Question

The error in the measurement of the length of the simple pendulum is $0.2\%$ and the error in time period is $4\%$. The maximum possible error in measurement of $\frac{L}{T^2}$ is :

• A. $4.2\%$
• B. $3.8\%$
• C. $7.8\%$
• D. $8.2\%$

Answer 1

Answer: (D)

$8.2\%$

Here given , $\frac{\Delta L}{L}\times 100=0.2$% and $\frac{\Delta T}{T}\times 100=4$%
Let $y=\frac{L}{T^2}$
So, $\frac{\Delta y}{y}=(\frac{\Delta L}{L}+2\frac{\Delta T}{T})$
Thus, percentage error $\frac{\Delta y}{y}\times 100=(\frac{\Delta L}{L}\times 100+2\frac{\Delta T}{T}\times 100)=0.2+2\times 4=8.2$%