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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

A capillary tube of radius \(\mathrm{R}\) is immersed in water and water rises in it to a height \(\mathrm{H}\). Mass of water in the capillary tube is M. If the radius of the tube is doubled. Mass of water that will rise in the capillary tube will now be (A) \(\mathrm{M}\) (B) \(2 \mathrm{M}\) (C) \((\mathrm{M} / 2)\) (D) \(4 \mathrm{M}\)

A rubber ball when taken to the bottom of a \(100 \mathrm{~m}\) deep take decrease in volume by \(1 \%\) Hence, the bulk modulus of rubber is $\ldots \ldots \ldots . . .\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]$ (A) \(10^{6} \mathrm{~Pa}\) (B) \(10^{8} \mathrm{~Pa}\) (C) \(10^{7} \mathrm{~Pa}\) (D) \(10^{9} \mathrm{~Pa}\)

When \(100 \mathrm{~N}\) tensile force is applied to a rod of $10^{-6} \mathrm{~m}^{2}\( cross-sectional area, its length increases by \)1 \%$ so young's modulus of material is \(\ldots \ldots \ldots \ldots\) (A) \(10^{12} \mathrm{~Pa}\) (B) \(10^{11} \mathrm{~Pa}\) (C) \(10^{10} \mathrm{~Pa}\) (D) \(10^{2} \mathrm{~Pa}\)

Water raises in a vertical capillary tube upto a height of \(2.0\) \(\mathrm{cm}\). If tube is inclined at an angle of \(60^{\circ}\) with the vertical then the what length the water will rise in the tube. (A) \(2.0 \mathrm{~cm}\) (B) \(4.0 \mathrm{~cm}\) (C) \((4 / \sqrt{3}) \mathrm{cm}\) (D) \(2 \sqrt{2} \mathrm{~cm}\)

The specific heat at constant pressure and at constant volume for an ideal gas are \(\mathrm{C}_{\mathrm{p}}\) and \(\mathrm{C}_{\mathrm{v}}\) and adiabetic \(\&\) isothermal elasticities are \(E_{\Phi}\) and \(E_{\theta}\) respectively. What is the ratio of \(\mathrm{E}_{\Phi}\) and \(\mathrm{E}_{\theta}\) (A) \(\left(\mathrm{C}_{\mathrm{v}} / \mathrm{C}_{\mathrm{p}}\right)\) (B) \(\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)\) (C) \(\mathrm{C}_{\mathrm{p}} \mathrm{C}_{\mathrm{y}}\) (D) \(\left[1 / C_{p} C_{y}\right]\)
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