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Laws of Motion NEET Questions — Free MCQ Practice

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A balloon starts rising from the ground with an acceleration of 1.25 m/s² . After 8s, a stone is released from the balloon. The stone will (g = 10 m/s²)

A.

Reach the ground in 4 second

B.

Begin to move down after being released

C.

Have a displacement of 50 m

D.

Cover a distance of 40 m in reaching the ground

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Explanation

When the stone is released from the balloon. Its height $h = \frac{1}{2} a t^{2} = \frac{1}{2} \times 1 .25 \times {(8)}^{2} = 40 m$ and velocity

$v = a t = 1 .25 \times 8 = 10 m / s$

$T i m e t a k e n b y t h e s t o n e t o r e a c h t h e g r o u n d : h = u t + \frac{1}{2} a t^{2} - 40 = 10 t + \frac{1}{2} (- 10) t^{2} t^{2} - 2 t - 8 = 0 (t - 4) (t + 2) = 0 t = 4 s e c$

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A body is thrown vertically upwards with a velocity u. Find the true statement from the following

A.

Both velocity and acceleration are zero at its highest point

B.

Velocity is maximum and acceleration is zero at the highest point

C.

Velocity is maximum and acceleration is g downwards at its highest point

D.

Velocity is zero at the highest point and maximum height reached is $u^{2} / 2 g$

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Explanation

At highest point $v = 0$ and $H_{\max} = \frac{u^{2}}{2 g}$

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A man throws a ball vertically upward and it rises through 20 m and returns to his hands. What was the initial velocity (u) of the ball and for how much time (T) it remained in the air $[g = 10 m / s^{2}]$

A.

u = 10 m/s, T = 2s

B.

u = 10 m/s, T = 4s

C.

u = 20 m/s, T = 2s

D.

u = 20 m/s, T = 4s

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Explanation

$u = \sqrt{2 g h} = \sqrt{2 \times 10 \times 20} = 20 m / s$

and $T = \frac{2 u}{g} = \frac{2 \times 20}{10} = 4$ sec

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A particle when thrown, moves such that it passes from same height at 2s and 10s, the height is

A.

g

B.

2 g

C.

5 g

D.

10 g

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Explanation

let initial velocity of the particle= u

after time $t_{1}$ :

$h = u t_{1} - \frac{1}{2} g {t_{1}}^{2} . . . . . . (1)$

after time $t_{2}$ :

$h = u t_{2} - \frac{1}{2} g {t_{2}}^{2} . . . . . . (2)$

From (1) & (2) :

$u (t_{1} - t_{2}) = \frac{1}{2} g ({t_{1}}^{2} - {t_{2}}^{2}) u = \frac{1}{2} g (t_{1} + t_{2})$

Put it in eqution (1) -

$h = \frac{1}{2} g t_{1} (t_{1} + t_{2}) - \frac{1}{2} g {t_{1}}^{2} h = \frac{1}{2} g t_{1} t_{2}$

If $t_{1}$ and $t_{2}$ are the time, when body is at the same height then, $h = \frac{1}{2} g t_{1} t_{2} = \frac{1}{2} \times g \times 2 \times 10 = 10 g$

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If the body is moving in a circle of radius r with a constant speed v, its angular velocity is

A.

v²/r

B.

vr

C.

v/r

D.

r/v

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Explanation

$v = r ω$ ⇒ $ω = \frac{v}{r}$ = constant [As v and r are constant]

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Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁ and r₂ respectively. Their speeds are such that each makes a complete circle in the same duration of time t. The ratio of the angular speed of the first to the second car is

A.

m₁ : m₂

B.

r₁ : r₂

C.

1 : 1

D.

m₁ r₁ : m₂ r₂

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Explanation

As time periods are equal therefore the ratio of angular speeds will be one. $ω = \frac{2 π}{T}$

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If a particle moves in a circle describing equal angles in equal times, its velocity vector

A.

Remains constant

B.

Changes in magnitude

C.

Changes in direction

D.

Changes both in magnitude and direction

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Explanation

It is always directed in a direction of tangent to circle.

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A motor cyclist going round in a circular track at constant speed has

A.

Constant linear velocity

B.

Constant acceleration

C.

Constant angular velocity

D.

Constant force

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Explanation

Linear velocity, acceleration and force varies in direction.

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Two bodies of mass 10 kg and 5 kg moving in concentric orbits of radii R and r such that their periods are the same. Then the ratio between their centripetal acceleration is

A.

R/r

B.

r/R

C.

R²/r²

D.

r²/R²

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Explanation

$\frac{a_{R}}{a_{r}} = \frac{ω_{R}^{2} \times R}{ω_{r}^{2} \times r} = \frac{T_{r}^{2}}{T_{R}^{2}} \times \frac{R}{r} = \frac{R}{r}$ [As T_r = T_R]

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A particle is moving in a horizontal circle with constant speed. It has constant

A.

Velocity

B.

Acceleration

C.

Kinetic energy

D.

Displacement

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Explanation

K.E. $= \frac{1}{2} m v^{2}$ . Which is scalar, so it remains constant.

Laws of Motion MCQs for NEET — Physics Questions with Answers

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