JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Elasticity & SHM

A square aluminium (shear modulus is

25 \times 1 0^{9} Nm^{- 2}

) slab of side 60 cm and thickness 15 cm is subjected to a shearing force (on its narrow face) of

18.0 \times 1 0^{4} N

. The lower edge is riveted to the floor. The displacement of the upper edge is

____

μ m

Answer: 48

Solution

For shear deformation, the relation between shearing force

F

, shear modulus

η

, area

A

, displacement

x

, and height

l

is:

\frac{F}{A} = η \frac{x}{l}

Solving for displacement:

x = \frac{F l}{A η}

Given:

F = 18 \times 1 0^{4}

l = 60

= 0.6

m (height),

A = 60 \times 15

cm²

= 0.6 \times 0.15

m² (narrow face area),

η = 25 \times 1 0^{9}

N/m²

x = \frac{18 \times 1 0 ^{4} \times 0.6}{0.6 \times 0.15 \times 25 \times 1 0 ^{9}} = \frac{18 \times 1 0 ^{4}}{0.15 \times 25 \times 1 0 ^{9}} = 48 \times 1 0^{- 6}

= 48

μm

Therefore, the answer is 48.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer