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Physics · JEE | UNITS AND MEASUREMENT
हिन्दीThe position of a particle at time \( t \) is given by \( x = v_0 t + \frac{1}{2} a t^2 \). If the dimensions of \( x \) and \( t \) are \( [L] \) and \( [T] \) respectively, the dimensional formula of \( \frac{v_0^2}{a} \) is:
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Correct answer · Option B
\( [L] \)
Explanation
By the principle of homogeneity, each term has the same dimension as \( x \). Thus \( v_0 t = [L] \implies v_0 = [LT^{-1}] \), and \( a t^2 = [L] \implies a = [LT^{-2}] \). Therefore, \( \frac{v_0^2}{a} = \frac{[L^2 T^{-2}]}{[LT^{-2}]} = [L] \).
