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Physics MCQs for NEET — Practice Questions with Answers

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A bomb of 12 kg explodes into two pieces of masses 4 kg and 8 kg. The velocity of 8kg mass is 6 m/sec. The kinetic energy of the other mass is

48 J

32 J

24 J

288 J

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Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is

4 : 1

$\sqrt{2} : 1$

1 : 2

1 : 16

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Explanation

$P = \sqrt{2 m E} .$ If E are same then $P \propto \sqrt{m}$

⇒ $\frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

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If the K.E. of a body is increased by 300%, its momentum will increase by

100%

150%

$\sqrt{300} %$

175%

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Explanation

Let initial kinetic energy, $E_{1} = E$

Final kinetic energy, $E_{2} = E + 300 %$ of E = 4E

As $P \propto \sqrt{E}$ ⇒ $\frac{P_{2}}{P_{1}} = \sqrt{\frac{E_{2}}{E_{1}}} = \sqrt{\frac{4 E}{E}} = 2$ ⇒ $P_{2} = 2 P_{1}$

⇒ $P_{2} = P_{1} + 100 %$ of $P_{1}$

i.e. Momentum will increase by 100%.

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A light and a heavy body have equal kinetic energy. Which one has a greater momentum ?

The light body

The heavy body

Both have equal momentum

It is not possible to say anything without additional information

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Explanation

$P = \sqrt{2 m E}$ if E are equal then $P \propto \sqrt{m}$

i.e. heavier body will possess greater momentum.

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If the K.E. of a particle is doubled, then its momentum will

Remain unchanged

Be doubled

Be quadrupled

Increase $\sqrt{2}$ times

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Explanation

$P = \sqrt{2 m E}$ ∴ $P \propto \sqrt{E}$

i.e. if kinetic energy of a particle is doubled the its momentum will becomes $\sqrt{2}$ times.

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Two bodies of different masses m₁ and m₂ have equal momenta. Their kinetic energies E₁ and E₂ are in the ratio

$\sqrt{m_{1}} : \sqrt{m_{2}}$

$m_{1} : m_{2}$

$m_{2} : m_{1}$

$m_{1}^{2} : m_{2}^{2}$

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Explanation

$E = \frac{P^{2}}{2 m}$ if bodies possess equal linear momenta then

$E \propto \frac{1}{m}$ i.e. $\frac{E_{1}}{E_{2}} = \frac{m_{2}}{m_{1}}$

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Two bodies A and B having masses in the ratio of 3 : 1 possess the same kinetic energy. The ratio of their linear momenta is then

3 : 1

9 : 1

1 : 1

$\sqrt{3} : 1$

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Explanation

$P = \sqrt{2 m E}$ ∴ $\frac{P_{1}}{P_{2}} = \sqrt{\frac{m_{1}}{m_{2}}}$ (if E = constant)

∴ $\frac{P_{1}}{P_{2}} = \sqrt{\frac{3}{1}}$

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Two bodies with kinetic energies in the ratio of 4 : 1 are moving with equal linear momentum. The ratio of their masses is

1 : 2

1 : 1

4 : 1

1 : 4

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Explanation

$E = \frac{p^{2}}{2 m}$ ∴ $m \propto \frac{1}{E}$ (If momentum are constant)

$\frac{m_{1}}{m_{2}} = \frac{E_{2}}{E_{1}} = \frac{1}{4}$

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If the kinetic energy of a body becomes four times of its initial value, then new momentum will

Becomes twice its initial value

Become three times its initial value

Become four times its initial value

Remains constant

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Explanation

$P = \sqrt{2 m E}$

∴ $P \propto \sqrt{E}$ i.e. if kinetic energy becomes four time then new momentum will become twice.

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A bullet is fired from a rifle. If the rifle recoils freely, then the kinetic energy of the rifle is

Less than that of the bullet

More than that of the bullet

Same as that of the bullet

Equal or less than that of the bullet

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Explanation

$E = \frac{P^{2}}{2 m} .$ If P = constant then $E \propto \frac{1}{m}$

i.e. kinetic energy of heavier body will be less. As the mass of gun is more than bullet therefore it possess less kinetic energy.

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