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Physics MCQs for NEET — Practice Questions with Answers

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A parallel plate capacitor is made by stocking n equally spaced plates connected alter- nately. If the capacitance between any two plates is x, then the total capacitance is,

A. nx B. $ nx^2$ C. n/x D. (n-1) x

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Explanation

When 'n' equally spaced plates are connected alternately to form a parallel plate capacitor, the effective number of capacitors in series is (n-1). The total capacitance for capacitors in series is given by the formula: $C_{total} = (n-1) imes x$.

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The electric potential V at any point x, y, z (all in metre) in space is given by $V = 4x^2$ volt. The electric field at the point (1m, 0, 2m) in $Vm^{–1}$ is

$ + 8 \hat i $

$ - 8 \hat i $

$ - 16 \hat i $

$ +16 \hat i $

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Explanation

$ V = 4x^2 \Rightarrow \bar E = - { dv \over dr} = -8 x $ Now put the value of x

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A parallel plate condenser with dielectric of constant K between the plates has a capacity C and is charged to potential V volt. The dielectric slab is slowly removed from between the plates and reinserted. The network done by the system in this process is

A. 0 B. $ { 1 \over 2 } ( K-1) cv^2 $ C. $ ( K-1) cv^2 $ D. $ cv^2 { (K-1 ) \over K } $

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Explanation

Knowledge base question

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A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and work done by the battery will be

A. 1/2 B. 2/1 C. 1 D. 1/4

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Explanation

Use equation, $ E = { -dV \over dx } $

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Two spherical conductors A and B of radii 1mm and 2mm are separated by a distance of 5mm and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of sphere of A and B is

A. 1:2 B. 2:1 C. 4:1 D. 1:4

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Explanation

When spheres are connected by a conducting wire their potentials become equal $ { C_1 \over C_2 } = { r_1 \over r_2 } = { 1 \over 2 } {q_1 \over q_2 } = { C_1V \over C_2 V } = { C_1 \over C_2 } = {1 \over 2 } $ Now, $ { E_1 \over E_2 } = { { Kq_1 \over r_1^2 } \over {Kq_2 \over r_2^2 } } $ Now find out ratio

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A parrallel plate capacitor of capacitance 5 F and plate separation 6 cm is connected to a 1 V battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is intro- duced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is

A. 2c B. 5c C. 3c D. 10c

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Explanation

$ Q = CV = 5 \mu c $ $ = C^1 = { A\varepsilon_0 \over d-(t -{ t\over k } ) } = { A \varepsilon_0 /d \over 1-({ t -t/k \over d } ) } $ Now put the value

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64 identical drops of mercury are charged simultaneously to the same potential of 10 volt. Assuming the drops to be spherical, if all the charged drops are made to combine to form one large drop, then its potential will be

A. 100V B. 320V C. 640V D. 160V

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Explanation

$ V = { 64q \over 4 \pi \varepsilon R } = { 64q \over 4 \pi \varepsilon_0 (4r) } $ $ C = A \pi \varepsilon_0 / d $

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Two metal plate form a parallel plate capacitor. The distance between the plates is d. A metal sheet of thickness d/2 and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases ?

A. 4:1 B. 3:1 C. 2:1 D. 5:1

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Explanation

$ C^1 = { A \varepsilon_0 \over d-t(1 - {1 \over k })}$ use this equation

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A parallel plate capacitor has plate of area A and separation d. It is charged to a potential difference $V_o$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

A. $ { A \epsilon _o V_0 ^2 \over d } $ B. $ { A \epsilon _o V_0 ^2 \over 2d } $ C. $ { A \epsilon _o V_0 ^2 \over 3d } $ D. $ { A \epsilon _o V_0 ^2 \over 4d } $

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Explanation

Work done = Final energy - Initial energy = $ { Q^2 \over 2C^1} - { Q^2 \over 2C } $

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Two identical capacitors have the same capacitance C. one of them is charged to a potential $V_1$ and the other to $V_2$. 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. $ { 1 \over 4 } c ( v_1 ^2 - v_2^2 ) $ B. $ { 1 \over 4 } c ( v_1 ^2 + v_2^2 ) $ C. $ { 1 \over 4 } c ( v_1 - v_2 )^2 $ D. $ { 1 \over 4 } c ( v_1 + v_2)^2 $

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Explanation

$U_i = { 1 \over 2 } C ( V_1^2 + V_2^2 ) $ and $ V = { q_1 + q_2 \over c_1 + c_2 } ={V_1 + V_2 \over 2 } $ $ U_f = { 1 \over 2 } (2C) V^2 Now find U_i U_f $

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