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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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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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