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MP Board Class 12th Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance Important Questions Objective Type Questions

Question 1.
Choose the correct answer of the following:

Electrostatics Important Questions For Board Question 1.
The SI unit of electrical capacitance:
(a) Stat farad
(b) Farad
(c) Coulomb
(d) Stat coulomb.
Answer:
(b) Farad

Question 2.
The potential difference between the plates of a capacitor is constant. A dielectric medium is filled instead of air in between the plates. The intensity of electric field will:
(a) Decrease
(b) Remains unchanged
(c) Become zero
(d) Increase.
Answer:
(b) Remains unchanged

Buffer capacity is the measure of a buffer’s ability to resist pH change.

Class 12 Physics Important Questions Chapter 2 Question 3.
On replacing the air by an insulating material between the plates of a capacitor its capacity:
(a) Remains unchanged
(b) Increases
(c) Decreases
(d) Nothing can be said.
Answer:
(b) Increases

Question 4.
On increasing the separation between the plates of a parallel plate capacitor its capacitance :
(a) Remains unchanged
(b) Increases
(c) Decreases
(d) Nothing can be said.
Answer:
(c) Decreases

Equivalent Capacitance Questions Class 12 Question 5.
When two capacitors are joined in series each capacitor will have the same :
(a) Charge
(b) Potential
(c) Charge and potential
(d) Neither charge nor potential.
Answer:
(a) Charge

Question 6.
When two capacitors are joined in parallel each capacitor will have the same:
(a) Charge and potential
(b) Only charge
(c) Only potential
(d) Neither charge nor potential.
Answer:
(c) Only potential

Question 7.
Two capacitors of equal capacitance first connected in parallel then connected in series. What is the ratio of their capacities in both the cases:
(a) 2 : 1
(b) 1 : 2
(c) 4 : 1
(d) 1 : 4.
Answer:
(c) 4 : 1

Physics Important Questions Class 12 MP Board 2023 Question 8.
The formula of capacitance of a spherical conductor is:
(a) C = \(\frac { 1 }{ 4π{ £ }_{ 0 }R } \)
(b) C = 4πt£₀R
(c) C = 4πr£₀R²
(d) C = 4π£₀R³
Answer:
(b) C = 4πt£₀R

Question 2.
Fill in the blanks:

1. 1 farad = one coulomb/ ……………
2. 1 farad = …………… stat farad.
3. Dimensional formula of capacitance is ……………
4. is a device in which with or out changing in shape or size of a conductor its capacitance can be increased ……………
5. On increasing the distance between the plater of a parallel plate capacitor its capacity ……………
6. Three capacitor each of 3pF are joined in series their equivalent capacitance will be ……………
7. The dimensional formula of electric potential is ……………
8. The potential due to a point charge q at a distance r is given as ……………
9. The potential difference = Intensity of electric field × ……………
10. The increase in kinetic energy of a charge q when it is accelerated by a potential difference V is ……………
11. Due to presence of dielectric medium the potential ……………
12. The work done in moving a charge perpendicular to the electric field is ……………
13. The potential of earth is considered to be ……………

Answers:

1. 1 Volt
2. 9 × 10¹¹
3. [M^-1L^-2T⁴A²]
4. Capacitor
5.  Decreases
6. lµF
7. [ML²T^-3 A^-11]
8. V = \(\frac { 1 }{ 4π{ £ }_{ 0 }R } \) \(\frac { q }{ r}\)
9. Distance between the two point
10. qV
11. Decreases
12. Zero
13. Zero.

Class 12 Physics Important Questions MP Board Question 3.
Match the Column:

Answers:

1. (c) Q/V
2. (d) \(\frac { 1}{2}\) C/V²
3. (e) 4π£₀R
4. (a) £₀A/d
5. (b) 4π£₀ab / (b – a)

Question 4.
Write the answer in one word / sentence:

1. What is the potential of earth. Write SI units ?
2. What will be the electric field intensity inside a hallow sphere ?
3. In which direction of electric dipole, electric potential is zero ?
4. What is the net charge of a charge condenser ?
5. What quantity remains constant when the condenser are connected in series ?
6. What quantity remain constant when the conductor are connected in parallel ?

Answers:

1. Zero, volt
2. Zero
3. Broad-side-on position
4. Zero
5. Charge
6. Potential difference.

Electrostatic Potential and Capacitance Important Questions Very Short Answer Type Questions

Question 1.
What do you understand by equal potential surface ?
Answer:
The surface of the conductor where potential is in every point is called equal potential surface.

Imp Questions Of Physics Class 12 MP Board Question 2.
Write the name of the physical quantity whose SI unit in J/C. Is it a scalar or vector ?
Answer:
Electric potential, it is a vector quantity.

Question 3.
Draw a equi potential surface for a unit charge.
Answer:

Question 4.
Define farad.
Answer:
If the potential of a conductor increases by one volt when one coulomb of charges is given to it, then the capacity of the conductor is said to be one farad.

Physics Important Questions Class 12 MP Board Question 5.
On going in direction of electric lines of force, electric potential decreases or increases.
Answer:
Electric potential decreases.

Question 6.
Give an example in which electric field is non-zero but potential is zero.
Answer:
At broad-side-on position of an electric dipole electric field is non-zero and potential is zero.

Question 7.
Does electron try to go toward high potential area or low potential area ?
Answer:
Since electron is negatively charge so it tries to go toward high potential area.

MP Board Class 12th Physics Important Questions Question 8.
Potential between two parallel surface are same. The distance between them is R. If a charge q is bought from one surface to another, then what will be the work I done to do this ?
Answer:
Amount of work done will be zero on both the surface are equipotential.

Question 9.
If area of a plate of a parallel plate condenser in made half. Will it behave as condenser.
Answer:
When area of the plate if a parallel plate condenser is made half. Its capacity become half. Therefore it will not act as condenser.

Question 10.
A capacitor of capacity C is charged with potential difference V. What will be the magnitude of electric flux passing through the surface of it ?
Answer:
Zero.

MP Board Class 12 Physics Notes In English Question 11.
Why condenser are used in computer’s ?
Answer:
Condenser are used as memory chip in computer.

Question 12.
Write one use of capacitor ?
Answer:
To accumulate electric charge.

Electrostatic Potential and Capacitance Important Questions Short Answer Type Questions

Question 1.
What is potential ? Is it a vector or scalar quantity ?
Answer:
Work done in bringing a unit positive charge from infinity to a point in the I electric field is called potential at that point. If charge q is brought from infinity to a point and IT work is done.
∴ V = \(\frac {W}{p}\)
It is a scalar quantity.

Question 2.
Can same amount of charge be given 1 and a solid sphere of same radii, if they have same potential ?
Answer:
No, because capacities of both spheres of same radii are always equal. Therefore i both the spheres can hold same amount of charge at same potential.

Physics Important Questions Class 12 MP Board 2024 Question 3.
What is meant by capacity of a conductor ? Give its unit.
Answer:
The capacity of a conductor is defined by the charge given to the conductor, which increases its potential through unity.
Capacity = \(\frac {Charge}{Potential}\)
or C = \(\frac {q}{v}\)
Its SI unit is farad.

Question 4.
The surface of any conductor is always equipotential. Why ?
Or
The potential at every point on a charged conductor is same. Why ?
Answer:
All the points of the surface of a conductor are in electrical contact with one another. If the potential is not equal then the charges will flow from higher potential to lower potential till the potential of both the points on the surface becomes same. This will give rise to electrodynamics situations. Thus, the surface of a conductor is always equipotential.

MP Board Class 12th Physics Imp Questions 2023 Question 5.
What would be the work done if a point charge +q is taken from a point A to point B on the circumference of a circle with another point charge +q at the center:
Answer:
The points A and B are at same distance from the charge + q at the center, so V_A = V_B So, work done, W= q₀ (V_A – V_B) = 0.

Question 6.
Explain the meaning of capacity of a capacitor. What will be the effect on capacity of a parallel plate capacitor if a dielectric medium of dielectric constant k is filled in between the plates ?
Answer:
The capacity of a capacitor, is equal to charge given to one of its plates which produces unit potential difference across the plates. In this case capacity increases, it becomes k times its initial value.

Class 12 Physics Chapter 2 Notes Question 7.
What will be the change in the value of charge and potential difference between the plates of a parallel plate capacitor, if after charging its battery is removed and distance between its plates is reduced ?
Answer:
Charge remains same but potential difference decreases.

Question 8.
Two equipotential surface does not intersect each other, why ?
Answer:
Electric lines of forces are always perpendicular to equipotential surface. If two equipotential surface intersect each other then at the point of intersection there will be two direction of electric fields which is impossible. Therefore they does not intersect each other.

MP Board Physics Question 9.
Why must electrostatic field be normal to the surface at every point of a charged conductor ?
Answer:
If electric field is randomly directed, then it can be resolved, into two components. The horizontal component on this surface is E sin θ.
For electrostatic situation
£ sin θ = 0
⇒ sin θ = 0
⇒ θ = 0°

So, the electric field is normal to surface.

Question 10.
The potential at any point inside the hollow conductor remains same. Why ?
Answer:
When charge is given to a hollow conductor then the distribution of charge takes place on its upper surface. Therefore the intensity of electric field inside the conductor is zero. Hence, no work is done in moving unit positive charge inside it. Therefore potential at every point inside the conductor remains same.

Question 11.
Can the potential be zero where electric field is not zero ?
Answer:
Yes, the electric field on the equatorial line of a dipole is not zero but potential is zero.

Chapter 2 Physics Class 12 Important Questions Question 12.
What will be the effect on electric field, potential, difference, electric capacity and energy if a dielectric of dielectric constant K is filled between the plates of a capacitor ?
Answer:
The electric field will become \(\frac {1 }{ K }\)times, potential difference will become \(\frac {1 }{ K }\) times, electric field will become K time and energy will become \(\frac {1 }{ K }\) times.

Question 13.
Can 1 coulomb charge be given to a sphere of radius 1cm ?
Answer:
As we know that the formula of potential ¡s V = \(\frac { 1 }{ 4π { £ }_{ 0 } } \) \(\frac { q }{ r }\) …(1)
Given,q = lC, r = lcm = 10^-2m
Putting these values in eqn. (1)
V = 9 × 10⁹ × \(\frac{1}{10^{-2}}\) = 9 x 10¹¹volt
Where = \(\frac{1}{4 \pi \varepsilon_{0}}\) =  9 × 10⁹ in SI unit.
This value of potential is greater than barrier potential of air. Therefore IC charge cannot be given to a sphere of radius 1cm.

Chapter 2 Physics Class 12 Question 14.
In the shown figure what will be the work done to bring a z point charge from the point X to Y to Z?
Answer:
There the point Z and Y are situated on same equaipotential surface. Therefore work done to bring a point charge from A’to Zand from X to Z will be same.
.’. W_y= W_z.

Question 15.
Derive an expression for electric potential due to a point charge. Is it scalar or vector and why?
Answer:
Consider a point charge q placed at origin O. Potential at P has to be found out. Let the medium between charge ‘q’ and P has dielectric constant E_r.

Electric field at P due to charge q is
E = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \cdot \frac{q}{r^{2}}\)
The electric field \(\vec{E}\) points away from the charge q. A force \(\vec{F}\) = -q₀ \(\vec{E}\) has to be applied on the charge so that it can be brought near to q. The small work required to move the test charge q₀ from P to Q through a small distance dr is given by dW = Fdr
= -q₀ Edr
The total work done in moving the charge q0 from infinity to point P will be obtained by integrating the above equation as –

But electric potential is defined as work done per unit test charge.

Potential at P is V = \(\frac{1}{4 \pi \varepsilon_{0} \varepsilon_{r}} \cdot \frac{q}{r}\)
If medium between q and q₀ is vacuum then £_r = 1
Then , V =\(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)
This is the required expression.

MP Board Class 12th Physics Question 16.
What are the factors affecting the potential of a charged
Answer:
The factors affecting the potential of a charged conductor are:
1. Amount of charge on conductor:
By the formula V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) it is clear that
V ∝ q, hence more is the charge, more will be the potential of charged conductor.

2. Shape of conductor (Area of conductor):
If the charge is kept constant on a conductor and its surface area is decreased then the potential of conductor increases whereas on increasing the surface area its potential decreases. So the potential of a conductor is inversely proportional to the radius.

3. Presence of other conductor near the charged conductor:
If an uncharged conductor is brought near a charged conductor then the potential of the charged conductor decreases.

4. Medium surrounding the conductor:
Due to presence of insulating medium near the charged conductor its potential will decrease.

Question 17.
Define equipotential surface. Write its properties.
Answer:
Equipotential surface:
An equipotential surface is the locus of all those points at which the potential due to distribution of charge remains same.

Properties:

• Potentials on every point are equal
• No work is done in moving a positive charge from one point to another
• The electrical lines of force are normal to the equipotential surface
• Two equipotential surfaces do not intersect each other.
• All the points on the surface of a conductor are in electric contact. If the potentials are not same then the
• charge will flow from higher potential to lower potential till the potential of both the points become same.
• Thus the surface of a conductor is always equipotential.

Uniform Distribution Calculator … Calculate the probability density.

Class 12th Physics Chapter 1 Important Questions MP Board Question 18.
Obtain a relation between electric Held intensity and potential difference.
Or
Prove that E = \(\frac { dv }{ dr }\)where symbols have their usual meanings.
Answer:
Suppose A and B are two points in the electric field of charge q. The direction of electric field is radially outwards from A to B. Suppose the distance between A and B is very small (i. e., dr) then the electrie field between A and B can be taken as uniform. As the potential is inversely proportional to distance hence potential at A is more than that of B. Let the potential at B is V then that at A is V + dV.

Work done in bringing the test charge q0 from B to A is –
dW = q₀dV …….(1)
Force acting on q₀ will be
F = q₀ E
Work done in bringing the test charge against the repulsion force will be
dW = -q₀Edr
(Work = Force x Displacement) ………(2)
The negative sign shows that the direction of displacement and direction of force are opposite to each other.
From eqns. (1) and (2), we get,
q₀dV = q₀Edr
or dV = -Edr
E = – \(\frac { dv }{ dr }\)
This is required relation between intensity of electric field and potential difference.

Question 19.
Prove that capacity of an isolated spherical conductor is directly proportional to its radius.
Or
Derive an expression for the capacity of a spherical conductor.
Answer:
The capacity of a conductor is its ability to store electrical energy and it is equal
to that charge which increases its potential by unity.
∴ Capacity = \(\frac { Charge }{ Potential }\)

Capacity of an isolated spherical conductor:
Let us consider about a spherical conductor of radius r. The charge + Q is given to it. The charge will be distributed on its surface uni- formly. Therefore the lines of force will be emitted normally to the surface seem to becoming from its center. Hence, we can suppose that all the charges are kept at the centre.
∴ Potential on the surface V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}\)
But capacity C = \(\frac { Q }{ V }\)
Putting the value of V, we get
C = \(\frac{Q}{\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}}\)
C = 4πE₀r
C ∝ r.
Thus, the capacity is proportional to the radius of the spherical conductor.

Question 20.
What do you mean by a capacitor ? Explain its principle.
Answer:
The capacitor is a device by which the capacitance of a conductor is increased without changing its size or volume. Actually it stores electrical energy.
Principle of capacitor:
Let A be a charged conducting plate. Another uncharged conductor plate B is brought near to A, therefore due to induction negative charges will be induced on the front surface and positive charges on the other side of plate B.
Now, the negative charge reduces the potential while the positive charge increases.

As the negative charge is nearer therefore the potential of plate A decreases. Now, the plate B is earthed then the free positive charge will go to earth and hence the potential of A decreases by more value.
C = \(\frac { Q }{V}\)
As V decreases, C will increase. This arrangement is called capacitor or condenser.

Question 21.
Derive an expression for parallel plate capacitor.
Answer;
Let A and B be two plates of a parallel plate capacitor separated by a distance d apart. Area of each plate is A and dielectric constant of the medium between them is E_r Now, plate A is given + Q charge. Therefore, – Q charge will be induced on the nearer surface of the plate B and + Q charge on the other side. As B is connected to earth, + Q charge of B will go to earth. Let the charge density of A is cr, therefore that of B will be -σ.

Now, σ = \(\frac { Q }{A}\)
Intensity between the plates will be given by
E = \(\frac{\sigma}{\varepsilon_{0} \varepsilon_{\mathrm{r}}}\)
E = \(\frac{Q}{A \varepsilon_{0} \varepsilon_{r}}\)
But, potential difference between the plates A and B is
V = Electric field intensity ×
Distance between to plates = Ed
V = \(\frac{Q}{A \varepsilon_{0} \varepsilon_{s}} \cdot d\)
But, C = \(\frac{Q}{V}=\frac{Q}{\frac{Q d}{A \varepsilon_{0} \varepsilon_{r}}}\)
C = \(\frac{\varepsilon_{r} \varepsilon_{0} A}{d}\)
This is the required relation.
For air or vacuum, E_r = 1
C = \(\frac{\varepsilon_{0} A}{d}\)

Question 22.
Three capacitors of capacitance’s C₁ C₂ and C₃ are connected in series. Derive an expression for the equivalent capacitance.
Answer:
The given figure shows three capacitors of capacitances C₁ C₂ and C₃ con – nected in series. A potential difference of V is applied across the combination, charges of + Q and – Q are developed on the plates of the capacitor.
Potential difference across the individual capacitors will be

V₁ = \(\frac{Q}{C_{1}}\) , V₂ = \(\frac{Q}{C_{2}}\), V₃ = \(\frac{Q}{C_{3}}\) …….(1)
The sum of these must be equal to the applied potential difference V.
V = V₁ + V₂ + V₃ ………(2)
Let C be the equivalent capacitance of the series combination
∴ C = \(\frac { Q }{ V }\) or V = \(\frac { Q }{ C }\) ………..(3)
V₁ + V₂ + V₃ = \(\frac { Q }{ C }\) [from equ..(2)]
\(\frac{Q}{C_{1}}\) + \(\frac{Q}{C_{2}}\) + \(\frac{Q}{C_{3}}\) = \(\frac{Q}{C}\) [from equ..(1)]
Q(\(\frac{1}{C_{1}}\) + \(\frac{1}{C_{2}}\) + \(\frac{1}{C_{3}}\)) = \(\frac{Q}{C}\)
\(\frac { 1 }{ c }\) = \(\frac{1}{C_{1}}\) + \(\frac{1}{C_{2}}\) + \(\frac{1}{C_{3}}\)
This is the required expression.

Question 23.
Three capacitors of capacitance’s C₁ C₂ and C₃ are connected in parallel. Derive an expression for the equivalent capacitance C.
Answer:
Consider three capacitor of capacitance’s C₁ C₂ and C₃ connected in parallel. A potential difference V is applied across the combination. Charges set up in the individual capacitor will be.

Q₁ = C₁V, Q₂ = C₂V,Q₃ = C₃V …(1)
Total charge stored in the parallel combination is
Q = Q₁ + Q₂ + Q₃ ……….(2)
If C is the equivalent capacitance of the combination
Then, C = \(\frac { Q }{ V }\) Q = CV …………(3)
Q₁ + Q₂ + Q₃ = CV [from eq. (2)]
C₁V + C₂V + C₃V = CV [from eq. (1)]
V(C₁ + C₂ + C₃) = CV
C = C₁ + C₂ + C₃
This is the required expression.

Question 24.
Derive an expression for the energy of a charged conductor.
Or
Prove that energy of a charge conductor is directly proportional to its square of potential.
Answer:
The work done in charging a conductor is stored as energy in it. This energy is called electrostatic potential of conductor.

Formula derivation:
Let us consider about a conductor of capacity C which is given charge +Q due to which its potential becomes V. As the charge increases work done also increases. Let at any instant the potential of conductor be V due to charge q.
∴ C = \(\frac { q }{ v}\)
or v = \(\frac { q }{ C}\)
Now, at potential Kthe work done in giving the charge dq will be dW
∴ dw = Vdq
or dw = \(\frac { q }{ C}\)dq
Work done in charging the conductor from 0 to Q

This work done is stored as potential energy on the conductor. Energy of a charge conductor
U = \(\frac{1}{2} \frac{Q^{2}}{C}\)
But Q = CV
∴ U = \(\frac{1}{2} \frac{C^{2} V^{2}}{C}\)
or U = \(\frac{1}{2} C V^{2}\)
∴ U ∝ V² because C is constant

Question 25.
Prove that on connecting two charged conductors, charges distribute on them according to their capacities.
Answer:
When two isolated charged conductors A and B are connected by a thin wire, charge flows from the conductors at high potential to the conductor at low potential till the potential of both A and B became equal. The phenomenon involved is called distribution of charges and the total charge of the entire system remains conserved. Let the capacitance of A and B be C₁ and C₂, the charges be Q₁ and Q₂ respectively. Then the potentials are V₁ and V₂ respectively.
∴ Initially, Q₁ = C₁V₁ and Q₂ = C₂V₂

The conductors are joined by a wire of negligible capacitance, the charges flow from- the conductor at higher potential to the conductor at lower potential till the potentials on each conductor become equal.
The net charge on the system,
Q = Q₁ + Q₂
The common potential, V = \(\frac { Total charge }{ Total capacitance }\)

After the potential becomes equal let the charge on A ₁ be Q₁ and charge on A₂ be Q₂.

Dividing eqn. (2) by eqn. (3), we get

When the conductors are joined, then the charges get distributed in the ratio of their capacities.

Question 26.
Obtain an expression for potential due to a group of point charges.
Or
Derive the expression for potential energy.
Answer:
Consider a group of point charges q₁,q₂,q₃……..q_n which are situated at a dis-tance of r₁, r₂, r₃…….. _nn respectively from the point P. The potential due to these point charges is to be obtained at P. Now potential at P due to q₁ is

potential due to q₂,q₃, ……… q_netc

Total potential at P will be V = V₁ + V₂ + V₃ + ……….. + V_n

This is the required expression.

Electrostatic Potential and Capacitance Important Questions Long Answer Type Questions

Question 1.
Derive the expression for the capacity of a parallel plate capacitor, when the medium between the plates is partially filled by a dielectric medium.
Answer:
Let A and B are parallel plates of a capacitor. The distance between the plates is d and plate of thickness t and dielectric constant Er is introduced.

Now, plate A is given charge +Q.
Let the charge density be σ.
∴ σ = \(\frac { Q}{ A}\)
Intensity of field in air ,
E_{0}=\(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\)
If the intensity of field inside the dielectric medium be E, then
Dielectric constant = \(\frac {Electric field in vacuum}{ Electric field in medium}\)
or \(\varepsilon_{r}=\frac{E_{0}}{E}\)
or E = \(\frac{E_{0}}{\varepsilon_{r}}=\frac{Q}{\varepsilon_{0} \varepsilon_{r} A}\)
Now, potential difference between A and B,
V=E₀ (d – t) + Et, [(d – t) is vacuum distance]

This is the required expression.
Metal is a conductor. When metal is used in place of the dielectric, it will conduct electricity and the potential difference will become zero. So, capacitor will not work.

Question 2.
Calculate the loss of energy, when two charged conductors are connected.
Or
The capacities of two conductors are C₁ and C₂, Q₁ and Q₂ charges are given to them so that their potentials become V₁ and V₂ respectively. If they are connected by a wire, then calculate the following:

• Common potential
• Loss of energy.

or
prove that when two charged conductors are connected, there will be a loss of energy
Or
In redistribution of charges, is there a loss of energy ? Deduce an expression to confirm the answer.
Answer:
Let A and B be two conductors of capacities C₁ and C₂ respectively. When charges Q₁ and Q₂ are given separately the potentials become V₁ and V₂ respectively.Total charges, Q = Q₁ + Q₂ ………..(1)
But, Q₁ = C₁V₁ and Q₂ = C₂V₂
By eqn. (1), we get
Q = C₁V₁ + C₂V₂
Total capacity, C = C₁ + C₂

(1) Common potential:
Let the conductors are connected by a wire and the common potential becomes-V.
Q₁ + Q₂ = (C₁ + C₂)V
V = \(\frac{Q_{1}+Q_{2}}{C_{1} + C_{2}}\)
V = \(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\) ……..(2)
This is the expression for the common potential.

(2) Loss of energy: Total energy of the conductors before connection:

and total energy after connection,

Putting the value of V from eqn. (2) in eqn. (4), we get

Hence, difference of energy’,

(V₁ – V₂)² is positive, hence (V₁ – V₂) is positive. Hence, during redistribution, there will be always loss of energy.
i.e., U₁ – U₂>0 ⇒ U₁>U₂
i.e., energy before joining is greater than energy after joining.
The loss in energy,

Question 3.
Explain the construction and working of Van de Graaff generator. Write its uses.
Answer:
Van de Graaff generator is a machine which produces electricity of about 107 V or more potential difference.
Construction:
It consists of a large metallic sphere S of diameter 5 m, mounted on high insulating support PP about 15 m high. An endless insulating belt made up of rubber passes over the pulleys p₁ and P₂. A motor rotates p₁ C₁ and C₂ are two metallic combs called spray comb and collecting comb respectively. C₁ is connected to S. To prevent the leakage of charge, the generator is put inside a large enclosure filled with gas at 15 atm. pressure. This iron enclosure is connected to earth.

Working:
The comb C₁ is connected to the positive terminal of the battery, therefore the surface density of the points becomes very high which causes the wind present nearby it to get charged. Thus, the spray comb sprays the charge on the belt. Now, the electric wind moves up to the collector comb C₂ When it reaches in front of the collector comb C₂ opposite charge induces on the tip to neutralize the same type of charge. The negative charge wind of C₂, cancels the positive charge of the belt. Thus, by the repeated actions more and more positive charge is induced on sphere, hence its potential increases to about 10⁷volts or more.

