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WBJEE · Physics · Rotational Motion

The velocity of a car travelling on a straight road is \(3.6 \mathrm{kmh}^{-1}\) at an instant of time. Now travelling with uniform acceleration for \(10 \mathrm{s}\), the velocity becomes exactly double. If the wheel radius of the car is \(25 \mathrm{cm}\), then which of the following is the closest to the number of revolutions that the wheel makes during this 10 s?

  1. A 84
  2. B 95
  3. C 126
  4. D 135
Verified Solution

Answer & Solution

Correct Answer

(B) 95

Step-by-step Solution

Detailed explanation

We have, \(\theta=2 \pi n=\frac{\left(\frac{v_{f}^{2}}{r^{2}}-\frac{v_{1}^{2}}{r^{2}}\right)}{\left(2 \frac{a}{r}\right)}\) \[ n=\frac{v_{f}^{2}-v_{i}^{2}}{(2 a r)(2 \pi)}=95 \]