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Physics MCQs for NEET — Practice Questions with Answers

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If $V_{H}, V_{N}$ and $V_{O}$ denote the root-mean square velocities of molecules of hydrogen, nitrogen and oxygen respectively at a given temperature, then

$V_{N} > V_{O} > V_{H}$

$V_{H} > V_{N} > V_{O}$

$V_{O} > V_{N} > V_{H}$

$V_{O} > V_{H} > V_{N}$

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Explanation

$2 . v_{rms} \propto \frac{1}{\sqrt{M}} \Rightarrow V_{H} > V_{N} > V_{O} (∵ M_{H} < M_{N} < M_{O})$

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The r.m.s. speed of a group of 7 gas molecules having speeds $(6, 4, 2, 0, - 2, - 4, - 6) m / s$ is

$1.5 m / s$

$3.4 m / s$

$9 m / s$

$4 m / s$

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Explanation

$4 . v_{rms} = \sqrt{\frac{{(6)}^{2} + {(4)}^{2} + {(2)}^{2} + {(0)}^{2} + {(- 2)}^{2} + {(- 4)}^{2} + {(- 6)}^{2}}{7}} = 4 m / s$

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If the ratio of vapour density for hydrogen and oxygen is $\frac{1}{16}$ , then under constant pressure the ratio of their rms velocities will be

$\frac{4}{1}$

$\frac{1}{4}$

$\frac{1}{16}$

$\frac{16}{1}$

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Explanation

$1 . v_{rms} = \sqrt{\frac{3 P}{ρ}} \Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{ρ_{2}}{ρ_{1}}} = \sqrt{\frac{16}{1}} = \frac{4}{1}$

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Molecular motion shows itself as

Temperature

Internal Energy

Friction

Viscosity

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Explanation

1. Temperature

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What is the velocity of wave in monoatomic gas having pressure 1 kilopascal and density $2.6 kg / m^{3}$ ?

$3.6 m / s$

$8.9 \times 10^{3} m / s$

Zero

None of these

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Explanation

$4 . v = \sqrt{\frac{γP}{ρ}} = \sqrt{\frac{\frac{5}{3} \times 10^{3}}{2.6}} = 25 m / \sec$

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If the pressure in a closed vessel is reduced by drawing out some gas, the mean free path of the molecules

Is decreased

Is increased

Remains unchanged

Increases or decreases according to the nature of the gas

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Explanation

2. The distance travelled by a molecule in two consecutive collision is called free path. If pressure is reduced, a molecule will travel longer distance before collision, so mean free path is increased.

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At constant volume, temperature is increased. Then

Collision on walls will be less

Number of collisions per unit time will increase

Collisions will be in straight lines

Collisions will not change

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Explanation

2. As the temperature increases, the average velocity increases. So the collisions are faster.

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A cubical box with porous walls containing an equal number of $O_{2}$ and $H_{2}$ molecules is placed in a large evacuated chamber. The entire system is maintained at constant temperature T. The ratio of $v_{rms}$ of $O_{2}$ molecules to that of the $v_{rms}$ of $H_{2}$ molecules, found in the chamber outside the box after a short interval is

$\frac{1}{2 \sqrt{2}}$

$\frac{1}{4}$

$\frac{1}{\sqrt{2}}$

$\sqrt{2}$

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Explanation

2. Temperature remain constant so

$v_{rms} \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{v_{O_{2}}}{v_{H_{2}}} = \sqrt{\frac{M_{H_{2}}}{M_{O_{2}}}} = \sqrt{\frac{1}{16}} = \frac{1}{4}$

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The value of C_V for one mole of neon gas is

$\frac{1}{2} R$

$\frac{3}{2} R$

$\frac{5}{2} R$

$\frac{7}{2} R$

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Explanation

2. Neon gas is monoatomic and for monoatomic gases $C_{v} = \frac{3}{2} R$

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At constant volume, for different diatomic gases the molar specific heat is

Same and 3 cal/mole/°C approximately

Exactly equal and its value is 4 cal/mole/°C

Will be totally different

Approximately equal and its value is 5 cal/mole/°C

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Explanation

$4 . {(C_{V})}_{dia} = \frac{5}{2} R Joule / mol / ° C = \frac{5}{2} \frac{R}{J} \approx 5 \frac{cal}{mol - ° C}$

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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.