JEE Main 2025 Jan 28 Shift 1, Physics Q23: Units And Dimensions

JEE Main2025Jan 28, Shift 1Physics

In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of $[M^{a} L^{b} T^{c}]$ . If $b = - 3$ , the value of $c$ is _____

Your answer

Solution

Begin with the dimensions of each quantity.

Modulus of elasticity $=$ stress/strain $\to$ has the dimensions of pressure: $[M L^{- 1} T^{- 2}]$ .

Torque has dimensions of energy: $[M L^{2} T^{- 2}]$ .

Ratio (modulus of elasticity per unit torque):

$\frac{[ M L ^{- 1} T ^{- 2} ]}{[ M L ^{2} T ^{- 2} ]} = [M^{0} L^{- 3} T^{0}]$ .

Concept behind this question

Units And DimensionsNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 28 Shift 1, Physics Q23: Units And Dimensions

Concept behind this question

More previous year questions on Units And Dimensions

Practice more Physics