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Physics NEET Questions — Free MCQ Practice

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A parallel plate condenser has a uniform electric

field E(V/m) in the space between the plates. If

the distance between the plates is d(m) and area

of each plate is $A (m^{2})$ , the energy (joule) stored

in the condenser is

A.

$\frac{1}{2} ε_{0} E^{2}$

B.

$ε_{0} EAd$

C.

$\frac{1}{2} ε_{0} E^{2} Ad$

D.

$E^{2} Ad / ε_{0}$

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Explanation

The energy stored in the condenser

$U = \frac{1}{2} {CV}^{2} U = \frac{1}{2} (\frac{{Aε}_{0}}{d}) (Ed)^{2} (so, C = \frac{{Aε}_{0}}{d} and V = Ed) U = \frac{1}{2} ε_{0} E^{2} Ad$

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A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 4V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it, as the first combination has. The value of C₂, in terms of C₁, is then

A.

$\frac{2 C_{1}}{n_{1} n_{2}}$

B.

16 $\frac{n_{2}}{n_{1}}$ C₁

C.

2 $\frac{n_{2}}{n_{1}}$ C₁

D.

$\frac{16 C_{1}}{n_{1} {n_{2}}_{}}$

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Explanation

Case I. When the capacitors are joined in series

U_series = $\frac{1}{2}$ $\frac{C_{1}}{n_{1}}$ (4V)²

Case II. When the capacitors are joined in parallel

U_parallel= $\frac{1}{2}$ (n₂C₂)V²

Given, U_series= U_parallel

or $\frac{1}{2}$ $\frac{C_{1}}{n_{1}}$ (4V)² = $\frac{1}{2}$ (n₂C₂)V²

$\Rightarrow$ C₂ = $\frac{16 C_{1}}{n_{2} n_{1}}$

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Three concentric spherical shells have radii a, b and c (a<b<c) and have surface charge densities $σ, - σ$ and $σ$ respectively. If $V_{A}, V_{B}$ and $V_{C}$ denote the potential of the three shells, if c=a+b, we have

A.

$V_{C} = V_{A} \neq V_{B}$

B.

$V_{C} = V_{B} \neq V_{A}$

C.

$V_{C} \neq V_{B} \neq V_{A}$

D.

$V_{C} = V_{B} = V_{A}$

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Explanation

$Here, Potential at the surface of A : V_{A} = \frac{1}{4 {πε}_{0}} \frac{σ 4 {πa}^{2}}{a} - \frac{1}{4 {πε}_{0}} \frac{σ 4 {πb}^{2}}{b} + \frac{1}{4 {πε}_{0}} \frac{σ 4 {πc}^{2}}{c} = \frac{σ}{ε_{0}} (a - b + c) = \frac{σ}{ε_{0}} (2 a) (∵ c = a + b)$

$Potential at the surface of B : V_{B} = \frac{1}{4 {πε}_{0}} \cdot \frac{σ 4 {πa}^{2}}{b} - \frac{1}{4 {πε}_{0}} \frac{σ 4 {πb}^{2}}{b} + \frac{1}{4 {πε}_{0}} \frac{σ 4 {πc}^{2}}{c} = \frac{σ}{ε_{0}} (\frac{a^{2}}{b} - b + c) = \frac{σ}{ε_{0}} (\frac{a^{2}}{b} + a) (∵ c = a + b) and V_{C} = \frac{1}{4 {πε}_{0}} \cdot \frac{σ 4 {πa}^{2}}{c} - \frac{1}{4 {πε}_{0}} \frac{σ 4 {πb}^{2}}{c} + \frac{1}{4 {πε}_{0}} \frac{σ 4 {πc}^{2}}{c} = \frac{σ}{ε_{0}} (\frac{a^{2}}{c} - \frac{b^{2}}{c} + c) = \frac{σ}{ε_{0}} (\frac{a^{2} - b^{2} + c^{2}}{c}) = \frac{σ}{ε_{0}} (\frac{a^{2} - b^{2} + {(a + b)}^{2}}{c}) = \frac{σ}{ε_{0}} (2 a) (∵ c = a + b) Hence, V_{A} = V_{C} \neq V_{B}$

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The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section A such that the uniform electric field between the plates is E, is

A.

$\frac{1}{2} ε_{0} E^{2} / A d$

B.

$ε_{0} E^{2} / A d$

C.

$ε_{0} E^{2} A d$

D.

