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Laws of Motion MCQs for NEET — Physics Questions with Answers

Practice free Laws of Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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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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A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity ω₀. If the length of the string and angular velocity are doubled, the tension in the string which was initially T₀ is now

T₀

T₀/2

4 T₀

8 T₀

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Explanation

Tension in the string $T_{0} = m R ω_{0}^{2}$

In the second case $T = m (2 R) (4 ω_{0}^{2}) = 8 m R ω_{0}^{2} = 8 T_{0}$

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In a circus stuntman rides a motorbike in a circular track of radius R in the vertical plane. The minimum speed at highest point of track will be

$\sqrt{2 g R}$

2gR

$\sqrt{3 g R}$

$\sqrt{g R}$

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Explanation

Minimum speed at the highest point of vertical circular path $v = \sqrt{g R}$

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A block of mass m at the end of a string is whirled round in a vertical circle of radius R. The critical speed of the block at the top of its swing below which the string would slacken before the block reaches the top is

(Rg)²

R/g

$\sqrt{R g}$

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Explanation

At highest point $\frac{m v^{2}}{R} = m g$

⇒ $v = \sqrt{g R}$

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A bucket tied at the end of a 1.6 m long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position (Take g = 10 m/s²)

4 m/sec

6.25 m/sec

16 m/sec

None of the above

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Explanation

Critical velocity at highest point $= \sqrt{g R} = \sqrt{10 \times 1 .6}$ = 4 m/s

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