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Two rods A and B, made of different metals, are held between two rigid walls. Ratios of cross sectional areas of the rods, Young's modulus and coefficient of linear expansion of their materials are 1:2, 3:4 and 9:8 respectively. When temperature of the rods is increased by 20oC, compressive forces FA and FB are induced in them then FA/FB is

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a

3:8

b

1:3

c

27:64

d

2:3

answer is C.

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Detailed Solution

F = Induced compressive force = Cross sectional area x thermal stress.=a×γαΔθ∴FAFB=aAaB×YAYB×αAαB=12×34×98=2764

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