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 $m / s^{2}$ using the formula, $Y = \frac{4 M g L}{π D^{2} l}$ the maximum permissible error in Y is

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6.50%

7.96%

8.42%

4.56%

answer is C.

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Detailed Solution

$Y = \frac{4 M g L}{π D^{2} l}$ so maximum permissible error in Y is

$\frac{∆ Y}{Y} \times 100 = [\frac{∆ M}{M} + \frac{∆ g}{g} + \frac{∆ L}{L} + 2 \frac{∆ D}{D} + \frac{∆ l}{l}] \times 100$

= $(\frac{1}{300} + \frac{1}{981} + \frac{1}{2820} + 2 \times \frac{1}{41} + \frac{1}{87}) \times 100$

= 0.065 $\times$ 100 = 6.5%

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