In an experiment, the following observation's 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 = \(\frac{4MgL}{\pi D^{2}l}\) , the maximum permissible error in Y is
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\(\gamma = \frac{4MgL}{\pi D^{2} l}\) so maximum permissible error in Y = \(\frac{\Delta Y}{Y} \times 100 = \left( \frac{\Delta M}{M} + \frac{\Delta g}{g} + \frac{\Delta L}{L} + \frac{2 \Delta D}{D} + \frac{\Delta l}{l} \right) \times 100\)
\(= \left(\frac{1}{300} + \frac{1}{981} + \frac{1}{2820} + 2 \times \frac{1}{41} + \frac{1}{87}\right) \times 100\) \(= 0.065 \times 100 = 6.5\%\)
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