Uses:

• To generate high potential.
• To accelerate the positive particles such as protons, Deuteronomy, are particle etc. and used in nuclear disintegration.

Question 4.
When Anil opened the cap of the tap, then he found no water in coming out of it. Then he opened the cap of the water tank and found no water in the tank. To fill up water in the water tank he switch on the switch of the motor and found motor is not starting. Then he called the electric technician. The technician said him on checking that the condenser of the motor is not functioning. On replacing capacitor, the motor start working.

Answer the following questions:

1. What values does Anil exhibits ?
2. What is the function of condenser ?
3. What is total charge on a charged condenser ?
4. The capacity of a capacitor is 3pF. If it is charged up to 100 V potential difference, then what will be charged stored in it ?

Answer:

1. Anil exhibited his presence of mind.
2. It accumulate charge and hence it conserved energy.
3. Net charge on a condenser is zero.
4. C = 3µF = 3 × l0^-6 F, V = 100V

∴ By formula Q = CV = 3 × l0^-6 × l00
or Q = 3 × 10^-4C.

Electrostatic Potential and Capacitance Important Questions Numerical Questions

Question 1.
Can 1 coulomb charge be given to a sphere of radius 1cm ?
Answer:
No.
As we know that the formula of potential is V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) …(1)
Given, q = 1C, r = 1cm = 10^-2m
Putting these values in eqn. (1)
V = 9 × 10⁹ x \(\frac{1}{10^{-2}}\) = 9 × 10¹¹volt
Where = \(\frac{1}{4 \pi \varepsilon_{0}}\) = 9 × 10⁹ in SI unit.
This value of potential is greater than barrier potential of air. Therefore lC charge cannot be given to a sphere of radius 1cm.

Question 2.
You are given three capacitor of 4pF each. How they will be combined to obtain resultant capacity of 6pF ?
Solution:
Given : Q = C₂ = C₃ = 4µF
When two capacitor is joined in series and third capacitor joined parallel with them, then resultant capacity is obtained as 6µF.

C₁ ,C₂ is in series, its resultant (C’) is
\(\frac { 1 }{ c}\) = \(\frac{1}{C_{1}}+\frac{1}{C_{2}}\)
or \(\frac { 1 }{ c}\) = \(\frac { 1 }{ 4}\) + \(\frac { 1 }{ 4}\) = \(\frac { 2 }{ 4}\)
or C = \(\frac { 4 }{ 2}\) = 2µF.
C and C₃ is in parallel combination,
Its resultant C is C = C + C₃
C = 2 + 4 or C = 6µF.

Question 3.
A hollow metallic sphere of radius 0-1 m is given 6pC. Calculate its potential:

1. At the surface of sphere
2. At the center.

Solution
Given, r = 01 m, q = 6µC = 6 x 10^-6 C
Formula: V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\)

1. Potential at the surface:
V = 9 × 10⁹ × \(\frac{6 \times 10^{-6}}{0 \cdot 1}\)
V = 54 × 10⁴
V- 5.4 × 10⁵ volt.

2. At the center:
Inside the sphere the potential remains same and equal to that on the surface hence V = 5 . 4 x 10⁵volt.

Question 4.
A test charge is moved from A to B, B to C and A to C in an electric field E as shown in the figure :

Find (1) Potential difference between A and C
(2) At which point electric potential will be high and why ?
Solution:
1. In right angled ∆ABC
AB² = AC² – BC² = 5² – 3²
∴ AB = 4 = dr
BC is perpendicular to electric field, therefore potential will be same at B and C.
V_A – V_C = V_A – V_B = -Edr = -4E
2. Therefore potential at point C will be more than potential of point A.

Question 5.
Identical water droplets, having equal charge on each are combined to form a big drop. Compare the capacity of bigger drop with that of a small drop.
Solution:
Let radius of small droplet = r
Radius of the big drop = R
Volume of big drop = Volume of 27 droplets
\(\frac { 4 }{ 3 }\) πR³ = 27 × \(\frac { 4 }{ 3 }\) πR³
or R³ = 27r³
or R³ = (3r)³
or R = 3r
or \(\frac { R }{ r }\) = \(\frac { 3}{ 1 }\)
Since, C ∝ radius
or \(\frac{C_{1}}{C_{2}}=\frac{r_{1}}{r_{2}}\)
or \(\frac{C_{1}}{C_{2}}\) = \(\frac { 3}{ 1}\) = 3
or C₁ = 3C₂
The capacity of bigger drop is three times that of smaller one.

Question 6.
How three capacitor of 3pF each can be combined such that their resultant capacity is :

1. 9pF,
2. 4.5pF.

Solution:
1. When the three capacitor is joined in parallel, then
C = C₁ + C₂ + C₃
= 3 + 3 + 3 = 9µF.

2. When two capacitor are joint in series, then resultant C’ is
\(\frac { 1 }{ c}\) = \(\frac{1}{C_{1}}+\frac{1}{C_{2}}\)
= \(\frac { 1 }{ 3}\) + \(\frac { 1 }{ 3}\) = \(\frac { 2 }{ 3}\)
C = \(\frac { 3 }{2}\) = 1.5 µF
Now C is joined in parallel with C₃
C = C + C₃ = 1.5 + 3 = 4.5µF.

Question 7.
The potential difference between two points is 10V. How much work is required to move a charge 100 pC from a point to the other ?
Solution:
Given, V= 10 volt, q = l00µC = l00 × l0^-6C
Formula : w = qV
= 100 × 10^-6 × 10
= 10^-3 joule.

Question 8.
Find the area of the plate of a 2F parallel plate capacitor, if the separation between the plates is 0.5 cm ?
Solution:
As C = \(\frac{\varepsilon_{0} A}{d}\)
A = \(\frac{C d}{\varepsilon_{0}}\)
Here, C = 2F, d= 0.5cm = 0.5 x 10^-2m
A = \(\frac{2 \times 0 \cdot 5 \times 10^{-2}}{8.85 \times 10^{-12}}\)
= 1.13 × 10⁹ m² = 1130Km²

Engineering Physics MCQ Electrical Engineering.

Question 9.
Two charges 5 x 10^-8C and -3 x 10^-8C are located 16 cm apart At what point, on the line joining the two charges, is the electric potential zero ? Take the potential at infinity to be zero. (NCERT)
Solution:
Case I.
Let electric potential be zero at point C lying at distance x from the positive charge.

Given, q₁ = 5 × 10^-8 C;
q₂ = -3 × l0^-8 C
AC = x cm : CB = (16 – x) cm
Now, Potential at C is zero i. e.,
V₁ + V₂ = 0

-8x + 80 = 0
8x = 80
x = 19 cm
i.e., electric potential at a distance of 10 cm from positive charge will be zero.

Case II.
The other possibility is that the point C may also lie on produced AB.

Now, V₁ + V₂ = 0

5(x – 16) – 3x = 0
5x – 80 – 3x = 0
2x – 80 = 0
x = 40 cm from the positive charge

Question 10.
Determine the equivalent capacitance between A and B in the following circuits:

Solution:
(i). Mark the junctions as C and D.

(But C will be A and D will be B)
Draw the equivalent network, which is given below

Equivalent capacitance,
C = C₁ + C₂ + C₃
or C = 1 + 1 + 1 = 3µF

(ii). To move from A to B, there are two paths P -1 and P – II.

(As A and B, the path is dislocated temporarily)
The capacitors in P -II are in series. So, the equivalent becomes

The resultant capacity of series combination is
\(\frac { 1 }{ C’ }\) = \(\frac { 1}{ 3 }\) + \(\frac { 1 }{ 3}\) + \(\frac { 1 }{3 }\)
= \(\frac { 3 }{ 3 }\) = 1µF = C’ = 1µF
The equivalent further becomes

Total capacity C = 3 +1 = 4µF.

MP Board Class 12th Physics Important Questions

In this article, we will share MP Board Class 10th Sanskrit Solutions Chapter 13 महाभारते विज्ञानम् Pdf, Surbhi Class 8, These solutions are solved subject experts from the latest edition books.

MP Board Class 8th Sanskrit Solutions Surbhi Chapter 18 सत्कर्म एव धर्मः

MP Board Class 8th Sanskrit Chapter 18 अभ्यासः

Class 8 Sanskrit Chapter 18 MP Board प्रश्न 1.
एकपदेन उत्तरं लिखत(एक शब्द में उत्तर लिखो-)
(क) विक्रमादित्यः नगरभ्रमणसमये किं दृष्टवान्? (विक्रमादित्य ने नगर भ्रमण के समय क्या देखा?)
उत्तर:
रुग्णम्। (रोगी को)

(ख) विक्रमादित्यः महामन्त्रिणि राज्यभारं समर्प्य कुत्र अगच्छत्? (विक्रमादित्य महामन्त्री पर राज्यभार को समर्पित करके कहाँ गये?)
उत्तर:
वनम्। (वन में)

(ग) महात्मा कस्य समीपे तपस्यारतः आसीत्? (महात्मा किसके पास तपस्यारत थे?)
उत्तर:
विक्रमादित्यस्य। (विक्रमादित्य के)

(घ) महात्मा योगबलेन तत्र कस्य दृश्यं दर्शितवान्? (महात्मा ने योग के बल से वहाँ किसका दृश्य दिखाया?)
उत्तर:
यमलोकस्य। (यमलोक का)

(ङ) कर्मणां लेखनं कस्य पार्वे अस्ति? (कर्मों का लेखा किसके पास है?)
उत्तर:
चित्रगुप्तस्य। (चित्रगुप्त के)

(च) श्रेष्ठाचरणस्य प्रतिज्ञां कृत्वा विक्रमादित्यः कुत्रः आगतवान्? (श्रेष्ठ आचरण की प्रतिज्ञा करके विक्रमादित्य कहाँ आये?)
उत्तर:
उज्जयिनीम्। (उज्जयिनी में)

Class 8 Sanskrit Chapter 18 प्रश्न 2.
एकवाक्येन उत्तरं लिखत(एक वाक्य में उत्तर लिखो-)
(क) महात्मा विक्रमादित्यं किम् उपदिष्टवान्? (महात्मा ने विक्रमादित्य को क्या उपदेश दिया?)
उत्तर:
महात्मा विक्रमादित्यं उपदिष्टवान् यत्-“राजन्! भवान् तु यथानीति राजधर्मस्य पालनं करोतु। धर्मानुकूलं शासनम् अपि धर्म एव भवति” इति। (महात्मा ने विक्रमादित्य को उपदेश दिया कि-“हे राजन्! आप तो नीति के अनुसार राजधर्म का पालन करो। धर्म के अनुसार शासन भी धर्म ही होता है।)

(ख) कर्म-तपस्योः कः भेदः? (कर्म और तपस्या में क्या भेद है?)
उत्तर:
कर्मणः स्थानम् भिन्नम् परं तपस्या स्वर्गप्राप्तेः साधनम्।” इति कर्म-तपस्ययोः भेदः। (“कर्म का स्थान भिन्न है परन्तु तपस्या स्वर्ग प्राप्ति का साधन है।” ऐसा कर्म और तपस्या का भेद है।)

(ग) विक्रमादित्यः अन्ते किं ज्ञातवान्? (विक्रमादित्य ने अन्त में क्या जाना?)
उत्तर:
विक्रमादित्यः अन्ते ज्ञातवान् यद्-‘सदाचारः एव तपस्या, सत्कर्म एव धर्म’ इति। (विक्रमादित्य ने अन्त में यह जाना कि-“सदाचार ही तपस्या है, सत्कर्म ही धर्म है”।)

(घ) विक्रमादित्यः लोके कथं प्रसिद्धः? (विक्रमादित्य संसार में कैसे प्रसिद्ध हैं?)
उत्तर:
विक्रमादित्यः लोके सत्कर्मणा एव प्रसिद्धः। (विक्रमादित्य संसार में सत्कर्म से ही प्रसिद्धः हैं।)

(ङ) विक्रमादित्यस्य ध्येयवाक्यं किम् आसीत्? (विक्रमादित्य का ध्येय वाक्य क्या था?)
उत्तर:
विक्रमादित्यस्य ध्येयवाक्यं ‘सत्कर्म एव धर्मः’ इति आसीत्। (विक्रमादित्य का ध्येय वाक्य ‘सत्कर्म ही धर्म है’ यह था।)

(च) विक्रमादित्यस्य मनसि केन वैराग्यम् उद्भूतम्? (विक्रमादित्य के मन में किससे वैराग्य उत्पन्न हुआ?)
उत्तर:
विक्रमादित्यस्य मनसि रुग्णदर्शनेन वैराग्यम् उद्भूतम्। (विक्रमादित्य के मन में रोगी को देखने से वैराग्य उत्पन्न हुआ।)

(छ) “सत्कर्म एव धर्मः” कथायाः सारः कः? (“सत्कर्म ही धर्म है” कथा का सारांश क्या है?)
उत्तर:
अस्याः कथायाः सारः यत् ‘मनुष्यः उत्तमानि कर्माणि कृत्वा अपि परलोकं साधयितुम् शक्नोति’ इति। (इस कथा का सारांश है कि ‘मनुष्य अच्छे कार्य करके भी परलोक सिद्ध कर सकता है।)

MP Board Class 8 Sanskrit Chapter 18 प्रश्न 3.
रिक्तस्थानानि पूरयत(रिक्त स्थान भरो-)
(क) विक्रमादित्यस्य समीपे एकः ……….. तपस्यारतः आसीत्। (महात्मा/दुरात्मा)
(ख) तपस्या ……… साधनम्। (स्वर्गप्राप्ते/राज्यप्राप्तेः)
(ग)यमराजः …………. आदिष्टवान्। (विक्रमादित्यम्/दूतेभ्यः)
(घ) वने सः कठिनां …………. आरब्धवान्। (दिनचर्यां/तपस्याम्)
(ङ) प्रजापालनं धर्म …….. अस्ति। (तपस्वीनाम्/राज्ञाम्)
उत्तर:
(क) महात्मा
(ख) स्वर्गप्राप्तेः
(ग) दूतेभ्यः
(घ) तपस्याम्
(ङ) राज्ञाम्।।

Class 8 Sanskrit Chapter 18 Question Answer प्रश्न 4.
उचितं मेलयत(उचित को मिलाओ-)

उत्तर:
(क) → (v)
(ख) → (iv)
(ग) → (ii)
(घ) → (iii)
(ङ) → (i)

MP Board Class 8 Sanskrit प्रश्न 5.
नामोल्लेखपूर्वकं समासविग्रहं कुरुत(नाम का उल्लेख करते हुए समास विग्रह करो-)
(क) ध्येयवाक्यम्
(ख) तपस्यारतः
(ग) प्रजापालनम्
(घ) योगबलेन
(ङ) श्रेष्ठाचरणेन।
उत्तर:

Class 8 Sanskrit Chapter 17 MP Board प्रश्न 6.
नामोल्लेखपूर्वकं सन्धिविच्छेदं कुरुत(नामे का उल्लेख करते हुए सन्धि विच्छेद करो-)
(क) धर्मानुकूलम्
(ख) नेति
(ग) सदाचारः
(घ) अद्यापि
(ङ) विक्रमादित्यः।
उत्तर:

Class 7 Sanskrit Chapter 18 प्रश्न 7.
रेखाङ्कितशब्दानाम् आधारेण प्रश्ननिर्माणं कुरुत (रेखांकित शब्दों के आधार पर प्रश्न निर्माण करो-)
(क) विक्रमः अवदत्। (विक्रम बोला।)
उत्तर:
कः अवदत्? (कौन बोला?)

(ख) तपस्या तु महात्मनां कर्म इति। (तपस्या तो महात्माओं का काम है।)
उत्तर:
तपस्या तु केषाम् कर्म इति? (तपस्या तो किनका काम है?

(ग) तत्रैव यमलोकस्य दृश्यं दर्शितवान्। (वहीं यमलोक का दृश्य दिखाया?)
उत्तर:
कुत्र यमलोकस्य दृश्यं दर्शितवान्? (कहाँ यमलोक का दृश्य दिखाया?)

(घ) यमराजः दूतान् पृच्छति? (यमराज दूतों से पूछते हैं।)
उत्तर:
यमराजः कान् पृच्छति? (यमराज ने किनसे पूछा?)

(ङ) राज्ञः धर्म प्रजापालनम्। (राजा का धर्म प्रजा का पालन है।)
उत्तर:
कस्य धर्मः प्रजापालनम्? (किसका धर्म प्रजा का पालन है?)

सत्कर्म एव धर्मः हिन्दी अनुवाद

एकदा विक्रमादित्यः नगरभ्रमणसमये एकम् मरणासन्नं रुग्णं दृष्टवान्। तस्य दर्शनेन मनसि उद्भूतम्। अतः मायामोहमयं संसारं ज्ञात्वा सः महामन्त्रिणि राज्यभारं समर्प्य वनम् अगच्छत्।

अनुवाद :
एक बार विक्रमादित्य ने नगर में भ्रमण के समय एक मरणासन्न (मरने के निकट) रोगी को देखा। उसको देखने से मन में वैराग्य उत्पन्न हुआ। इसलिए माया मोह से भरे संसार को जानकर वह महामन्त्री को राज्यभार सौंपकर वन चले गये।

वने सः कठिना तपस्याम् आरब्धवान्। तस्य समीपे एव एकः महात्मा अपि तपस्यारतः आसीत्। महात्मा तम् अवदत्, “राजन्! भवान् तु यथानीति राजधर्मस्य पालनं करोतु। धर्मानुकूलं शासनम् अपि धर्म एव भवति” इति स महात्मा विक्रमादित्यम् उपदिष्टवान्। विक्रमः अवदत्-“नहि महात्मन्! तपसा एव परलोकः साध्यते, कर्मणा नेति।” महात्मा अवदत्, “राजन! राज्ञः धर्मः प्रजापालनं, शासनम् एव अस्ति तपस्या तु महात्मानां कर्म इति।” इत्युक्त्वा महात्मा राजानाम् पृष्टवन्-कर्म-तपस्ययोः कः भेदः?

अनुवाद :
वन में उन्होंने कठिन तपस्या आरम्भ कर दी। उनके पास में एक महात्मा भी तपस्यारत (तपस्या में लगे) थे। महात्मा ने उनसे कहा, “हे राजन्! आप तो नीति के अनुसार राजधर्म का पालन करो। धर्म के अनुकूल शासन भी धर्म ही होता है।” इस प्रकार उस महात्मा ने विक्रमादित्य को उपदेश दिया। विक्रम ने कहा-“नहीं महात्मन्! तपस्या से ही परलोक प्राप्त होता है, कर्म से नहीं।” महात्मा ने कहा, “हे राजन्! राजा का धर्म प्रजा का पालन और शासन ही है, तपस्या तो महात्माओं का काम है।” ऐसा कहकर महात्मा ने राजा से पूछा-कर्म और तपस्या में क्या भेद है?

राजा अवदत्-कर्मणः स्थानम् भिन्नम् परं तपस्या स्वर्गप्राप्तेः साधनम्। एतच्छ्रुत्वा महात्मा हसन् अवदत्-“राजन् ! मनुष्यः उत्तमानि कर्माणि कृत्वा अपि परलोकं साधयितुम् शक्नोति।” अनन्तरम् महात्मा योगबलेन तत्रैव यमलोकस्य दृश्यं विक्रमं दर्शितवान्। दृश्ये यमराजः दूतान् पृच्छति-“एतस्य कर्म कीदृशम् ?” एकः दूतः अवदत्-“कर्मणां लेखनं तु चित्रगुप्तस्य पार्वे अस्ति।” क्षणं विचार्य यमराजः दूतान् आदिष्टवान् यत्-“यदि एतस्य कर्माणि उत्तमानि सन्ति तर्हि स्वर्गस्य द्वारम् उद्घाटयतु यदि कर्माणि अधमानि, तदा बलात् नरके पातयतु।”

अनुवाद :
राजा ने कहा-कर्म का स्थान भिन्न है परन्तु तपस्या स्वर्ग प्राप्ति का साधन है। ऐसा सुनकर महात्मा हँसते हुए बोले-“हे राजन्! मनुष्य अच्छे कर्म करके भी परलोक सिद्ध कर सकता है।” इसके बाद महात्मा ने योग के बल से वहीं यमलोक का दृश्य विक्रम को दिखाया। दृश्य में यमराज दूतों से पूछ रहे हैं-“इसका कर्म कैसा है? एक दूत ने कहा-“कर्मों का लेखा तो चित्रगुप्त के पास है।” कुछ देर विचार करके यमराज ने दूतों को आदेश दिया कि-यदि इसके कर्म अच्छे हैं तो स्वर्ग का द्वार खोल दो यदि कर्म बुरे हैं तो जबरदस्ती नरक में डाल दो।”

इदं दृश्यं दृष्ट्वा विक्रमः ज्ञातवान् यद्-‘सदाचारः एव तपस्या, सत्कर्म एव धर्म’ इति। अनन्तरं सः तम् महात्मानम् प्रणम्य, श्रेष्ठाचरणस्य प्रतिज्ञां कृत्वा राजधानीम् उज्जयिनीम् आगतवान्। आगत्य धर्मानुकूलंनीतिपूर्वकम् प्रजापालनपुरस्सरं शासनं कृतवान्। सः अद्यापि लोके सत्कर्मणा एव प्रसिद्धः। तस्य जीवनस्य ध्येयवाक्यम् आसीत्-‘सत्कर्म एव धर्मः।’

अनुवाद :
इस दृश्य को देखकर विक्रम जान गये कि-“सदाचार (अच्छा व्यवहार) ही तपस्या है और सत्कर्म (अच्छे कर्म) ही धर्म।” इसके बाद वह उन महात्मा को प्रणाम करके, श्रेष्ठ आचरण की प्रतिज्ञा करके राजधानी उज्जयिनी आ गये। आकर धर्म के अनुसार नीतिपूर्वक प्रजा पालन को प्रमुखता देते हुए शासन किया। वह आज भी संसार में अच्छे कर्म से ही प्रसिद्ध हैं। उनके जीवन का ध्येय वाक्य था-‘सत्कर्म ही धर्म है।’

सत्कर्म एव धर्मः शब्दार्थाः

बलात् = बलपूर्वक। साधयितुम् = सिद्ध करने लिए। पार्वे = पास में। प्रजापालनपुरस्सरम् = प्रजापालन को प्रमुखता देते हुए। मरणासन्न = मरने के निकट। उद्भूतम् = उत्पन्न हुआ। पातयतु = गिराओ। एतच्छ्रुत्वा = ऐसा सुनकर।

MP Board Class 12th Physics Important Questions Chapter 3 Current Electricity

Current Electricity Important Questions

Current Electricity Objective Type Questions

Question 1.
Choose the correct answer of the following:

Class 12 Current Electricity Important Questions Question 1.
The flow of current through a conduction is due to:
(a) Protons
(b) Positive ions
(c) Free electrons
(d) Positive and negative ions.
Answer:
(c) Free electrons

Question 2.
The specific resistance of a wire depends upon:
(a) Length
(b) Diameter
(c) Mass
(d) Material.
Answer:
(b) Diameter

Physics Important Questions Class 12 MP Board 2023 Question 3.
A wire is stretched redouble its length. Its resistance will be:
(a) Halved
(b) Doubled
(c) One fourth
(d) Four times.
Answer:
(d) Four times.

Question 4.
The unit of specific resistance is :
(a) Ohm
(b) Ohm^-1
(c) Ohm meter
(d) Ohm^-1
Answer:
(c) Ohm meter

Question 5.
Which of the following is an oh-mic resistance:
(a) Junction transistor
(b) Transistor
(c) LED
(d) Copper wire.
Answer:
(d) Copper wire.

Current Electricity Class 12 Important Questions Question 6.
On increasing temperature the resistance of the which of the following decreases :
(a) Semiconductor
(b) Metal
(c) Electrolyte
(d) Alloy.
Answer:
(a) Semiconductor

Question 7.
The conductance of super conductor is :
(a) Infinite
(b) Very high
(c) Very low
(d) Zero.
Answer:
(a) Infinite

Question 8.
In the series combination of two resistance which quantity remains same :
(a) Only potential difference
(b) Only current
(c) Current and potential difference both
(d) Neither current nor potential difference.
Answer:
(b) Only current

12th Physics Chapter 3 Important Questions Question 9.
Electric cell is source of:
(a) Electrons
(b) Electrical energy
(c) Electric charge
(d) Electric current.
Answer:
(b) Electrical energy

Question 10.
The e.m.f. of a cell depends upon:
(a) Quantity of electrolyte filled in it
(b) Distance between the electrodes
(c) Size of electrodes
(d) Nature of electrolyte and electrodes.
Answer:
(d) Nature of electrolyte and electrodes.

