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: Write the formula for \(g\)
From \(T=2\pi\sqrt{l/g}\) we get \(g=\frac{4\pi^2l}{T^2}\).
Errors add in a product or quotient, and a power counts that many times: \(\frac{\Delta g}{g}=\frac{\Delta l}{l}+2\frac{\Delta T}{T}\).

Step 2: Error in \(l\)
\(\frac{\Delta l}{l}=\frac{0.1}{80}=0.00125\), that is \(0.125\%\).

Step 3: Error in \(T\)
The time for \(50\) oscillations is \(50\times1.5=75\) s, measured with a least count of \(0.1\) s. So \(\frac{\Delta T}{T}=\frac{0.1}{75}=0.00133\), that is \(0.133\%\). Timing many oscillations keeps this error small.

Step 4: Combine
\(\frac{\Delta g}{g}\times100=0.125+2\times0.133=0.39\%\), which is nearly \(0.4\%\). Option (B).

Final Answer:
The percentage error in \(g\) is nearly \(0.4\), option (B). \[ \boxed{\text{(B) }0.4} \]