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Physics NEET Questions — Free MCQ Practice

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In a region, the electric field intensity E is given by $E = 100 / x^{2}$ where x is in metre. The potential difference between the points at $x = 10 m$ and $x = 20 m$ will be :

A.

$1 V$

B.

$2 V$

C.

$5 V$

D.

$10 V$

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Explanation

$3 . \vec{E} = \frac{100}{x^{2}} \hat{x} V_{B} - V_{A} = - \int_{10}^{20} \vec{E} . d \vec{a} = - \int_{10}^{20} \frac{100}{x^{2}} dx = + 100 {[\frac{1}{x}]}_{10}^{20} = 100 [\frac{1}{20} - \frac{1}{10}] = - 5 V ∴ |V_{B} - V_{A}| = 5 V$

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An uncharged capacitor with a solid dielectric is connected to a similar air capacitor charged to a potential of $V_{0}$ . If the common potential after sharing of charges becomes $V$ , then the dielectric constant of the dielectric must be –

A.

$\frac{V_{0}}{V}$

B.

$\frac{V}{V_{0}}$

C.

$\frac{V_{0} - V}{V}$

D.

$\frac{V_{0} - V}{V_{0}}$

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Explanation

3. Let $K$ be the dielectric constant of solid capacitor and if $C_{0}$ is capacitance of air capacitor then

solid capacitor will have capacitance ${KC}_{0}$ .

After charge sharing the common potential becomes $V$ .

$\Rightarrow V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} \Rightarrow V = \frac{{CV}_{0} + KC (0)}{C + KC} \Rightarrow K = \frac{V_{0} - V}{V}$

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Two parallel conducting plates $5 mm$ apart are held horizontally one above the other. The upper plate is maintained at a positive potential of $15 kV$ while the lower plate is earthed. If a small oil drop of relative density $0.92$ and of radius $5 µ m$ remains stationary between the plates, then the charge on the drop will be.

A.

$10 e$

B.

$8 e$

C.

$5 e$

D.

$3 e$

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Explanation

1. Electric field intensity between plates $E = \frac{V}{d} = \frac{15 \times 10^{3}}{5 \times 10^{- 3}} = 3 \times 10^{6} V / m$

Weight of the oil drop $= \frac{4}{3} {πa}^{3} ρg$

= $\frac{4}{3} π {(5 \times 10^{- 6})}^{3} \times (0.92 \times 10^{3}) (9.8)$

For equilibrium $qE = mg .$

$∴ q = \frac{mg}{E} = \frac{4 \times π \times 125 \times 0.92 \times 9.8 \times 10^{- 15}}{3 \times 3 \times 10^{6}} C = \frac{4 \times π \times 125 \times 0.92 \times 9.8 \times 10^{- 15}}{9 \times 10^{6} \times 1.6 \times 10^{- 19}} = 10 e$

electronic charge.

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An infinite number of charges (each of magnitude $1 µ c$ ) are placed along the $X$ -axis at x = 1, 2, 4, 8....metre. If the charges are alternately of opposite sign, then the potential at the point x = 0 due to these charges will be.

A.

$6 \times 10^{3} V$

B.

$9 \times 10^{3} V$

C.

$1.8 \times 10^{4} V$

D.

$1.2 \times 10^{4} V$

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Two charged conducting sphere of radii $R_{1}$ and $R_{2}$ , separated by a large distance, are connected by a long wire. The ratio of the charges on them is –

A.

$\frac{R_{1}}{R_{2}}$

B.

$\frac{R_{2}}{R_{1}}$

C.

$\frac{R_{1}^{2}}{R_{2}^{2}}$

D.

$\frac{R_{2}^{2}}{R_{1}^{2}}$

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Explanation

1. When connected by a long wire then

$\frac{Q_{1}}{Q_{2}} = \frac{C_{1} V}{C_{2} V} = \frac{4 {πε}_{0} R_{1} V}{4 {πε}_{0} R_{2} V} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{R_{1}}{R_{2}}$

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A proton and an $α$ -particle are at a distance r from each other. After letting them free if they move to infinity, the kinetic energy of the proton will be -

A.

