Question

If the units of $ML$ are doubled, then the unit of kinetic energy will become:
(A) 8 times
(B) 16 times
(C) 4 times
(D) 2 times

Answer 1

We can recall the dimension of the kinetic energy of a body. If it exists, double the dimension $M$ and $L$, then compare to the original dimension.
Formula used: In this solution we will be using the following formulae;
$KE = \dfrac{1}{2}m{v^2}$ where $KE$ is the kinetic energy of a body, $m$ is the mass of the body and $v$ is the speed of the body.

Complete Step-by-Step solution:
The question explains us to find the unit of kinetic energy if the units of the M and L are doubled.
Now, to get the dimension of kinetic energy, we recall the formula which is given as
$KE = \dfrac{1}{2}m{v^2}$ where $KE$ is the kinetic energy of a body, $m$ is the mass of the body and $v$ is the speed of the body.
Hence, the dimension, which neglects constant, can be given as
$\left[ {KE} \right] = M{L^2}{T^{ - 2}}$ where the bracket signifies dimension of…., $M$ is the dimension of mass, $L$ is the dimension of length and $T$ is the dimension of time. m
Now, let us double the dimension M and L and let's call that $K{E_2}$, hence,
$\left[ {K{E_2}} \right] = 2M{\left( {2L} \right)^2}{T^{ - 2}}$
By simplifying, we get
$\left[ {K{E_2}} \right] = 2M\left( {4{L^2}} \right){T^{ - 2}} = 8ML{T^{ - 2}}$
Hence, by comparing with first kinetic energy, we have
$\left[ {K{E_2}} \right] = 8\left[ {KE} \right]$
Hence, the kinetic energy becomes 8 times the initial one.

Thus, the correct option is A.

Note: For clarity, we can derive the dimension of kinetic energy as follows
From $KE = \dfrac{1}{2}m{v^2}$
The dimension of mass is simply $M$
Now, as known, velocity is distance over time or length over time, hence, the dimension will be $\dfrac{L}{T} = L{T^{ - 1}}$
Now, we square the velocity, hence we get
${\left( {L{T^{ - 1}}} \right)^2} = {L^2}{T^{ - 2}}$
Then we multiply this by the dimension of mass
$\left[ {KE} \right] = {L^2}{T^{ - 2}} \times M = M{L^2}{T^{ - 2}}$ just as written above.
The dimensions of constants are given as 1.