A cylindrical rod of length 1 m and radius 4 cm is mounted vertically. It is subjected to a shear force of $ 10^5 $ N at the top. Considering infinitesimally small displacement in the upper edge, the angular displacement $ \theta $ of the rod axis from its original position would be: (shear moduli $ G = 10^{10} $ N/m$^2$)
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- \( \frac{1}{160\pi} \)
- \( \frac{1}{4\pi} \)
- \( \frac{1}{40\pi} \)
- \( \frac{1}{2\pi} \)
The Correct Option is A
Approach Solution - 1
To determine the angular displacement \( \theta \) of the rod axis from its original position, we need to use the formula for angular displacement due to shear force in a cylindrical body:
\(\theta = \frac{F \cdot L}{G \cdot A}\)
Where:
- \(F\) is the shear force applied at the top of the rod.
- \(L\) is the length of the rod.
- \(G\) is the shear modulus of the material.
- \(A\) is the cross-sectional area of the rod.
Given values are:
- \(F = 10^5\) N
- \(L = 1\) m
- \(G = 10^{10}\) N/m2
- The radius of the rod is 4 cm, which is \(0.04\) m
First, calculate the cross-sectional area \(A\) of the cylindrical rod:
\(A = \pi r^2 = \pi (0.04)^2 = 0.0016\pi\) m2
Substitute these values into the formula:
\(\theta = \frac{10^5 \cdot 1}{10^{10} \cdot 0.0016\pi}\)
Now, simplify:
\(\theta = \frac{10^5}{10^7 \cdot 0.0016\pi} = \frac{10^5}{1.6 \times 10^4 \pi}\)
\(\theta = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{10^5}{1.6 \times 10^4 \pi} = \frac{1}{160\pi}\)
Thus, the angular displacement \( \theta \) is \(\frac{1}{160\pi}\). Hence, the correct answer is:
\( \frac{1}{160\pi} \)
Approach Solution -2
The angular displacement \( \theta \) due to a shear force is given by: \[ \theta = \frac{F L}{G A} \] where:
- \( F = 10^5 \, \text{N} \) is the shear force, - \( L = 1 \, \text{m} \) is the length of the rod,
- \( G = 10^{10} \, \text{N/m}^2 \) is the shear modulus,
- \( A = \pi r^2 = \pi (0.04)^2 = 5.027 \times 10^{-3} \, \text{m}^2 \) is the cross-sectional area of the rod.
Substitute the values into the formula: \[ \theta = \frac{10^5 \times 1}{10^{10} \times 5.027 \times 10^{-3}} = \frac{10^5}{5.027 \times 10^7} = \frac{1}{160\pi} \]
Thus, the angular displacement is \( \frac{1}{160\pi} \), and the correct answer is (1).
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