In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of [Mᵃ Lᵇ Tᶜ]. If b=3, the value of c is _________.

Solution

Physics sir · 4 steps

Write both dimensions down; the T⁻² in stress and in torque cancel exactly.

1. Modulus of elasticity. Any modulus is stress divided by strain. Strain has no dimensions, so the modulus has the dimensions of force per area: [ML⁻¹T⁻²].
2. Torque. Torque is force times distance, so it has dimensions [MLT⁻²] × [L] = [ML²T⁻²].

   [ML⁻¹T⁻²]/[ML²T⁻²] = [M⁰L⁻³T⁰]

3. Read off powers. Mass cancels and time cancels, so a = 0 and c = 0. Length comes out with a power of size 3, which matches the given b. Its sign only depends on which quantity sits on top, and the power of T is 0 either way.
4. So. The power of time is c = 0.

The idea to keep. For a ratio of two quantities, subtract the powers of M, L and T one at a time. Both stress and torque carry T⁻², so any ratio of the two has no time dependence at all.

Answer from the NTA final answer key. Solution by PrepWiser.