Q. 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
A) 7.96%
B) 4.56%
C) 6.50%
D) 8.42%
Explanation
$Y = \frac{4MgL}{\pi D^{2}I}$ so maximum permissible error in $Y$ $= \frac{\mathrm{\Delta}Y}{Y} \times 100 = \left( \frac{\mathrm{\Delta}M}{M} + \frac{\mathrm{\Delta}g}{g} + \frac{\mathrm{\Delta}L}{L} + \frac{2\mathrm{\Delta}D}{D} + \frac{\mathrm{\Delta}l}{l} \right) \times 100$ $= \left( \frac{1}{300} + \frac{1}{981} + \frac{1}{2820} + 2 \times \frac{1}{41} + \frac{1}{87} \right) \times 100$ $= 0.065 \times 100 = 6.5\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.