Question 11.
Three cells each having e.m.f. E and internal resistance r are joined in series. One cell by mistake is joined in reverse order. The resultant e.m.f. and internal resistance will be:
(a) 3E, 3r
(b) E,3r
(c) E, r
(d) 3E,r
Answer:
(b) E,3r

Class 12 Physics Chapter 3 Important Questions With Answers Question 12.
The internal resistance of a cell can be decreased by :
(a) Decreasing the size of electrodes
(b) Increasing the distance between electrodes
(c) Decreasing the distance between electrodes
(d) None of these.
Answer:
(c) Decreasing the distance between electrodes

Question 13.
Two cells of e.m.f. and internal resistance E₁r₁ and E₂, r₂ respectively are joined in parallel. Their equivalent e.m.f. will be:
(a) E₁ + E₂
(b) E₁ – E₂
(c) \(\frac{\mathrm{E}_{1} r_{2}+\mathrm{E}_{2} r}{r_{1}+r_{2}}\)

(d) \(\frac{\mathrm{E}_{1}+\mathrm{E}_{2}}{r_{1}+r_{2}} \times r_{1} \cdot r_{2}\)
Answer:

(c) \(\frac{\mathrm{E}_{1} r_{2}+\mathrm{E}_{2} r}{r_{1}+r_{2}}\)

Current Electricity Imp Questions Question 14.
Potentiometer measures:
(a) Terminal voltage of cell
(b) Current in circuit
(c) e.m.f. of cell
(d) None of these.
Answer:
(c) e.m.f. of cell

Question 15.
In the null deflection position:
(a) No current flows through the galvanometer
(b) Current flows due to primary circuit through the galvanometer
(c) Current flows through galvanometer due to secondary circuit
(d) Nothing can be said.
Answer:
(a) No current flows through the galvanometer

Question 16.
The SI unit of the potential gradient is:
(a) ohm/cm
(b) volt
(c) volt cm
(d) volt/cm.
Answer:
(d) volt/cm.

Current Electricity Class 12 Numericals Question 17.
To increase the sensitivity of potentiometer:
(a) The potential difference across its wire should be high
(b) The length of its wire should be less
(c) The current through the wire should be high
(d) potential difference across its wire should be low and length of wire should be large.
Answer:
(d) potential difference across its wire should be low and length of wire should be large.

Question 18.
In the balance condition of potentiometer its resistance is :
(a) Zero
(b) Infinite
(c) Very small
(d) Very high.
Answer:
(d) Very high.

Question 2.
Fill in the blanks:

1. According to ……….. law, a if the physical conditions of a conductor remains unchanged then current flowing through it is directly proportional to the applied potential difference.
2. On increasing the length of a conductor its resistance ………..
3. On decreasing the area of cross-section of a conductor its resistance ………..
4. On increasing the temperature of a metallic conductor its resistance ………..
5. KirchhofFs first law is in accordance with the law of conservation of ………..
6. In series grouping of resistors the resistance but in parallel grouping the resistance ………..
7. With increase in length of potentiometer its sensitivity ………..
8. Meter bridge works on the principle of ………..
9. The specific resistance of alloys is ……….. and temperature coefficient of resistance is ………..
10. The Kirchhoff’s first law is in accordance with the law of conservation of ……….. while the second law is in accordance with law of conservation of ………..
11. On increasing the distance between electrodes of a cell, its internal resistance ……….. but on increasing area its internal resistance ………..
12. The phenomenon in which at low temperature the resistivity of a substance becomes zero is celled ………..
13. In Ohm’s law the V-I graph is a ………..

Answers:

1. Ohm’s
2. Increases
3. Increases
4. Increases
5. Charge
6. Increases, decreases,
7. Increases
8. Wheatstone bridge
9. High, low
10. Charge, energy
11. Increases, decreases
12. Super conductivity
13. Straight line.

Class 12 Physics Important Questions MP Board Question 3.
Match the Column:

Answer:

1. (e) Metre Bridge
2. (c) Law of conservation of charge
3. (b) Ohmic conductor
4. (a) Law of energy conservation
5. (d) Non-ohmic conductor

Answers:

1. (e) coulomb/sec. (or ampere).
2. (d) ampere/metre²
3. (b) joule/coulombx ampere
4. (c) ohm x meter
5. (a) ohm^-1

Answers:

1. (e) I²Rt
2. (a) ohm
3. (b) volt
4. (c) Electrical energy
5. (d) VI

Answer:

1. (c) Potential difference
2. (e) e.m.f.
3. (b) Vector quantity
4. (a) l₁/l₂
5. (d) \(\mathrm{R}\left(\frac{l_{1}}{l_{2}}-1\right)\)

Imp Questions Of Physics Class 12 MP Board Question 4.
Write the answer in one word / sentence:

1. What is the direction of electric current ?
2. What is the unit of specific resistance ?
3. Kirchhoffs first law is based upon which law ?
4. Kirchhoffs second law is based upon which law ?
5. On which principle does f&etre bridge works ?
6. Does Kirchhoffs law can be applied for both a.c. and d.c. currents ?
7. In a carbon resistance, there is green, violet, red and silver strip’s. What will be the resultant resistance.
8. What is the relation between 1 kWh and joule.
9. When the potentiometer is in equilibrium, what will be its resistance ?
10. The algebraic sum of current at a point is zero. Then what will be that point ?

Answer:

1. Opposite to flow of electrons,
2. Unit is ohm x meter
3. Laws of conservation of charges
4. Laws of conservation of energy
5. Wheatstone bridge
6. Yes
7. 57 x 10² ± 10% ohm
8. 1 kWh = 3.6 x 106 joule
9. Infinity
10. Junction point.

Current Electricity Very Short Answer Type Questions

Question 1.
To flow current in a conducting wire, how much charges is present in it ?
Answer:
On flowing current through the conducting wire, charges becomes zero.

Question 2.
Why Meter bridge is known by these name ?
Answer:
Because in Meter bridge, a wire of one meter length is used.

MP Board Class 12th Physics Important Questions Question 3.
Two bulb are marked as 25 W and 100 W. Whose resistance will be more ?
Answer:
R ∝ \(\frac {1}{V}\), Resistance of 25 W bulb will be more.

Question 4.
What do you mean by electric power. Write its unit ?
Answer:
In any electrical circuit the rate of energy decay is called as power. Its unit is watt.

Question 5.
On which factor relaxation time depend on ?
Answer:
It depends on nature of the material of the conductor.

Physics Important Questions Class 12 MP Board Question 6.
Why there is internal resistance in a cell ?
Answer:
Because, inside the cell, the motion of ions get obstacled by the collision of electrolytes molecules.

Question 7.
Why wire of Meter bridge is not made of copper wire ?
Answer:
Since the resistance of copper wire is very less and its temperature coefficient is more, therefore it is not used in Meter bridge.

Current Electricity Short Answer Type Questions

Question 1.
What is current ? Write its unit
Answer:
The electric current is defined as the rate of flow of charge through any section of a
Electric current = \(\frac {Total charge flowing}{ Time taken }\)
l = \(\frac {q}{ t}\) The SI unit of electric current is ampere (A).

MP Board Class 12 Physics Notes In English Question 2.
Is electric current a scalar or a vector quantity ? Give reason.
Answer:
Electric current is a scalar quantity. This is because laws of ordinary algebra are used to add electric current and laws of vector addition are not applied for the addition of electric current.

Question 3.
Define current density. Is it scalar or vector ? Write its unit.
Answer:
Current density at a point in a conductor is defined as the amount of current flowing per unit area of the conductor held perpendicular to the flow of current.
Current density J = \(\frac {Current (l)}{ Area (S)}\) (or) \(\frac {l}{ S}\)
Current density is a vector quantity. The SI unit of current density is ampere / meter².

Question 4.
What is drift velocity ? What is its value ?
Answer:
Drift velocity is defined as the average velocity with which the velocity free electrons with which they get drifted towards the positive terminal of the conductor under the influence of the external electric field. Its value is 10^-5ms^-1

Physics Important Questions Class 12 MP Board 2024 Question 5.
Is Ohm’s law applicable to all conductors ? Write conditions for its application
Or
Write the condition or under which Ohm’s law is not obeyed.
Answer:
No, Ohm’s law is not applicable for non – ohmic conductors, e.g., semiconductor diode, discharge tube, etc. It is valid only when there is no change in physical conditions of conductor i.e., temperature, length or mechanical strain etc.

Question 6.
Define specific resistance or resistivity. Write its unit and dimensional formula.
Answer:
Specific resistance or resistivity is numerically equal to the resistance offered by a conductor of unit length and unit cross-sectional area. Its unit is ohm – meter. It depends upon temperature. a’ ‘ The dimensional formula of specific resistance is [ML³T³A^-2].

Current Electricity Class 12 Question 7.
If you are given two wires of same material, having same length but different diameters, then which wire will have higher resistance and which will have high specific resistance ?
Answer:
The thin wire (having less diameter) will have higher resistance than that of thicker one as R ∝ \(\frac { 1 }{ A }\) Both wires will have same specific resistance as they are made of same material and specific resistance is characteristic of the material of wire.

Question 8.
It is easier to start motor car during summer rather than in winter. Why ?
Answer:
The internal resistance of the battery during summer is less as compared to winter, therefore more current can be drawn from the battery.

Question 9.
When a high power heater is connected to mains, bulbs become dim. Why ?
Answer:
All electrical appliances are connected in parallel at home, therefore when power heater is used, very high current passes through it. So, potential difference falls on wires connected through mains (V = E – lr), which results in decreasing intensity of bulbs used in circuit.

MP Board Class 12th Physics Imp Questions 2023 Question 10.
What do you measure by meter bridge ? When is its sensitivity maximum ?
Answer:
Resistances are measured by meter bridge. Sensitivity of meter bridge is maximum when the null point is obtained at the center of wire. Moreover, all the resistances must be of same order.

Question 11.
What is the wire of meter bridge made up of ?
Answer:
The wire of meter bridge is made of manganin or constantan because it has low temperature coefficient of resistance and high specific resistance.

Question 12.
How can the sensitivity of potentiometer be increased ?
Answer:
The sensitivity of the potentiometer will be more if the potential gradient is less. For this, V should be less but more than the emf of E. The length of potentiometer should be large.

Question 13.
Write two possible reasons for obtaining deflection on one side in the experiment of potentiometer.
Answer:
The two possible reasons are as follows:

• If the emf of secondary cell in primary circuit is less than the emf of primary cell in the secondary circuit.
• If positive terminals of all the cells are not connected to the same point.

Question 14.
What do you understand by potential gradient of a potentiometer ? Give its unit
Answer:

• Fall in potential per unit length is called potential gradient. If the length of wire be / and potential difference across the wire is v, then
• Potential gradient, K =\(\frac {V}{l}\) Its unit is volt / m.

Current Electricity Important Questions Question 15.
The potential gradient of potentiometer wire is doubled, what will happen to its null point ?
Answer:
Let emf of cell E is balanced at the length of –
E = Kl = l = \(\frac {E}{ K}\) = l ∝ \(\frac {1}{ K}\)
Hence, the null point will be at the half of the length.

Question 16.
Why the length of wire in potentiometer is more ?
Answer:
For greater sensitivity, the length of null point should be greater, for which the potential gradient should be smaller.
Potential gradient, K =\(\frac {V}{ l}\) To decrease K the value of l should be greater and F should be smaller. But, the value of V can be decreased up to a certain limit, hence the length is increased sufficiently.

Question 17.
Prove that potentiometer is an ideal voltmeter.
Answer:
When the potential difference between two points is measured by the voltmeter, then same current passes through the voltmeter, hence it measures little less than actual potential difference. But, when potentiometer measures the potential difference between two points, no current flows through it. Hence, it measures the accurate potential difference.

Physics Notes Class 12th MP Board Question 18.
A carbon resistance has bands blue, red and green respectively. What is its resistance ?
Answer:

.’. The value of carbon resistance is 62 x 105 ohm with tolerance ± 20% .

Question 19.
A carbon resistance has bands blue, green and red respectively. Write its resistance.
Answer:

The value of carbon resistance is 65 x 102 ohm with tolerance ± 20% .

Question 20.
A carbon resistance has bands red, blue and green. What is its resistance ?
Answer:

The value of carbon resistance is 26 x 105 ohm with tolerance ±20% .

Question 21.
In high tension battery internal resistance should be high. Why ?
Answer:
If high tension battery is used in a circuit having low external resistance, current may go beyond its safety limit. If internal resistance of the battery is high, then current cannot go beyond its safety limit. So, internal resistance of high tension battery must be high.

Class 12 Physics Chapter 3 Notes Question 22.
What is cause of end error in a meter bridge ?
Answer:
The end error in meter bridge is due to following reasons:
The zero mark of the scale provided along the wire may not start from the position where the bridge wire leaves the copper strip and 100 cm mark of the scale may not end at the position where the wire touches the copper strip. The resistance of the copper wire and metal strip of meter bridge has not been taken in account.

Question 23.
What is Ohm’s law ? On what factors does the resistance of a conductor depend upon ?
Answer:
If all the physical conditions of any conductor as length, temperature, etc. remain constant, then the current which flows through it is proportional to the potential difference applied across the ends of the conductor. If I is the current in conductor and V is the potential difference, then
V ∝ l = V=RI
Where, R is a constant, called resistance of conductor.
For factors affecting resistance of conductor:

• Length: The resistance of conductor is directly proportional to the length of the conductor i.e., R ∝ l
• Area of cross – section: The resistance of a conductor is inversely proportional to the area of cross – section of the conductor i.e., R ∝ \(\frac{l}{A}\)
• Temperature: Increase in temperature, increases the resistance.

Combining above two laws, we get,
R ∝ \(\frac{l}{A}\)
R = p \(\frac{l}{A}\)
Where, p= a constant, called specific resistance of the material of the conductor.
Specific resistance:
We have,
R = p \(\frac{l}{A}\)
Let l = 1 And A = l, then
R = P

Thus, the specific resistance of a material is defined by the resistance of unit length and unit area of cross-section of that material.
Unit: Now, p = \(\frac{R.A}{l}\)
= \(\frac{Unit of R x Unit of A}{Unit of l}\)
= \(\frac{\mathrm{ohm} \times \mathrm{m}^{2}}{\mathrm{m}}\) = Ohm x m = Ω x m

Question 24.
What are the possible errors of a meter bridge and how can they be removed ?
Answer:
The possible errors and their removal methods are:
1. It might happen that the wire is not uniform. To remove this error, balance point should be obtained at the-middle.

2. During’the experiment, it is assumed that the resistance of L shaped plates are negligible, but actually it is not so. The error created due to this is called end error. To remove this error, the resistance box and the unknown resistance must be interchanged and then the mean reading should be taken.

3. If the jockey is pressed for a long period of time, then it gets heated and its resistance changes. Hence, jockey must not be pressed for a long interval.

Current Electricity Class 12 Hsc Important Questions Question 25.
Prove Ohm’s law on the basis of free electron theory.
or
Determine the formula for specific resistance in terms of free electron density and relaxation time.
Answer:
Consider a conductor of length l, area of cross-section A having ‘n’ No. of free electron per unit volume. If potential difference V is applied between its opposite ends then,
\(\frac { V }{ l }\)
Force experienced by each free electron will be
F = eE or F = \(\frac { eV }{ l }\)
The acceleration produced on the electron of mass m is
a = \(\frac { F }{ m}\) = \(\frac { eV }{ ml }\)
If the time interval between two successive collisions (relaxation time) is r then drift velocity is
V_d = \(\frac {eEτ }{ m }\)
= \(\frac {eVτ }{ ml }\)
But cuurent l = neAv_d
l = neA\(\left(\frac{e V \tau}{m l}\right)=\frac{n e^{2} \tau}{m} \frac{A}{l}V\)
\(\frac { V }{ l }\) = \(\frac{m}{n e^{2} \tau} \cdot \frac{l}{A}\)
If temperature remains constant, then all terms on R.H.S. will be constant, so
\(\frac { V }{ l }\) = R
This is Ohm’s law. Where R is constant called resistance.
R = \(\frac{m}{n e^{2} \tau} \cdot \frac{l}{A}\) ……….(1)
But resistance is directly proportional to length / and inversely proportional to area of cross-section A of conductor
R ∝ \(\frac { l }{A }\)
R = p\(\frac { l }{A }\) ………..(2)
From eqns. (1) and (2),
p = \(\frac{m}{n e^{2} \tau}\)

Question 26.
Obtain a relationship between current and drift velocity.
Answer:
Consider a conductor of length /and area of cross-section A. Let n be the number of electrons per unit volume.
Volume of conductor = Al
Total number of electrons = nAl.
Total charge of conductor, Q = nAle.
The conductor is now joined to a cell of potential difference V.
So, the electrons get drifted towards the positive terminal of the cell. Let the drift velocity be v_d.
Time taken to cover length l is –

t = \(\frac{l}{v_{d}}\)
l = \(\frac { Q }{t }\)
= \(\frac { nAel }{t }\)
l = nAev_d,
( v_d = \(\frac { distance}{time }\) = \(\frac { l }{t }\)
This is the required relationship.

Question 27.
Write four differences between electromotive force and potential difference.
Or
Define e.m.f. of a cell and potential difference. Give differences between them.
Answer:
e.m.f.: Work done in flowing 1 coulomb charge through a circuit is called e.m.f. of a cell When the terminals are not connected to an external circuit, the maximum potential difference between the terminals is equal to the e.m.f. of the cell. Unit of e.m.f is volt.

Potential difference:
The difference of potentials of any two points of the circuit is called potential difference.
Or
Work done in bringing unit positive charge from one point to another is called potential difference between these two points.

Differences between e.m.f. and potential difference:
e.m.f.:

• The maximum potential difference bet-ween the terminals of a cell is called its e.m.f., when the cell is in open circuit.
• This term is used for the electric sources as generator, cell, battery, dynamo, etc.
• It is established even the circuit is off.
• e.m.f. does not depend upon the resistance of the circuit.

Potential difference:

• Difference of potential of any two points of the circuit is called potential difference.
• This term is used for any two points of the circuit.
• It is established till the current is flowing through the circuit.
• It depends upon the resistance of the circuit

Physics Chapter 3 Class 12 Important Questions Question 28.
What do you understand by internal resistance of a cell ? On what factors does It depend and how ?
Or
What do you mean by internal resistance of a cell ? Write the factors affecting it.
Answer:
The resistance offered by the electrolyte of the cell during the flow of current inside the cell is called its internal resistance.
The following factors affect the internal resistance:

• Distance between the electrodes:
  As the distance increases, the internal resistance increases.
• Area of the immersed electrodes:
  As the area increases, the internal resistance decreases.
• Concentration of the electrolyte:
  As the concentration is more, the internal resistance is more.
• Temperature:
  The increase of temperature, decreases the internal resistance.

Question 29.
Establish the relationship between internal resistance of a cell, e.m.f. and extecoafresistance.
Or
Establish the relationship between the electromotive force, potential difference and internal resistance of a cell.
Answer:

Let e.m.f. of a cell be E and its internal resistance is r. If current / is flowing through a resistance R, then by Ohm’s law,
l = \(\frac { V}{R }\) ……………(1)
Where, V is the potential difference across the resistance. Now, total e.m.f. of the circuit = E and total resistance of the circuit = R + r
l = \(\frac {E}{R+r }\)
By eqns. (1) and (2), we get
\(\frac {V}{R}\) = \(\frac {E}{R+r }\)
VR + Vr = ER
Vr = ER – VR
r = \(\frac {R(E – V)}{V}\)
r = R(\(\frac {E}{V}\) – 1)
This is the required relationship.

Question 30.
Explain Kirchhoff’s laws of distribution of current through the different conduct^£
Or
Write down the Kirchhoff’s laws relating to the distribution of electric current.
Or
State and explain Kirchhoff’s laws.
Answer:
Kirchhoff’s laws:
(i) The algebraic sum of current meeting at any junction in a circuit is zero. In this law, the currents flowing towards the junction are considered as positive and those flowing  way from the junction as negative. As shown in the figure, we have

i₁ – i₂ – i₃ – i₄ – i₅ = 0
i₁ + i₄ = i₂ + i₃ + i₅

(ii) in any closed mesh (or loop) of an electrical circuit, the algebraic sum of the product of the currents and resistances is equal to the total e.m.f. of the mesh. If we go along the direction of conventional current, the potential difference will be taken as negative and opposite to it will be positive. Inside the cell, if we move from low to high potential, along the direction of conventional current, the e.m.f. will be positive.

For loop 1,
E₂ – i₂R₂ – (i₁ + i₂ )R₃ = 0
or E₂ = i₂R₂ + (i₁ + i₂ )R<sub>3

For loop 2,
i2R2 – E2 – i1R1 + E1 = 0
or E1 – E2 = i1R1 – i2R2</sub>

Question 31.
Derive the principle of Wheatstone bridge by Kirchhoffs law.
Or
Explain the principle of Wheatstone bridge and obtain the expression for balance condition by Kirchhoff s laws.
Answer:
Principle of Wheatstone bridge:
Four resistances P, Q, R and S are connected to form a quadrilateral ABCD. A cell E is connected across the diagonal AC and a galvanometer across BD. When the current is flown through the circuit and galvanometer does not give any deflection, then the bridge is said to be balanced. In this condition,
\(\frac { p }{ Q }\) = \(\frac { R }{ S }\)
This is the principle of Wheatstone bridge.

Formula derivation:
Let the current i is divided into two parts and i2, flowing through P, Q and R, S respectively. In the position of equilibrium, the galvanometer shows zero deflection, i.e,, the potential of B and D will be equal. In the closed mesh ABDA, by Kirchhoffs second law, we get

i₁P – i₂R = 0
or i₁ P = i₂R ……….(1)
Similarly, in the closed mesh BCDB, we have
i₁Q – i₂S = 0
or i₁Q = i₂S ……….(2)
Dividing eqn. (1) by eqn. (2), we get
\(\frac{i_{1} P}{i_{1} Q}=\frac{i_{2} R}{i_{2} S}\)
\(\frac{P}{Q}\) = \(\frac{R}{S}\)
This is Wheatstone bridge principle or principle of balance.

Question 32.
Explain the principle of potentiometer.
Answer:
Let AB be the wire of potentiometer of length L. A storage cell C, a key K and a rheostat Rh are connected in series with the resistance wire.
When the key is inserted, the current starts flowing through the circuit and a potential difference is established between A and B, let it is V.
∴ Potential gradient, P = \(\frac{V}{L}\)

Now, the +ve terminal of the experimental cell is connected to A and -ve to galvanometer which is connected to jockey J. When jockey is touched near A, we get the deflection in one side and if it is touched near B, the deflection is in other side. Thus, a point on AB is found such that galvanometer gives no deflection. Hence, no current flows through the galvanometer and get the null point J.

In this balanced position,
e.m.f. of the cell = Potential difference between A and J
If the length of AJ = l.
∴ E = pl
Thus, knowing p and E can be calculated.
This is the principle of potentiometer.

Question 33.
On what factors the resistance of a wire or conductor depends ? Define specific resistance and write its unit.
Or
On what factors the resistance of a conductor depends and how ?
Answer:
The factors affecting the resistance are:

• Length: The resistance of conductor is directly proportional to the length of the conductor i.e., R ∝ l
• Area of cross – section: The resistance of a conductor is inversely proportional to the area of cross – section of the conductor i.e., R ∝ \(\frac{l}{A}\)
• Temperature: Increase in temperature, increases the resistance.

Combining above two laws, we get,
R ∝ \(\frac{l}{A}\)
R = p \(\frac{l}{A}\)
Where, p= a constant, called specific resistance of the material of the conductor.
Specific resistance:
We have,
R = p \(\frac{l}{A}\)
Let l = 1 And A = l, then
R = P

Thus, the specific resistance of a material is defined by the resistance of unit length and unit area of cross-section of that material.
Unit: Now, p = \(\frac{R.A}{l}\)
= \(\frac{Unit of R x Unit of A}{Unit of l}\)
= \(\frac{\mathrm{ohm} \times \mathrm{m}^{2}}{\mathrm{m}}\) = Ohm x m = Ω x m

Question 34.
What do you understand by specific resistance or resistivity of a conductor ? Give its unit and dimensional formula.
Answer:
For specific resistance of a conductor: Refer Short Answer: Type Q. No. 33(iii). Dimensional formula of specific resistance is determined as:
As p = \(\frac{RA}{l}\)
[p] = \(\frac{[R][A]}{[l]}\)
[R] = \(\frac{[V]}{[l]}\) = \(\frac{W/P}{[l]}\) = \(\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-2}\right] /[\mathrm{AT}]}{[\mathrm{A}]}\)
= \(\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-3} \mathrm{A}^{-1}\right]}{[\mathrm{A}]}\) = [ML²T^-3A^-2
[p] = \(\frac{\left[\mathrm{ML}^{2} \mathrm{T}^{-3} \mathrm{A}^{-2}\right]\left[\mathrm{L}^{2}\right]}{[\mathrm{L}]}\)
= \(\left[\mathrm{ML}^{3} \mathrm{T}^{-3} \mathrm{A}^{-2}\right]\)

Question 35.
Give differences between specific resistance and resistance.
Answer:
Differences between specific resistance (resistivity) and resistance:

Specific resistance:

• It is defined by the resistance offered by the conductor of unit length and unit area of cross – ection.
• Its unit is ohm x m.
• It does not depend upon the length and area of cross – ection.

Resistance:

• The hindrance offered by a conductor to the flow of current is called electrical resistance.
• Its unit is ohm.
• It depends upon the length and area of cross-section of the conductor.

Question 36.
Evaluate It and I₂ as shown in the figure.

Solution:
Equivalent figure of fig. (a) is fig. (b)
In closed current BEFC, by Kirchhoff s second law,
2l₁ – 6I₂ = 0
or I₁ = 3I₂ …….(1)
In closed path ABCD
2I₁ + l(l₁ + l₂) = 10
3I₁ + I₂ = 10 …….(2)

From eqns. (1) and (2),
3 x 3I₂ + I₂ = 10
or 9I₂ + I₂ = 10
or I₂ = 1 amp.
Putting the value of 12 in eqn. (1) we get,
I₁ = 3 x l = 3 amp

Question 37.
Three resistors R₁ R₂ and R₃ are connected in series. Obtain the expression for the equivalent resistance.
Answer:
The series combination of three resistors having resistances R₁ R₂ and R₃ are shown in figure.

