JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 19)

A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the other end of spring is fixed at point O. If the body moves about O in a circular path on a smooth horizontal surface with constant angular speed 5 rad/s. Then the ratio of extension in the spring to its natural length will be :

1 : 2

2 : 3

2 : 5

1 : 1

Explanation

JEE Main 2023 (Online) 24th January Evening Shift Physics - Circular Motion Question 24 English Explanation

$\because k x=m \omega^{2}(\ell+x)$

$12.5(x)=\frac{1}{5}(5)^{2}(\ell+x)$

$\Rightarrow \frac{5}{2} x=\ell+x$

$\Rightarrow \frac{3}{2} x=\ell$

$\Rightarrow \frac{x}{\ell}=\frac{2}{3}$

JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 19)

Explanation

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