JEE PHYSICS MCQ: UNITS & MEASUREMENTS

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Analyzing +4 | -1

P=P_0 e^{-\left[\frac{\alpha t}{x+\beta}\right]}

, where

t

is time and

x

is displacement. Then dimension of

\left[\frac{\alpha^2}{\beta}\right]

is same as

View answer and explanation

Correct option: B

Acceleration

Option B is correct. By analyzing the dimensions of the given equation, it's found that the dimension of

\frac{{\alpha^2}}{{\beta}}

is the same as acceleration.

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