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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?
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The correct answer is:
95
We have, $\theta=2 \pi n=\frac{\left(\frac{v_{f}^{2}}{r^{2}}-\frac{v_{1}^{2}}{r^{2}}\right)}{\left(2 \frac{a}{r}\right)}$
$$
n=\frac{v_{f}^{2}-v_{i}^{2}}{(2 a r)(2 \pi)}=95
$$
$$
n=\frac{v_{f}^{2}-v_{i}^{2}}{(2 a r)(2 \pi)}=95
$$
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