Two parallel and opposite forces each 5000 N are applied tangentially to the upper and lower faces of a cubical metal block of side 25 cm. The angle of shear is (The shear modulus of the metal is 80 GPa.)

Question

Two parallel and opposite forces each 5000 N are applied tangentially to the upper and lower faces of a cubical metal block of side 25 cm. The angle of shear is (The shear modulus of the metal is 80 GPa.) (A) 10-4 rad (B) 10-5 rad (C) 10-6 rad (D) 10-7 rad

Tardigrade Admin · Accepted Answer

Here, F= 5000 N L=25 cm =25×10-2 m A =(25×10-2m)2 =625×10-4m2 η=80 Gpa =80×109pa Angle of shear, θ =(F/Aη) =(5000/625×10 -4×80×109) =10-6 rad

Q. Two parallel and opposite forces each $5000 N$ are applied tangentially to the upper and lower faces of a cubical metal block of side $25 c m$ . The angle of shear is
(The shear modulus of the metal is $80 GP a$ .)

$1 0^{- 4} r a d$

$1 0^{- 5} r a d$

$1 0^{- 6} r a d$

$1 0^{- 7} r a d$

Here, $F = 5000 N$
$L = 25 c m$ $= 25 \times 1 0^{- 2} m$
$A$ $= (25 \times 1 0^{- 2} m)^{2}$ $= 625 \times 1 0^{- 4} m^{2}$
$η = 80 Gp a$ $= 80 \times 1 0^{9} p a$
Angle of shear, $θ$ $= \frac{F}{A η}$ $= \frac{5000}{625 \times 1 0 ^{- 4} \times 80 \times 1 0 ^{9}}$ $= 1 0^{- 6} r a d$

Q. Two parallel and opposite forces each 5000N are applied tangentially to the upper and lower faces of a cubical metal block of side 25cm. The angle of shear is (The shear modulus of the metal is 80GPa.)

10−4rad

10−5rad

10−6rad

10−7rad