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Physics MCQs for NEET — Practice Questions with Answers

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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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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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