A stone tied to $180 \mathrm{~cm}$ long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is $\frac{1936}{x} ms^{-2}$. The value of $x$ ________. (Take $\pi=\frac{22}{7}$ )
Answer (integer)
125
Solution
Acceleration of stone $a=\frac{v^{2}}{r}=\omega^{2} R$
<br/><br/>$$
\begin{aligned}
& a=\left(\frac{28 \times 2}{60} \times \frac{22}{7}\right)^{2} \times 1.8 \\\\
& =\frac{1936}{125}
\end{aligned}
$$
<br/><br/>So, $x=125$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Uniform Circular Motion
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