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Physics MCQs for NEET — Practice Questions with Answers

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The electric field lines due to a single negative charge are represented by

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Explanation

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At the mid point of a line joining an electron and a proton, the values of E and V will be.

$E = 0, V \neq 0$

$E \neq 0, V = 0$

$E \neq 0, V \neq 0$

$E = 0, V = 0$

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Explanation

$2 . E = \frac{{Kq}^{2}}{x^{2}} + \frac{{Kq}^{2}}{x^{2}} = \frac{2 {Kq}^{2}}{x^{2}} towards electron V = \frac{Kq}{x} - \frac{Kq}{x} = 0 ∴ At mid point, E \neq 0, V = 0$

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A charge of $10 µ C$ is kept at the origin of $X - Y$ coordinate system. The potential difference in volts between two points (a, 0) and $(a / \sqrt{2}, a / \sqrt{2})$ will be.

Zero

$9 \times 10^{4}$

$\frac{9 \times 10^{4}}{a}$

$\frac{9 \times 10^{4}}{\sqrt{2}}$

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Explanation

1. The distance of both these points from the origin is a. Hence potential at these points will be

equal and potential difference will be zero.

$V = \frac{Kq}{a} - \frac{Kq}{a} = 0$

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Two point charge of $8 μC$ and $12 μC$ are kept in air at a distance of 10 cm from each other. The work required to change the distance between them to 6 cm will be.

$5.8 J$

$4.8 J$

$3.8 J$

$2.8 J$

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Explanation

1. Work = Increase in potential energy

$= {Kq}_{1} q_{2} (\frac{1}{r_{2}} - \frac{1}{r_{1}}) = 9 \times 10^{9} \times 8 \times 10^{- 6} \times 12 \times 10^{- 6} (\frac{1}{6 \times 10^{- 2}} - \frac{1}{10 \times 10^{- 2}}) = 9 \times 8 \times 12 (\frac{1}{6} - \frac{1}{10}) \times 10^{- 1} = 5.8 J$

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Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant K. The potential difference across the capacitors now becomes –

$\frac{3 V}{K + 2}$

$KV$

$\frac{V}{K}$

$\frac{3}{KV}$

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Explanation

1. Initial charge on first capacitor is $CV = Q_{1}$ .

Initial charge on second capacitor is $2 CV = Q_{2}$ .

Final capacitance of first capacitor is $KC$ .

If $V'$ is the common potential then

$V' = \frac{Q_{1} + Q_{2}}{C'_{1} + C_{2}} \Rightarrow V' = \frac{CV + 2 CV}{KC + 2 C} = \frac{3 V}{2 + K}$

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Maximum charge stored on a metal sphere of radius $15$ cm may be $7.5 µ C$ . The potential energy of the sphere in this case is :

$9.67 J$

$0.25 J$

$3.25 J$

$1.69 J$

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Explanation

$4 . Potential energy = \frac{1}{2} QV = \frac{1}{2} Q (\frac{KQ}{R}) U = \frac{1}{2} \frac{7.5 \times 10^{- 6} \times 7.5 \times 10^{- 6} \times 9 \times 10^{9}}{15 \times 10^{- 2}} = 1.687 J \approx 1.69 J$

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A parallel plate capacitor is charged to a certain potential difference. A slab of thickness 3 mm is inserted between the plates and it becomes necessary to increase the distance between the plates by 2.4 mm to maintain the same potential difference. The dielectric constant of the slab is–

2.5

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Explanation

$2 . \frac{σ}{ε_{0}} d = \frac{σ}{ε_{0}} (d + 2.4 - 3 + \frac{3}{k}) \Rightarrow d = d - 3 + 2.4 + \frac{3}{k} \Rightarrow 3 - 2.4 = \frac{3}{k} \Rightarrow 0.6 = \frac{3}{k} \Rightarrow k = 5$

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A charge $Q$ is distributed over two concentric hollow spheres of radii r and $R (R > r)$ such that the surface densities are equal. The potential at the common centre is $\frac{1}{4 {πε}_{0}}$ times –

$Q (\frac{r + R}{r^{2} + R^{2}})$

$\frac{Q}{2} (\frac{r + R}{r^{2} + R^{2}})$

$2 Q (\frac{r + R}{r^{2} + R^{2}})$

Zero

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Explanation

$1 . q_{r} + q_{R} = Q . . . . . . . . (1) \frac{q_{r}}{q_{R}} = \frac{4 {πr}^{2} σ}{4 {πR}^{2} σ} = \frac{r^{2}}{R^{2}} \Rightarrow \frac{q_{r}}{q_{R}} = \frac{r^{2}}{R^{2}} . . . . . . . . (2) From (1) and (2) q_{r} = \frac{{Qr}^{2}}{R^{2} + r^{2}} and q_{R} = \frac{{QR}^{2}}{R^{2} + r^{2}} So, V = \frac{q_{r}}{4 {πε}_{0} r} + \frac{q_{R}}{4 {πε}_{0} R} \Rightarrow V = \frac{Q}{4 {πε}_{0}} (\frac{r + R}{r^{2} + R^{2}})$

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A charge $+ Q$ is uniformly distributed over a thin ring of radius $R$ , velocity of an electron at the moment it passes through the centre $O$ of the ring, if the electron was initially at rest at a point $A$ which is very far away from the centre and on the axis of the ring is

$\sqrt{\frac{2 kQe}{mR}}$

$\sqrt{\frac{kQe}{m}}$

$\sqrt{\frac{kme}{QR}}$

$\sqrt{\frac{kQe}{mR}}$

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Explanation

$1 . U_{\infty} = 0, U_{0} (centre) = - \frac{KQe}{R} loss in P . E . = \frac{KQe}{R}, Gain in K . E . = \frac{1}{2} {mV}^{2} = \frac{KQe}{R} V = \sqrt{\frac{2 KQe}{Rm}}$

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The electric potential V as a function of distance x (in metre) is given by: $V = (5 x^{2} + 10 x - 9) V .$ The value of the electric field of x = 1m would be -

$20 V / m$

$6 V / m$

$11 V / m$

$- 23 V / m$

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Explanation

$1 . \vec{E} = - \frac{dV}{dx} \hat{i} \Rightarrow \vec{E} = (- 10 x - 10) \hat{i} At x = 12 m, \vec{E} = - 20 \hat{i} \Rightarrow |\vec{E}| = 20 {Vm}^{- 1}$

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