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JEE Main
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JEE Main
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Physics
Chapter
Mechanical Properties of Solids

JEE Main · Physics · Mechanical Properties of Solids

A solid metal cylinder is fixed at its bottom end. A tangential...

A solid metal cylinder is fixed at its bottom end. A tangential force of 360π N360 \pi \text{ N} is applied to its top face, producing an angular displacement of 10−5 rad10^{-5} \text{ rad} in the cylinder. If the shear modulus of the material is 1010 N/m210^{10} \text{ N/m}^2, the diameter of the cylinder is :

Options

  1. A12 cm12 \text{ cm}
  2. B6 cm6 \text{ cm}
  3. C3.6 cm3.6 \text{ cm}
  4. D1.2 cm1.2 \text{ cm}

Correct answer

A. 12 cm12 \text{ cm}

Step-by-step solution

Given tangential force F=360π NF = 360 \pi \text{ N}, angular displacement θ=10−5 rad\theta = 10^{-5} \text{ rad}, and shear modulus G=1010 N/m2G = 10^{10} \text{ N/m}^2.

The shear modulus is given by G=FAθG = \dfrac{F}{A \theta}, where AA is the cross-sectional area.

Rearranging for AA:
A=FGθ=360π1010×10−5=36π×10−4 m2A = \dfrac{F}{G \theta} = \dfrac{360 \pi}{10^{10} \times 10^{-5}} = 36 \pi \times 10^{-4} \text{ m}^2

Since A=πr2A = \pi r^2, we have:
πr2=36π×10−4⇒r=6×10−2 m=6 cm\pi r^2 = 36 \pi \times 10^{-4} \Rightarrow r = 6 \times 10^{-2} \text{ m} = 6 \text{ cm}

The diameter of the cylinder is D=2r=12 cmD = 2r = 12 \text{ cm}.

Answer: 12 cm12 \text{ cm}
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