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Chapter 7: Problem 969

Which is the dimensional formula for modulus of rigidity? (A) \(\mathrm{M}_{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (B) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\) (C) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-1}\) (D) \(\mathrm{M}^{1} \mathrm{~L}^{-2} \mathrm{~T}^{-2}\)

Short Answer

Expert verified
The dimensional formula for modulus of rigidity is (B) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\).

Step by step solution

01

Define the formula for shear stress

Shear stress is defined as the force applied per unit area. Mathematically, it is given by the formula: \[ \text{Shear stress} = \frac{\text{Force}}{\text{Area}} \] The dimensional formula for force is given by \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\), and the dimensional formula for the area is given by \(\mathrm{L}^{2}\).
02

Find the dimensional formula for shear stress

Using the formula for shear stress, we can substitute the dimensional formulas for force and area: \[ \text{Shear stress} = \frac{\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}}{\mathrm{L}^{2}} \] Now, simplify the expression: \[ \text{Shear stress} = \mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2} \]
03

Define the formula for shear modulus

The formula for shear modulus or modulus of rigidity is given by the following equation: \[ \text{Shear modulus} = \frac{\text{Shear stress}}{\text{Shear strain}} \] Shear strain is dimensionless, as it is the ratio of displacement or deformation to the initial length. Therefore, it does not have any dimensions.
04

Find the dimensional formula for modulus of rigidity

Since shear strain is dimensionless, the dimensional formula for shear modulus or modulus of rigidity will be the same as the dimensional formula for shear stress derived in Step 2: \[ \text{Modulus of rigidity} = \mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2} \] Comparing the result with the given options, we find that: The dimensional formula for modulus of rigidity is (B) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\).

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Most popular questions from this chapter

The amount of work done in blowing a soap bubble such that its diameter increases from \(d\) to \(D\) is \((T=\) Surface tension of solution) (A) \(4 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (B) \(8 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (C) \(\pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\) (D) \(2 \pi\left(\mathrm{D}^{2}-\mathrm{d}^{2}\right) \mathrm{T}\)

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