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…

Given: shear force F = 10^5 N, length L = 1 m, shear modulus G = 10^10 N/m^2, radius r = 4 cm = 0.04 m. Find: angular displacement of the rod axis. For a cylindrical rod under shear force, = F LG A where A is the cross-sectional area. First calculate the area: A = r^2 = (0.04)^2 = 0.0016 m^2 Substitute the values: = 10^5 110^10 0.0016 = 10^51.6 10^7 = 1160 Therefore, the angular displacement is 1160. The correct option is A. The solution also writes the area as A = r^2 = (0.04)^2 = 5.027 10^-3 m^2 which is numerically equivalent to 0.0016 m^2. Then, = 10^5 110^10 5.027 10^-3 = 10^55.027 10^7 = 1160 Hence the angular displacement remains 1160, so the correct option is A.

A cylindrical rod of length 1 m and radius 4 cm is mounted vertic… — Solution | JEE Physics

MCQEasyJEE 2025Young's Modulus, Bulk & Rigidity Modulus

A cylindrical rod of length $1 \, \text{m}$ and radius $4 \, \text{cm}$ is mounted vertically. It is subjected to a shear force of $10^5 \, \text{N}$ 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 $G = 10^{10} \, \text{N/m}^2$ )

A
$\frac{1}{160\pi}$
B
$\frac{1}{4\pi}$
C
$\frac{1}{40\pi}$
D
$\frac{1}{2\pi}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: shear force $F = 10^5 \, \text{N}$ , length $L = 1 \, \text{m}$ , shear modulus $G = 10^{10} \, \text{N/m}^2$ , radius $r = 4 \, \text{cm} = 0.04 \, \text{m}$ .

Find: angular displacement $\theta$ of the rod axis.

For a cylindrical rod under shear force,

\theta = \frac{F L}{G A}

where $A$ is the cross-sectional area.

First calculate the area:

A = \pi r^2 = \pi (0.04)^2 = 0.0016\pi \, \text{m}^2

Substitute the values:

\theta = \frac{10^5 \cdot 1}{10^{10} \cdot 0.0016\pi}

\theta = \frac{10^5}{1.6 \times 10^7 \pi}

\theta = \frac{1}{160\pi}

Therefore, the angular displacement is $\frac{1}{160\pi}$ . The correct option is A.

Using cross-sectional area explicitly

The solution also writes the area as

A = \pi r^2 = \pi (0.04)^2 = 5.027 \times 10^{-3} \, \text{m}^2

which is numerically equivalent to $0.0016\pi \, \text{m}^2$ .

Then,

\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}

Hence the angular displacement remains $\frac{1}{160\pi}$ , so the correct option is A.

Common mistakes

Using $r = 4 \, \text{m}$ instead of converting $4 \, \text{cm}$ to $0.04 \, \text{m}$ is incorrect because SI units must be consistent in the formula. Convert the radius to metres before calculating the area.
Taking the cross-sectional area as $A = \pi r$ instead of $A = \pi r^2$ is wrong because the rod has a circular cross section. Always use the full area formula for a circle.
Confusing shear strain relation and substituting an incorrect formula for $\theta$ leads to an incorrect result. Here the solution uses $\theta = \frac{F L}{G A}$ , so apply that relation directly.

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