The metal cube of side 10 cm is subjected to a shearing stress of 104N m-2. The modulus of rigidity if the top of the cube is displaced by 0.05 cm with respect to its bottom is

Question

The metal cube of side 10 cm is subjected to a shearing stress of 104N m-2. The modulus of rigidity if the top of the cube is displaced by 0.05 cm with respect to its bottom is (A) 2×106 N m-2 (B) 105N m-2 (C) 1×107 N m-2 (D) 4×105 N m-2

Tardigrade Admin · Accepted Answer

Here, l=10 cm, Δ l=0.05 cm Shearing stress=104 N m-2 Shearing strain=(Δ l/l) =(0.05/10) ∴ η =(Shearing stress/Shearing strain) =(104/0.005) =2×106 N m-2

Q. The metal cube of side $10 c m$ is subjected to a shearing stress of $1 0^{4} N$ $m^{- 2}$ . The modulus of rigidity if the top of the cube is displaced by $0.05 c m$ with respect to its bottom is

$2 \times 1 0^{6} N$ $m^{- 2}$

$1 0^{5} N$ $m^{- 2}$

$1 \times 1 0^{7} N$ $m^{- 2}$

$4 \times 1 0^{5} N$ $m^{- 2}$

Here, $l = 10 c m$ , $Δ l = 0.05 c m$
Shearing stress $= 1 0^{4} N$ $m^{- 2}$
Shearing strain $= \frac{Δ l}{l}$ $= \frac{0.05}{10}$
$∴$ $η$ $= \frac{S h e a r in g s t ress}{S h e a r in g s t r ain}$ $= \frac{1 0 ^{4}}{0.005}$ $= 2 \times 1 0^{6} N m^{- 2}$

Q. The metal cube of side 10cm is subjected to a shearing stress of 104N m−2. The modulus of rigidity if the top of the cube is displaced by 0.05cm with respect to its bottom is

2×106N m−2

105N m−2

1×107N m−2

4×105N m−2