Question

The velocity of a car travelling on a straight road is 3.5kmh−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?
$\begin{align}
  & \text{A}\text{. 84} \\
 & \text{B}\text{. 95} \\
 & \text{C}\text{. 126} \\
 & \text{D}\text{. 135} \\
\end{align}$

Answer 1

Hint: The given question is a question of kinematic circular motion. Use the kinematic equation of linear motion and kinematic equation of rotational motion. Unit of angular velocity in rad/sec. and unit of angular acceleration in rad/sec square.

Complete step by step solution:
A car travelling on a straight road is $3.5km/hr$ . when velocity gets double, the acceleration is uniform for 10 seconds. The radius of the wheel is 25cm. Then in that 10 seconds, what is the number of revolution is done by car.
Aim: which of the following is the closest to the number of revolutions that the wheel makes during this 10s?
So the initial velocity of a car is $3.5km/hr$
$u=3.5km/hr$
Time to travel car is 10second
$t=10s$
$\Rightarrow t=\dfrac {10}{3600}hr$
$\Rightarrow t=\dfrac {1}{360}hr$
Now it is given that acceleration is uniform when velocity get double so,
$v=7km/hr$
By kinematic equation, we have
$v=u+at$
Put values of u, v and t in the above equation,
$7=3.5+\dfrac{a}{360}$
$\Rightarrow 7-3.5=\dfrac{a}{360}$
$\Rightarrow 3.5=\dfrac{a}{360}$
$\Rightarrow a=3.5 \times 360 km/{{hr}^{2}}$
To calculate angular acceleration $\alpha$, we know that $\alpha =\dfrac{a}{r}$
Given that $r=25cm=25\times 10^{-5}km$
$\Rightarrow \alpha =\dfrac {3.5\times 360}{25\times {10}^{-5}}$
Converting hours square into seconds square we get,
$\Rightarrow \alpha =\dfrac {3.5\times 360}{25\times {10}^{-5}}\times \dfrac{1}{3600\times 3600}
= 3.88 rad /{{s}^{2}}........(1)$
To calculate angular velocity, $\omega =\dfrac{v}{r}$
Substituting, we have
$\Rightarrow \omega=\dfrac{3.5}{2.5}\times \dfrac{{{10}^{5}}}{3600}=38.8rad/s$
Similar to kinetic linear motion, for kinetic circular motion we have,
$\theta=\omega\times t+\dfrac{1}{2}\times \alpha \times t^{2}$
Where $\theta$ is the angle swept during the time $t$
Substituting, we have,
$\Rightarrow \theta =38.8\times 10+\dfrac{1}{2}3.88\times {{10}^{2}}=582rad$
No of revolution $N$ is then given as
$N=\dfrac{\theta }{2\pi }=\dfrac{582}{2\pi }=92.6$
And 92.6 is nearest to 95 revolutions.
Therefore option (B) is the correct option.

Note: To convert km into cm multiply with and to convert sec into hour divide with to equation (1).formula of a number of revolution is given by, No of revolution $=\dfrac{\theta }{2\pi }$ . When you calculate the revolution number might not be an exact whole number. So observe options then allocate the nearest matching value given options.