JEE MAIN - Physics (2023 - 31st January Evening Shift - No. 17)
A stone of mass $1 \mathrm{~kg}$ is tied to end of a massless string of length $1 \mathrm{~m}$. If the breaking tension of the string is $400 \mathrm{~N}$, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is :
$20 \mathrm{~ms}^{-1}$
$40 \mathrm{~ms}^{-1}$
$400 \mathrm{~ms}^{-1}$
$10 \mathrm{~ms}^{-1}$
Explanation
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$$ \begin{aligned} & T \sin \theta=\frac{m v^{2}}{l \sin \theta} \\\\ & \cos \theta=\frac{m g}{T} ........(1) \\\\ & \sin^2 \theta=\frac{m v^{2}}{T l} ........(2) \end{aligned} $$
From (1) and (2),
$1=\left(\frac{m g}{T}\right)^{2}+\frac{m v^{2}}{T l}$
$$ \Rightarrow $$ $1=\left(\frac{10}{400}\right)^{2}+\frac{v^{2}}{400}$
$$ \Rightarrow $$ $v^{2}=399.78$
$$ \Rightarrow $$ $v=20 \mathrm{~m} / \mathrm{s}$
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