JEE Physics 2023 Question with Solution

Given: Mass of the particle is $m = 10 \, \text{g} = 10^{-2} \, \text{kg}$ and retardation is $a = -2x$.

Find: The value of $n$ if the loss of kinetic energy is $(10/x)^{-n} \, \text{J}$.

Using

$$a = v\frac{dv}{dx}$$

we get

$$v\frac{dv}{dx} = -2x$$

So,

$$v \, dv = -2x \, dx$$

Integrating from initial velocity $v_1$ to final velocity $v_2$, and displacement from $0$ to $x$,

$$\int_{v_1}^{v_2} v \, dv = -2 \int_0^x x \, dx$$ $$\frac{v_2^2}{2} - \frac{v_1^2}{2} = -\frac{2x^2}{2}$$ $$\frac{v_2^2}{2} - \frac{v_1^2}{2} = -x^2$$

Multiplying by $m$,

$$\frac{mv_1^2}{2} - \frac{mv_2^2}{2} = mx^2$$

Hence, loss of kinetic energy is

$$\Delta KE_{\text{loss}} = mx^2 = 10^{-2}x^2$$

Now,

$$10^{-2}x^2 = \left(\frac{10}{x}\right)^{-2}$$

Comparing with the given form $(10/x)^{-n}$, we get

$$n = 2$$

Therefore, the value of $n$ is $2$.

Given: The retardation varies with displacement as $a = -2x$.

Find: The exponent $n$ in the loss of kinetic energy expression.

By the work-energy idea, the retardation causes a decrease in kinetic energy. First rewrite acceleration as

$$a = v\frac{dv}{dx}$$

Thus,

$$v\frac{dv}{dx} = -2x$$ $$v \, dv = -2x \, dx$$

Integrating,

$$\int_{v_1}^{v_2} v \, dv = -2\int_0^x x \, dx$$ $$\frac{v_2^2 - v_1^2}{2} = -x^2$$

So,

$$\frac{1}{2}m(v_2^2 - v_1^2) = -mx^2$$

This means the change in kinetic energy is

$$KE_2 - KE_1 = -mx^2$$

Therefore, the loss of kinetic energy is

$$KE_1 - KE_2 = mx^2$$

Now substitute $m = 10 \, \text{g} = 0.01 \, \text{kg} = 10^{-2} \, \text{kg}$:

$$KE_1 - KE_2 = 10^{-2}x^2 \, \text{J}$$

Also,

$$\left(\frac{10}{x}\right)^{-2} = \left(\frac{x}{10}\right)^2 = \frac{x^2}{100} = 10^{-2}x^2$$

Hence the given expression matches when

$$n = 2$$

Therefore, the required value is $2$.