Q88 / 90 · JEE Main 2022 (27 Jul Shift 1) · Physics

A square aluminium (shear modulus is $25 \times 10^{9} {Nm}^{- 2}$ ) slab of side $60 cm$ and thickness $15 cm$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^{4} N$ . The lower edge is riveted to the floor. The displacement of the upper edge is _____ $μ m$ .

Correct answer: 48

Solution

Shearing stress is given by $\frac{F}{A} = η \frac{x}{l}$
Thus, displacement is $x = \frac{F l}{A η}$
$\Rightarrow x = \frac{18 \times 10^{4} \times 60 \times 10^{- 2}}{60 \times 10^{- 2} \times 15 \times 10^{- 2} \times 25 \times 10^{9}}$
$= 48 \times 10^{- 6} m = 48 μ m$

Hence, the correct answer is 48.

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A square aluminium (shear modulus is 25×109Nm-2) slab of side 60 cm and thickness 15 cm is subjected to a shearing force (on its narrow face) of 18.0×104 N. The lower edge is riveted to the floor. The displacement of the upper edge is _____ μm.

Solution

More Mechanical Properties of Solids Questions

A square aluminium (shear modulus is $25 \times 10^{9} {Nm}^{- 2}$ ) slab of side $60 cm$ and thickness $15 cm$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^{4} N$ . The lower edge is riveted to the floor. The displacement of the upper edge is _____ $μ m$ .