Question

The velocity of a car travelling on a straight road is $3.5km{h}^{-1}$ at an instant of time. Now travelling with uniform acceleration for $10s$ the velocity becomes exactly double. If the wheel radius of the car is $25cm$, then which of the following is the closest to the number of revolutions that the wheel makes during this $10s$?

• A. $84$
• B. $95$
• C. $126$
• D. $135$

Answer 1

The correct option is B $95$

Initial velocity is $3.5$ $\frac{km}{h}$

Time taken $=10s$, Final velocity $=7\frac{km}{h}$

Radius of the wheel (r) $=25cm$

Using, $v=u+at$

$\Rightarrow 7=3.5+a\times \frac{10}{3600}$

$\Rightarrow a=3.5\times 360km/h$

Angular acceleration $=\frac{a}{r}$

Angular Velocity $=\frac{u}{r}$

From: $\theta =\omega \times t+\frac{1}{2}\times \alpha \times t^2$

Net rotation of the wheel will be calculated and $\frac{\theta }{2\pi }$ will give total no. of revolutions 95.