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Physics MCQs for NEET — Practice Questions with Answers

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The energy stored in a condenser of capacity C which has been raised to a potential V is given by -

$\frac{1}{2} C V$

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

$\frac{1}{2 V C}$

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Explanation

$U = \int_{0}^{V} C V d V = \frac{1}{2} C V^{2}$

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If two conducting spheres are separately charged and then brought in contact -

The total energy of the two spheres is conserved

The total charge on the two spheres is conserved

Both the total energy and charge are conserved

The final potential is always the mean of the original potentials of the two spheres

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Explanation

Law of conservation of charge.

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Eight drops of mercury of equal radii possessing equal charges combine to form a big drop. Then the capacitance of bigger drop compared to each individual small drop is

8 times

4 times

2 times

32 times

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Explanation

Volume of 8 small drops = Volume of the big drop

$8 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 2 r As capacity is 4 {πε}_{0} r, hence capacity becomes 2 times .$

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A condenser of capacity 50 μF is charged to 10 volts. Its energy is equal to

2.5 × 10^–3Joule

2.5 × 10^–4Joule

5 × 10^–2Joule

1.2 × 10^–8Joule

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Explanation

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 50 \times 10^{- 6} \times {(10)}^{2} = 2 .5 \times 10^{- 3} J$

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A parallel plate condenser has a capacitance 50 μF in air and 110 μF when immersed in an oil. The dielectric constant ‘k’ of the oil is

0.45

0.55

1.10

2.20

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Explanation

$C_{m e d i u m} = K C_{a i r}$

⇒ $K = \frac{C_{m e d i u m}}{C_{a i r}} = \frac{110}{50} = 2 .20$

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Separation between the plates of a parallel plate capacitor is d and the area of each plate is A. When a slab of material of dielectric constant k and thickness t(t < d) is introduced between the plates, its capacitance becomes -

$\frac{ε_{0} A}{d + t (1 - \frac{1}{k})}$

$\frac{ε_{0} A}{d + t (1 + \frac{1}{k})}$

$\frac{ε_{0} A}{d - t (1 - \frac{1}{k})}$

$\frac{ε_{0} A}{d - t (1 + \frac{1}{k})}$

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The capacity of parallel plate condenser depends on

The type of metal used

The thickness of plates

The potential applied across the plates

The separation between the plates

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Explanation

$C = \frac{K ε_{0} A}{d}$

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The capacity of a parallel plate condenser is C. Its capacity when the separation between the plates is halved will be

4 C

2 C

$\frac{C}{2}$

$\frac{C}{4}$

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Explanation

$C = \frac{ε_{0} A}{d} . C' = \frac{ε_{0} A}{d / 2}$

⇒ C’ = 2C

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Eight small drops, each of radius r and having same charge q are combined to form a big drop. The ratio between the potentials of the bigger drop and the smaller drop is

8 : 1

4 : 1

2 : 1

1 : 8

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Explanation

$V o l u m e o f l a r g e d r o p = V o l u m e o f n s m a l l d r o p s \Rightarrow \frac{4}{3} {πR}^{3} = n \times \frac{4}{3} {πr}^{3} \Rightarrow R = n^{\frac{1}{3}} r Charge on large drop, Q = nq Potential at the centre of small drop, V_{small} = \frac{kq}{r} Potential at the centre of large drop, V_{Big} = \frac{KQ}{R} = \frac{nkq}{n^{\frac{1}{3}} r} = n^{\frac{2}{3}} \frac{kq}{r} = n^{\frac{2}{3}} V_{small} Put n = 8 - \frac{V_{B i g}}{V_{s m a l l}} = {(8)}^{2 / 3} = \frac{4}{1}$

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1000 small water drops each of radius r and charge q coalesce together to form one spherical drop. The potential of the big drop is larger than that of the smaller drop by a factor of

1000

100

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Explanation

$Potential of small drop : ‐ V = \frac{kq}{r} Charge on larger drop : ‐ Q_{net} = 1000 q Let radius of larger drop = R . \Rightarrow \frac{4}{3} {πR}^{3} = 1000 \times \frac{4}{3} {πr}^{3} R = 10 r Then potential of larger drop : ‐ V_{net} = \frac{kQ}{R} = \frac{K \times 1000 q}{10 r} V_{net} = \frac{kq}{r} \times 100 V_{net} = 100 V$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.