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A motor cyclist moving with a velocity of 72 km/hour on a flat road takes a turn on the road at a point where the radius of curvature of the road is 20 meters. The acceleration due to gravity is 10 m/sec². In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than

$θ = \tan^{- 1} 6$

$θ = \tan^{- 1} 2$

$θ = \tan^{- 1} 25 .92$

$θ = \tan^{- 1} 4$

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Explanation

$v = 72 k m / h o u r = 20 m / \sec$

$θ = \tan^{- 1} (\frac{v^{2}}{r g}) = \tan^{- 1} (\frac{20 \times 20}{20 \times 10}) = \tan^{- 1} (2)$

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A particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance ‘r’. The speed of the particle is

Proportional to r²

Independent of r

Proportional to r

Proportional to 1/r

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Explanation

$\frac{m v^{2}}{r} \propto \frac{K}{r}$ ⇒ v ∝ r°

i.e. speed of the particle is independent of r.

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Two masses M and m are attached to a vertical axis by weightless threads of combined length l. They are set in rotational motion in a horizontal plane about this axis with constant angular velocity ω. If the tensions in the threads are the same during motion, the distance of M from the axis is

$\frac{M l}{M + m}$

$\frac{m l}{M + m}$

$\frac{M + m}{M} l$

$\frac{M + m}{m} l$

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Explanation

When two masses M and m are rotating with the same angular velocity Ï‰ about a vertical axis, their centripetal accelerations must be equal. This means that M(Ï‰r)^2 = m(Ï‰(l-r))^2, where r is the distance of M from the axis. Solving this equation gives r = ml/(M+m).

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A ball of mass 0.25 kg attached to the end of a string of length 1.96 m is moving in a horizontal circle. The string will break if the tension is more than 25 N. What is the maximum speed with which the ball can be moved

14 m/s

3 m/s

3.92 m/s

5 m/s

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Explanation

$T = \frac{m v^{2}}{r}$ ⇒ $25 = \frac{0 .25 \times v^{2}}{1 .96}$ ⇒ v = 14 m/s

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A body of mass 5 kg is moving in a circle of radius 1m with an angular velocity of 2 radian/sec. The centripetal force is

10 N

20 N

30 N

40 N

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Explanation

Centripetal force $= m r ω^{2} = 5 \times 1 \times {(2)}^{2} = 20 N$

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A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal circle. The maximum tension the string can withstand is 16 Newton. The maximum velocity of revolution that can be given to the stone without breaking it, will be

20 ms^–1

16 ms^–1

14 ms^–1

12 ms^–1

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Explanation

Maximum tension = $\frac{m v^{2}}{r} = 16 N$

⇒ $\frac{16 \times v^{2}}{144} = 16$ ⇒ v = 12 m/s

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Find the maximum velocity for skidding for a car moved on a circular track of radius 100 m. The coefficient of friction between the road and tyre is 0.2

0.14 m/s

140 m/s

1.4 km/s

14 m/s

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Explanation

$v_{\max} = \sqrt{μ r g} = \sqrt{0 .2 \times 100 \times 9 .8} = 14 m / s$

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An unbanked curve has a radius of 60m. The maximum speed at which a car can make a turn if the coefficient of static friction is 0.75, is

2.1 m/s

14 m/s

21 m/s

7 m/s

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Explanation

$v_{\max} = \sqrt{μ r g} = \sqrt{0 .75 \times 60 \times 9 .8} = 21 m / s$

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A ball of mass 0.1 Kg. is whirled in a horizontal circle of radius 1 m. by means of a string at an initial speed of 10 R.P.M. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is

5 r.p.m.

10 r.p.m.

20 r.p.m.

14 r.p.m.

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Explanation

$T = m ω^{2} r$ ⇒ $ω \propto \sqrt{T}$

∴ $\frac{ω_{2}}{ω_{1}} = \sqrt{\frac{1}{4}} \Rightarrow ω_{2} = \frac{ω_{1}}{2} = 5 r p m$

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A cyclist riding the bicycle at a speed of $14 \sqrt{3}$ ms^–1 takes a turn around a circular road of radius $20 \sqrt{3}$ m without skidding. Given g = 9.8 ms^–2, what is his inclination to the vertical

30^o

90^o

45^o

60^o

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Explanation

$θ = \tan^{- 1} (\frac{v^{2}}{r g}) = \tan^{- 1} [\frac{{(14 \sqrt{3})}^{2}}{20 \sqrt{3} \times 9 .8}] = \tan^{- 1} [\sqrt{3}]$ $= 60 °$

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