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JEE Main 2025 (28 Jan, Shift 1) · Units & Dimensions

Units & Dimensions question from JEE Main 2025 (28 Jan, Shift 1)

Q22JEE Main 2025 · 28 JAN · SHIFT 1Medium

Units & Dimensions

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 _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 0

  1. Modulus of elasticity =[ML−1T−2]= [\mathrm{ML^{-1}T^{-2}}]; torque =[ML2T−2]= [\mathrm{ML^2T^{-2}}].
  2. Yτ=[ML−1T−2][ML2T−2]=[M0L−3T0]\dfrac{Y}{\tau} = \dfrac{[\mathrm{ML^{-1}T^{-2}}]}{[\mathrm{ML^2T^{-2}}]} = [\mathrm{M^0L^{-3}T^0}]
  3. So a=0a = 0, b=−3b = -3, c=0c = 0.
  4. Hence c=0c = 0.
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