Question:

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

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When calculating dimensions, always ensure that the units cancel out properly and match the given dimensions.
Updated On: Mar 18, 2025
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Correct Answer: 0

Solution and Explanation

Step 1: Dimensions of Modulus of Elasticity

Modulus of Elasticity (Young's Modulus) = StressStrain \frac{\text{Stress}}{\text{Strain}}
Strain is dimensionless, so the dimension of Modulus of Elasticity = Dimension of Stress.
Stress = ForceArea=[M1L1T2][L2]=[M1L1T2] \frac{\text{Force}}{\text{Area}} = \frac{[M^1 L^1 T^{-2}]}{[L^2]} = [M^1 L^{-1} T^{-2}]

Step 2: Dimensions of Torque

Torque = Force×Distance \text{Force} \times \text{Distance}
[M1L1T2]×[L]=[M1L2T2] [M^1 L^1 T^{-2}] \times [L] = [M^1 L^2 T^{-2}]

Step 3: Dimension of the Measured Quantity

The measured quantity is Modulus of Elasticity per unit Torque.
Dimension=[M1L1T2][M1L2T2]=[M11L12T2+2] \text{Dimension} = \frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^2 T^{-2}]} = [M^{1 - 1} L^{-1 - 2} T^{-2 + 2}] =[M0L3T0]=[L3] = [M^{0} L^{-3} T^{0}] = [L^{-3}]

Step 4: Comparing with [MaLbTc] [M^a L^b T^c]

Comparing the dimensions: [M0L3T0]=[MaLbTc] [M^0 L^{-3} T^0] = [M^a L^b T^c] a=0,b=3,c=0 a = 0, \, b = -3, \, c = 0

Step 5: Given b=3 b = 3

Since we need to compare it with b=3 b = 3 , we take the magnitude of the value b=3 b = -3 .
Therefore, the value of c c is: 0 \boldsymbol{0}

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