A 5.0 cm cube of substance ; has its upper face displaced by 0.65 cm, by a tangential force of 0.25 N . Calculate the modulus of rigidity of the substance.

Question

A 5.0 cm cube of substance ; has its upper face displaced by 0.65 cm, by a tangential force of 0.25 N . Calculate the modulus of rigidity of the substance. (A) 769.2 Nm-2 (B) 754.3 Nm-2 (C) 735.2 Nm-2 (D) 714.3 Nm-2

Tardigrade Admin · Accepted Answer

η = (FL/Al) , A = L2 , η = (FL/L2l) = (F/Ll) Here, L = 5.0 × 10-2 m l = 0.65 × 10-2 m , F = 0.25 N η = (0.25/5.0 × 10-2 × 0.65 × 10-2) = (0.25 × 104/3.25) = 769.2 Nm-2

Q. A 5.0 cm cube of substance ; has its upper face displaced by 0.65 cm, by a tangential force of 0.25 N . Calculate the modulus of rigidity of the substance.

769.2 $N m^{- 2}$

754.3 $N m^{- 2}$

735.2 $N m^{- 2}$

714.3 $N m^{- 2}$

$η = \frac{F L}{A l}, A = L^{2}, η = \frac{F L}{L ^{2} l} = \frac{F}{L l}$
Here, $L = 5.0 \times 1 0^{- 2} m$
$l = 0.65 \times 1 0^{- 2} m, F = 0.25 N$
$η = \frac{0.25}{5.0 \times 1 0 ^{- 2} \times 0.65 \times 1 0 ^{- 2}} = \frac{0.25 \times 1 0 ^{4}}{3.25}$
= 769.2 $N m^{- 2}$

Q. A 5.0 cm cube of substance ; has its upper face displaced by 0.65 cm, by a tangential force of 0.25 N . Calculate the modulus of rigidity of the substance.

769.2 Nm−2

754.3 Nm−2

735.2 Nm−2

714.3 Nm−2

η=AlFL​,A=L2,η=L2lFL​=LlF​ Here, L=5.0×10−2m l=0.65×10−2m,F=0.25N η=5.0×10−2×0.65×10−20.25​=3.250.25×104​ = 769.2 Nm−2

769.2 $N m^{- 2}$

754.3 $N m^{- 2}$

735.2 $N m^{- 2}$

714.3 $N m^{- 2}$

$η = \frac{F L}{A l}, A = L^{2}, η = \frac{F L}{L ^{2} l} = \frac{F}{L l}$
Here, $L = 5.0 \times 1 0^{- 2} m$
$l = 0.65 \times 1 0^{- 2} m, F = 0.25 N$
$η = \frac{0.25}{5.0 \times 1 0 ^{- 2} \times 0.65 \times 1 0 ^{- 2}} = \frac{0.25 \times 1 0 ^{4}}{3.25}$
= 769.2 $N m^{- 2}$