Question

The period of oscillation of a simple pendulum is given by \(T = 2π\sqrt{\frac{l}{g}}\) where \(l\) is about \(80\) cm and is known to have \(0.1\) cm accuracy. The period is about \(1.5\) s. The time of \(50\) oscillations is measured by a stop watch of least count \(0.1\) s. The percentage error in \(g\) is nearly

• A. \(0.8\)
• B. \(0.4\)
• C. \(0.1\)
• D. \(1\)

Hint:

For $g=4\pi^2l/T^2$, $\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}$.

Answer 1

The Correct Option is B

Step 1: Use maximum error idea
The largest possible error in $g$ is the sum of the relative error of $l$ plus twice the relative error of $T$.

Step 2: Numbers
Length part: $\frac{0.1}{80}\times100=0.125\%$. Period part: for $50$ swings, $t=75$ s and the error in $t$ is $0.1$ s, so the relative error is $\frac{0.1}{75}\times100=0.133\%$ and doubled it is $0.267\%$.
Total $=0.125+0.267=0.392\%\approx0.4\%$.

Final Answer:
Option (B). \[ \boxed{\text{(B)}} \]