Question

In an experiment, the following observations were recorded :$L=2.820m,M=3.00kg,l=0.087cm,$ Diameter $D=0.041cm$. Taking $g=9.81m/s^2$ 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. $8.42\%$

Answer 1

Answer: (C)

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

$\frac{\Delta Y}{Y}\times 100=\left(\frac{\Delta M}{M}+\frac{\Delta g}{g}+\frac{\Delta L}{L}+\frac{2\Delta D}{D}+\frac{\Delta l}{l}\right)\times 100$

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

$=0.065\times 100=6.5\%$