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Physics MCQs for NEET — Practice Questions with Answers

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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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If a cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m, then coefficient of friction between the cycle tyres and road is

0.41

0.51

0.61

0.71

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Explanation

$μ = \frac{v^{2}}{r g} = \frac{{(4 .9)}^{2}}{4 \times 9 .8} = 0 .61$

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A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be

Double

Half

4 times

$\frac{1}{4}$ times

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Explanation

$F = \frac{m v^{2}}{r}$ ⇒ $F \propto v^{2}$ i.e. force will become 4 times.

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Two bodies of equal masses revolve in circular orbits of radii R₁ and R₂ with the same period. Their centripetal forces are in the ratio

${(\frac{R_{2}}{R_{1}})}^{2}$

$\frac{R_{1}}{R_{2}}$

${(\frac{R_{1}}{R_{2}})}^{2}$

$\sqrt{R_{1} R_{2}}$

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Explanation

$F = m (\frac{4 π^{2}}{T^{2}}) R$ . If masses and time periods are same then $F \propto R$

∴ $F_{1} / F_{2} = R_{1} / R_{2}$

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