JEE Physics 2023 Question with Solution

Given: A car accelerates from rest to $u$ and the energy spent is $E_1$. The energy required to accelerate it from $u$ to $2u$ is compared with $E_1$.

Find: The relation between the two energies and the correct option.

The kinetic energy is

$$E_k = \frac{1}{2}mv^2$$

For motion from rest to $u$,

$$E_1 = \frac{1}{2}mu^2$$

For speed $2u$, the kinetic energy is

$$E_k(2u) = \frac{1}{2}m(2u)^2 = 2mu^2$$

Hence the energy required to go from $u$ to $2u$ is

$$E_2 = 2mu^2 - \frac{1}{2}mu^2$$ $$E_2 = \frac{3}{2}mu^2$$

Using $E_1 = \frac{1}{2}mu^2$,

$$E_2 = 3E_1$$

Therefore, the required value is $3$, so the correct option is C. The answer key disagrees with the solution, and the solution has been followed.

Given: Kinetic energy is proportional to $v^2$.

Find: The ratio of energy intervals.

Since

$$E_k \propto v^2$$

energy from rest to $u$ is proportional to

$$u^2 - 0 = u^2$$

and energy from $u$ to $2u$ is proportional to

$$(2u)^2 - u^2 = 4u^2 - u^2 = 3u^2$$

So,

$$E_2 : E_1 = 3 : 1$$

Hence,

$$E_2 = 3E_1$$

Therefore, the correct option is C.