ISRO Mechanical & Aerospace 2011 - Question 47

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A circular rod of 100 mm diameter and 500 mm length is subjected to a tensile force of 1000 kN. Determine the modulus of rigidity (G) if $E = 2 \times 10^5$ N/mm$^2$ and Poisson's = 0.3

  • A. $0.335 \times 10^5$ N/mm$^2$
  • B. $0.521 \times 10^5$ N/mm$^2$
  • C. $0.7692 \times 10^5$ N/mm$^2$
  • D. $0.2256 \times 10^5$ N/mm$^2$

Answer: C. $0.7692 \times 10^5$ N/mm$^2$

Explanation: $G = E / 2(1+\mu) = 2\times 10^5 / 2(1.3) = 10^5 / 1.3 = 0.7692 \times 10^5$.

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