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Physics MCQs for NEET — Practice Questions with Answers

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The average force necessary to stop a bullet of mass 20 g moving with a speed of 250 m/s, as it penetrates into the wood for a distance of 12 cm is

2.2 × 10³ N

3.2 × 10³ N

4.2 × 10³ N

5.2 × 10³ N

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Explanation

$u = 250 m / s$ , $v = 0$ , $s = 0 .12 m e t r e$

$F = m a = m (\frac{u^{2} - v^{2}}{2 s}) = \frac{20 \times 10^{- 3} \times {(250)}^{2}}{2 \times 0 .12}$

∴ $F = 5 .2 \times 10^{3} N$

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A man measures time period of a pendulum (T) in stationary lift. If the lift moves upward with acceleration $\frac{g}{4},$ then new time period will be

$\frac{2 T}{\sqrt{5}}$

$\frac{\sqrt{5} T}{2}$

$\frac{\sqrt{5}}{2 T}$

$\frac{2}{\sqrt{5} T}$

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Explanation

$T = 2 π \sqrt{\frac{l}{g}}$ ⇒ $\frac{T^{'}}{T} = \sqrt{\frac{g}{g^{'}}} = \sqrt{\frac{g}{g + \frac{g}{4}}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$

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If rope of lift breaks suddenly, the tension exerted by the surface of lift (a = acceleration of lift)

m(g + a)

m(g – a)

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Explanation

If rope of lift breaks suddenly, acceleration becomes equal to g so that tension, $T = m (g - g) = 0$

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A lift of mass 1000 kg is moving with an acceleration of 1 m/s² in upward direction. Tension developed in the string, which is connected to the lift, is

9,800 N

10,000 N

10,800 N

11,000 N

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Explanation

$T = m (g + a) = 1000 (9 .8 + 1) = 10800 N$

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A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively

g, g

g – a, g – a

g – a , g

a, g

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Explanation

Due to relative motion, acceleration of ball observed by observer in lift = (g – a) and for man on earth the acceleration remains g.

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A monkey of mass 20kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can climb up along the rope (g = 10 m/s²)

10 m/s²

25 m/s²

2.5 m/s²

5 m/s²

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Explanation

Tension the string $= m (g + a) =$ Breaking force

⇒ $20 (g + a) = 25 \times g$ ⇒ $a = g / 4 = 2 .5 m / s^{2}$

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A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of a. What will be the angle of inclination with vertical

tan^–1(a/g)

tan^–1(g/a)

cos^–1(a/g)

cos^–1(g/a)

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A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be

$m g \cos θ$

$m g \sin θ$

$m g / \cos θ$

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Explanation

When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right.

For the condition of equilibrium

$m g \sin θ = m a \cos θ$ ⇒ $a = \frac{g \sin θ}{\cos θ}$

∴ Force exerted by the wedge on the block

$R = m g \cos θ + m a \sin θ$

R $= m g \cos θ + m (\frac{g \sin θ}{\cos θ}) \sin θ = \frac{m g (\cos^{2} θ + \sin^{2} θ)}{\cos θ}$

R $= \frac{m g}{\cos θ}$

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The linear momentum p of a body moving in one dimension varies with time according to the equation p = a + bt² where a and b are positive constants. The net force acting on the body is

A constant

Proportional to t²

Inversely proportional to t

Proportional to t

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Explanation

$F = \frac{d p}{d t} \equiv \frac{d}{d t} (a + b t^{2}) = 2 b t$

∴ $F \propto t$

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A man of weight 80 kg is standing in an elevator which is moving with an acceleration of 6 m/s² in upward direction. The apparent weight of the man will be (g = 10 m/s²)

1480 N

1280 N

1380 N

None of these

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Explanation

The apparent weight of man,

$R = m (g + a) = 80 (10 + 6) = 1280 N$

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