JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Work, Power & Energy

WhatsApp Email

A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface.

The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be

\frac{1}{2} xgh m / s

. The value of x is ______。

Answer: 3

Solution

Using energy conservation between positions P and Q:

Net loss in PE = Gain in KE

The 12 kg wheel descends height

h

, while the 3 kg mass rises height

h

12 g h - 3 g h = \frac{1}{2} (3) v^{2} + \frac{1}{2} (12) v^{2} + \frac{1}{2} I ω^{2}

For a disc/wheel,

I = \frac{1}{2} m r^{2} = \frac{1}{2} (12) r^{2}

and

ω = \frac{v}{r}

9 g h = \frac{1}{2} [3 + 12 + 12] v^{2} = \frac{27 v ^{2}}{2}

v^{2} = \frac{18 g h}{27} = \frac{2 g h}{3}

v = \frac{2 g h}{3} = \frac{1}{2} \frac{8 g h}{3}

Comparing with

v = \frac{1}{2} xg h

x = \frac{8}{3} \approx 2.67 \approx 3

Therefore,

x = 3

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs on Work, Power & Energy →

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Work, Power & Energy

Solution

More JEE Main questions on Work, Power & Energy