As shown in the figure, force of $105 N$ each are applied in opposite directions, on the upper and lower force of a cube of side $10 c m$ ,shifting the upper face parallel to itself by $0.5 c m$ . If the side of another cube of the same material is $20 c m$ , then under similar condition as shown as above, the displacement will be:

0.25 c m

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0.37 c m

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0.75 c m

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1.00 c m

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Solution

The correct option is A $0.25 c m$
$T h e c u b e i n t h e b o t h c a s e s i s i n e q u i l i b r i u m S h e a r s t r e s s τ = S h e a r m o d u l u s (G) \times s h e a r s t r a i n i n t h e f i r s t c a s e s h e a r s t r a i n = \frac{0.5}{10} = 0.05 S h e a r s t r e s s = F o r c e / A r e a = \frac{105 N}{10 \times 10 {c m}^{2}} G = 21 N / {c m}^{2} I n t h e s e c o n d c a s e : Δ l = (20 \times S h e a r s t r e s s) / G {S h e a r f o r c e = \frac{105 N}{20 \times 20 {c m}^{2}}} Δ l = .25 c m$

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