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physics ncert exemplar solutions class 12th chapter two

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Electrostatic Potential and Capacitance

A capacitor $C$ is fully charged with voltage $V_{0}$ . After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance. $\frac{C}{2}$ . The energy loss in the process after the charge is distributed between the two capacitor is:

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alok kumar singh

Contributor-Level 10

Sol. $g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Units and Measurements

The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be:

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Payal Gupta

Contributor-Level 10

S. I. unit Pascal – second $= \frac{N}{m^{2}} - s e c$

= $\frac{M L T^{- 2} T}{L^{2}}$

= $M L^{- 1} T^{- 1}$

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physics ncert exemplar solutions class 12th chapter two Units and Measurement

The distance of the Sun from earth is 1.5 * 10¹¹ m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be

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alok kumar singh

Contributor-Level 10

1° ® 60'

Þ 60' ® 10°

L = 1.5 * 10¹¹ m

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3 months ago

physics ncert exemplar solutions class 12th chapter two Class 12th

A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations.

A. Key K is kept closed and plates of capacitors are moved apart using insulating handle.

B. Key K is opened and plates of capacitors are moved apart using insulating handle.

Choose the correct option(s).

(a) In A, Q remains the same but G changes

(b) In B, V remains the same but C changes

(c) In A, V remains the same hence Q changes

(d) In B ,Q remains the same hence V changes

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c), (d) Case A: When key K is kept closed and plates of capacitors are moved apart using insulating handle.
The battery maintains the potential difference across connected capacitor in every circumstance. The separation between two plates increases which in turn decreases its capacitance (C=? ₀A/d)and potential difference across connected capacitor continue to be the same as capacitor is still connected with battery. So the charge stored decreases as Q = CV.

Case B: When key K is opened and plates of capacitors are moved apart using insulating handle.The charge s

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physics ncert exemplar solutions class 12th chapter two Class 12th

If a conductor has a potential V≠0 and there are no charges anywhere else outside, then

(a) There must be charges on the surface or inside itself

(b) There cannot be any charge in the body of the conductor

(c) There must be charges only on the surface

(d) There must be charges inside the surface

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) The potential of a body is due to charge of the body and due to the charge of surrounding. If there are no charges anywhere else outside, then the potential of the body will be due to its own charge. If there is a cavity inside a conducting body, then charge can be placed inside the body. Hence there must be charges on its surface or inside itself. Also The charge resides on the outer surface of a closed charged conductor. Hence there cannot be any charge in the body of the conductor.

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physics ncert exemplar solutions class 12th chapter two Class 12th

In the circuit shown in figure initially key K₁is closed and key K₂ is open. Then K₁ is opened and K₂ is closed (order is important). [Take Q'₁ and Q'₂ as charges on C₁ and C₂ and V₁ and V₂ as voltage respectively.] Then, E

(a) Charge on C, gets redistributed such that V₁ = V₂

(b) Charge on C₁ gets redistributed such that Q'₁ = Q'₂

(c) Charge on C₁ gets redistributed such that C₁V₁ + C₂V₂ = C₁E

(d) Charge on C₁ gets redistributed such that Q'₁ + Q'₂=Q

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) Initially key K₁ is closed and key K₂ is open, the capacitor C₁ is charged by battery and capacitor C₂ is still uncharged. Now K₁ is opened and K₂ is closed, the capacitors C₁ and C₂ both are connected in parallel. The charge stored by capacitor C₁, gets redistributed between C₁ and C₂ till their potentials become same, i.e., V₂ = V₁.
By law of conservation of charge, the charge stored in capacitor Cx is equal to sum of charges on capacitors C₁ and C₂ when K₁ is opened and K₂ is closed, i.e., Q

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3 months ago

physics ncert exemplar solutions class 12th chapter two Class 12th

In the circuit shown in figure initially key K₁is closed and key K₂ is open. Then K₁ is opened and K₂ is closed (order is important). [Take Q’₁ and Q’₂ as charges on C₁ and C₂ and V₁ and V₂ as voltage respectively.] Then, E

(a) Charge on C, gets redistributed such that V₁ = V₂

(b) Charge on C₁ gets redistributed such that Q’₁ = Q’₂

(c) Charge on C₁ gets redistributed such that C₁V₁ + C₂V₂ = C₁E

(d) Charge on C₁ gets redistributed such that Q’₁ + Q’₂=Q

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3 months ago

physics ncert exemplar solutions class 12th chapter two Class 12th

In a region of constant potential

(a) The electric field is uniform

(b) The electric field is zero

(c) There can be no charge inside the region

(d) The electric field shall necessarily change if a charge is placed outside the region

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) We know, the electric field intensity E and electric potential V are dV related as E =- dV/dr or we can write |E|=ΔV/Δr

The electric field intensity E and electric potential V are related as E = 0 and for V = constant, dV/dr=0 this imply that electric field intensity E = 0.

If some charge is present inside the region then electric field cannot be zero at that region, for this V = constant is not valid.

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3 months ago

physics ncert exemplar solutions class 12th chapter two Class 12th

The work done to move a charge along an equipotential from A to B

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) If potential of two point is same then work done is also zero according to relation As W=qdv=q (final potential-initial potential).

Also W=-qdv= -q $\int_{1}^{2} E . d l$

For equipotential surface V₂-V₂=0 and W=0

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physics ncert exemplar solutions class 12th chapter two Class 12th

Equipotential surfaces

(a) are closer in regions of large electric fields compared to regions of lower electric fields

(b) will be more crowded near sharp edges of a conductor

(c) will be more crowded near regions of large charge densities

(d) will always be equally spaced

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (b), (c) Properties of equipotential surfaces

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