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Question: The displacement of a body at a particular second n is given by the expression \({S_n}^{th} = u + \l...

The displacement of a body at a particular second n is given by the expression Snth=u+(2n−1){S_n}^{th} = u + \left( {2n - 1} \right). The dimensional formula of Snth{S_n}^{th} in this equation is
(A) [M2L0T1]\left[ {{M^2}{L^0}{T^1}} \right].
(B) [M0L1T0]\left[ {{M^0}{L^1}{T^0}} \right].
(C) [M0L1T−1]\left[ {{M^0}{L^1}{T^{ - 1}}} \right].
(D) [M0L0T0]\left[ {{M^0}{L^0}{T^0}} \right].

Explanation

Solution

In the given expression Snth=u+(2n−1){S_n}^{th} = u + \left( {2n - 1} \right), Snth{S_n}^{th} is represented as the displacement of the body at a particular second n. To find the dimensional formula of Snth{S_n}^{th} in the given equation, formula of displacement should be use, also by using the dimensions of all the symbols that used in the displacement equation for the nth second, we can find the dimensional formula of Snth{S_n}^{th} for the given equation.

Complete step by step answer: Given: Snth=u+(2n−1){S_n}^{th} = u + \left( {2n - 1} \right)
Snth{S_n}^{th} = displacement of the body at nth second .
a= acceleration of the body.
u= initial velocity of the body.
Now taking L.H.S, we get,
Snth{S_n}^{th} =initialvelocity×time = initial velocity \times time =[M0L1T−1]×[M0L0T1]=[M0L1T0] = [{M^0}{L^1}{T^{ - 1}}] \times [{M^0}{L^0}{T^1}] = [{M^0}{L^1}{T^0}].
Now taking R.H.S , we get,
RHS=  u+a2(2n−1)RHS = \;u + \dfrac{a}{2}\left( {2n - 1} \right)
u= initial velocity = [M0L1T−1][{M^0}{L^1}{T^{ - 1}}]
a=acceleration= [M0L1T−2][{M^0}{L^1}{T^{ - 2}}]
Time=(2n-1)=n= [M0L0T1][{M^0}{L^0}{T^1}]
1/2 is the dimensionless quantity.
Therefore, R.H.S= [M0L1T−1][{M^0}{L^1}{T^{ - 1}}] ++ [M0L1T−2][{M^0}{L^1}{T^{ - 2}}] ×\times [M0L0T1][{M^0}{L^0}{T^1}] == [M0L1T0][{M^0}{L^1}{T^0}]
Hence, L.H.S=R.H.S
Therefore, Snth{S_n}^{th} dimensional formula is [M0L1T0][{M^0}{L^1}{T^0}].
Hence, option (B) is the correct option.

Note: Whenever we come up with this type of problem, you must remember the formula that the displacement travelled by a body during n seconds is Snth=  u+a2(2n−1){S_n}^{th} = \;u + \dfrac{a}{2}\left( {2n - 1} \right). Here, we first found the dimensional formula of the L.H.S of the equation, then found the dimensional formula R.H.S of the equation. After, equating LHS and RHS side we will get the dimensional formula of Snth{S_n}^{th}. of the given equation.