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Physics MCQs for NEET — Practice Questions with Answers

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An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms^-1 and the second part of mass 2kg moves with 8 ms^-1 speed. If the third part flies off with 4 ms^-1 speed, then its mass is

3kg

5kg

7kg

17kg

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Explanation

(b) We have p₁+p₂+p₃=0 [∴ P=mv]
∴ 1x12i+2x8j+p₃=0

=> 12i+16j+p₃=0

=> p₂=-(12i+16j)

p₃=√(12)²+(16)²=144+256=20 kg-m/s

Now, p₃=m₃v₃

=> m₃=p₃/v₃=20/4=5kg

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A small object of uniform density rolls up a curved surface with an initial velocity v'. It reaches up to a maximum height of 3v²/4g with respect to the initial position. The object is

ring

solid sphere

hollow sphere

disc

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Explanation

(d) As V=√2gh/1+k²/r²
Given h=3v²/4g

v²=2gh/1+k²/r²=2g3v²/4g(1+k²/r²)=6gv²/4g(1+u²/v²)

1=3/2(1+k²/v²)

or 1+k²/r²=3/2 or k²/r²=3/2-1=1/2

k²=1/2r² (Equation of disc)

Hence, the object is a disc

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When a mass is rotating in a plane about a fixed point, its angular momentum is directed along

a line perpendicular to the plane of rotation

the line making an angle of $45 °$ to the plane of rotation

the radius

the tangent to the orbit

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Explanation

We know L= $m (r \times v)$

So here, angular momentum is directed along a line perpendicular to the plane of rotation.

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Two persons of mass 55 kg and 65 kg respectively, are at the opposite ends of a boat.The length of the boat is 3.0m and weighs 100 kg.The 55 kg man walks up to the 65 kg man and sits with him.If the boat is in still water the centre of mass of the system shifts by

3.0m

2.3m

zero

0.75m

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Explanation

Here on the entire system net external force on the system is zero hence centre of mass remains unchanged.

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A solid cylinder of mass 3kg is rolling on a horizontal surface with velocity 4 $m s^{- 1} .$ It collides with a horizontal spring of force constant 200 $N m^{- 1}$ . The maximum compression produced in the spring will be

0.5 m

0.6 m

0.7 m

0.2 m

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Explanation

Loss in KE = Gain in spring energy

$\frac{1}{2} m v^{2} [1 + \frac{K^{2}}{R^{2}}] = \frac{1}{2} k x_{m a x}^{2}$

where k is the force contant.

Given,v=4m/s,m=3kg,k=200N/m

For solid cylinder, $\frac{K^{2}}{R^{2}} = \frac{1}{2}$

$∴ \frac{1}{2} \times 3 \times {(4)}^{2} [1 + \frac{1}{2}] = \frac{1}{2} \times 200 \times x_{m a x}^{2}$

The maximum compression in the spring

$x_{m a x} = 0.6 m$

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Two spheres A and B of masses $m_{1} a n d m_{2}$ respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction.The mass A moves after collision in the direction

same as that of B

opposite to that of B

$θ = \tan^{- 1} (\frac{1}{2}) t o t h e x ‐ a x i s$

$θ = \tan^{- 1} (\frac{- 1}{2}) t o t h e x ‐ a x i s$

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Three masses are placed on the x-axis:

300 g at origin, 500 g at x= 40 cm and 400g

at x=70 cm. The distance of the center of

mass from the origin is

40 cm

45 cm

50 cm

30 cm

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Explanation

$x_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}} = \frac{300 (0) + 500 (40) + 400 (70)}{300 + 500 + 400} = \frac{20000 + 28000}{1200} = 40 cm$

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The instantaneous angular position of a point on

a rotating wheel is given by the equation

$θ (t) = 2 t^{3} - 6 t^{2}$

The torque on the wheel becomes zero at

t=0.5 s

t=0.25 s

t=2 s

t=1 s

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Explanation

According to question, torque, $τ = 0$

It means that, $α = 0$

$α = \frac{d^{2} θ}{{dt}^{2}}$

Given $θ (t) = 2 t^{3} - 6 t^{2}$

so, $\frac{dθ}{dt} = 6 t^{2} - 12 t \frac{d^{2} θ}{{dt}^{2}} = 12 t - 12 (since, α = \frac{d^{2} θ}{{dt}^{2}} = 0) 12 t - 12 = 0 t = 1 s$

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The moment of inertia of a thin uniform rod of

mass M and length L about an axis passing

through its mid-point and perpendicular to its

length is $I_{0}$ . Its moment of inertia about an

axis passing through one of its ends and

perpendicular to its length is

$I_{0} + {ML}^{2} / 4$

$I_{0} + 2 {ML}^{2}$

$I_{0} + {ML}^{2}$

$I_{0} + {ML}^{2} / 2$

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Explanation

The theorem of parallel axis for moment of

inertia.

$I = I_{CM} + {Mh}^{2} I = I_{0} + M {(\frac{L}{2})}^{2} I = I_{0} + \frac{{ML}^{2}}{4}$

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A body of mass M hits normally a rigid wall with velocity v and bounces back with the same velocity. The impulse experienced by the body is

1.5 Mv

2 Mv

zero

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Explanation

Impulse $| J | = | ∆ p |$

$= M v - (- M v) = 2 M v$

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