Question

What is the unit of $ k $ in the relation $ U = \dfrac{{ky}}{{{y^2} + {a^2}}} $ where $ U $ represents potential energy, $ y $ represents the displacements and ‘$a$’ represents the amplitude
(A) $ m{s^{ - 1}} $
(B) $ m\;s $
(C) $ J\;m $
(D) $ J{s^{ - 1}} $

Answer 1

Hint: To solve this question we need to understand all parameters given in the question. We then evaluate the dimensional formula of all the given parameters comparing them and then rearranging them using the transposition method to evaluate the value of $ k $ .

Complete step by step solution:
Here given that $ U $ is the potential energy which can be defined as the stored energy that depends on the relative position of the various parts of a system. The SI unit of potential energy is given as joules. The joules can be defined as $ 1 $ newton's force expanded over $ 1 $ meter and hence it can be given as the product of force and distance.
The unit of potential energy is given as $ kg{m^2}{s^{ - 2}} $ . Hence its dimensional formula is given as
 $ \Rightarrow U = \left[ {{M^1}{L^2}{T^{ - 2}}} \right] $ ………. $ (1) $
Where $ M $ represents mass in kilogram, $ L $ represents the length in meters, and $ T $ represents time in seconds.
As given $ y $ represents displacement which is in length, hence its dimensional formula of $ {y^2} $ is given as
 $ \Rightarrow {y^2} = \left[ {{M^2}} \right] $ ………. $ (2) $
Similarly $ a $ is also in meters because it is amplitude, hence the dimensional formula of $ {a^2} $ is also denoted by
 $ \Rightarrow {a^2} = \left[ {{M^2}} \right] $ ………. $ (3) $
Now substituting the values of dimensional formula from the equation $ (1) $ , $ (2) $ and $ (3) $ in the given equation
 $ \Rightarrow U = \dfrac{{ky}}{{{y^2} + {a^2}}} $
 $ \Rightarrow \left[ {{M^1}{L^2}{T^{ - 2}}} \right] = \dfrac{{k\left[ {{L^1}} \right]}}{{\left[ {{L^2}} \right]}} $
Now for evaluating the value of $ k $ we will use the transposition method
 $ \Rightarrow k = \dfrac{{\left[ {{L^2}} \right] \times \left[ {{M^1}{L^2}{T^{ - 2}}} \right]}}{{\left[ {{L^1}} \right]}} $
 $ \Rightarrow k = \left[ {{L^1}} \right] \times \left[ {{M^1}{L^2}{T^{ - 2}}} \right] $
 $ \therefore k = \left[ {{M^1}{L^3}{T^{ - 2}}} \right] $
Hence the value of $ k $ is given as $ \left[ {{M^1}{L^3}{T^{ - 2}}} \right] $ now converting it in units we get it is equal to Joules meter, so
  $ \therefore k = Jm $
Therefore the option (C) is the correct answer.

Note:
While calculating the dimensional formula we have to express the units of quantities in the KGS system. The same method can be applied to find the dimensional formula of any quantity by using the units of quantities in the KGS system and representing them in terms of parameters $ L $ , $ M $ , $ T $ , etc.