JEE MAIN - Physics (2022 - 27th July Morning Shift - No. 24)
A square aluminum (shear modulus is $$25 \times 10^{9}\, \mathrm{Nm}^{-2}$$) slab of side $$60 \mathrm{~cm}$$ and thickness $$15 \mathrm{~cm}$$ is subjected to a shearing force (on its narrow face) of $$18.0 \times 10^{4}$$ $$\mathrm{N}$$. The lower edge is riveted to the floor. The displacement of the upper edge is ____________ $$\mu$$m.
Answer
48
Explanation
_27th_July_Morning_Shift_en_24_1.png)
$$Y = {{Fl} \over {A\Delta l}}$$
$$\Delta l = {{Fl} \over {YA}}$$
$$ = {{18 \times {{10}^4} \times 60 \times {{10}^{ - 2}}} \over {25 \times {{10}^9} \times 60 \times 15 \times {{10}^{ - 4}}}}$$
$$ = 48 \times {10^{ - 6}}$$ m
Comments (0)
