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Q.

In Searle's experiment, which is used to find Young's modulus of elasticity, the diameter of experimental wire is D=0.05 cm (measured by a scale of least count 0.001 cm) and length is L=110 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of l=0.125 cm (measured by a micrometer of least count 0.001 cm ). Find maximum possible error in the value of Young's modulus. Screw gauge and meter scale are free from error.

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a

1.09×1010 N/m2

b

0.09×1010 N/m2

c

10.09×1010 N/m2

d

1.19×1010 N/m2

answer is A.

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Detailed Solution

Young's modulus of elasticity is given byY= stress strain =F/Al/L=FLLA=FLlπd24Substituting the values, we getY=50×1.1×41.25×10-3×π×5.0×10-42=2.24×1011 N/m2ΔYY=ΔLL+Δll+2Δdd=0.1110+0.0010.125+20.0010.05=0.0489ΔY=(0.0489)Y=(0.0489)×2.24×1011 N/m2=1.09×1010 N/m2

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In Searle's experiment, which is used to find Young's modulus of elasticity, the diameter of experimental wire is D=0.05 cm (measured by a scale of least count 0.001 cm) and length is L=110 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of l=0.125 cm (measured by a micrometer of least count 0.001 cm ). Find maximum possible error in the value of Young's modulus. Screw gauge and meter scale are free from error.