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Physics MCQs for NEET — Practice Questions with Answers

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An electric fan has blades of length 30 cm as measured from the axis of rotation. If the fan is rotating at 1200 r.p.m, the acceleration of a point on the tip of the blade is about

1600 m/sec²

4740 m/sec²

2370 m/sec²

5055 m/sec²

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Explanation

$ω^{2} r = 4 π^{2} n^{2} r$ = $4 π^{2} {(\frac{1200}{60})}^{2} \times \frac{30}{100} = 4740 m / s^{2}$

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The angular speed of seconds needle in a mechanical watch is

$\frac{π}{30}$ rad/s

2π rad/s

π rad/s

$\frac{60}{π}$ rad/s

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Explanation

$ω = \frac{2 π}{T} = \frac{2 π}{60} = \frac{π}{30} r a d / s$

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A particle moves with constant speed v along a circular path of radius r and completes the circle in time T. The acceleration of the particle is

$2 π v / T$

$2 π r / T$

$2 π r^{2} / T$

$2 π v^{2} / T$

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Explanation

Acceleration $= ω^{2} r = \frac{v^{2}}{r} = ω v = \frac{2 π}{T} v$

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In uniform circular motion

Both the angular velocity and the angular momentum vary

The angular velocity varies but the angular momentum remains constant

Both the angular velocity and the angular momentum stay constant

The angular momentum varies but the angular velocity remains constant

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Explanation

$v = R ω .$ In U.C.M. ω = constant ∴ v = constant.

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If a_r and a_t represent radial and tangential accelerations, the motion of a particle will be uniformly circular if

a_r = 0 and a_t = 0

a_r = 0 but $a_{t} \neq 0$

$a_{r} \neq 0$ but a_t = 0

$a_{r} \neq 0$ and $a_{t} \neq 0$

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Explanation

In uniform circular motion tangential acceleration remains zero but magnitude of radial acceleration remains constant.

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A stone ties to the end of a string 1m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, what is the magnitude and direction of acceleration of the stone

$\frac{π^{2}}{4} m s^{- 2}$ and direction along the radius towards the centre

$π^{2} m s^{- 2}$ and direction along the radius away from the centre

$π^{2} m s^{- 2}$ and direction along the radius towards the centre

$π^{2} m s^{- 2}$ and direction along the tangent to the circle

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Explanation

a $= \frac{v^{2}}{r} = ω^{2} r$ $= 4 π^{2} n^{2} r = 4 π^{2} {(\frac{22}{44})}^{2} \times 1 = π^{2} m / s^{2}$

and its direction is always along the radius and towards the centre.

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A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In the first 2 sec, it rotates through an angle θ₁. In the next 2 sec, it rotates through an additional angle θ₂. The ratio of θ₂/θ₁ is

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Explanation

Using relation,

$θ = ω_{0} t + \frac{1}{2} α t^{2}$

$θ_{1} = \frac{1}{2} (α) {(2)}^{2} = 2 α$ …(i) (As $ω_{0} = 0, t = 2 \sec$ )

Now using same equation for t = 4 sec, ω₀ = 0

$θ_{1} + θ_{2} = \frac{1}{2} α {(4)}^{2} = 8 α$ …(ii)

From (i) and (ii), $θ_{1} = 2 α$ and $θ_{2} = 6 α$

∴ $\frac{θ_{2}}{θ_{1}} = 3$

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If the equation for the displacement of a particle moving on a circular path is given by $(θ) = 2 t^{3} + 0 .5$ , where θ is in radians and t in seconds, then the angular velocity of the particle after 2 sec from its start is

8 rad/sec

12 rad/sec

24 rad/sec

36 rad/sec

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Explanation

$ω = \frac{d θ}{d t} = \frac{d}{d t} (2 t^{3} + 0 .5) = 6 t^{2}$

at t = 2 s, $ω = 6 \times {(2)}^{2} = 24 r a d / s$

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For a particle in a non-uniform accelerated circular motion

Velocity is radial and acceleration is transverse only

Velocity is transverse and acceleration is radial only

Velocity is radial and acceleration has both radial and transverse components

Velocity is transverse and acceleration has both radial and transverse components

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Explanation

In a non-uniform circular motion, the particle possesses both centripetal as well as tangential acceleration.

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The coordinates of a moving particle at any time ‘t’ are given by x = αt³ and y = βt³. The speed of the particle at time ‘t’ is given by

$\sqrt{α^{2} + β^{2}}$

$3 t \sqrt{α^{2} + β^{2}}$

$3 t^{2} \sqrt{α^{2} + β^{2}}$

$t^{2} \sqrt{α^{2} + β^{2}}$

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Explanation

$x = α t^{3}$ and $y = β t^{3}$ (given)

$v_{x} = \frac{d x}{d t} = 3 α t^{2}$ and $v_{y} = \frac{d y}{d t} = 3 β t^{2}$

Resultant velocity $= v = \sqrt{v_{x}^{2} + v_{y}^{2}} = 3 t^{2} \sqrt{α^{2} + β^{2}}$

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