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Q.

In an experiment, the following observations were recorded : L = 2.820 m, M = 3.00 Kg, l = 0.087 cm. Diameter D = 0.041 cm. Taking g = 9.81 m/s2 using the formula, Y = 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 is∆YY×100 = [∆MM+∆gg+∆LL+2∆DD+∆ll]×100= (1300+1981+12820+2×141+187)×100= 0.065×100 = 6.5%

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