Question 2
A force is represented by $$F = ax^2 + bt^{1/2}$$, where $$x$$ = distance and $$t$$ = time. The dimensions of $$\frac{b^2}{a}$$ are :
Solution
$$F = ax^2 + bt^{1/2}$$. $$[F] = MLT^{-2}$$.
$$[a] = [F]/[x^2] = MLT^{-2}/L^2 = ML^{-1}T^{-2}$$.
$$[b] = [F]/[t^{1/2}] = MLT^{-2}/T^{1/2} = MLT^{-5/2}$$.
$$[b^2/a] = M^2L^2T^{-5}/(ML^{-1}T^{-2}) = ML^3T^{-3}$$.
The answer is Option (1): $$[ML^3T^{-3}]$$.
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