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Q.

A block B of mass 20 Kg is suspended by a metallic rod of length 2 m, mass 10 Kg cross sectional area 4 mm2 and young's modulus 2×1010 N/m2. Then elongation produced in the rod is

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a

8.25 mm

b

12.15 mm

c

4.85 mm

d

6.25 mm

answer is D.

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Detailed Solution

δ=100×14×10−6×2×1010+200×24×10−6×2×1010m⇒δ=1800+1200m=6.25 mm
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