When this combination is connected to a cell (C) of e.m.f. E volt, then I current flows through combination.
Let the potential difference across R₁ R₂ and R₃ be V₁, V₂ and V₃ respectively. Then, by Ohm’s law,
V₁= IR₁ V₂ = IR₂ and V₃ = IR₃ .
If potential difference across A and B be V, then
V = V₁ + V₂ + V₃
or V = IR₁ + IR₂ + IR₃
V = I(R₁ + R₂ + R₃) ……(1)
If equivalent resistance of this combination be R, then
V = IR …..(2)
From eqns. (1) and (2), we get
1R = I(R₁ + R₂ + R₃)
or R = R₁ + R₂ + R₃ …(3)
Thus, when a number of resistances are connected in series with each other, the equivalent resistance of the combination is equal to the sum of their individual resistances.

Question 38.
Find out equivalent resistance of the three resistances R₁ ,R₂ and R₃ connected in parallel combination.
Answer:
The resistances R₁ ,R₂ and R₃ are connected in parallel as shown in adjacent figure. As all the resistances are connected between two points A and B, hence the potential difference between two points will be same for all. Let it be V. Let I be the total current. This current is divided into three parts at point A. If through R₁, R₂ and R₃; currents I₁ I₂ and I₃ are respectively flowing. Then,

I = I₁ + I₂ + I₃
By Ohm’s law,
I₁ = \(\frac{V}{R_{1}}\), I₂ = \(\frac{V}{R_{2}}\), I₃ = \(\frac{V}{R_{3}}\)
I = \(\frac{V}{R_{1}}\) + \(\frac{V}{R_{2}}\) + \(\frac{V}{R_{3}}\) ……….(1)
If the equivalent resistance of this combination be R, then
I = \(\frac{V}{R}\) ……….(2)
From eqns. (1) and (2), we have
\(\frac{V}{R}\) = \(\frac{V}{R_{1}}\) + \(\frac{V}{R_{2}}\) + \(\frac{V}{R_{3}}\)
\(\frac{1}{R}\) = \(\frac{1}{R_{1}}\) + \(\frac{1}{R_{2}}\) + \(\frac{1}{R_{3}}\) …………(3)
Thus, if a number of resistances are connected in parallel, the reciprocal of equivalent resistance of the combination is equal to the sum of the reciprocals of their individual resistances.

Current Electricity Long Answer Type Questions

Question 1.
Describe an experiment to compare the e.m.f. of two cells by potentiometer onTne following points :

1. Circuit diagram
2. Derivation of formula
3. Two precautions.

Or
Describe an experiment to compare the e.m.f. of two cells using a potentiometer under the following heads :

1. Labelled diagram of electric circuit
2. Formula used
3. Observation table
4. Two main precautions.

Answer:
1. Circuit diagram:
AB → Potentiometer wire
B₁ → Lead accumulator
K₁ → Plug key
Rh → Rheostat
E₁ E₂ → Experimental cells
K₂ → Two ways key
G → Galvanometer
J → Jockey.

2. Derivation of formula:
Let the first cell is having e.m.f. E, and the balancing point is obtained at distance l₁. Then by the principle of potentiometer,
E₁ = pl₁ ……(1)
Where, p is potential gradient.
Let E₂ is the e.m.f. of second cell whose balancing point is at l₂, then
E₂ = pl₂ ……(2)
Dividing eqn. (1) by eqn. (2), we get,
\(\frac{E_{1}}{E_{2}}=\frac{pl_{1}}{pl_{2}}=\frac{l_{1}}{l_{2}}\)

3. Observation tab:

4. Precautions:
The e.m.f. of lead accumulator should be greater than that of experimental cells. All the positive terminals should be connected to one point

Question 2.
Determine the internal resistance of a cell by potentiometer on the following points:

1. Labelled circuit diagram
2. Derivation of formula
3. Precautions.

Or
Determine an experiment to find out the internal resistance of a cell by potentiometer under the following points :

1. Labelled circuit diagram
2. Formula derivation
3. Observation table
4. Precautions (any two).

Answer:
1. Circuit diagram:
AB → Potentiometer wire
C → Storage cell
K → Plug key
Rh → Rheostat
E → Experimental cell
R.B. → Resistance box
G → Galvanometer
J → Jockey.

2. Formula derivation:
Let the e.m.f. of the cell is E and its internal resistance is r. When it is connected to external resistance R, its potential difference is V, then

r = R (\(\frac{E}{V}\) – 1) ……..(1)
When R is not connected, then let the balance point is found at a distance l₁ from A.
∴ E = pl₁
Similarly, when R is also connected, then let the balanced point is obtained at a distance l₂ from A.
∴ V = pl₂
Now, putting these values in eqn. (1), we get
r = R \(\left(\frac{\rho l_{1}}{\rho l_{2}}-1\right)\) = R \(\left(\frac{l_{1}}{l_{2}}-1\right)\)
This is the required formula

3. Observation table :

4. Precautions:

• The e.m.f. of the storage cell (C) must be greater than that of experimental cell (£).
• All the positive terminals must be connected to a single point

Question 3.
How are the cells connected in mixed combination ? Derive an expression for current flowing through external resistance.
Or
Prove that the current in the external circuit for a mixed combination of cells is maximum, when the internal resistance of the combination equals the external resistance.
Answer:
In this combination, series connections of equal number of cells are connected in parallel and finally providing one +ve and one -ve terminal as shown in Fig. (a). These two terminals are connected to the ends of an external resistance R so that current starts flowing through the resistance.

Suppose n cells are connected in series combination and let m be the number of such series connected in parallel. Let the e.m.f. of each cell be E and internal resistance be r. Total e.m.f. of each series combination i.e, for a row, will be nE and total internal resistance will be nr. Hence, the equivalent becomes as given in Fig. (b).

As all the cells are now in parallel, so total e.m.f. of the combination will also be nE and total internal resistance will be r’. This is shown in Fig. (c). As each resistor is of value nr and nr such resistors are connected in parallel, hence total internal resistance r’ will be calculated as:

\(\frac{1}{r’}\) = \(\frac{1}{nr}\) + \(\frac{1}{nr}\) + ………. + m terms = \(\frac{m}{nr}\) or r’ = \(\frac{nr}{m}\)
As R is the external resistance, then r’ and R are in series. Hence, total resistance of combination becomes,
R_total = R + r’ = R + \(\frac{nr}{m}\)
The current through the circuit will be given by Ohm’s law as

l = \(\frac{Total e.m.f.of battery }{Total resistance of circuit}\)

l = \(\frac{n \mathrm{E}}{\mathrm{R}+\frac{n r}{m}}\) = \(\frac{m n \mathrm{E}}{m \mathrm{R}+n r}\) ……….(1)

This is the expression for the current. For the current to be maximum, mR+nr must be minimum

mR + nr = \((\sqrt{m \mathrm{R}})^{2}+(\sqrt{n r})^{2}-2 \sqrt{m \mathrm{R}} \sqrt{n r}+2 \sqrt{m \mathrm{R}} \sqrt{n r}\) (as subtracting and adding \(2 \sqrt{m \mathrm{R}} \sqrt{n r}\))

= \((\sqrt{m \mathrm{R}}-\sqrt{n r})^{2}+2 \sqrt{m \mathrm{R}} \times \sqrt{n r}\)

But \((\sqrt{m \mathrm{R}}-\sqrt{n r})^{2}\) cannot be negative, as it is a perfect square.

mR + nr will be minimum, only when \((\sqrt{m \mathrm{R}}-\sqrt{n r})^{2}\) = 0

or \(\sqrt{m \mathrm{R}}-\sqrt{n r}\) = 0

\(\sqrt{m \mathrm{R}}\) = \(\sqrt{n r}\) ⇒ mr = nr ………(2)
∴ R = \(\frac{n r}{m}\) ………..(3)

Hence, the external resistance should be equal to the internal resistance of the battery. Under this condition the current supplied by the battery to the resistance R is maximum. The value of maximum current can be obtained by putting eqn. (2) in eqn. (1).

∴ I_max = \(\frac{mnE}{mR + mR}\)

= \(\frac{mnE}{2mR}\) = \(\frac{nE}{2R}\)
On substituting mR = nr, we get

I_max = \(\frac{mnE}{nr + nr}\) = \(\frac{mnE}{2nr}\)

= \(\frac{mE}{2r}\) .

Question 4.
explain the experiment of determining the unknown resistance of a wire with he help of meter bridge on the following points:

1. Electrical circuit
2. Principle.

Or
Describe an experiment to determine the unknown resistance by meter bridge on following points :

1. Circuit diagram
2. Description of apparatus
3. Observation table
4. Precautions.

Or
What are the possible errors in performing the experiment with metre bridge and how they can be removed ?
Answer:
1. Circuit diagram:

Where ,
AC→ wire
E → cell
R → resistance box
S → unknown resistance,
G → galvanometer
K → key.

2. Description of apparatus:
It consists of 1 meter long manganin or constant resistance wire, fixed on a wooden plank. Thick steel or brass strips are fixed as shown in the figure with two gaps. In one gap, resistance box R and in other, unknown resistance S are connected. A, C and D are terminals. A jockey slides on the wire.

3. Formula derivation or principle:
Meter bridge consists of 1 m long constant-an wire AC, fixed on a wooden plank. In two gaps resistance R and unknown resistance S are connected. Galvanometer G is connected between B and D, where B is sliding point i. e., jockey.
A cell E, with a plug key K is connected between A and C.
Let null point is obtained at a distance l from A.
∴ AB = l cm
and BC = (100 – l)cm
If x be the resistance per unit length, then Resistance offered by AB is P = lx
and resistance offered by BC is Q = (100 – l)x
Now, by the principle of Wheatstone bridge,

\(\frac{P}{Q}\) = \(\frac{R}{S}\)

\(\frac{lx}{(100 – l)}\)

S = \(\frac{R(100 – l)}{l}\)
Wheatstone bridge is sensitive when all the four resistances are of same order. Hence, metre bridge is also suitable for the resistance of same order.

4. Observation table:

5. Precautions:

• The connection should be tight.
• The plugs of R.B. should not be loose.
• The current should be passed only when readings are to be taken.
• Jockey should not be rubbed with the wire.

Possible errors and their removal:
1. It might happen that the wire is not uniform. To remove this error, balance point should be obtained at the-middle.

2. During’the experiment, it is assumed that the resistance of L shaped plates are negligible, but actually it is not so. The error created due to this is called end error. To remove this error, the resistance box and the unknown resistance must be interchanged and then the mean reading should be taken.

3. If the jockey is pressed for a long period of time, then it gets heated and its resistance changes. Hence, jockey must not be pressed for a long interval.

Question 5.
How are cells connected in series ? Derive an expression for current flowing through outer circuit. When is this combination useful ?
Answer:
In this combination, the -ve terminal of one cell is connected to the + ve terminal of second cell, whose -ve terminal is connected to the +ve terminal of third one and so on as shown in Fig. (a). Let n cells each of e.m.f. E and internal resistance r be connected in series through an external resistance R, then

Total e.m.f. of this combination = n E ’ and total internal resistance
r’ = r + r + r +…………….+ n times
= nr
The series combination of n cells is equivalent to a single cell having e.m.f. «E and internal resistance nr. This is shown in Fig. (b).
Now nr and external resistance R are in series. So, the equivalent resistance of circuit becomes R + nr Applying Ohm’s law, current flowing through the circuit is given by –
l = \(\frac{e.m.f. of battery }{Total resistance}\)
= \(\frac{nE }{nr + R}\)
This is the expression for current flowing through the external resistance R. If r « R, then nr + R * R.
Hence, from eqn. (1), we get
l = n. \(\frac{E }{R}\)
= n x Current flowing through each cell.

Utility:
Thus, if the internal resistance of each cell is negligible as compared to external resistance, the current flowing through the battery is n times the current supplied by each cell.
So, the cells should be connected in series only when the internal resistance of each cell is much less than the external resistance.

Question 6.
n cells are connected in parallel combination with internal resistance. Derive an expression for current flowing through external resistance. When is this combination useful?
Answer:
Let n cells be connected in parallel and e.m.f. of each cell be E and internal resistance be r. Since, all the positive terminals of cells connected to point A and -ve terminals connected to the point B, the total e.m.f. of battery will be E (because in parallel combination potential difference remains same).

Since the cells are connected in parallel, therefore their internal resistance will also be in parallel combination. Let the equivalent internal resistance be r’
∴ \(\frac{1 }{r’}\) = \(\frac{E }{R}\) + \(\frac{E }{R}\) + …………. + n times \(\frac{n }{r}\)
or r’ = \(\frac{r }{n}\)
Hence, the equivalent circuit becomes

Where, r’ is the total resistance of the combination of cells. As R and r’ are in series, so the total resistance of the circuit becomes
= R + r’ = R + \(\frac{r }{n}\)
Current through external resistance R will be given by Ohm’s law as
l = \(\frac{Total e.m.f.of battery }{Total resistance of circuit}\)
l = \(\frac{E}{R+\frac{r}{n}}\) = \(\frac{n E}{n R+r}\)
This is the required expression.
If R << r, then nR + r≈r
Now, from eqn. (1), we have
I = n \(\frac{E}{r}\) = n x Current given by one cell .

Utility:
Thus, if internal resistance of each cell is much greater than the external resistance, then the current flowing through the battery is n times the current through each cell i.e., maximum current is obtained. So, the cells should be connected in parallel when internal resistance of each cell is much greater than the external resistance.

Current Electricity Numerical Questions

Question 1.
The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery? (NCERT)
Solution:
Given: E = 12 V; r – 0.4 Ω
Current l = \(\frac{E}{R+r}\)
For current to be maximum R = 0
I_max = \(\frac{E}{r}\) = \(\frac{12}{0.4}\) = 30A

Question 2.
A battery of emf 10V and internal resistance 3) is connected to a resistor. If the current in the circuit ¡s 0.5A. What is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed? (NCERT)
Solution:
Given:
E =10V; r = 3Ω l = 0.5A

l = \(\frac{E}{R + r}\)
R+r = \(\frac{E}{l}\)
R = \(\frac{E}{l}\) – r = \(\frac{10}{0.5}\) – 3 = 20 – 3 = 17Ω
Now, V = E – lr = 10 – 0.5 x 3 = 10 – 1.5 = 8.5V

Question 3.
(a) Three resistance IΩ, 2Ω and 3Ω are combined in series. What is the total resistance of the combination ?
(b) If the combination is connected to a battery of emf 12V and negligible internal resistance, obtain the potential drop across each resistor. (NCERT)
Solution:
(a) Given:
R₁ = IΩ; R₂ = 2Ω; R₃ = 3Ω
R₂ = R₂ + R₂ + R₂ = 1 + 2 + 3 = 6Ω .
(b) E = 12V; r = 0
l = \(\frac{E}{R+r}\) = \(\frac{E}{R_{S}+0} = \frac{12}{6}\)
I = 2A
Potential difference across (R₁)
V₁ = IR₁ = 2 x l = 2V
Potential difference across (R₂)
V₂ = IR₂ = 2 x 2 = 4V
Potential difference across (R₃)
V₃ = IR₃ = 2 X 3 = 6V.

Question 4.
(a)Three resistors 2Ω, 4Ω and 5Ω are combined in parallel. What is the total resistance of the combination ?
(b) If the combination is connected to a battery of emf 20V and negligible internal resistance, determine the current through each resistor and the total current drawn from the battery. (NCERT)
Solution?:
(a) Given:
R₁ = 20; R₂ = 40; R₃ = 50
Equivalent resistance:
\(\frac{1}{R_{p}}\) = \(\frac{1}{R_{1}}\) + \(\frac{1}{R_{2}}\) + \(\frac{1}{R_{3}}\)
= \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{5}\) = \(\frac{10+5+4}{20}\) = \(\frac{19}{20}\)
∴ \(\frac{1}{R_{p}}\) = \(\frac{20}{19}\) Ω

(b) E = 20V; r = 0
Current through different resistors,
l₁ = \(\frac{E}{R_{1}}\) = \(\frac{20}{2}\) = 10A
l₂ = \(\frac{E}{R_{2}}\) = \(\frac{20}{4}\) = 5A
l₃ = \(\frac{E}{R_{3}}\) = \(\frac{20}{5}\) = 4A
Total current drawn
l = l₁ + l₂ + l₃ = 10 + 5 + 4 = 19A

Question 5.
At room temperature (27°C) the resistance of a heating element is 100Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that temperature coefficient of the resistor material is 1.70 x 10^-4 °C^-1 ? (NCERT)
Solution:
Given:
t₁ = 27 °C
R₁ = 1000
R₂ = 1170
α = 1.70 x 10^-4 °C^-1
We know that, α = \(\frac{R_{2}-R_{1}}{R_{1}\left(t_{2}-t_{1}\right)}\)
t₂ – t₁ = \(\frac{R_{2}-R_{1}}{R_{1} \alpha}\)
= \(\frac{117-100}{170 \times 10^{-4} \times 100}\)
t₂ – t₁ = 1000
t₂ = 1000 + t₁ = 1000 + 27 = 1027 °C

Question 6.
A negligible small current is passing through a wire of length 15 m and uniform cross – section 6.0 x 10^-7 m² and its resistance is measured to be 5.0ΩWhat is the resistivity of the material at the temperature of the experiment ? (NCERT)
Sol. Given:
l = 15m; A = 6.0 x l0^-7m²; R = 5.0Ω
using R = \(\frac { pl }{ A }\)
p = \(\frac {RA}{ l }\) = \(\frac{5 \times 6 \times 10^{-7}}{15}\) = 2.0 x l0^-7Ωm

Question 7.
Determine the current in each branch of the network shown in figure. (NCERT)

solution:
\(\frac {P}{ Q }\) = \(\frac {10}{5}\) = \(\frac {2}{ 1 }\) and \(\frac {R}{S }\) = \(\frac {5}{ 10 }\) = \(\frac {1}{ 2 }\).
i.e \(\frac {P}{ Q}\) = \(\frac {R}{ S}\)

Therefore, bridge is not in equilibrium. Current flowing through different branches is as shown in fig. Now applying Kirchhoffs loop rule.

For loop ABDA
10I₁ + 5I₂ – 5(I – I₁) = 0
3I₁ + l₂ – I = 0 ……(1)

For loop BCDB.
5(I₁ – I₂) – 10 (I – l₁ + I₂) – 5I₂ = 0
or I₁ – I₂ – 2 (I – I₁ + I₂) I₁ = O
or 3I₂ – 4I₂ – 2I = 0 …..(2)
By eqns. (1) and (2), 5I₂ + I = 0
I = – 5I₂
I₂ = – \(\frac {1}{ 5 }\) I …….(3)
And I₁ = \(\frac {2}{ 5 }\) I …….(4)

For loop ADCA,
5(I – I₁) + 10 (1 – l₁ + l₂) + 10I = 10
5I – 3I + 2I₁ = 2

Question 8.
Find out effective resistance between points A and B for circuit given below:

Solution:
\(\frac {1}{ R}\) = \(\frac {1}{ 2 + 2 + 2 }\) + \(\frac {1}{ 3 }\)

\(\frac {1}{ R}\) = \(\frac {1}{6}\) + \(\frac {1}{3}\)

\(\frac {1}{ R }\) = \(\frac {1 + 2}{ 6 }\) = \(\frac {3}{6}\)
R = \(\frac {6}{3}\) = 2Ω

Question 9.
Find out effective resistance between points A and B for circuit given below:

\(\frac{1}{\mathrm{R}_{1}}\) = \(\frac {1}{ 4}\) + \(\frac {1}{ 4}\) = \(\frac{1}{\mathrm{R}_{1}}\) = \(\frac {2}{ 4}\)

R₁ = \(\frac {4}{2}\) = 2Ω

\(\frac{1}{\mathrm{R}_{2}}\) = \(\frac {1}{ 6}\) + \(\frac {1}{ 6}\) =\(\frac{1}{\mathrm{R}_{2}}\) = \(\frac {2}{ 6}\)

R₂ = \(\frac {6}{ 2}\) = 3Ω
Effective resistance R = R₁ + R₂
R = 2 + 3 = 5Ω

Question 10.
10⁹ electrons flows from point A to B in 10^-3 second. Find out magnitude and direction of electric current.
Solution:
Formula: I = \(\frac {ne}{t}\)
Given: n = 10⁹ t = 10^-3 sec.
Putting the given value in the formula, we get
I = \( \frac{10^{9} \times 16 \times 10^{-19}}{10^{-3}}\)
or l = 1.6 x l 0^-7 ampere.
Direction of electric current will be from B to A. Ans.

Question 11.
150 m A current flows through a conductor. How many electrons will flow through it in 20 sec ?
Solution:
Given:
I = 150 mA = 0.15 A, t = 20 sec.
Formula:
n = \(\frac {It}{e}\) = \(\frac{0.15 \times 20}{1 \cdot 6 \times 10^{-19}\)

n = \(\frac{3}{1 \cdot 6} \times 10^{19}\) = 1.875 x 10¹⁹

Question 12.
200 mA current flows through a conductor. How many electrons will flow through it in 5 sec ?
Solution:
Given:
I = 200 mA = 0.2 A, t = 5 sec
Formula:
n = \(\frac {It}{e}\) = \(\frac{0\cdot 2\times 5}{1\cdot 6\times 10^{-19}}\)

n = \(\frac{1}{1 \cdot 6} \times 10^{19}\)
n = 0.625 x 10¹⁹

Question 13.
The length of a wire becomes twice when stretched. How many times resistance will increase ?
Solution:
On stretching the wire, radius of the wire will decrease but volume will remain same. Let l₁ and r₁ be initial length and radius of wire respectively. On stretching, the length becomes l₂ say and radius r₂
Initial volume = Final volume

From the formula, R = p \(\frac {1}{A}\)

Putting the value of \(\left(\frac{r_{2}}{r_{1}}\right)^{2}\) from eqn. (1) in eqn. (2), we get

As per the question, l₂ = 2l₁
\(\frac{R_{1}}{R_{2}}\) = \(\left(\frac{l_{1}}{2 l_{1}}\right)^{2}\) = \(\frac {1}{4}\)
R₂ = 4R₁
Hence, the resistance of wire is increased four times.

Question 14.
The ratio of area of cross-section of two wires made of same metal and equal length is 2 : 1. It applied potential difference between their ends are same then what will be the ratio of current among them ?
Solution:

Question 15.
The ratio of length of two wire, made of same metal and equal area of cross-section is 2 : 1. If the applied potential difference between their ends are same, then what will be the ratio of current following among them ?
Solution:

Question 16.
125 mA current flows through a lamp in 10 sec. How many electrons will flow across it ?
Solution:
Given:
I = 125 mA = 0.125 A. t = 10 sec.

