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JEE Main 202528 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Morning Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2025 · 28 January · Morning Shift · Physics · Units and Dimensions

In a measurement, it is asked to find modulus of elasticity per...

In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of [MaLbTc]\left[M^a L^b T^c\right]. If b=−3b=-3, the value of cc is ________

Correct answer

0

Step-by-step solution

[Modulus of elasticity] =ML−1 T−2=\mathrm{ML}^{-1} \mathrm{~T}^{-2} [Torque] = ML 2 T−2^2 \mathrm{~T}^{-2} [Modulus of elasticity per unit torque] =ML−1 T−2ML2 T−2=L−3=\frac{\mathrm{ML}^{-1} \mathrm{~T}^{-2}}{\mathrm{ML}^2 \mathrm{~T}^{-2}}=\mathrm{L}^{-3}
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