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Q.

In an experiment, the following observation's were recorded: L=2.820 m, M=3.00kg, l=0.087cm, Diameter D=0.041 cm. Taking g=9.81 m/s2,using the formulaY=4MgLπD2l, the maximum permissible error in Y is

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a

7.96%

b

4.56%

c

6.50%

d

8.42%

answer is C.

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Detailed Solution

Y=4MgLπD2l so maximum permissible error in Y=ΔYY×100=ΔMM+Δgg+ΔLL+2ΔDD+Δll×100 =0.013.00+0.019.81+0.0012.820+2×0.0010.041+0.0010.087×100 =0.0649×100=6.50%
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