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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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An object with a mass 10 kg moves at a constant velocity of 10 m/sec. A constant force then acts for 4 second on the object and gives it a speed of 2 m/sec in opposite direction. The acceleration produced in it, is

3 m/sec²

–3 m/sec²

0.3 m/sec²

–0.3 m/sec²

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Explanation

$\vec{a} = \frac{\vec{v_{2}} - \vec{v_{1}}}{t} = \frac{(- 2) - (+ 10)}{4} = \frac{- 12}{4} = - 3 m / s^{2}$

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The force acting on the object of mass 1 kg moving with acceleration $3 m / s^{2}$ is -

30 N

– 30 N

3 N

– 3 N

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Explanation

$F = m a = 1 \times (3) = 3 N$

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A machine gun is mounted on a 2000 kg car on a horizontal frictionless surface. At some instant the gun fires bullets of mass 10 gm with a velocity of 500 m/sec with respect to the car. The number of bullets fired per second is ten. The average thrust on the system is

550 N

50 N

250 N

250 dyne

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Explanation

u = velocity of bullet

$\frac{d m}{d t} =$ Mass thrown per second by the machine gun

= Mass of bullet × Number of bullet fired per second

$= 10 g \times 10 b u l l e t / \sec = 100 g / \sec = 0 .1 k g / \sec$

∴ Thrust $= \frac{u d m}{d t} = 500 \times 0 .1 = 50 N$

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