JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 22)
In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of $\left[M^a L^b T^c\right]$. If $b=3$, the value of $c$ is _________.
Answer
0
Explanation
Given, measured quantity $$ = {{Modulus\,of\,elasticity} \over {torque}}$$
$$ = {\sigma \over {\varepsilon \tau }} = {F \over {A\varepsilon \tau }}$$ where, $\sigma$ = stress, $$\varepsilon $$ = strain, $\tau$ = torque
$$ = {F \over {A\varepsilon FL}}$$
So, dimension $$ = {1 \over {[{L^2}][L]}}$$ ($$\varepsilon $$ is dimensionless)
$$ = [{L^{ - 3}}] = [{M^0}{L^{ - 3}}{T^0}] = [{M^a}{L^b}{T^c}]$$
$$ \Rightarrow c = 0$$
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