A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing modulus of cube is :

Question

A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing modulus of cube is : (A)  2× 109 Nm-2  (B)  4× 109 Nm-2  (C)  8× 109 Nm-2  (D)  16× 109 Nm-2

Tardigrade Admin · Accepted Answer

Shearing modulus of cube η =(Fl/AΔ l)=(8× 103× 40× 10-3/(40× 10-3)× (0.1× 10-3)) =2× 109 Nm-2

Q. A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing modulus of cube is :

$2 \times 10^{9} N m^{- 2}$

$4 \times 10^{9} N m^{- 2}$

$8 \times 10^{9} N m^{- 2}$

$16 \times 10^{9} N m^{- 2}$

Shearing modulus of cube $η = \frac{Fl}{A Δ l} = \frac{8 \times 10 ^{3} \times 40 \times 10 ^{- 3}}{( 40 \times 10 ^{- 3} ) \times ( 0.1 \times 10 ^{- 3} )}$ $= 2 \times 10^{9} N m^{- 2}$

Q. A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8 kN. The shearing modulus of cube is :

2×109Nm−2

4×109Nm−2

8×109Nm−2

16×109Nm−2

Shearing modulus of cube η=AΔlFl​=(40×10−3)×(0.1×10−3)8×103×40×10−3​ =2×109Nm−2

$2 \times 10^{9} N m^{- 2}$

$4 \times 10^{9} N m^{- 2}$

$8 \times 10^{9} N m^{- 2}$

$16 \times 10^{9} N m^{- 2}$

Shearing modulus of cube $η = \frac{Fl}{A Δ l} = \frac{8 \times 10 ^{3} \times 40 \times 10 ^{- 3}}{( 40 \times 10 ^{- 3} ) \times ( 0.1 \times 10 ^{- 3} )}$ $= 2 \times 10^{9} N m^{- 2}$