Question

Question: Given that force F is given by $F = P{t^{ - 1}} + Qt$, where t is the time. The unit of P is same as that of
A. Velocity
B. Acceleration
C. Displacement
D. Momentum

Answer 1

On the basis of homogeneity of dimensions principle, we can say that for any equation to be dimensionally correct, the dimensions of the terms on the left hand side and right hand side of that equation should be the same. In the solution of this problem, we will be comparing the dimensions of the terms on the right hand and left hand side to calculate the dimensions of unknown quantities.

Complete step by step answer:
The unit of force F is given by $Kg . m/s^2$ and its dimension is expressed as $\left[ {ML{T^{- 2}}} \right]$.
For the given equation $F = P{t^{ - 1}} + Qt$ to be dimensionally correct, the dimensions of the units on the right hand side and left hand side must be equal. This statement is drawn on the basis of the principle of homogeneity of dimensions so dimension of the term $\left( {P{t^{ - 1}}} \right)$ must be equal to dimension of the force F.

$$\left[ {P{T^{ - 1}}} \right] = \left[ {ML{T^{ - 2}}} \right]\\\ \Rightarrow\dfrac{{\left[ P \right]}}{{\left[ T \right]}} = \left[ {ML{T^{ - 2}}} \right]\\\ \therefore\left[ P \right] = \left[ {ML{T^{ - 1}}} \right]$$

We know that the unit momentum is $(Kg . m/s)$ and its dimension is $\left[ {ML{T^{ - 1}}} \right]$.
Therefore, the unit of P is the same as that of the unit of momentum and option (D) is correct.

Note: Alternate method: We know that the dimensions of two terms added or subtracted must be the same. So we can again find the dimension of $\left( {Qt} \right)$ by comparing it with the dimension of force F. By knowing the dimension of $\left( {Qt} \right)$ we can compare it with the dimension of $\left( {P{t^{ - 1}}} \right)$ to evaluate the dimension of P which will come out to be equal to dimension of momentum. Also, take extra care while rearranging the dimensions and express non-dimensional terms in the power rise to the power zero form.