MP Board Class 12th Physics Important Questions

In this article, we will share MP Board Class 11th Biology Solutions Chapter 3 वनस्पति जगत Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 11th Biology Solutions Chapter 3 वनस्पति जगत

वनस्पति जगत NCERT प्रश्नोत्तर

Shaival Ke Vargikaran Ka Kya Aadhar Hai प्रश्न 1.
शैवालों के वर्गीकरण का क्या आधार है?
उत्तर:
शैवालों का वर्गीकरण निम्नलिखित लक्षणों के आधार पर किया गया है –

1. प्रकाश – संश्लेषी वर्णक का प्रकार
2. अन्य वर्णक के प्रकार
3. संचित भोज्य पदार्थ का प्रकार
4. सूकाय का प्रकार।

प्रश्न 2.
लिवरवर्ट, मॉस, फर्न, जिम्नोस्पर्म तथा एंजियोस्पर्म के जीवन – चक्र में कहाँ और कब निम्नीकरण विभाजन होता है ?
उत्तर:
लिवरवर्ट, मॉस, ‘फर्न एवं जिम्नोस्पर्म में निम्नीकरण विभाजन (Reduction division) बीजाणु (Spore) निर्माण की प्रक्रिया के समय बीजाणु मातृ कोशिका (Spore mother cell) में होता है। एंजियोस्पर्म में निम्नीकरण विभाजन पुंकेसर के परागकोष में परागण (Pollen grain) निर्माण के समय तथा भ्रूणकोष (Embryo sac) के निर्माण में अंडप (Ovule) के द्वारा किया जाता है।

Vanaspati Jagat In Hindi प्रश्न 3.
पौधों के तीन वर्गों के नाम लिखिए, जिनमें स्त्रीधानी होती है। इनमें से किसी एक के जीवन-चक्र का संक्षिप्त वर्णन कीजिए।
उत्तर:
पौधों के निम्नलिखित तीन वर्गों में स्त्रीधानी (Archegonia) पायी जाती है –

1. ब्रायोफाइटा
2. टेरिडोफाइटा एवं
3. जिम्नोस्पर्म।

टेरिडोफाइटा (Pteridophyta) का जीवन – चक्र:
टेरिडोफाइटा के जीवन-चक्र में द्विगुणित तथा अगुणित अवस्थाएँ स्पष्ट रूप से आती हैं, इस क्रिया को पीढ़ी का एकान्तरण (Alternation of generation) कहते हैं। पीढ़ी एकान्तरण करने वाले जीवों में युग्मकोद्भिद (Gametophytic phase) तथा बीजाणुद्भिद (Sporophytic phase) क्रम से बनते रहते हैं। उदाहरण – साइकस। साइकस का मुख्य पौधा स्पोरोफाइट होता है। यह एकलिंगी पौधा है, जिसमें दो प्रकार की जड़ें, शल्कपत्र व बीजाणुधानियाँ पायी जाती हैं।

नर पौधे में नर शंकु तथा मादा पौधे में मादा शंकु के स्थान पर मेगास्पोरोफिल पाया जाता है। मेगास्पोरोफिल में 4-6 बीजाण्ड होते हैं, इनमें महाबीजाणु मातृकोशिका में अर्धसूत्री विभाजन के फलस्वरूप मादा युग्मकोद्भिद बनता है। नर बीजाणुधानी में अर्द्धसूत्री विभाजन में परागकण बनते हैं, जो बीजाणुधानी के फटने से हवा में मुक्त हो जाते हैं। साइकस में युग्मकोद्भिद पूर्णतः स्पोरोफाइट पर आश्रित होता है यह अल्पकालिक व परजीवी होता है। स्पोरोफाइट ही जीवन-चक्र में विकसित रूप में होता है।

प्रश्न 4.
निम्नलिखित की सूत्रगुणता (Ploidy) बताइए-मॉस के प्रथम तन्तुक कोशिका, द्विबीजपत्री के प्राथमिक भ्रूणपोष का केन्द्रक, मॉस की पत्तियों की कोशिका, फर्न के प्रोथैलस की कोशिकाएँ, मार्केन्शिया की जेमा कोशिका, एकबीजपत्री की मेरिस्टेम कोशिका, लिवरवर्ट के अण्डाशय तथा फर्न के युग्मनज।
उत्तर:

1. मॉस की प्रथम तंतुक कोशिका (Protonemal cell) – अगुणित (Haploid)
2. द्विबीजपत्री के प्राथमिक भ्रूणपोष (Endosperm) का केन्द्रक – त्रिगुणित (Triploid)
3. मॉस की पत्तियों की कोशिका – द्विगुणित (Diploid)
4. फर्न के प्रोथैलस की कोशिकाएँ – अगुणित (Haploid)
5. मार्केन्शिया की जेमा (Gemma) कोशिकाएँ – अगुणित (Haploid)
6. एकबीजपत्री की मेरिस्टेम कोशिका – द्विगुणित (Diploid)
7. लिवरवर्ट के अण्डाशय (Ovum) – अगुणित (Haploid)
8. फर्न के युग्मनज (Zygote) – द्विगुणित (Diploid)।

Bryophyte Mein Bijanu Matra Koshika Hoti Hai प्रश्न 5.
शैवाल तथा जिम्नोस्पर्म के आर्थिक महत्व पर टिप्पणी लिखिए।
उत्तर:
1. शैवाल का आर्थिक महत्व:
(a) लाल शैवाल का आर्थिक महत्व –

1. औद्योगिक रूप से अगर – अगर का उत्पादन लाल शैवालों द्वारा ही किया जाता है।
2. इनका उपयोग लक्जेटिव एवं इमल्सीफाइंग कारक के रूप में सीरप (दवाइयाँ) बनाने में किया जाता है।
3. इससे प्राप्त अगर-अगर का उपयोग चॉकलेट उद्योग में किया जाता है।
4. कुछ लाल शैवालों का उपयोग खाद्य के अलावा आइसक्रीम, चीज़ व सलाद के रूप में किया जाता है।

(b) भूरे शैवाल का आर्थिक महत्व –

1. यह जापानी लोगों का एक प्रमुख भोजम है।
2. भूरे शैवाल के शुष्क भार में लगभग 30% पोटैशियम क्लोराइड होता है। अत: इनसे पोटैशियम लवण प्राप्त किया जाता है।
3. इनसे विटामिन, आयोडीन, ऐसीटिक अम्ल प्राप्त किया जाता है।
4. इनका उपयोग सौन्दर्य प्रसाधनों, आइसक्रीम तथा कपड़ा उद्योग में भी किया जाता है।

2. जिम्नोस्पर्म का आर्थिक महत्व –

1. जिम्नोस्पर्म मृदा के कणों को पकड़ कर रखती है, अतः मृदा अपरदन (Soil erosion) से बचाव होता है।
2. बाग-बगीचों में जिम्नोस्पर्म का उपयोग सजावटी पौधों के रूप में किया जाता है। उदा.- साइकस, थूजा, ऑरोकेरिया आदि।
3. जिम्नोस्पर्म के तनों से एक प्रकार का स्टार्च प्राप्त किया जाता है, जिसकी सहायता से साबूदाना बनाया जाता है। इसी प्रकार पाइनस के बीज का उपयोग सूखे मेवे (चिलगोजे) के रूप में किया जाता है।
4. इसी प्रकार जिम्नोस्पर्म के विभिन्न भागों जैसे-इनकी लकड़ियों से पेंसिल, कलम, सिगार दान बनाये जाते हैं। सूखी पत्तियों एवं उसके डंठल से टोकरी, झाड़ आदि बनता है।
5. जिम्नोस्पर्म की विभिन्न प्रजातियों से खाद्य तेल प्राप्त किया जाता है।

प्रश्न 6.
जिम्नोस्पर्म तथा एंजियोस्पर्म दोनों में बीज होते हैं, फिर भी उनका वर्गीकरण अलग-अलग क्यों है?
उत्तर:
जिम्नोस्पर्म तथा एंजियोस्पर्म दोनों में बीज होते हैं फिर भी उनका वर्गीकरण अलग-अलग किया जाता है। इसका कारण है कि जिम्नोस्पर्म के बीज नग्न (Nacked) होते हैं अर्थात् इनका बीज फल के अन्दर पाया जाता है, जबकि एंजियोस्पर्म के बीज फलों के अन्दर ढंके हुए होते हैं।

Bryophyta Pteridophyta Mein Antar प्रश्न 7.
विषम बीजाणुकता क्या है ? इसकी सार्थकता पर संक्षिप्त टिप्पणी लिखिए। इसके दो उदाहरण दीजिए।
उत्तर:
विषम बीजाणुकता (Heterospory) दो प्रकार के बीजाणु (Spores) बनने की प्रक्रिया है, जैसे कि छोटा लघुबीजाणु (Microspore) तथा बड़ा दीर्घबीजाणु (Megaspore) विषम बीजाणुकता को सर्वप्रथम टेरिडोफाइट्स सिलैजिनेला (Selaginella) में देखा गया। साल्विनिया में भी विषम बीजाणुकता पायी जाती है। बड़े दीर्घबीजाणु (मादा) तथा छोटे लघु बीजाणु (नर) से क्रमशः मादा और नर युग्मकोद्भिद बन जाते हैं। ऐसे पौधों में मादा युग्मकोद्भिद अपनी आवश्यकताओं की पूर्ति करने के लिए पैतृक स्पोरोफाइट से जुड़ा रहता है।.

प्रश्न 8.
उदाहरण सहित निम्नलिखित शब्दावली का संक्षिप्त वर्णन कीजिए –

1. प्रथम तंतु
2. पुंधानी
3. स्त्रीधानी
4. द्विगुणितक
5. बीजाणुपर्ण
6. समयुग्मकी।

उत्तर:
1. प्रथम तंतु (Protonema):
यह मॉस के जीवन-चक्र की प्रारंभिक अवस्था है। प्रथम तंतु का निर्माण बीज (Spore) के अंकुरण से होता है। प्रथम तंतु या प्रोटोनिमा विकसित होकर हरे रंग की, विसी, शाखित तथा तन्तुमय हो जाती है।

2. पुंधानी (Antheridium):
ब्रायोफाइटा एवं टेरिडोफाइटा के नर जनन अंग को पुंधानी कहा जाता है। पुंधानी के द्वारा द्विकशाभिक पुमंग (Antherozoids) उत्पन्न किया जाता है।

3. स्त्रीधानी (Archegonium):
ब्रायोफाइटा के मादा जनन अंग को स्त्रीधानी कहा जाता है। यह बहुकोशिकीय तथा फ्लास्क के आकार की होती है। इनके द्वारा अण्ड का निर्माण होता है।

4. द्विगुणितक (Diplontic):
यह जीवन-चक्र की एक अवस्था है, जिसमें द्विगुणित (Diploid) बीजाणुद्भिद (Sporophyte) प्रभावी होता है। इस अवस्था में पौधे स्वतंत्र होते हैं, तथा प्रकाश-संश्लेषण की क्रिया द्वारा भोजन का निर्माण करते हैं । युग्मकोद्भिद इसके विपरीत प्रायः एककोशिकीय तथा अगुणित (Haploid) होता है। जीवन-चक्र की इस अवस्था को द्विगुणितक (Diplointic) कहा जाता है।

5. बीजाणुपर्ण (Sporophyll):
जिम्नोस्पर्म की जिन संरचना में बीजाणुधानी (Sporangium) पाया जाता है, उन्हें बीजाणुपर्ण कहा जाता है।

6. समयुग्मकी (Isogamy):
समान युग्मकों (Gametes) के आपस में संलयित होने की क्रिया को समयुग्मकी (Isogamy) कहा जाता है। उदाहरण-क्लैमाइडोमोनास।

Sporophyte Class 11 प्रश्न 9.
निम्नलिखित में अंतर कीजिए

1. भूरेशैवाल तथा लाल शैवाल।
2. लिवरवर्ट तथा मॉस।
3. विषम बीजाणुक तथा समबीजाणुक टेरिडोफाइटा।
4. युग्मक संलयन तथा त्रिसंलयन।

उत्तर:
(i) भूरा शैवाल तथा लाल शैवाल में अंतर –
1.भूरा शैवाल-ये भूरे रंग के समुद्रीय विकसित शैवाल हैं, जिनमें निम्नलिखित लक्षण पाये जाते हैं –

•  फ्यूकोजैन्थिन की उपस्थिति के कारण ये भूरे रंग के दिखाई देते हैं। इसके अलावा इनमें क्लोरोफिल a, c, और -कैरोटीन भी पाया जाता है।
•  इनके प्रजनन में द्विकशाभिकीय चलजन्यु बनते हैं। इनकी एक कशाभिका बड़ी तथा एक छोटी होती है।
• इनके भोज्य पदार्थों का संग्रहण मैनिटॉल एवं लैमीनेरिन के रूप में किया जाता है।
• इनका शरीर अचल, बहुकोशिकीय, सूकायवत् होता है तथा इनकी कोशिका भित्ति में एल्जिनिक एवं फ्यूसिनिक अम्ल का जमाव होता है। उदाहरण-लैमिनेरिया, सारगासम।

2. लाल शैवाल-लाल रंग के इन शैवालों में निम्नलिखित लक्षण देखे जा सकते हैं –

• इनका लाल रंग इनकी कोशिकाओं में पाये जाने वाले वर्णक फाइकोइरीथिन के कारण होता है। इसके अलावा इनमें क्लोरोफिल – a, d तथा फाइकोसायनिन वर्णक भी पाये जाते हैं।
•  जनन अवस्था में अचल कोशिकाएँ ही पायी जाती हैं। जल कोशिकाओं का पूर्णतः अभाव होता है।
• उनमें संचित भोजन फ्लोरिडियन स्टार्च के रूप में होता है।।
• इनका लैंगिक जनन ऊगैमस प्रकार का होता है। उदाहरण-पोरफायरा, पॉलिसाइफोनिया।

(ii) लिवरवर्ट और मॉस में अंतर –
1. लिवरवर्ट (Liver wort) –

1. ये ब्रायोफाइटा वर्ग के हिपैटीकोप्सिडा (Hepaticopsida) के सदस्य हैं।
2. इनका थैलस अधर और पृष्ठ से चपटे होते हैं तथा इनकी आकृति यकृत की पालियों (Lobe) के सदृश्य दिखाई पड़ती हैं।
3. लिवरवर्ट में अलैंगिक जनन थैलस के विखण्डन अथवा विशिष्ट संरचना जेमा द्वारा होता है।
4. इसका मूलाभ (Rhizoids) एककोशिकीय होता है। उदाहरण – रिक्सिया।

2. मॉस (Moss):

• ये ब्रायोफाइटा वर्ग ब्रायोप्सिडा (Bryopsida) का सदस्य है।
• इनका थैलस पत्तीनुमा तथा त्रिज्या सममिति (Radially-symmetrical) होता है।
• मॉस में कायिक जनन विखंडन तथा मुकुलन द्वारा होता है।
• इनके मूलाभ (Rhizoids) बहुकोशिकीय होते हैं। उदाहरण- फ्यूनेरिया।

(iii) विषम बीजाणुक तथा समबीजाणुक टेरिडोफाइटा –
1. विषम बीजाणुक टेरिडोफाइटा (Heterosporous pteridophyte):
ये दो प्रकार के बीजाणु (Spores) उत्पन्न करते हैं, वृहत या दीर्घबीजाणु (Megaspore) तथा लघुबीजाणु (Microspore)। उदाहरण – सिलैजिनेला।

2. समबीजाणुक टेरिडोफाइटा (Homosporous pteridophyte):
ऐसे टेरिडोफाइट्स, जो केवल एक ही प्रकार के बीजाणु (Spores) उत्पन्न करते हों, उन्हें समबीजाणुक कहा जाता है। उदाहरण – लाइकोपोडियम।

(iv) युग्मक संलयन तथा त्रिसंलयन –
1. युग्मक संलयन (Syngamy):
यह प्रथम नर युग्मक का मादा के अंडकोशिका से संलयित होकर युग्मनज (Zygote) बनाने की क्रिया है।

2. त्रिसंलयन (Triple Fusion):
यह द्वितीय नर युग्मक का द्वितीयक केन्द्रक से युग्मित होकर त्रिगुणित भ्रूणपोष (Endosperm) बनाने की घटना को त्रिसंलयन (Triploid fusion) कहा जाता है।

Aavritabiji Aur Anavrtabiji Mein Antar प्रश्न 10.
एकबीजपत्री को द्विबीजपत्री से किस प्रकार विभेदित करेंगे?
उत्तर:
एकबीजपत्री एवं द्विबीजपत्री पादपों में अन्तर –

प्रश्न 11.

उत्तर:

1. (c) शैवाल
2. (d) जिम्नोस्पर्म
3. (b) टेरिडोफाइट
4. (a) मॉस

Class 11th Biology Chapter 3 Question Answer In Hindi प्रश्न 12.
जिम्नोस्पर्म के महत्वपूर्ण अभिलक्षणों का वर्णन कीजिए।
उत्तर:
जिम्नोस्पर्म के महत्वपूर्ण अभिलक्षण –

1. बीजाण्ड आवरणहीन होते हैं।
2. भ्रूणपोष निषेचन से पहले बन जाते हैं।
3. जनन अंग शंकु रूप में होता है।
4. ये पौधे प्राय: एकलिंगी होते हैं।
5. जाइलम में वाहिकाएँ व फ्लोएम में कैस्पोरियन सेल व सीव ट्यूब नहीं पाई जातीं।
   नोट-उपर्यक्त प्रश्न क्रमांक 2 भी देखें।

वनस्पति जगत अन्य महत्वपूर्ण प्रश्नोत्तर

वनस्पति जगत वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –
1. रोडोफाइटा में थैलस का रंग लाल दिखाई देता है, इसकी उपस्थिति से –
(a) R – फाइकोइरीथ्रिन
(b) R – फाइकोबिलीन
(c) दोनों साथ हों तब
(d) इनमें से कोई नहीं।
उत्तर:
(a) R – फाइकोइरीथ्रिन

2. ताजे जल में पाये जाने वाले शैवाल हैं –
(a) लाल शैवाल
(b) भूरी शैवाल
(c) हरी शैवाल
(d) पीली शैवाल।
उत्तर:
(d) पीली शैवाल।

3. ब्रायोफाइट्स की प्रभावी अवस्था है –
(a) अगुणित
(b) गुणित
(c) दोनों समानः
(d) दोनों नहीं।
उत्तर:
(a) अगुणित

4. जूस्पोर्स निम्न में से किसके परिणामस्वरूप होते हैं –
(a) अलैंगिक जनन
(b) लैंगिक जनन
(c) असूत्री विभाजन
(d) समसूत्री विभाजन।
उत्तर:
(a) अलैंगिक जनन

5. फीते के समान हरितलवक किस पौधे में पाया जाता है –
(a) यूलोथ्रिक्स
(b) स्पाइरोगायरा
(c) ऊडोगोनियम
(d) मार्केन्शिया।
उत्तर:
(b) स्पाइरोगायरा

6. स्पाइरोगाइरा के जीवन-चक्र में मिओसिस की क्रिया कब होती है –
(a) युग्मकों के निर्माण के समय
(b) चल बीजाणुओं के निर्माण के समय
(c) युग्मनज के अंकुरण के समय
(d) युग्मनज के निर्माण के समय।
उत्तर:
(c) युग्मनज के अंकुरण के समय

7. मॉसों का आवास है –
(a) शुष्क स्थान
(b) स्वच्छ जल
(c) अलवणीय जल
(d) नम तथा छायादार स्थान।
उत्तर:
(d) नम तथा छायादार स्थान।

8. मॉस का शुक्राणु होता है –
(a) एककशाभिकीय
(b) द्विकशाभिकीय
(c) बहुकशाभिकीय
(d) अचल।
उत्तर:
(b) द्विकशाभिकीय

9. मॉसों का नम वातावरण में पैदा होने का कारण है –
(a) संवहनी ऊतकों की अनुपस्थिति
(b) निषेचन के लिए जल की आवश्यकता
(c) सम्पूर्ण सतह पर जल अवशोषण
(d) उपर्युक्त सभी।
उत्तर:
(d) उपर्युक्त सभी।

10. फ्यूनेरिया का मादा जनन अंग कहलाता है –
(a) बीजाणुधानी
(b) शुक्राणुधानी
(c) आर्कीगोनिया
(d) बीजाणुपर्ण।
उत्तर:
(c) आर्कीगोनिया

11. शुक्राणु का आर्कीगोनियम तक पहुँचना किस माध्यम में होता है –
(a) वायु
(b) जल
(c) जन्तु
(d) कीट।
उत्तर:
(b) जल

12. मॉस बड़े आकार के नहीं हो सकते क्योंकि इनमें –
(a) सम्वहनी तन्त्र अनुपस्थित होता है
(b) संवहनी तन्त्र उपस्थित होता है
(c) पुष्प नहीं पाया जाता
(d) बीज नहीं पाया जाता।
उत्तर:
(a) सम्वहनी तन्त्र अनुपस्थित होता है

13. फर्न प्रोथैलस होता है –
(a) अगुणित
(b) द्विगुणित
(c) त्रिगुणित
(d) स्पोरोफाइट।
उत्तर:
(a) अगुणित

14. प्रोथैलस का आकार होता है –
(a) गोलाकार
(b) लम्बा
(c) हृदय के समान
(d) बहुभुजीय।
उत्तर:
(c) हृदय के समान

15. फर्न, मॉस की अपेक्षा ज्यादा विकसित है, क्योंकि –
(a) इसमें संवहन ऊतक होता है
(b) बीजाणुद्भिद प्रभावी होता है
(c) जड़, तना और पत्तियाँ विकसित होती हैं
(d) उपर्युक्त सभी।
उत्तर:
(d) उपर्युक्त सभी।

16. प्रोथैलस होता है –
(a) एक स्वतन्त्र रचना
(b) एक आश्रित रचना
(c) गैमीटोफाइट पर आश्रित
(d) स्पोरोफाइट पर आश्रित।
उत्तर:
(a) एक स्वतन्त्र रचना

17. स्पाइरोगाइरा में होता है –
(a) निकटवर्ती कोशिकाओं के मध्य लिंगी-जनन
(b) भिन्न द्विपक्ष्मी पुमंग एवं ऊगोनियम
(c) लिंगी प्रजनन एक कोशिका के विभिन्न तत्वों के मध्य
(d) केवल अलिंगी प्रजनन।
उत्तर:
(a) निकटवर्ती कोशिकाओं के मध्य लिंगी-जनन

18. फ्यूनेरिया में होता है –
(a) एककोशिकीय सरल मूलाभास
(b) गुलिकीय मूलाभास
(c) भिन्न शाखित मूलाभास
(d) बहुकोशिकीय तिर्यक।
उत्तर:
(a) एककोशिकीय सरल मूलाभास

19. फ्यूनेरिया के पेरिस्टोम में कितने दाँत होते हैं –
(a) एक मण्डल में 4.
(b) दो मण्डल में 32
(c) एक मण्डल में 16
(d) दो मण्डलों में 16
उत्तर:
(b) दो मण्डल में 32

20. स्पाइरोगाइरा के लैंगिक जनन में किस तरह के युग्मकों का समेकन होता है –
(a) दो समान गतिशील युग्मकों का
(b) दो समान गतिहीन युग्मकों का
(c) एक गतिशील व एक गतिहीन युग्मकों का
(d) दो असमान गतिहीन युग्मकों का।
उत्तर:
(d) दो असमान गतिहीन युग्मकों का।

21. फ्यूनेरिया के किस कोशिका में न्यूनकारी विभाजन होता है –
(a) पुंधानी कोशिका
(b) प्रसूतक कोशिका
(c) युग्मकजीवीय कोशिका
(d) बीजाणु मातृ कोशिका में
उत्तर:
(d) बीजाणु मातृ कोशिका में

22. निम्नलिखित में से कौन-सी दो शैवाल अगर-अगर का उत्पादन करती हैं –
(a) स्पाइरोगायरा व नॉस्टॉक
(b) ग्रैसिलेरिया व डोलेडियम
(c) वॉलवॉक्स व वाउचेरिया
(d) रिवुलेरिया व यूलोथ्रिक्स।
उत्तर:
(b) ग्रैसिलेरिया व डोलेडियम

23. यूलोथ्रिक्स पाया जाता है –
(a) प्रवाही ताजे जल में
(b) प्रवाही लवण जल में
(c) शुद्ध ताजे जल में
(d) शुद्ध लवण जल में।
उत्तर:
(a) प्रवाही ताजे जल में

24. फ्यूनेरिया में रन्ध्र कहाँ स्थित होते हैं –
(a) पत्ती पर
(b) तने पर
(c) थीका पर
(d) एपोफाइसिस पर।
उत्तर:
(d) एपोफाइसिस पर।

25. अनावृतबीजी में जननांग कहलाते हैं –
(a) बीजाणुधर
(b) स्त्री केसर
(c) शंकु या स्ट्रॉबिलस
(d) इनमें से कोई नहीं।
उत्तर:
(c) शंकु या स्ट्रॉबिलस

26. साइकस की पत्ती होती है –
(a) सरल
(b) पक्षवत् संयुक्त
(c) संयुक्त
(d) अभिमुख।
उत्तर:
(b) पक्षवत् संयुक्त

27. रसकस, स्माइलेक्स, यूक्का, उदाहरण है –
(a) पामी के
(b) ग्रैमिनी के
(c) म्यूजेसी के
(d) लिलियेसी के।
उत्तर:
(d) लिलियेसी के।

28. शाक, झाड़ी तथा वृक्ष के प्रकार का आधार है –
(a) शारीरिकी
(b) आकारिकी
(c) (a) तथा (b) दोनों
(d) इनमें से कोई नहीं।
उत्तर:
(b) आकारिकी

29. युग्मनज में अर्द्धसूत्री विभाजन होता है –
(a) सिलेजिनेला में
(b) स्पाइरोगाइरा में
(c) पाइनस में
(d) इक्वीसिटममें।
उत्तर:
(b) स्पाइरोगाइरा में

30. स्पाइरोगाइरा का जीवन-चक्र होता है –
(a) अगुणित
(b) द्विगुणित
(c) अगुणितीय जीवी
(d) द्विगुणितीय जीवी।
उत्तर:
(a) अगुणित

Seval Ke Vargikaran Ka Kya Aadhar Hai प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिए –

1. प्रभावी तथा आत्मनिर्भर स्पोरोफाइट ……………. पादप समूह में होता है।
2. नग्न बीज वाले पौधों के समूह का नाम ……………. है।
3. ……….. समूह में वे जीव आते हैं जिनका जनन अंग बहुकोशिकीय होता है।
4. …………. में बहुभ्रूणता पायी जाती है जिसका कारण है स्त्रीधानियों का होना।
5. फ्यूनेरिया का जीवन-चक्र एक ………… से शुरू होता है।
6. ………….. शरीर में संवहनी ऊतक नहीं पाया जाता।
7. समुद्री खरपतवार ………….. को कहते हैं।
8. हृदय के समान अगुणित रचना …………… कहलाता है।
9. ………….. में बीज फल के अन्दर विकसित होता है।
10. द्वि-जगत वर्गीकरण में शैवाल व कवक को …………… में रखा जाता है।
11. कोशिकाओं की कॉलोनी के रूप में पाये जाने वाले शैवाल …………… हैं।
12. स्पाइरोगाइरा के वर्ग का नाम ……………. है।
13. रिक्सिया, मार्केन्शिया के प्रतिपृष्ठ सतह पर मूल रोम समान रचना ………….. कहलाती है।
14. हरा सूकायवत् शाखित ब्रायोफाइट ……………. कहलाते हैं।

उत्तर:

1. ट्रैकियोफाइटा
2. जिम्नोस्पर्म
3. एम्ब्रियोफाइटा
4. साइकस
5. अगुणित बीजाणु
6. शैवाल
7. भूरा शैवाल
8. प्रोथैलस
9. आवृतबीजियों
10. थैलोफाइटा
11. वॉलवॉक्स
12. क्लोरोफाइसी
13. मूलाभास
14. लिवरवर्ट।

Gymnosperm Ke Mahatvpurn Abhilakshan Ka Varnan Kijiye प्रश्न 3.
उचित संबंध जोडिए –

उत्तर:

1. (c) प्रथम स्थलीय पौधे
2. (d) काँटेदार पौधे
3. (a) उभयचर पौधे
4. (b) जड़, पत्तीविहीन पौधे

उत्तर:

1. (d) थैलोफाइट
2. (c) अपुष्पीय पौधे
3. (b) स्पोरोफाइट गैमिटोफाइट की कड़ी
4. (a) पुष्पीय पौधे

उत्तर:

1. (c) लाल शैवाल
2. (e) कवक
3. (d) माइसिलियम
4. (b) पीढ़ियों का एकान्तरण
5. (a) ब्रायोफाइटा

लिवरवर्ट तथा मास में अंतर बताइए प्रश्न 4.
सत्य / असत्य बताइए –

1. ब्रायोफाइट्स केवल स्थल में पाये जाते हैं, जल में नहीं।
2. सिलैजिनेला में विषमबीजाणुकता नहीं पाई जाती है।
3. म्यूकर एक विषमजालिक कवक है इसे ब्रेड मोल्ड भी कहते हैं।
4. स्पाइरोगायरा को तालाब का रेशम कहते हैं।
5. पाइनस जिरार्डियाना के बीज से चिरौंजी प्राप्त होता है।

उत्तर:

1. असत्य
2. असत्य
3. सत्य
4. सत्य
5. असत्य।

फ्यूनेरिया का जीवन चक्र प्रश्न 5.
एक शब्द में उत्तर दीजिए –

1. लाल शैवाल में कौन-सा वर्णक होता है ?
2. कवक जंतुओं के करीब किस गुण के कारण हैं ?
3. हाइफा के समूह को क्या कहते हैं ?
4. युग्मकोद्भिद का स्पोरोफाइट के साथ नियमित क्रम क्या कहलाता है ?
5. ब्रायोफाइटा में पोषण के आधार पर युग्मकोद्भिद कैसा होता है ?

