JEE MAIN - Physics (2021 - 25th July Evening Shift - No. 12)

The force is given in terms of time t and displacement x by the equation

F = A cos Bx + C sin Dt

The dimensional formula of $${{AD} \over B}$$ is :
$$[{M^0}L{T^{ - 1}}]$$
$$[M{L^2}{T^{ - 3}}]$$
$$[{M^1}{L^1}{T^{ - 2}}]$$
$$[{M^2}{L^2}{T^{ - 3}}]$$

Explanation

$$[A] = [ML{T^{ - 2}}]$$

$$[B] = [{L^{ - 1}}]$$

$$[D] = [{T^{ - 1}}]$$

$$\left[ {{{AD} \over B}} \right] = {{[ML{T^{ - 2}}][{T^{ - 1}}]} \over {[{L^{ - 1}}]}}$$

$$\left[ {{{AD} \over B}} \right] = [M{L^2}{T^{ - 3}}]$$

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