$8 {Ke}^{2} / 5 r$

B.

$2 {Ke}^{2} / 5 r$

C.

$8 {Ke}^{2} / r$

D.

${Ke}^{2} / 5 r$

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Explanation

1. The potential energy in the stationary state $= \frac{K (2 e) (e)}{r}$

Let the velocity of proton be $v_{P}$ and velocity of the $α$ particle be $v_{α}$ then for the conservation of momentum

${mv}_{P} = - 4 {mv}_{α} ∴ v_{α} = \frac{- v_{P}}{4}$

Increase in kinetic energy = Loss in potential energy

$∴ \frac{1}{2} {mv}_{P}^{2} + \frac{1}{2} (4 m) v_{α}^{2} = K \frac{2 e^{2}}{r} or {mv}_{P}^{2} + 4 m (\frac{v_{P}^{2}}{16}) = K \frac{4 e^{2}}{r} or \frac{5 {mv}_{P}^{2}}{4} = K \frac{4 e^{2}}{r}$

Thus the kinetic energy of the proton $= \frac{1}{2} {mv}_{P}^{2} = \frac{8 {Ke}^{2}}{5 r}$

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$10^{3}$ small water drops, each of radius r and each carrying charge q, combine to form one bigger drop. The potential of a bigger drop as compared to that of a smaller drop will be.

A.

$10^{5} V$

B.

$10^{3} V$

C.

$10 V$

D.

$10^{2} V$

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Explanation

$4 . Volume of n small drops = Volume of 1 big drop \Rightarrow n \times \frac{4}{3} {πr}^{3} = \frac{4}{3} {πR}^{3} \Rightarrow R = n^{\frac{1}{3}} r V' = \frac{kQ}{R} = \frac{knq}{n^{\frac{1}{3}} r} = n^{2 / 3} V V = {(10^{3})}^{2 / 3} V V = 10^{2} V$

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Infinite charges, each of q coulomb are lying on the x-axis at x = 1m, 2m, 4m, 8m, --------.The electric potential due to these charges at x = 0 will be –

A.

$Kq$

B.

$\frac{Kq}{2}$

C.

$2 Kq$

D.

$\frac{Kq}{3}$

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Explanation

$3 . V = Kq (\frac{1}{1} + \frac{1}{2} + \frac{1}{4} + . . . . . \infty) = Kq (\frac{a}{1 - r}) = Kq (\frac{1}{1 - \frac{1}{2}}) = 2 Kq$

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The plates of a parallel plate capacitor have an area of $90 {cm}^{2}$ each and are separated by $2.5 mm$ . The capacitor is charged by a 400 volt supply. How much electrostatic energy is stored by the capacitor?

A.

$2.55 \times 10^{- 6} J$

B.

$1.55 \times 10^{- 6} J$

C.

$8.15 \times 10^{- 6} J$

D.

$5.5 \times 10^{- 6} J$

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Explanation

$Here, A = 90 {cm}^{2} = 90 \times 10^{- 4} m^{2}; d = 2.5 mm = 2.5 \times 10^{- 3} m; V = 400 volt C = \frac{ε_{0} A}{d} = \frac{8.854 \times 10^{- 12} \times 90 \times 10^{- 4}}{2.5 \times 10^{- 3}} = 3.187 \times 10^{- 11} F W = \frac{1}{2} {CV}^{2} = \frac{1}{2} \times 3.187 \times 10^{- 11} \times {(400)}^{2} = 2.55 \times 10^{- 6} J$

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From a supply of identical capacitors rated $8 mF$ , $250 V$ , the minimum number of capacitors required to form a composite $16 mF$ , $1000 V$ is :

A.

$2$

B.

$4$

C.

$16$

D.

$32$

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Explanation

Let 'n' such capacitors are in series and 'm' such branches are in parallel.

$∴ 250 \times n = 1000 ∴ n = 4 . . . . . . (i)$

Also $\frac{8}{n} \times m = 16$

$m = \frac{16 \times n}{8} = 8 . . . . . . (ii)$

$∴$ No. of capacitor $= 8 \times 4 = 32$

Physics MCQs for NEET — Practice Questions with Answers

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