उत्तर:

1. फाइकोइरीथ्रिन
2. ग्लाइकोजन (संचित भोज्य पदार्थ)
3. माइसिलियम
4. पीढ़ियों का एकांतरण
5. स्वपोषी।

वनस्पति जगत अति लघु उत्तरीय प्रश्न

प्रश्न 1.
उस शैवाल का नाम बताइए जिससे अगर-अगर नामक पदार्थ प्राप्त किया जाता है।
उत्तर:
ग्रेसिलेरिया एवं गेलिडियम नामक शैवाल से अगर-अगर प्राप्त किया जाता है।

वनस्पति जगत प्रश्न 2.
स्पाइरोगायरा को जल रेशम क्यों कहते हैं ?
उत्तर:
स्पाइरोगांयरा चिकने तथा लसलसेदार होते हैं, इसलिए इन्हें जल रेशम या तालाबी रेशम (Pond silk) कहते हैं।

प्रश्न 3.
दो खाद्य शैवालों के नाम बताइए।
उत्तर:

1. लेमिनेरिया
2. पोरफायरा।

प्रश्न 4.
चाय में रस्ट रोग किस शैवाल से होता है ?
उत्तर:
सिफैल्यूरस नामक शैवाल से।

प्रश्न 5.
भारत में शैवाल के जनक के रूप में कौन जाने जाते हैं ?
उत्तर:
एम.ओ.पी. आयंगर नामक वैज्ञानिक भारत में शैवाल के जनक कहे जाते हैं।

वनस्पति जगत के नोट्स प्रश्न 6.
सूकाय किसे कहते हैं ?
उत्तर:
वह पादप शरीर जो जड़, तना तथा पत्ती में विभेदित न किया जा सके सूकाय कहलाता है। शैवालों का शरीर सूकायवत् होता है, जैसे-स्पाइरोगायरा तथा यूलोथ्रिक्स।

प्रश्न 7.
मॉस के एक स्त्रीधानी का नामांकित चित्र बनाइए।
उत्तर:

अध्याय 3 वनस्पति जगत प्रश्न 8.
किस वर्ग के स्थलीय पौधे सबसे आदिम हैं ?
उत्तर:
टेरिडोफाइटा वर्ग के स्थलीय पौधे सबसे आदिम हैं। उदाहरण – राइनिया, सायलोटम।

प्रश्न 9.
फर्न में बीजाणुधानी पुंज कहाँ पाये जाते हैं ?
उत्तर:
फर्न में बीजाणुधारी पुंज विशिष्ट पत्ती सदृश रचनाओं, बीजाणुपर्ण पर पाये जाते हैं।

प्रश्न 10.
साइकस नामक जिम्नोस्पर्म में पाये जाने वाले किन्हीं दो फर्न के लक्षणों को लिखिए।
उत्तर:

1. तरुण कली पत्तियों में सर्सिनेट वर्नेशन (अन्दर की ओर मुड़ा होना)।
2. कशाभिकीय लघुबीजाणु का होना।

प्रश्न 11.
किस पौधे का पुष्य सबसे बड़ा होता है ?
उत्तर:
रैफ्लेशिया नामक पादप का पुष्प सबसे बड़ा होता है।

वनस्पति जगत के प्रश्न उत्तर प्रश्न 12.
भारत में पाये जाने वाले किन्हीं तीन जिम्नोस्पर्स के उदाहरण दीजिए।
उत्तर:

1. पाइनस (चिनार)
2. सिड्रस (देवदार)
3. साइकस।

प्रश्न 13.
ऐन्जियोस्पर्म के किन्हीं दो अतिमहत्वपूर्ण लक्षणों को लिखिए।
उत्तर:

1. बीज का फल के अन्दर स्थित होना।
2. अण्डाशय की उपस्थिति।

प्रश्न 14.
शैवाल के कोई दो औषधीय महत्व लिखिये।
उत्तर:
शैवाल के औषधीय महत्व –

1. लेमीनेरिया जेपोनिका, कोंडस क्रिस्पस आदि में Vitamin. B पाया जाता है।
2. भूरे शैवालों से आयोडीन प्राप्त की जाती है।

वनस्पति जगत लघु उत्तरीय प्रश्न

प्रश्न 1.
मॉस कैप्स्यूल के अनुलम्ब काट का नामांकित चित्र बनाइए।
अथवा
मॉस के कैप्स्यूल का नामांकित चित्र बनाइए।
उत्तर:

प्रश्न 2.
क्लोरोफाइटा के चार लक्षण तथा चार उदाहरण लिखिए।
उत्तर:
क्लोरोफाइटा के लक्षण – ये हरे रंग के शैवाल हैं, जिनमें निम्नलिखित लक्षण पाये जाते हैं –

1. इनका हरा रंग क्लोरोफिल-a एवं b की अधिकता के कारण होता है, इसके अलावा इनमें कैरोटीन एवं जैन्थोफिल भी पाया जाता है।
2. इनकी कोशिकाभित्ति सेल्युलोज की बनी होती है।
3. इनमें बराबर कशाभिकाओं वाले चल बीजाणु बनते हैं।
4. ये भोजन का संग्रह मण्ड के रूप में करते हैं।
5. इनमें लैंगिक जनन समयुग्मक प्रकार का होता है। उदाहरण – स्पाइरोगायरा, यूलोथ्रिक्स, जिग्निमा, क्लैडोफोरा।

प्रश्न 3.
आवृतबीजी (ऐन्जियोस्पर्म) के प्रमुख लक्षण लिखिए।
उत्तर:
आवृतबीजी के प्रमुख लक्षण –

1. इनके बीज के चारों तरफ आवरण पाया जाता है तथा इनका मुख्य पौधा स्पोरोफाइट (2x) होता है।
2. पौधा जड़, तना व पत्तियों में विभाजित रहता है। प्रजनन के लिए इनमें पूर्ण विकसित पुष्प बनता है।
3. इनका संवहन पूल पूर्ण विकसित तथा जाइलम एवं फ्लोएम में विभेदित होता है। जाइलम के अन्दर वाहिकाएँ तथा फ्लोएम के अन्दर सह-कोशिकाएँ पायी जाती हैं।
4. इनमें वातावरण के प्रति बहुत अधिक अनुकूलन पाया जाता है।
5. इनमें प्रजनन अंग पुष्प के रूप में पाया जाता है।
6. ये परजीवी, मृतोपजीवी, सहजीवी, उपरिरोही या जन्तु समभोजी होते हैं।

प्रश्न 4.
ब्रायोफाइटा के प्रमुख लक्षण लिखिए।
उत्तर:
ब्रायोफाइट्स उभयचर जीव हैं, जिनमें निम्नलिखित लक्षण पाये जाते हैं –

1. ये असंवहनी, हरितलवक युक्त, नम भूमि या छालों इत्यादि पर पाये जाने वाले ऐसे पौधे हैं, जिनमें निषेचन के बाद भ्रूण बनता है।
2. मुख्य पौधा युग्मकोद्भिद होता है, बीजाणुद्भिद युग्मकोद्भिद के ऊपर आश्रित रहता है।
3. इनके जनन अंग बहुकोशिकीय होते हैं तथा बंध्य आवरण से घिरे रहते हैं। नर अंग को ऐन्थीरीडिया तथा मादा अंग को आर्कीगोनियम कहते हैं।
4. निषेचन के लिए जल आवश्यक होता है।
5. इनके जीवन – चक्र में दो बहुकोशिकीय रचनाएँ पायी जाती हैं-एक अगुणित, दूसरी द्विगुणित होती है। इन अगुणित और द्विगुणित रचनाओं में स्पष्ट विषमरूपी पीढ़ी रूपान्तरण, पीढ़ी एकान्तरण पाया जाता है।
6.  इनका थैलसनुमा (सूकायवत्) शरीर प्रायः द्विपार्श्व सममित होता है।

प्रश्न 5.
जिम्नोस्पर्म तथा ऐन्जियोस्पर्म में समानताएँ बताइए।
उत्तर:

1. दोनों पौधे बहुवर्षीय वृक्ष होते हैं तथा जनन अंग एक विशेष रचना में लगते हैं।
2. दोनों के परागकण अंकुरित होकर पराग नलिका बनाते हैं।
3. दोनों ही में जीवन की मुख्य रचना स्पोरोफाइटिक होती है तथा गैमीटोफाइट, स्पोरोफाइट पर आश्रित रहता है।
4. दोनों में बीज तथा बीजाण्ड निर्माण की क्रिया समान होती है।

वनस्पति जगत दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
जिम्नोस्पर्म ( अनावृतबीजी) एवं ऐन्जियोस्पर्म (आवृतबीजी) पौधों में पाँच अन्तर बताइए।
उत्तर:

प्रश्न 2.
ब्रायोफाइट्स तथा टेरिडोफाइट्स में अन्तर बताइए।
उत्तर:
ब्रायोफाइट्स तथा टेरिडोफाइट्स में अन्तर –

MP Board Class 9th Science Solutions Chapter 3 Atoms and Molecules

Atoms and Molecules Intext Questions

Atoms and Molecules Intext Questions Page No. 32 – 33

MP Board Class 9th Science Chapter 3 Question 1.
In a reaction 5.3g of sodium carbonate reacted with 6g of ethanoic acid. The products were 2.2g of carbon dioxide. 0. 9g water and 8.2g of sodium ethanoate. Show that these observations are in agreement with the law of co serration of mass.
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Answer:
The reaction is,
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
5.3g + 6g → 8.2g + 2.2g + 0.9g
Now,
Total mass of reactants = (5.3 + 6)g = 11.3g
And, total mass of products = (8.2 + 2.2 + 0.9)g = 11.39g
So, Mass of reactants = Mass of product
It shows the law of conservation of mass.

Molar Mass Calculator with steps for Chemistry Problems – determine the molar mass of a compound based on its formula.

Question 2.
Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3g of hydrogen gas?
Answer:
Ratio of hydrogen and oxygen in water = 1 : 8
So, oxygen is 8 times that of hydrogen by mass.
Let, xgrams of oxygen will react with 3g of hydrogen.
Then,
1 : 8 = 3 : x
x = 8 × 3
x = 24g
∴ 24g of oxygen gas required.

MP Board Class 9 Science Chapter 3 Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:
Postulate of Dalton’s theory based on law of conservation of mass is “Atoms are indivisible particles which can neither be created nor be destroyed in a chemical reaction.”

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
Postulate is the relative number and kinds of atoms remain constant in a given compound.

The mole to mass calculator tool calculates the mass of compound required to achieve a specific molar concentration and volume.

Atoms and Molecules Intext Questions Page No. 35

Question 1.
Define the atomic mass unit.
Answer:
Atomic mass unit is the mass unit equal to \(\frac { 1 }{ 12 }\)th mass of one carbon-12 atom.

Ch 3 Science Class 9 Question 2.
Why is it not possible to see an atom with naked eyes?
Answer:
The size of an atom is very very small that we can see with naked eyes. The size of an atoms lies in nano meters (nm).

Atoms and Molecules Intext Questions Page No. 39

Question 1.
Write down the formulae of:
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide.
Answer:
(i) Sodium Oxide

• Symbol → NaO
• Charge → +1-2
• Formula → Na₂O

(ii) Aluminium Chloride

• Symbol → AlCl
• Charge → +3-1
• Formula → AlCl₃

(iii) Sodium Sulphate

• Symbol → NaS
• Charge → +1-2
• Formula → Na₂S

(iv) Magnesium Hydroxide

• Symbol → MgOH
• Charge → +2-1
• Formula → Mg(OH)₂

Class 9 Subject Science Chapter 3 Question Answer Question 2.
Write down the names of compounds represented by the following formulae:

1. Al₂(SO₄)₃
2. CaCl₂
3. K₂SO₄
4. KNO₃
5. CaCO₃

Answer:

1. Al₂(SO₄)₃ → Aluminium sulphate
2. CaCl₂ → Calcium chloride
3. K₂SO₄ → Potassium sulphate
4. KNO₃ → Potassium nitrate
5. CaCO₃ → Calcium carbonate

Question 3.
What is meant by the term chemical formula?
Answer:
It is the representation of composition of a compounds in the form of symbols of elements present in it.

Class 9 Science Ch 3 Solutions Question 4.
How many atoms are present in a:

1. H₂S molecule and
2. PO₄^3- ion?

Answer:

1. H₂S Molecule -2 atoms of H + 1 atom of S = Total 3 atoms.
2. PO₄^3- 1 atom of phosphorus + 4 atoms of oxygen total 5 atoms.

Atoms and Molecules Intext Questions Page No. 40

Question 1.
Calculate the molecular masses of:

1. H₂
2. O₂
3. Cl₂
4. CO₂
5. CH₄
6. C₂H₆
7. C₂H₄
8. NH₃
9. CH₃OH.

Answer:

1. H₂ = (2 × 1)u = 2u
2. O₂ = (2 × 16)u = 32u
3. Cl₂ = (2 × 35.5)u = 71u
4. CO₂ = (1 × 12 + 2 × 16)u = (12 + 32)u = 44u
5. CH₄ = (1 × 12 + 4 × 1)u = (12 + 4)u = 16u
6. C₂H₆ = (2 × 12 + 6 × 1)u = (24 + 6)u = 30u
7. C₂H₄= (2 × 12 + 4 × 1)u = (24 + 4)u = 28u
8. NH₃ = (1 × 14 + 3 × 1)u = (14 + 3)u = 17u
9. CH₃OH = (1 × 12 + 3 × 1 + 1 × 16 + 1 × 1)u = (12 + 3 + 16 + 1)u = 32u.

Class 9th Science Chapter 3 Question 2.
Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u and O = 16u.
Answer:
Formula unit mass of

1. ZnO = (1 × 65 + 1 × 16)u = (65 + 16)u = 81u
2. Na₂O = (2 × 23 + 1 × 16)u = (46 + 16)u = 62u
3. K₂CO₃ = (2× 39 + 1 × 12 + 3 × 6)u = (78 + 12 + 48)u = 138u

Atoms and Molecules Intext Questions Page No. 42

Question 1.
If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
Answer:
1 mole of carbon atoms = 6.022 × 10²³ atoms
Also, 1 mole of carbon atoms = 12g
6.022 × 10²³ atoms of carbon weigh = 12g
1 atom of carbon weigh = 1.99 × 10²³g.

Chapter 3 Science Class 9 Question 2.
Which has more number of atoms, 100 grams of sodium or 100 grams of ion (given atomic mass of Na = 23u, Fe = 56u)?
Answer:
1 mole of sodium = 23g of Na
Atoms = 6.022 × 10²³ atoms
23g of Na = 6.022 × 10²³ atoms
100g of Na = \(\frac { 100 }{ 23 }\) × 6.022 × 10²³ = 2.617 × 10²⁴ atoms 8 23
Now, 1 mole of iron atoms = 56g of Fe = 6.022 × 10²³ atoms
56g of Fe = 6.022 × 10²³ atoms
100g of Fe = \(\frac { 100 }{ 56 }\) × 6.022 × 10²³ = 1.075 × 10²⁴ atoms
So, 100 g of Na contains more atoms.

Atoms and Molecules NCERT Textbook Exercises

Question 1.
A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:
Percentage composition of boron

Percentage composition of oxygen

Class 9 Science Chapter 3 Question Answer Question 2.
When 3.0g of carbon is burnt in 8.00g oxygen, 11.00g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00g of oxygen? Which law of chemical combination will govern your answer?
Answer:
We know that 3g of carbon is burnt in 8g of oxygen to form 11g of carbon dioxide.
When 3g of carbon is burnt in 50g of oxygen, then 11g of carbon dioxide will be formed.
And oxygen remain unreacted, (50 – 8)g = 42g.
Law of constant proportion governs here.

Question 3.
What are polyatomic ions? Give examples.
Answer:
The ions which contain more than one type of atoms (same kind or different kinds) as a single unit are called polyatomic ions.
Examples:

• Sulphate ion (SO₄^-2)
• Nitrate ion (NO₃^-2)
• Carbonate ion (CO₃^-2).

Class 9 Chapter 3 Science Question 4.
Write the chemical formulae for the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Answer:
(a) Magnesium chloride – MgCl₂
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO₃)₂
(d) Aluminium chloride – AlCl₃
(e) Calcium carbonate – CaCO₃

Question 5.
Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer:
(a) Quick lime (calcium oxide): Elements: Calcium and oxygen.
(b) Hydrogen bromide: Elements: Hydrogen and bromide.
(c) Baking powder (sodium hydrogen carbonate) Elements: Sodium, hydrogen, carbon, oxygen.
(d) Potassium sulphate: Elements: Potassium, sulphur, oxygen.

Science Ch 3 Class 9 Question 6.
Calculate the molar mass of the following substances:
(a) Ethyne, C₂H₂
(b) Sulphur molecule, S₈
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO₃.
Answer:
(a) Ethyne, C₂H₂ = (2 × 12 + 2 × 1)g = (24 + 2)g = 26g
(b) Sulphur molecules, S₈ = (8 × 32)g = 256g
(c) Phosphorus molecule, P₄ = (4 × 31)g = 124g
(d) Hydrochloric acid, HCl = (1 × 1 + 1 × 35.5)g = (1 + 35.5)g = 36.5g
(e) Nitric acid, HNO₃ = (1 × 1 + 1 × 14 + 3 × 16)g = (1 + 14 + 48)g = 63g

Question 7.
What is the mass of:
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na₂SO₃)?
Answer:
(a) 1 Mole of nitrogen atoms = 14g.

(b) Mass of 1 mole of aluminium atoms = 27g.
So, mass of 4 moles of aluminium atoms = (4 × 24)g = 108g.

(c) Mass of 1 mole of sodium sulphide
= (2 × 23 + 32 × 1 + 3 × 16)g = (46 + 32 + 48)g = 126g
∴ Mass of 10 moles of Na₂SO₃ = (126 × 10)g = 1260g.

Question 8.
Convert into mole:
(a) 12g of oxygen gas
(b) 20g of water
(c) 22g of carbon dioxide.
Answer:
(a) Mass of oxygen gas = 12g
Now, 1 mole of oxygen gas = 32g moles mass
So, 32g molar mass of oxygen = 1 moles
∴ 1g of oxygen gas = \(\frac { 1 }{ 32 }\) moles
and 12 g of oxygen gas = \(\frac { 1 }{ 32 }\) moles × 12 moles = 0.375 moles.

(b) 1 Mole of water = 18g molar mass
So, 18g moles mass of water = 1 mole
∴ 1g of water = \(\frac { 1 }{ 18 }\) moles
and 20 g of water = \(\frac { 1 }{ 18 }\) moles × 20 moles = 1.11 moles.

(c) 1 Mole of carbon dioxide = 44g molar mass
So, 44g molar mass of = 1 mole carbon dioxide
∴ 22g of carbon dioxide = \(\frac { 22 }{ 44 }\) moles = 0.5 mole.

Class 9 Science Chapter 3 Questions And Answers Question 9.
What is the mass of
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer:
(a) 1 mole of oxygen atoms = 16g
∴ 0.2 moles of oxygen atoms = (16 × 0.2)g = 3.2g

(b) 1 mole of water molecules = 18g
∴ 0.5 mole of water molecule = (18 × 0.5)g = 9g.

Question 10.
Calculate the number of molecules of sulphur (S₈) present in 16g of solid sulphur.
Answer:
256g of sulphur = 1 mole of sulphur molecules
So, 16g of sulphur = \(\frac { 1 }{ 16 }\) × 16 moles of sulphur molecules
= 0.0625 mole of sulphur molecule
Also, 1 mole of sulphur molecules = 6.023 × 10²³ molecules
So, 0.0625 moles of sulphur molecules = 6.025 × 10²³ × 0.0625 molecules
= 3.76 × 10²² molecules.

Class 9th Science Ch 3 Question 11.
Calculate the number of aluminium ions present in 0.051g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27u).
Answer:
Molar mass of aluminium oxide (Al₂O₃) = (2 × 27)g + (3 × 16)g
= (54 + 48)g = 102g
∴ 10²g of aluminium = 6.022 × 10²³ oxide contains aluminium ions
So, 0.051g Al₂O₃ contains aluminium ions = 6.022 × 10²⁰ aluminium ions.

Atoms and Molecules Additional Questions

Atoms and Molecules Multiple Choice Questions

Question 1.
Atomic theory of matter was proposed by.
(a) Newton
(b) John Dalton
(c) Rutherford
(d) Lavoisier.
Answer:
(b) John Dalton

Question 2.
Law of Conservation of mass was given by.
(a) Lavoisier
(b) Dalton
(c) Kennedy
(d) Faraday.
Answer:
(a) Lavoisier

Question 3.
The term mole was first introduced by.
(a) Grahm
(b) Dalton
(c) Ostwald
(d) Boyle.
Answer:
(c) Ostwald

Class 9th Science Ch 3 Question Answer Question 4.
Atomic radius of an atom is measured in.
(a) Micrometre
(b) Millimetre
(c) Nanometre
(d) Centimetre.
Answer:
(c) Nanometre

Question 5.
Law of constant proportions was proposed by.
(a) Dalton
(b) Bezelius
(c) Proust
(d) Lavoisier.
Answer:
(c) Proust

Ch 3 Class 9 Science Question 6.
Latin name of an atom is argentum. The English name of this element is.
(a) Argon
(b) Gold
(c) Silver
(d) Mercury.
Answer:
(c) Silver

Question 7.
Phosphorus molecule is.
(a) Diatomic
(b) Triatomic
(c) Tetra – atomic
(d) Mono – atomic.
Answer:
(c) Tetra – atomic

Question 8.
The atom chosen for reference for measuring atomic masses is.
(a) C-14
(b) C-12
(c) H-2
(d) O-12.
Answer:
(b) C-12

Class 9 Science Chapter 3 Question Answer In English Question 9.
IUPAC is.
(a) Indian Union of Pacific and Applied Chemistry.
(b) International Union of Permanent and Applied Chemicals.
(c) International Union of Pure and Applied Chemistry.
(d) Indian Union of Pure and Applied Chemistry.
Answer:
(c) International Union of Pure and Applied Chemistry.

Question 10.
Atomic mass is measured in.
(a) Grams
(b) Atomic mass radius
(c) Centigrams
(d) Atomic mass unit.
Answer:
(d) Atomic mass unit.

Question 11.
1 A.M.U. is equal to.
(a) 1.65 × 10^-23g
(b) 1.63 × 10^-25g
(c) 1.66 × 10^-24g
(d) 1.66 × 10^-22g.
Answer:
(c) 1.66 × 10^-24g

Science Class 9th Chapter 3 Question 12.
The naming of elements from first or first and second letter was introduced by.
(a) Berzellius
(b) Dalton
(c) Proust
(d) Lavoisier.
Answer:
(a) Berzellius

Question 13.
The combining capacity of an element is called.
(a) Atomicity
(b) Valency
(c) Reactivity
(d) None of these.
Answer:
(b) Valency

9th Class Science Chapter 3 Question Answer Question 14.
The cation of element has.
(a) More electrons than normal atom
(b) Equal electrons than normal atom
(c) Less electrons than normal atom
(d) Equal proton than normal atom.
Answer:
(c) Less electrons than normal atom

Question 15.
The formula of a compound is A₅B₄. Then valency of A and B will be.
(a) 5 and 4
(b) 5 and 9
(c) 4 and 9
(d) 4 and 5.
Answer:
(d) 4 and 5.

