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

Hint: 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.