JEE Physics 2023 Question with Solution

Given: Mass of the body is $m = 200 \, \text{g} = 0.2 \, \text{kg}$, spring constant is $k = 12.5 \, \text{N/m}$, and angular speed is $\omega = 5 \, \text{rad/s}$.

Find: The ratio of extension $x$ in the spring to its natural length $L_0$.

Let the natural length of the spring be $L_0$ and the extension be $x$. Then the radius of circular motion is $L_0 + x$. The spring force provides the required centripetal force, so

Substituting the given values,

Now simplify,

Therefore, the ratio of extension to natural length is $2:3$.

The solution working gives $x/L_0 = 2/3$, which corresponds to option C. The solution incorrectly labels the correct option as B.

Given: The body moves in a circle on a smooth horizontal surface, so no other horizontal force contributes to the centripetal force except the spring force.

The circular radius is not the natural length alone. It is the stretched length of the spring, that is $L_0 + x$.

Hence the centripetal-force equation is

Using $m = 0.2$ and $\omega = 5$,

Since $k = 12.5$,

So the correct option should be C.