Question 16.
1 Mole has.
(a) 6.012 × 10²³ particles
(b) 6.022 × 10²³ particles
(c) 6.022 × 10²² particles
(d) 6.022 × 10²⁵ particles
Answer:
(b) 6.022 × 10²³ particles

Class 9th Science Chapter 3 Intext Questions Question 17.
Which of the following has the maximum number of atoms?
(a) 18g of CH₄
(b) 18g of H₄O
(c) 18g of CO₂
(d) 18g of O_2.
Answer:
(a) 18g of CH₄

Question 18.
Atomic mass of C₆H₁₂O₆ is.
(a) 24
(b) 80
(c) 100
(d) 12
Answer:
(a) 24

Question 19.
1 atomic mass unit is equal to.
a) \(\frac { 1 }{ 14 }\) mass of a C – 12 atom
(b) \(\frac { 1 }{ 16 }\)mass of a C – 12 atom
(c) \(\frac { 1 }{ 18 }\) mass of a C – 12 atom
(d) \(\frac { 1 }{ 12 }\)mass of a C – 12 atom.
Answer:
(d) \(\frac { 1 }{ 22 }\)mass of a C – 12 atom.

Class 9th Chemistry Chapter 3 Question Answer Question 20.
Atomicity of Sulphate (SO₄^2-) ion.
(a) Mono – atomic
(b) Tri – atomic
(c) Poly – atomic
(d) Tetra – atomic
Answer:
(c) Poly – atomic

Question 21.
Formula of number of moles of a substance is.

Answer:

Atoms and Molecules Very Short Answer Type Questions

Question 1.
Name the scientist who gave atomic theory of matter.
Answer:
John Dalton.

Question 2.
Name the scientist who gave the law of conservation of mass.
Answer:
Antoine Lavoisier

Class 9 Sub Science Chapter 3 Question 3.
Name the scientist who gave the law of constant proportions.
Answer:
Joseph Proust.

Question 4.
How many metres are in 1nm?
Answer:
1nm = 10^-9m.

Question 5.
Give two examples of polyatomic molecules of elements.
Answer:
P₄ (Phosphorus) and S₈ (Sulphur).

Question 6.
Write full form of IUPAC.
Answer:
International Union of Pure and Applied Chemistry.

Question 7.
An element A and B has valency of 3 and 4. Write its chemical formula.
Answer:
A₄B_3.

9th Class Science 3 Chapter Question Answer Question 8.
Name the scientist who introduced the term ‘mole’.
Answer:
Wilhem Ostwald.

Question 9.
How many particles exist in 1 mole of atom?
Answer:
1 mole = 6.022 × 10²³ particles.

Question 10.
Write the formula of aluminium sulphate.
Answer:
Al₂(SO₄)₃.

Question 11.
Name the charged particles formed by gaining of electrons.
Answer:
Anion.

Class 9th Chapter 3 Question Answer Question 12.
Name the charged particles formed by loosening of electrons.
Answer:
Cation.

Question 13.
What is numerical value of Avogadro number or Avogadro’s constant?
Answer:
6.022 × 10²³.

Question 14.
Write the Latin names of iron, gold and copper.
Answer:

1. Iron – Ferrum
2. Gold – Aurum
3. Copper – Cuprum.

Question 15.
What is 1 amu?
Answer:
1 amu = \(\frac { 1 }{ 2 }\)th mass of a carbon-12 atom.

Question 16.
Calculate the number of atoms of oxygen present in its 3.5 moles.
Answer:
1 mole of oxygen atoms = 6.022 × 10²³
∴ 3.5 moles of oxygen atoms = 3.5 × 6.022 × 10²³ atoms
= 2.10 × 10²⁴ atoms.

Atoms and Molecules Short Answer Type Questions

Question 1.
Define:
(a) Law of conservation of mass.
(b) Law of constant proportions.
(c) Atomic mass.
(d) Molecules of element.
(e) Molecules of compound.
(f) Atomicity.
(g) Ion.
(h) Chemical formula.
(i) Valency.
(j) Molecular mass.
(k) Formula unit mass.
(l) Mole.
Answer:
(a) Law of conservation of mass: Matter is neither created nor destroyed during a chemical reaction i.e., total mass of reactants is equal to the total mass of products in a chemical reaction.

(b) Law of constant proportions: In a chemical compound, the elements are always present in a definite proportion by mass.
e.g.:

• In CO₂ the ratio of mass of carbon to the mass of oxygen is always 7 : 16.

(c) Atomic mass: The Atomic mass of an element is defined as the relative mass of its atom as compared with the mass of a carbon 12 atom taken as 12 units.

(d) Molecules of element: It contain two or more similar kinds of atoms chemically combined together.
e.g.:

• O₂, N₂, P₄.

(e) Molecules of compound: It contains two or more different kinds of atoms chemically combined together.
e.g.:

• SO₄, CO₂, H₂O.

(f) Atomicity: It is the total number of atoms present in a molecule.
e.g.:

• Atomicity of H₂ is 2
• Atomicity of P₄ is 4 and
• Atomicity of CO₂ is 3.

(g) Ion: It is the positively or negatively charged atom or group of atoms. It is formed by either loosening or gaining of electrons. Positively charged ion is called cation and negatively charged ion is called anion.
e.g.:

• Cation – Na⁺, Mg²⁺, NH₄⁺
• Anion – Cl^–, Br^–,OH^–, SO₄^2-

(h) Chemical formula: Chemical formula of a compound is representation in the form of symbols of elements present in it.
e.g.:

• Sodium chloride (NaCl)
• Calcium oxide (CaO).

(i) Valency: It is the combining power (or capacity) atom or group of atoms. It tells number of electrons lost or gained by the atom during the Chemical reaction.
e.g.:

(j) Molecular mass: It is the sum of atomic masses of all the atoms in a molecule of the substance.
e.g.:
Molecular mass of CO₂ is = Atomic mass of C + 2 × atomic mass of O
= 12 + 2 × 16
= 12 + 32 = 44u
It is expressed in atomic mass unit (u).

(k) Formula unit mass: It is the sum of atomic masses of all atoms in a formula unit of the compound containing constituents as ions. It is also expressed in atomic mass unit (u).
e.g.:
Formula unit mass of NaCl = Atomic mass of Na + atomic mass of Cl.
= (1 × 23 + 1 × 35.5)u
= (23 + 35.5)u = 58.5u.

(l) Mole: One mole of any substance like atoms or molecules or ions is that quantity which has mass equal to its atomic or molecular mass in grams and also contains 6.022 x 10²³ particles of that substance.
e.g.:
1 mole of H₂ molecule = (1 × 2)g = 2g and also,
1 Mole of H₂ molecule = 6.022 × 10²³ molecules of hydrogen.

Question 2.
Write the postulates of Dalton’s atomic theory.
Answer:
Postulates:

1. All matter is made up of very tiny particles called atoms.
2. Atoms are indivisible.
3. Atoms can neither be created nor be destroyed in a chemical reaction.
4. Atoms of a given element are identical in mass and chemical properties.
5. Atoms combine in the ratio of small whole numbers to form compounds.
6. The number of atoms and kind of atoms is fixed in a given compound.

Question 3.
What were the drawbacks of Dalton’s atomic theory?
Answer:
Drawbacks:

1. According to theory, atoms were indivisible but they can be divided in ions in electrons, protons and neutrons under special conditions.
2. According to theory, atoms of an element have masses but it is found that atoms of some elements can have some different masses.
3. According to theory, atoms of different elements have different masses but it is found that atoms of different elements can have same masses.

Question 4.
Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide. When 20g calcium carbonate is decomposed completely then 11.2g of calcium oxide is formed. Calculate the mass of carbon dioxide formed. Which law of chemical combination will be applied there?
Answer:
Law of conservation of mass is applied here.
Now,

Let, x grams of CO₂ is formed
Now, according to law of conservation of mass.
Total mass of reactants = Total mass of product
⇒ 20 = 11.2 + x
⇒ x = (20 – 11.2) = 8.89
So, Mass of CO₂ formed is 8.8g.

Question 5.
In an experiment 9.8g of copper oxide was obtained from 7.84g of copper. In another experiment 9.1g of copper oxide was obtained on reduction 7.28g of copper. Show with the help of calculations that these figures verify the law of constant proportions.
Answer:
Case I:
Mass of copper oxide = 9.8g
Mass of copper = 7.84g
Mass = (9.8 – 7.84)g = 1.96g
So, Ratio of copper to oxygen = \(\frac { 7.84 }{ 1.96 }\) = \(\frac { 4 }{ 1 }\) = 4 : 1

Case II:
Mass of copper oxide = 9.1g
Mass of copper = 7.28g
∴ Mass of oxygen = (9.1 – 7.28)g = 1.82g
So, Ratio of copper to oxygen = \(\frac { 7.28 }{ 1.82 }\) = \(\frac { 4 }{ 1 }\) = 4 : 1
Since, ratio of copper and oxygen in the two samples is same.
The law of constant proportions is verified.

Question 6.
Write the atomicity of following compounds.
(a) H₂O
(b) CaCO₃
(c) H₂SO₄
(d) C₆H₁₂O₆
(e) S_8.
Answer:

Question 7.
Write the chemical formula of:
(a) Hydrogen oxide
(b) Calcium carbonate
(c) Magnesium chloride
(d) Aluminium sulphate
Answer:

Question 8.
Calculate the molecular ma
(a) CH₃OH
(b) C₆H₁₂O₆
(c) NH₃
(d) H₂SO₄
Atomic masses:
C = 12u
H = 1u
O = 16u
N = 14u
S = 32u
Answer:
(a) Molecular mass of
CH₃OH = (1 × 12 + 3 × 1 + 1 × 16 + 1 × 1)u
= (12 + 3 + 16 + 1 )u = 32u.

(b) Molecular mass of
C₆H₁₂O₆ = (6 × 12 + 12 × 1 + 6 × 16)u
= (72 + 12 + 96)u = 180u.

(c) Molecular mass of
NH₃ = (1 × 14 + 3 × 1)u
= (14 + 3)u = 17u.

(d) Molecular mass of
H₂SO₄ = (1 × 2 + 1 × 32 + 4 × 16)u
= (2 + 32 + 64)u
= 98u.

Question 9.
Calculate the number of molecules of:
(a) SO₂ in its 3 moles.
(b) O₂ in its 2.5 moles.
Answer:
(a) 1 mole of SO₂ molecule
= 6.022 × 10²³ molecules of SO₄
= 3 moles of SO₄ molecule
= 3 × 6.022 × 10²³ molecules
= 18.066 × 10²³ molecules.

(b) 1 mole of O₂ molecule
= 6.022 × 10²³ molecules of O₂
= 2.5 moles of O₂ molecule
= 2.5 × 6.022 × 10²³ molecules of O₂
= 15.055 × 10²³ molecules
= 1.5055 × 10²⁴ molecules.

Question 10.
Calculate the number of moles in:
(a) 12.044 × 10²³ molecules of CO₂.
(b) 24.012 × 10²⁴ molecules of H₂
Answer:
(a) 1 mole of CO₂ = 6.022 × 10²³ particles of CO₂.
⇒ 6.022 × 10²³ particles = 1 mole of CO_2.
So, 1 particle of CO₂ molecule
= 1 / 6.022 × 10²³ mole and,
12.044 × 10²³ particles of CO₂ molecules
= 1 / 6.022 × 10²³ × 12.044 × 10²³
= 12.044 × 10²³ × 1 / 6.022 × 10²³ moles = 2 moles.

(b) 1 mole of H₂ contains = 6.022 × 10²³ particles of H₂ molecule
or 6.022 × 10²³ particles of H₂ = 1 mole of H₂
1 particle of H₂ molecule = 1 / 6.022 × 10²³ mole of H₂.
∴ 24.012 × 10²³ particles of H₂ = 1 / 6.022 × 10²³ × 24.012 × 10²³ moles of H₂
= 3.98 moles of H2.

Atoms and Molecules Long Answer Type Questions

Question 1.
Calculate the ratio of following elements by mass in the given compound:
(a) Hydrogen and oxygen in water (H₂O).
(b) Carbon and oxygen in carbon dioxide (CO₂).
(c) Carbon and hydrogen in ethene (C₂H₄).
Answer:
(a) In H₂O molecule,
Mass of H = 2 × 1g = 2g
Mass of O = 1 × 16g = 16g
So, Ratio of hydrogen and oxygen by mass = \(\frac { 2 }{ 16 }\) =\(\frac { 1 }{ 8 }\) = 3 : 8

(b) In CO₂ molecule,
Mass of C= 1 × 12g = 12g
Mass of O = 2 × 16g = 32g
.;. Ratio of carbon and oxygen by mass = \(\frac { 12 }{ 32 }\) = 3 : 8

(c) In ethene (C₂H₄) molecule,
Mass of C = 2 × 12g = 24g
Mass of H = (4 × 1)g = 4g
Ratio of carbon and hydrogen by mass = \(\frac { 24 }{ 4 }\) = 6 : 1

Question 2.
Calculate the formula unit mass of the following Ionic Compounds:
(a) Sodium Chloride (NaCl)
(b) Calcium Oxide (CaO)
(c) Copper Sulphate (CuSO₄)
(d) Calcium Nitrate [Ca(NO₃)₂]
[Atomic masses: Na = 23u, Cl = 35.5u, Ca = 40u, O = 16u, Cu = 63.5u, S = 32u, N = 14u]
Answer:
(a) Formula unit mass of NaCl Molecule = (1 × 23 + 1 × 35.5)u
= (23 + 35.5)u = 58.5u.

(b) Formula unit mass of CaO molecule = (1 × 40 + 1 × 16)u
= (40 + 16)u = 56u.

(c) Formula unit mass of CuSO₄ = (1 × 63.5 + 1 × 32 + 4 × 16)u
= (63.5 + 32 + 64)u = 159.5u.

(d) Formula unit mass of [Ca(NO₃)₂]
= 1 × 40 + 2 [1 × 14 4 + 3 × 16]
= 40 + 2 [14 4 + 48]u
= (40 + 124)u
= 164u.
MOLE CONCEPT (Graus to uoles):

Question 3.
Find the number of moles in:
(a) 20g of H₂O
(b) 140g of CO₂
(c) 200g of CaCO₃.
Answer:
(a) 1 mole of H₂O = Molar mass of H₂O
= (2 × 1 + 16)g
So, 1 Mole = 18g of H₂O
or 18g of H₂O = 1 Mole
1g of H₂O = \(\frac { 1 }{ 18 }\) Mole
and, 20g of H₂O = \(\frac { 1 }{ 18 }\) × 20 = 1.11 Mole

(b) 1 mole of CO₂ = Molar mass of CO₂ = (1 × 12 + 2 × 16)g
= (12 + 32)g = 44g
So, 1 mole = 44g of CO₂
or 44g of CO₂ = 1 mole
1 g of CO₂ = \(\frac { 1 }{ 44 }\) mole
and, 140 g of CO₂ = \(\frac { 1 }{ 44 }\) × 140 moles
= 3.18 moles.

(c) 1 mole of CaCO₃ = Molar mass of CaCO₃
= (1 × 40 + 1 × 12 + 3 × 16)g
= (40 + 12 + 48)g = 100g
So, 1 mole = 100g of CaCO₃
or 100g of CaCO₃ = 1 mole
1g of CaCO₃ = \(\frac { 1 }{ 100 }\) mole
and, 200 g of CaCO₃ = \(\frac { 1 }{ 100 }\) × 200 moles = 2 moles.
Moles to Grams

Question 4.
Calculate the mass in grams of:
(а) 3 moles of H₂O
(b) 2 moles of H₂SO₄
(c) 1.5 moles of carbon atoms (C atom).
Answer:

(a) 1 mole of H₂O = Molar mass of H₂O
= (2 × 1 + 16)g of H₂O
= 18g of H₂O
∴ 3 mole of H₂O = (3 × 18)g of H₂O
H₂O = 54g of H₂O

(b)1 mole of H₂SO₄ = Molar mass of H₂SO₄
= (2 × 1 + 1 × 32 + 4 × 16)g = 98g
2 mole of H₂SO₄ = (2 × 98)g = 196g

(c) 1 mole of C atom = Molar mass of C atom = 12g
∴ 1.5 moles of C -atom = (1.5 × 12)g = 18g.
Moles to Number of Particles.

Question 5.
Calculate the number of particles in:
(a) 115g of H₂O.
(b) 60g of CO₂.
Answer:
(a) We know,
1 mole of H₂O = molar mass of H₂O
= (2 × 1 + 16)g = 18g
It means, 18g of H₂O = 1 mole of CO₂
1g of H₂O = \(\frac { 1 }{ 18 }\) mole = 6.38 moles
Now, 1 mole of H₂O = 6.022 × 10²³ particles
6.38 moles of H₂O = 6.38 x 6.022 × 10²³ particles
= 3.84 × 10²⁴ particles.

(b) We know,
1 mole of CO₂ = Molar mass of CO₂
= (12 + 2 × 16)g = (12 + 32)g = 44g
It means, 44g of CO₂ = 1 mole of CO₂
1g of CO₂= \(\frac { 1 }{ 44 }\) mole of CO₂
and, 60g of CO₂ = \(\frac { 1 }{ 44 }\) × 60 moles of CO₂
= 1.3636 moles
Number of Particles to Mass.

Question 6.
Calculate the mass of:
(a) O₂ in its 36.48 × 10²⁵ particles.
(b) NH₃ in its 4.012 × 10²⁴ particles.
Answer:
(a) We know,
1 mole of O₂ = 6.023 × 10²³ particles
or 6.023 × 10²³ particles = 1 mole of O₂

Now, We also know,
1 mole of O₂ contains = molar mass of O₂
So, 60.57 moles of O₂ contains = 60.57 × 32g = 1938.49g.

(b) We know,
1 mole of ammonia (NH₃) = 6.022 X 1023 particles
or 6.022 × 1023 particles of ammonia (NH₃) = 1 mole

Now, We also know that
1 mole NH₃ contains = Molar mass of NH₃
= (14 + 3 × 1)g = (14 + 3)g
= 17g
So, 6.66 moles of NH₃ contains = (6.66 × 17)g = 113.25g

Question 7.
(a) Define atomicity and poly atomic ions.
(b) Find out the atomicity of following:
CO₂, NH₃, S₈, CaCO₃, H₂SO₄, Ca(OH)₂, K₂SO₄, Al₂(SO₄)₃, NaCl.
(c) Write 2 divalent and 2 trivalent polyatomic ions.
Answer:
(a) Atomicity: It is total number of atoms present in a molecule.
e,g.,

• Atomicity of H2 is 2.

Polyatomic ions: Ions which are formed from group of atoms are called polyatomic ions.
e,g.,

• Carbonate (CO₃^2-), Sulphate (SO₂^2-).

(b)

(c) Divalent polyatomic ions:

• Carbonate ion (CO₃s^2-)
• Sulphate ion (SO₄^2-).

Trivalent polyatomic ions:

• Phosphate ion (PO₄^3-)
• Phosphite ion (PO₃³).

Atoms and Molecules Higher Order Thinking Skills (HOTS)

Question 1.
A thermos P contains 0.5 mole of oxygen gas. Another thermos Q contain 0.4 mole of ozone gas. Which of the two thermos contain greater number of oxygen atom?
Answer:
1 molecule of oxygen (O₂) = 2 atoms of oxygen
1 molecule of ozone (O₃) = 3 atoms of oxygen
In thermos P:
1 mole of oxygen gas = 6.022 × 10²³ molecules
0.5 mole of oxygen gas = 6.022 × 10²³ × 0.5 molecules
= 6.022 × 10²³ × 0.5 × 2 atoms
= 6.022 × 10²³ atoms

In thermos Q:
1 mole of ozone gas = 6.022 × 10²³ molecules
0.4 mole of oxygen gas = 6 .022 × 10²³ × 0.4 molecules
= 6.022 × 10²³ × 0.4 x 3 atoms
= 7.32 × 10²³ atoms.

Question 2.
On analysing an impure sample of sodium chloride, the percentage of chlorine was found to be 45.5. What is the percentage of pure sodium chloride in the sample?
Answer:
Molecular mass of pure NaCl
= Atomic mass of Na + Atomic mass of Cl
= 23 + 35.5 = 58.5u
Precentage of chlorine in pure NaCl
Now, if chlorine is 60.6 parts
NaCl =100 parts
If chlorine is 45.5 parts,
Thus, percentage of pure NaCl = 75%.

Question 3.
Write the chemical formulae of the following:

1. Ammonium phosphate
2. Iron sulphate
3. Calcium nitrate
4. Magnesium nitride
5. Ammonium sulphate
6. Aluminium chloride
7. Copper Nitrate
8. Aluminium sulphate
9. Sodium carbonate
10. Barium chloride
11. Calcium nitrate
12. Potassium chloride
13. Hydrogen sulphide
14. Magnesium hydroxide
15. Zinc sulphate.

Answer:

1. Ammonium phosphate – (NH₄)₃PO₄
2. Iron sulphate – -Fe₂(SO₄)₃
3. Calcium nitrate – Ca(NO₃)₂
4. Magnesium nitrate – Mg(NO₃)₂
5. Ammonium sulphate – (NH₄)₂SO₄
6. Aluminium chloride – AlCl₃
7. Copper nitrate – CU(NO₃)₂
8. Aluminium sulphate – Al₂(SO₄)₃
9. Sodium carbonate – Na₂CO₃
10. Barium chloride – BaCl₂
11. Calcium nitrate – Ca(NO₃)₂
12. Potassium chloride – KCl
13. Hydrogen sulphide – H₂S
14. Magnesium hydroxide – Mg(OH)₂
15. Zinc sulphate – ZnSO₄

Atoms and Molecules Value Based Question 

Question 1.
Jolly buys gold ornaments and she is told that the ornaments has 90% gold and the rest is copper. She has been given a bill which amounts 100% charges of gold. Jolly refused to pay
the bill for 100% gold but settles the bill for 90% gold?
(a) How many atoms of gold are present in 1 gram of gold?
(b) Find out the ratio of gold and copper in the ornaments.
(c) What value of Jolly is seen in the above discussion?
Answer:
(a) 1 gram of gold will contain \(\frac { 90 }{ 100 }\) = 0.9 g of gold.

∴ 0.046 mol of gold will contain = 0.046 × 6.022 × 10²³ = 2.77 × 10²¹ atoms

(b) Ratio of gold : Copper 90 : 10

(c) Value of responsible behaviour and self – awareness is seen.

MP Board Class 9th Science Solutions

MP Board Class 11th Special English Important Questions with Answers Guide Pdf Free Download are part of MP Board Class 11th Solutions. Here we have given Madhya Pradesh Syllabus MP Board Class 11 Special English Question Bank Solutions Pdf.

MP Board Class 11th Special English Important Questions with Answers

• Prose Important Questions
• Poems Important Questions
• Vocabulary Exercises Important Questions
• Important Exercises: From Work Book
• Grammar Important Questions
• Unseen Passages Important Questions
• Essay Writing Important Questions
• Writing Skill Important Questions
• Letter and Application Writing Important Questions

MP Board Class 11th Special English Syllabus and Marking Scheme

Latest Syllabus and Marks Distribution Special English Class XI for the academic year 2019 – 2020 Year Examination.

Class XI
Time: 3 Hours
Maximum Marks: 100

Unit 1 – Reading an unseen passage and Poem (15 Marks)

• Literary or discursive passage of about 500 – 600 words.
• A poem of about 15 lines.

Unit 2 – Texts for Detailed Study (40 Marks)
A. Prose (15 Marks)

• One passage for comprehension with Short Answer Questions.
• One out of two questions to be answered in about 150 words.
• Two Short Answer Questions to be answered in about 30 – 40words each.

B. Poetry (15 Marks)

• Two extracts from the prescribed poem for literary interpretation, comprehension.
• One question on the prescribed poem for the appreciation to be answered in 150 words.

C. English by Choice (10 Marks)

• Two out of three questions to be answered in 150 – 200 words.

Unit 3 – Drama (10 Marks)
One out of two questions to be answered in about 150 – 200 words.

Unit 4 – Fiction (10 Marks)

• One out of two questions to be answered in about 150 words.
• Two out of three Short Answer Type Questions to be answered in about (30 – 40).

Unit 5 – Writing (15 Marks)

• An Essay (250 – 300 words).
• To write a shorter composition such as an article, report, a statement of purpose (100 – 125 words).

Unit 6 – Functional Grammar (10 Marks)

• Tenses
• Modals
• Determiners
• Articles
• Voices
• Narration
• Prepositions
• Clauses.

We hope the given MP Board Class 11th Special English Important Questions with Answers Guide Pdf Free Download will help you. If you have any queries regarding Madhya Pradesh Syllabus MP Board Class 11 Special English Question Bank Solutions Pdf, drop a comment below and we will get back to you at the earliest.

MP Board Class 7th Maths Solutions Chapter 2 भिन्न एवं दशमलव Ex 2.1

Mp Board Class 7 Maths Solutions English Medium प्रश्न 1.
हल कीजिए-

हल:
(i) 2 – \(\frac { 3 }{ 5 }\)
यहाँ, ल. स. = 5

कक्षा 7 गणित अध्याय 2 प्रश्न 2.
निम्नलिखित को अवरोही क्रम में रखिए-

हल:

Mp Board Class 7 Maths Solutions Hindi Medium प्रश्न 3.
एक “जादुई वर्ग” में प्रत्येक पंक्ति, प्रत्येक स्तम्भ एवं प्रत्येक विकर्ण की संख्याओं का योग समान होता है। क्या यह एक जादुई वर्ग है ?

