JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 26)
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
125
Explanation
Acceleration of stone $a=\frac{v^{2}}{r}=\omega^{2} R$
$$ \begin{aligned} & a=\left(\frac{28 \times 2}{60} \times \frac{22}{7}\right)^{2} \times 1.8 \\\\ & =\frac{1936}{125} \end{aligned} $$
So, $x=125$
$$ \begin{aligned} & a=\left(\frac{28 \times 2}{60} \times \frac{22}{7}\right)^{2} \times 1.8 \\\\ & =\frac{1936}{125} \end{aligned} $$
So, $x=125$
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