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Laws of Motion MCQs for NEET — Physics Questions with Answers

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A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at constant speed of 4 m/sec. The tension in the string is 6 N, when the stone is at (g = 10 m/sec²)

Top of the circle

Bottom of the circle

Half way down

None of the above

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Explanation

$m g = 1 \times 10 = 10 N,$ $\frac{m v^{2}}{r} = \frac{1 \times {(4)}^{2}}{1} = 16$

Tension at the top of circle = $\frac{m v^{2}}{r} - m g = 6 N$

Tension at the bottom of circle = $\frac{m v^{2}}{r} + m g = 26 N$

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The tension in the string revolving in a vertical circle with a mass m at the end which is at the lowest position

$\frac{m v^{2}}{r}$

$\frac{m v^{2}}{r} - m g$

$\frac{m v^{2}}{r} + m g$

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Explanation

Tension = Centrifugal force + weight $= \frac{m v^{2}}{r} + m g$

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A coin, placed on a rotating turn-table slips, when it is placed at a distance of 9 cm from the centre. If the angular velocity of the turn-table is trippled, it will just slip, if its distance from the centre is

27 cm

9 cm

3 cm

1 cm

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Explanation

In the given condition friction provides the required centripetal force and that is constant. i.e. mω²r = constant

⇒ $r \propto \frac{1}{ω^{2}}$ ∴ $r_{2} = r_{1} {(\frac{ω_{1}}{ω_{2}})}^{2} = 9 {(\frac{1}{3})}^{2} = 1 c m$

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A bucket full of water is revolved in vertical circle of radius 2m. What should be the maximum time-period of revolution so that the water doesn't fall off the bucket

1 sec

2 sec

3 sec

4 sec

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Explanation

Balancing vertical force at highest point:

$m g = m r w^{2}$

Minimum angular velocity $ω_{\min} = \sqrt{g / R}$

$∴ T_{\max} = \frac{2 π}{ω_{\min}} = 2 π \sqrt{\frac{R}{g}}$ $= 2 π \sqrt{\frac{2}{10}} = 2 \sqrt{2} ≅ 3 s$

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A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is

$\frac{M L ω^{2}}{2}$

$M L ω^{2}$

$\frac{M L ω^{2}}{4}$

$\frac{M L^{2} ω^{2}}{2}$

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A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration $α$ . If the coefficient of friction between the rod and the bead is μ, and gravity is neglected, then the time after which the bead starts slipping is

$\sqrt{\frac{μ}{α}}$

$\frac{μ}{\sqrt{α}}$

$\frac{1}{\sqrt{μ α}}$

Infinitesimal

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A particle is moving with a constant speed along a straight line path. A force is not required to

Increase its speed

Decrease the momentum

Change the direction

Keep it moving with uniform velocity

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Explanation

Particle will move with uniform velocity due to inertia.

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Newton's second law gives the measure of

Acceleration

Force

Momentum

Angular momentum

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Explanation

$\vec{F} = m \vec{a}$

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A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then

t₁ > t₂

t₂ > t₁

t₁ = t₂

t₁ >> t₂

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Explanation

For stationary lift $t_{1} = \sqrt{\frac{2 h}{g}}$

and when the lift is moving up with constant acceleration $t_{2} = \sqrt{\frac{2 h}{g + a}} ∴ t_{1} > t_{2}$

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If the tension in the cable of 1000 kg elevator is 1000 kg weight, the elevator

Is accelerating upwards

Is accelerating downwards

May be at rest or accelerating

May be at rest or in uniform motion

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Explanation

Since T= mg, it implies that elevator may be at rest or in uniform motion.

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