Question

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=$\frac{4Mg}{\pi D^{2}l}$, the maximum permissible error in Y is

• A. 7.96%
• B. 4.56%
• C. 6.50%
• D. 8.42%

Answer 1

Answer: (C)

6.50%

Answer 2

$Y = \frac{4MgL}{\pi D^{2}l}$ so maximum permissible error in Y =

$\frac{\Delta Y}{Y} \times 100 = \left( \frac{\Delta M}{M} + \frac{\Delta g}{g} + \frac{\Delta L}{L} + \frac{2\Delta D}{D} + \frac{\Delta l}{l} \right) \times 100 = \left( \frac{1}{300} + \frac{1}{9.81} + \frac{1}{9820} + 2 \times \frac{1}{41} + \frac{1}{87} \right) \times 100$ $= 0.065 \times 100 = 6.5\%$