हल:

चूँकि प्रत्येक पंक्ति, स्तम्भ तथा विकर्ण के योग समान हैं। अत: यह एक जादुई वर्ग है।

Mp Board Class 7 Maths Solutions Chapter 2 प्रश्न 4.
एक आयताकार कागज की लम्बाई 12 \(\frac { 1 }{ 2 }\) सेमी और चौड़ाई 10 \(\frac { 2 }{ 3 }\) सेमी है। कागज का परिमाप ज्ञात कीजिए।
हल:
लम्बाई = 12, सेमी = 25 सेमी ;
चौड़ाई = 103 सेमी = 32 सेमी
∵ आयत का परिमाप = 2 (लम्बाई + चौड़ाई)
∴ आयताकार कागज के टुकड़े का परिमाप

Class 7 Maths Chapter 2 Exercise 2.1 Solutions Hindi Medium प्रश्न 5.
दी हुई आकृति में (i) ∆ABE, (ii) आयत BCDE, का परिमाप ज्ञात कीजिए। किसका परिमाप ज्यादा है ?

हल:
(i) ∆ABE का परिमाप = AB + BE + AE

(ii) आयत BCDE का परिमाप = 2(BE + DE)


अत: ∆ABE का परिमाप ज्यादा है।

भिन्न के सवाल Class 7 प्रश्न 6.
सलील एक तस्वीर को किसी फ्रेम (चौखट) में जड़ना चाहता है। तस्वीर 7\(\frac { 3 }{ 5 }\) सेमी चौड़ी है। चौखट में उचित रूप से जड़ने के लिए तस्वीर की चौड़ाई 7 \(\frac { 3 }{ 10 }\) सेमी से ज्यादा नहीं हो सकती। तस्वीर की कितनी काट-छाँट की जानी चाहिए ?
हल:
तस्वीर की चौड़ाई = 7\(\frac { 3 }{ 5 }\) सेमी = \(\frac { 38 }{ 5 }\) सेमी
तस्वीर की अभीष्ट चौड़ाई = 7\(\frac { 3 }{ 10 }\) = \(\frac { 73 }{ 10 }\) सेमी
∴ तस्वीर की काट-छाँट करनी चाहिए

अतः तस्वीर की काट-छाँट \(\frac { 3 }{ 10 }\) सेमी की जानी चाहिए।

Mp Board Class 7 Maths Solutions प्रश्न 7.
रीतू ने एक सेब का \(\frac { 3 }{ 5 }\) भाग खाया और शेष सेब उसके भाई सोमू ने खाया। सेब का कितना भाग सोमू ने खाया ? किसका हिस्सा ज्यादा था ? कितना ज्यादा था ?
हल:
रीतू ने सेब का खाया = \(\frac { 3 }{ 5 }\) भाग
तथा सोमू ने सेब का शेष भाग खाया अर्थात्

अतः रीतू का हिस्सा सोमू से \(\frac { 1 }{ 5 }\) भाग अधिक था।

Mp Board Class 7th Maths Solutions प्रश्न 8.
माइकल ने एक तस्वीर में रंग भरने का कार्य \(\frac { 7 }{ 12 }\) घण्टे में समाप्त किया। वैभव ने उसी तस्वीर में रंग भरने का कार्य \(\frac { 3 }{ 4 }\) घण्टे में समाप्त किया। किसने ज्यादा समय कार्य किया ? यह समय कितना ज्यादा था ?
हल:
तस्वीर में रंग भरने में माइकल \(\frac { 7 }{ 12 }\) घण्टे लेता है और वैभव \(\frac { 3 }{ 4 }\) घण्टे लेता है।

अतः वैभव ने = घण्टे ज्यादा कार्य किया।

पाठ्य-पुस्तक पृष्ठ संख्या # 33

Class 7th Maths Mp Board प्रश्न 1.
क्या आप बता सकते हैं कि निम्नांकित आकृति किसे निरूपित करेगी?

हल:
इस आकृति में, हम प्राप्त करते हैं :

Class 7 Maths 2.1 Hindi Medium प्रश्न 2.
क्या आप बता सकते हैं कि निम्नांकित आकृति किसे निरूपित करेगी?

हल:
इस आकृति में, हम प्राप्त करते हैं :

Mp Board Solution Class 7th Maths प्रश्न 3.
क्या आप बता सकते हैं
(i) 3 x \(\frac { 2 }{ 7 }\) = ?
(ii) 4 x \(\frac { 3 }{ 5 }\) = ?
हल:

पाठ्य-पुस्तक पृष्ठ संख्या # 34

प्रयास कीजिए
Mp Board Solutions Class 7 Maths प्रश्न 1.
ज्ञात कीजिए :
(a) \(\frac { 2 }{ 7 }\) x 3
(b) \(\frac { 9 }{ 7 }\) x 6
(c) 3 x \(\frac { 1 }{ 8 }\)
(d) \(\frac { 13 }{ 11 }\) x 6
यदि गुणनफल एक विषम भिन्न है, तो इसे मिश्रित भिन्न के रूप में व्यक्त कीजिए।
हल:

Mp Board 7th Class Maths Syllabus प्रश्न 2.
2 x \(\frac { 2 }{ 5 }\) = \(\frac { 4 }{ 5 }\) को सचित्र निरूपित कीजिए।
हल:
2 x 2 = \(\frac { 4 }{ 5 }\) को हम निम्नांकित रूप में निरूपित कर सकते हैं :

प्रयास कीजिए

Ex 2.1 Class 7 Hindi Medium प्रश्न 1.
ज्ञात कीजिए-
(i) 5 x 2\(\frac { 3 }{ 7 }\),
(i) 1\(\frac { 4 }{ 9 }\) x – 6
हल:

पाठ्य-पुस्तक पृष्ठ संख्या # 35
प्रयास कीजिए

Exercise 2.1 Class 7 प्रश्न 1.
क्या आप बता सकते हैं कि (i) 10 का \(\frac { 1 }{ 2 }\), (ii) 16 का \(\frac { 1 }{ 4 }\), (iii) 25 का \(\frac { 2 }{ 5 }\), क्या है ?
हल:

MP Board Class 7th Maths Solutions

MP Board Class 8th Science Solutions Chapter 12 Friction

MP Board Class 8 Science Chapter 12 Question 1.
Fill in the blanks:
(a) Friction opposes the …………. between the surfaces in contact with each other.
(b) Fraction depends on the …………… of surfaces.
(c) Friction produces …………….. .
(d) Sprinkling of powder on the carrom board …………. friction.
(e) Sliding friction is ……………. than the static friction.
Answer:
(a) relative motion
(b) nature
(c) heat
(d) reduces
(e) less.

MP Board Class 8 Science Chapter 9 Friction Question 2.
Four children were asked to arrange forces due to rolling, static and sliding frictions in a decreasing order. Their arrangements are given below. Choose the correct arrangement.
(a) Rolling, static, sliding.
(b) Rolling, sliding, static
(c) Static, sliding, rolling
(d) Sliding, static, rolling.
Answer:
(c) Static, sliding, rolling

Class 8 Science Chapter 12 MP Board Question 3.
Alida runs her toy car on dry marble floor, wet marble floor, newspaper and towel spread on the floor. The force of friction acting on the car on different surfaces in increasing order will be:
(a) Wet marble floor, dry marble floor, newspaper and towel.
(b) Newspaper, towel, dry marble floor, wet marble floor.
(c) Towel, newspaper, dry marble floor, wet marble floor.
(d) Wet marble floor, dry marble floor, towel, newspaper.
Answer:
(a) Wet marble floor, dry marble floor, newspaper and towel.

Class 8 Science Chapter 12 Question Answer In English Question 4.
Suppose your writing desk is tilted a little. A book kept on it starts sliding down. Show the direction of frictional force acting on it.

Answer:
The frictional force will act parallel to the inclined surface, opposite to the direction of the sliding of book.

Class 8 Subject Science Chapter Friction Question 5.
You spill a bucket of soapy water on a marble floor accidentally. Would it make it easier or more difficult for you to walk on the floor? Why?
Answer:
It is difficult to walk on a soapy floor because layer of soap makes floor smooth. The coating of soap reduces the friction and the foot cannot make a proper grip on the floor and it starts getting to slip on the floor.

Science Ch 12 Class 8 Question 6.
Explain why sportsmen use shoes with spikes?
Answer:
Sportsmen use shoes with spikes to increase the friction so that their shoes do not slip while they run or play.

Friction Class 8 Question 7.
Iqbal has to push a lighter box and Seema has to push a similar heavier box on the same floor. Who will have to apply a larger force and why?
Answer:
Seema will experience more frictional force because the heavy object will be pressed hard against the opposite surface and produces more friction.

Class 8 Science Ch 12 Question 8.
Explain why sliding friction is less than the static friction.
Answer:
When the box starts sliding, the contact points on the surface of the box do not get enough time to lock into the contact points of the floor. Hence, the sliding friction is slightly less than the static friction. This is why the sliding friction is always less than the static friction.

Class 8 Chapter Friction Question 9.
Give examples to show that friction is both a friend and a foe.
Answer:
Friction is a friend because:

1. We cannot imagine to be able to walk without friction.
2. We cannot write with pen or pencil in absence of friction.
3. The automobiles could not be started or stopped or turned to change the direction of motion if there were no friction between the tyres of the vehicle and the surface of the road.
4. We cannot fix a nail in the wall or tie a knot if there is no friction.
5. No building could be constructed without friction.

Friction is a foe because:

1. It wears out the materials such as shoes, screws and ball bearings.
2. Friction causes heat. When a machine. is operated, heat generated causes much wastage of energy. Thus, Friction acts as a friend as well as a foe. Also, friction is a necessary evil.

Class 8 Friction Numericals Question 10.
Explain why objects moving in fluids must have special shapes.
Answer:
When objects move through fluids, they have to overcome the friction acting on them. Efforts are therefore made to minimize the friction so objects are given special shapes. Idea of such shapes come from the body structure of birds and fishes which have to move about in fluids all the time. Such shapes are called streamlined. Giving such shapes to the ships, planes and cars is called streamlining.

MP Board Class 8th Science Friction NCERT Extended Learning – Activities and Projects

Question 1.
What role does friction play in the sport of your choice? Collect some pictures of that sport in action where friction is either supporting it or opposing it. Display these pictures with proper captions on the bulletin board of your classroom.
Answer:
Do yourself.

Numericals On Friction Class 8 Question 2.
Imagine that friction suddenly vanishes. How would the life be affected. List ten such situations.
Answer:
Do yourself.

Lesson 12 Science Class 8 Question 3.
Visit a shop which sells sports shoes. Observe the soles of shoes meant for various sports. Describe your observations.
Answer:
Do yourself.

Class 8th Friction Question 4.
A toy to play with:
Take an empty matchbox. Take out its tray. Cut a used refill of a ball pen of the same width as the tray as shown in the figure below. Fix the refill with two pins on the top of the tray as shown in Fig. 12.2. Make two holes on the opposite sides of the tray. Make sure that the holes are large enough to allow a thread to pass through them easily. Take a thread about a metre long and pass it through the holes as shown. Fix beads at the two ends of the thread so that it does not come out. Insert the tray in the outer cover of the matchbox. Suspend the match box by the thread. Leave the thread loose. The match box will start falling down due to gravity. Tighten the thread now and observe what happens. Explain your observation. Can you relate it to friction?

Answer:
Do yourself.

MP Board Class 8th Science Friction NCERT Additional Important Questions

A. Short Answer Type Questions

Class 8 Ch Friction Science Question l.
What is friction?
Answer:
The force acting equal and opposite surfaces? Does it depend on the smoothness to the relative motion of two objects in of the surface?
contact is known as friction.

Friction Class 8 Extra Questions Question 2.
How is friction caused? the surfaces. Yes, it depends on the
Answer:
Friction is caused by the irregularities smoothness of the surface. It is the least on the two surfaces in contact. the smoothest surface.

Class 8 Science Chapter 9 Friction Question Answer Question 3.
In which direction does friction work? Smooth surfaces provide less friction, whereas
Answer:
Frictions works opposite to the motion, rough surfaces provide more friction.

Science Class 8 Chapter 12 Question Answer Question 4.
Is the friction the same for all the
Answer:
No, friction is not the same for all

B. Long Answer Type Questions

Class 8 Friction Question Answer Question 5.
Why does the friction between two
Answer:
Since the friction is due to the interlocking of irregularities in the two surfaces which slide with respect to each other, it is obvious that the force of friction will increase if the two surfaces are pressed harder. One can experience it by dragging a mat when nobody is sitting on it and when a person is sitting on it.

Class 8 Friction Chapter Question 6.
What would happen if there were no force of friction? Imagine and describe in your own words.
Answer:
If there were no friction, many problems could have been arise. One cannot able to walk if there were no friction at all. One could not write with pen or pencil, if there were no friction. When one is writing with chalk on the blackboard, its rough surface rubs off some chalk particles. So, we can se the writing on the black board. If any object just started moving, it would never stop. Had there been no friction between the tyres of the automobiles and the road, they could not be started or stopped or turned to change the direction of motion. One could not fix a nail in the wall or knot. Also, without friction no building could so constructed.

Physics Friction Class 8 Question 7.
It is easy to move an object on a sliding surface than the plane surface. Why?
Answer:
Every surface has some irregularities on it. On a plane surface these irregularities get interlocked easily and it becomes a little hard for the object to move. In case of sliding surfaces, the irregularities do not get enough time to interlock, so the object moves easily on the sliding surface.

Worksheet Of Friction Class 8 Question 8.
Why is friction called an evil?
Answer:
Friction is an evil. It causes a lot of problems for us. It wears out the objects or materials on which it acts. The soles of our shoes get worn due to friction. The floors, machines, their parts, clothes, metals etc., get weared due to the friction they bear. Friction spoils the parts of the machines. They are, therefore, regularly lubricated to minimise friction. Friction produces heat, which further damages the objects on which it is exerted. The grooves of the tyres are finished due to regular friction they bear, which is offered by their regular motion on roads. Except this wear and tear, friction utilize a large amount of energy spent to overcome it.

MP Board Class 8th Science Solutions

MP Board Class 7th Special English Solutions Revision Exercises 2

Comprehension

Class 7 English Revision Exercise 2 Question 1.
Where does the poet breathe a song?
Answer:
The poet breathe’s a song into the air.

Class 7th English Revision Exercise 2 Question 2.
Why is Shravan Kumar famous?
Answer:
Shravan kumar is famous for his devotion and obedience to his blind parents.

MP Board Class 7th English Revision Exercise 2 Question 3.
What is a talking machine? What is a ‘talkie’?
Answer:
A talking machine is one that can talk. The modem gramophone is simply an improvement on this talking machine. A ‘talkie’ is a talking picture. This is the modern cinema.

Revision Exercise 2 Class 7 English Question 4.
Why did the Americans turn off their electric lights for a while when Edison died?
Answer:
Edison had invented the electric light. Therefore, when Edison died, the Americans turned off their electric lights for a while as a mark of respect to the wonderful inventor.

Class 7 Revision Test 2 Question 5.
What kind of life did Sri Ramakrishna live as a temple priest?
Answer:
As a temple priest Sri Ramakrishna spent all his time in the service of ‘Kali’ the Divine Mother whom he worshipped out of religious duty. He lived a happy and a deeply religious life. He spent days and nights in prayer. He led a very hard life. He often went without food and water and ruined his health.

Class 5 English Revision Exercise 2 MP Board Question 6.
Why is quarrelling in the name of religion foolish?
Answer:
All the religions lead to the same goal. They also teach the same truth. In fact, people worship the same God under different names. Therefore, quarrelling in the name of religion is foolish.

Class 7 English Revision Lesson Question 7.
What do you understand by the lines “till …….began No, time to wait?
Answer:
The smile appears in the eyes before it reaches the mouth. It takes a few moments. But the poet has no time even to view that smile.

Revision Test 2 Class 7 Question 8.
What words and phrases did Anand fail to understand?
Answer:
The words and phrases which Anand failed to understand are = Cross, boiling of blood, get, novel.

Class 5 English Revision Exercise 2 Question 9.
Whom do the villagers give credit for their prosperity?
Answer:
The villagers give credit to God (who sends rain)

Class 7th Revision Exercise 2 Question 10.
What is the best of all for the country people?
Answer:
The country faith is the best of all for the country-people.

Word Power

A. Choose the correct words to complete the sentences.

1. She is very ………. to me. (deer/dear)
2. He has a very bad eye ………… (sight/site)
3. The ………… was bright that day. (Son/sun)
4. He went inside the temple to ………… to god. (prey/pray)
5. He saved the ………. from the hunter. (dear/deer).

Answer:

1. dear
2. sight
3. sun
4. pray
5. deer.

B. Match the words in column (A) with the meaning given in column (B)

Answer:
1. → (b)
2. → (d)
3. → (a)
4. → (e)
5. → (c)

Grammar in Use

A. Rearrange the words in the following sentences and rewrite them to make meaningful sentences:

1. the/bet/is/What/?/
2. surprised/very/am/I/./
3. committed/a/blunder/he/./
4. archer/great/a/was/Dashrath/./
5. jobs/lost/Edison/many/./

Answer:

1. What is the bet?
2. I am very surprised.
3. He committed a blunder.
4. Dashrath was a great archer.
5. Edison lost many jobs.

B. Use the prefix ‘dis’ in the words given below and fill in the blank space, trust obey honest similar like

1. We should never ………. our elders.
2. People of ……….. nature never come together.
3. I ………. eating kachori.
4. Mr. Natawarlal is a ………… man.
5. Don’t ………… a priest.

Answer:

1. disobey
2. dissimilar
3. dislike
4. dishonest
5. distrust

Let’s Write

1. Write five sentences about the wood.
2. Write ten sentences on Shravan Kumar. Take the help of the clues, obedient, devoted, dutiful, pilgrimage, forest, put aside, thirsty, old and blind, river.
3. Write a paragraph on gramophone.
4. You are a cricket player of your school team. Give a set of three or four rules of the game using ought to and ought not to.
5. Write a message to your friend stating that as you didn’t attend the school that day and you want him to send the homework for English given by the teacher.
Answer:
1. The wood :
The wood is a place of beasts. We find many trees in the wood. The trees provide us fuel, food and fruit. Rishis live in the wood. They practise penance there.

2. Shravan Kumar was a devoted and obedient son. He used to carry his old and blind parents for pilgrimage. One day he was passing through the forest around Ayodhya. His parents asked him to bring water. Shravan took a pot. He went to Saryu river to bring water. His pot produced a gurgling sound. King Dashrath drought him to be a big animal. He shot an arrow. It pierced shravan’s chest. Shravan asked Dashrath to give water to his parents just then he died.

3. Edison invented the first talking machine. It looked very funny. However, it could talk. It was a new machine. The modern gramophone or record player is an improvement on the talking machine. The present day telephones and wireless sets are also talking machines. We can play the records on the gramophone of our choice and hear songs. The modem cinema is called a talkie due to the same.

4.Three or four rules of the game-

1. The bowler ought not to cross the crease- line while bowling.
2. The bowler ought not to throw the ball above chest height of the batsman.
3. We ought to obey the orders of the umpire.
4. We ought to play a fair game.

5. Class Room Activity in School

MP Board Class 7th English Solutions

MP Board Class 8th Special English Solutions Quarterly Evaluation

Question 1.
Do as directed :

(i) Write the opposite words of the following :
(a) poet
(b) cow
(c) priest
(d) woman
Ans.
(a) poetess
(b) bull
(c) Priestess
(d) man.

(ii) Change the number :
(a) mice
(b) city
(c) life
(d) deer
Answer:
(a) mouse
(b) cities
(c) lives
(d) deer

(iii) Use the apporopriate prepositions in the balnk spaces:
(a) The colour the but is blue, (form/of/for)
(b) The bus goes from Mahow. pithampur. (to/into/in)
Answer:
(a) of
(b) to.

(iv) Use the correct form of the verbs in the sentences given below.
1. She has already. her home work. (finish)
2. Ravi always his work regularly, (do)
Answer:
1. finished
2. does

(v) Insert suitable articles in the following sentences :
(a) ………….. umbrella is used in the rainy season.
(b) The students join ………….. university after completily the schooling.
Answer:
(a) An
(b) a.

(vi) Change into interrogative :
(a) He write an essay.
(b) The boys are playing football.
Answer:
(a) Does he write an essay?
(b) Are the boys playing football.

How Did The Parents Nature Influence Mohan Question 2.
Give one word for :
(i) One who flies an aeroplane is called a …………..
(ii) One who goes in space in called an …………..
(iii) The doctor who takes care of your teeth is called a …………..
(iv) A person who tooks after the patients in a hospital is called a
Answer:
(i) pilot
(ii) astronant
(iii) dentist
(iv) doctor.

Class 8th English MP Board Question 3.
(A) Answer any six of the following questions in two to three sentences each:

Question 1
Who is the misssile man?
Answer:
Dr. APJ Abdul Kalam is the Missile Man.

Class 8th Quarterly Exam Paper MP Board Question 2
What was the plan of shekh chilli?
Answer:
Shekh Chilli’s plan was to buy a cow when I got some money.

Question 3
Why does narmada feel proud?
Answer:
The Narmada say that she feels proud because she flows in India where a river is considered as a living being and is worshipped as goddess.

Question 4
What is Maheshwar famous for?
Answer:
The humble palace of Devi Ahilya Bai is located at the ghat of Maheshwar.

MP Board Class 8 Supplementary Reader Solutions Question 5
Where did the part of wolves live?
Answer:
The pair of wolves lived in a cave in the Seeonee Hills.

Question 6
What did Abdul Kalam do to support his family?

Question 7
Where was the pack council meeting held?

B. Answer any two of the following questions in three to five sentences each:

8th English Quarterly Question Paper Question 1.
How did Abdul Kalam learn that persemceamce makes miracles?
Answer:
Once Kalam was invited his teacher Mr. Subramania Iyer. Iyer’s wife was an orthodox Hindu lady. So she refused to serve food to Kalam. Iyer, without showing any adverse reaction, served the food himself. It was his patience and gravity that prompted his wife another time to serve food to Kalam. This taught Kalam the lesson of perseverance.

Class 8 MP Board English Book Question 2.
How did Shrad become disabled?
Answer:
Sharad lost his right hand in a train ride. He left his seat in an express train after having supper to wash his hands. As he finished and turned, the train suddenly jerked throwing him off balance. He was flung night out of the door to lose his right hand.

Question 3.
How do you know about the origin of the Narmada?
Answer:
The Narmada has his origin from the mackal. ranges. There is a beautiful place name Amarbeanlake near anooppur. At amarkantal the Narmada appears from a source (Kand).

MP Board Solutions Class 8 English Supplementary Question 4.
Answer any of the following questions :

1. How can we get success?
Answer:
We can get success by fighting with all the difficulties that come in our life now and then.

2. Which lines of the poem ‘Don’t Quit’ do you like most and why?
Ans.
Life is queer with its twists and turns,

Question 5
Write the summary of any one of the following poems.
(a) ‘Don’t Quit’
(b) ‘Tree are the Kindest things I know’

Class 8 English MP Board Question 6.
write a short note on any one of the following topics.
(a) The Narmada
(b) Cleanliness.
Answer:
Do yourself.

Class 8th MP Board English Paper Question 7
Read the passage carefully and answer the questions that follow :

One day a little Indian girl was shut up in a room alone for a whole day by her father because she would not learn English. The child was Sarojini who later on. As Sarojini Naidu, became one of the most distinguished poets in the very language she had refused to learn.

Sarojini, like the rest of her family, was educated in her early years under her father’s own care. Her father wanted her to become a great mathematician or scientist, but she did not like doing sums or conducting experiments. She loved to dream and write poems. She wrote her first poem when she was hardly eleven. At the age of thirteen she wrote a long poem in the manner of the famous English poets. In this way began her poetic career.

After passing the Matriculation Examination in India, and before she was fifteen she went to England for higher education. It was during her stay in London that she met Sir Edmund Gosse, a distinguished man of letters of the time. He was very impressed by her knowledge, intelligence and learning. When he came to know from her that she wrote English verse, he asked her to show him some of her poems. On reading her poems, he was surprised to find that they were written in almost faultless English, but was disappointed that they had nothing of the East in them. They were all about English sights and this day on wards, she devoted herself to writing verses about India.

Class 8 MP Board Supplementary Reader Question 1.
Why was Sarojini shut up in a room?
Answer:
She was shut up in a room because she would not learn English.

Question 2
Where did Sarojini receive the heigher education?
Answer:
She received the higher education in England.

Question 3
What qualities of Sarojini’s character does the passage reveal?
Answer:
She devoted herself to writing verses about India.

MP Board Class 8 English Question 4
How were Sarojini’s early poems different from her later ones?
Answer:
Sarojini’s early poems were all about English sights but her lats poems were about India.

Question 5
What does the phrase ‘man of letters’ mean?
Answer:
Learned person.

Question 6
Give a suiable title to the above passage.

MP Board Solutions Class 8 English Question 8.
write an application to the Headmas-ter of your school requesting him to issue you a character certificate.
Or
You are Radhika. Write a letter to your friend Reena, inviting her to spend Diwali vacation with you.

Question 9.
Write an essay on any one of the following topics :

1. Teachers Day
2. Any Great Leader
3. Uses and Abuses of Science

Class 8 English Reader Question 10.
Answer any four of the following questions from the supplementary Reader:

1. What effect had the book, Shrawan-Pitra-Bhakti, on Mohan Das?
2. How did the parents’ Natures influence mohan?
3. Who is a Shatavadhani?
4. During his ship journey why did Gandhi feel relieved at night?
5. Why was Gandhiji called to Africa for the third time?

MP Board Class 8th English Solutions