$\frac{1}{2} ε_{0} E^{2} A d$

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Explanation

Energy given by the cell

$E = \frac{1}{2} C V^{2}$

Here, C = capacitance of condenser = $\frac{A ε_{0}}{d}$

V = potential difference across the plates = Ed

Therefore, E = $\frac{1}{2} (\frac{A ε_{0}}{d}) {(E d)}^{2}$

= $\frac{1}{2} A ε_{0} E^{2} d$

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100 capacitors each having a capacity of 10 μF are connected in parallel and are charged by a potential difference of 100 kV. The energy stored in the capacitors and the cost of charging them, if electrical energy costs 108 paise per kWh, will be

A.

10⁷ joule and 300 paise

B.

5 × 10⁶ joule and 300 paise

C.

5 × 10⁶ joule and 150 paise

D.

10⁷ joule and 150 paise

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Explanation

Energy stored in the capacitor $= \frac{1}{2} C V^{2} \times 100$

$= \frac{1}{2} \times 10 \times 10^{^{- 6}} \times {(100 \times 10^{3})}^{2} \times 100 = 5 \times 10^{6} J$

Electric energy costs $= 108 P a i s e p e r k W H$ $= \frac{108 P a i s e}{3 .6 \times 10^{6} J}$

∴ Total cost of charging $= \frac{2 \times 5 \times 10^{6} \times 108}{3 .6 \times 10^{6}} = 300 P a i s e$

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A 10 μF capacitor and a 20 μF capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor

A.

$\frac{400}{9} V$

B.

$\frac{800}{9} V$

C.

400 V

D.

200 V

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Explanation

Initially potential difference a cross each capacitor

$V_{1} = \frac{20}{(10 + 20)} \times 200 = \frac{400}{3} V$

and $V_{2} = \frac{10}{(10 + 20)} \times 200 = \frac{200}{3} V$

Finally common potential $V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$

$V = \frac{10 \times \frac{400}{3} + 20 \times \frac{200}{3}}{(10 + 20)} = \frac{800}{9} V$

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Three capacitors of capacitance 3 μF, 10 μF and 15 μF are connected in series to a voltage source of 100V. The charge on 15 μF is

A.

50 μC

B.

100 μC

C.

200 μC

D.

280 μC

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Explanation

$\frac{1}{C_{e q}} = \frac{1}{3} + \frac{1}{10} + \frac{1}{15} \Rightarrow C_{e q} = 2 μ F$

Charge on each capacitor

Q = C_eq × V

$\Rightarrow 2 \times 100 = 200 μ C$

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A parallel plate capacitor has capacitance C. If it is equally filled with parallel layers of materials of dielectric constants K₁ and K₂ its capacity becomes C₁. The ratio of C₁ to C is

A.

K₁ + K₂

B.

$\frac{K_{1} K_{2}}{K_{1} - K_{2}}$

C.

$\frac{K_{1} + K_{2}}{K_{1} K_{2}}$

D.

$\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

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Two identical capacitors, have the same capacitance C. One of them is charged to potential V₁ and the other to V₂. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

A.

$\frac{1}{4} C (V_{1}^{2} - V_{2}^{2})$

B.

$\frac{1}{4} C (V_{1}^{2} + V_{2}^{2})$

C.

$\frac{1}{4} C {(V_{1} - V_{2})}^{2}$

D.

$\frac{1}{4} C {(V_{1} + V_{2})}^{2}$

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Explanation

Initial energy of the system

$U_{i} = \frac{1}{2} C {V_{1}}^{2} + \frac{1}{2} C {V_{2}}^{2}$

When the capacitors are joined, common potential $V = \frac{C V_{1} + C V_{2}}{2 C} = \frac{V_{1} + V_{2}}{2}$

Final energy of the system

$U_{f} = \frac{1}{2} (2 C) V^{2} = \frac{1}{2} 2 C {(\frac{V_{1} + V_{2}}{2})}^{2} = \frac{1}{4} C {(V_{1} + V_{2})}^{2}$

Decrease in energy = $U_{i} - U_{f} = \frac{1}{4} C {(V_{1} - V_{2})}^{2}$

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Three capacitors of capacitance 3 μF are connected in a circuit. Then their maximum and minimum capacitances will be

A.

9 μF, 1 μF

B.

8 μF, 2 μF

C.

9 μF, 0 μF

D.

3 μF, 2 μF

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Explanation

$C_{\max} = n C = 3 \times 3 = 9 μ F$ , $C_{\min} = \frac{C}{n} = \frac{3}{3} = 1 μ F$

Physics MCQs for NEET — Practice Questions with Answers

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