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Physics NEET Questions — Free MCQ Practice

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Two identical cylindrical vessels with their bases at the same level each contain a liquid of density; $ρ$ . The height of the liquid in one vessel is h₁ and that in the other vessel is h₂ $(h_{1} > h_{2})$ .The area of either base is A. The work done by gravity in equalizing the levels when the two vessels are connected is :

A.

$(h_{1} - h_{2}) g ρ$

B.

$(h_{1} - h_{2}) g A ρ$

C.

$\frac{1}{2} {(h_{1} - h_{2})}^{2} g A ρ$

D.

$\frac{1}{4} {(h_{1} - h_{2})}^{2} g A ρ$

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If the increase in the kinetic energy of a body is 22%, then the increase in the momentum will be

A.

22%

B.

44%

C.

10%

D.

300%

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Explanation

$P = \sqrt{2 m E} .$ If m is constant then

$\frac{P_{2}}{P_{1}} = \sqrt{\frac{E_{2}}{E_{1}}} = \sqrt{\frac{1 .22 E}{E}}$ ⇒ $\frac{P_{2}}{P_{1}} = \sqrt{1 .22} = 1 .1$

⇒ $P_{2} = 1 .1 P_{1}$ ⇒ $P_{2} = P_{1} + 0 .1 P_{1} = P_{1} + 10 % of P_{1}$

So the momentum will increase by 10%

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If a body of mass 200 g falls from a height 200 m and its total P.E. is converted into K.E. at the point of contact of the body with earth surface, then what is the decrease in P.E. of the body at the contact (g = 10 m/s²)

A.

200 J

B.

400 J

C.

600 J

D.

900 J

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Explanation

$Δ U = m g h = 0 .2 \times 10 \times 200$ = 400 J

∴ Gain in K.E. = decrease in P.E. = 400 J.

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If momentum is increased by 20%, then K.E. increases by

A.

44%

B.

55%

C.

66%

D.

77%

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Explanation

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

⇒ $\frac{E_{2}}{E_{1}} = {(\frac{P_{2}}{P_{1}})}^{2} = {(\frac{1 .2 P}{P})}^{2} = 1 .44$

⇒ $E_{2} = 1 .44 E_{1} = E_{1} + 0 .44 E_{1}$

$E_{2} = E_{1} + 44 %$ of $E_{1}$

i.e. the kinetic energy will increase by 44%

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The kinetic energy of a body of mass 2 kg and momentum of 2 Ns is

A.

1 J

B.

2 J

C.

3 J

D.

4 J

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Explanation

$E = \frac{P^{2}}{2 m} = \frac{{(2)}^{2}}{2 \times 2} = 1 J$

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The decrease in the potential energy of a ball of mass 20 kg which falls from a height of 50 cm is

A.

968 J

B.

98 J

C.

1980 J

D.

None of these

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Explanation

$Δ U = m g h = 20 \times 9 .8 \times 0 .5 = 98 J$

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A 0.5 kg ball is thrown up with an initial speed 14 m/s and reaches a maximum height of 8.0m. How much energy is dissipated by air drag acting on the ball during the ascent

A.

19.6 Joule

B.

4.9 Joule

C.

10 Joule

D.

9.8 Joule

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Explanation

If there is no air drag then maximum height

$H = \frac{u^{2}}{2 g} = \frac{14 \times 14}{2 \times 9 .8} = 10 m$

But due to air drag ball reaches up to height 8m only. So loss in energy

$= m g (10 - 8) = 0 .5 \times 9 .8 \times 2 = 9 .8 J$

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An ice cream has a marked value of 700 kcal. How many kilowatt- hour of energy will it deliver to the body as it is digested

A.

0.81 kWh

B.

0.90 kWh

C.

1.11 kWh

D.

0.71 kWh

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Explanation

It is based on conversion of units from kcal to KWh

$1 k c a l = 10^{3} C a l o r i e = 4200 J$ .................. (1)

$1 k W h = 1 k W X 1 h = 10^{3} \frac{J}{s} \times 3600 s = 3.6 \times 10^{6} J 1 J = \frac{1 k W h}{3.6 \times 10^{6}} ∴ 4200 J = \frac{4200 k W h}{3.6 \times 10^{6}} . . . . (2) F r o m (1) a n d (2), 1 k c a l = \frac{4200 k W h}{3.6 \times 10^{6}} 700 k c a l = \frac{700 \times 4200 k W h}{3.6 \times 10^{6}} = 0.81 k W h$

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A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1m/s so as to have same K.E. as that of the boy. The original speed of the man will be

A.

$\sqrt{2} m / s$

B.

$(\sqrt{2} - 1) m / s$

C.

$\frac{1}{(\sqrt{2} - 1)} m / s$

D.

$\frac{1}{\sqrt{2}} m / s$

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Explanation

$L e t, m a s s o f m a n = m t h e n, m a s s o f b o y = \frac{m}{2} v = v e l o c i t y o f b o y, V = v e l o c i t y o f m a n$

$K E o f t h e m a n = \frac{1}{2} K E o f t h e b o y \frac{1}{2} m V^{2} = \frac{1}{2} [\frac{1}{2} \frac{m}{2} v^{2}] - - - - - - - - - - (i)$

$W h e n m a n i n c r e a s e s h i s s p e e d b y 1 m / s t h e n, K E o f t h e m a n = K E o f t h e b o y \frac{1}{2} m {(V + 1)}^{2} = [\frac{1}{2} \frac{m}{2} v^{2}] - - - - - - - - - - - (i i)$

Solve both equations, $V = \frac{1}{\sqrt{2} - 1}$

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A particle of mass m at rest is acted upon by a force F for a time t. Its Kinetic energy after an interval t is

A.

$\frac{F^{2} t^{2}}{m}$

B.

$\frac{F^{2} t^{2}}{2 m}$

C.

$\frac{F^{2} t^{2}}{3 m}$

D.

$\frac{F t}{2 m}$

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Explanation

Kinetic energy $E = \frac{P^{2}}{2 m} = \frac{{(F t)}^{2}}{2 m} = \frac{F^{2} t^{2}}{2 m}$ [As P = F t]

Physics MCQs for NEET — Practice Questions with Answers

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