Question

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for $20$ oscillations is measured by using a watch of $1$ second least count. The mean value of time taken comes out to be $30\, s$. The length of pendulum is measured by using a meter scale of least count $1\, mm$ and the value obtained is $55.0\, cm$. The percentage error in the determination of $g$ is close to :

• A. 0.70%
• B. 0.20%
• C. 3.50%
• D. 6.80%

Answer 1

The Correct Option is D

To determine the percentage error in the calculation of acceleration due to gravity (\(g\)) using the simple pendulum experiment, we start by considering the formula for the period \(T\) of a simple pendulum:

T = 2\pi \sqrt{\frac{L}{g}}

where \(L\) is the length of the pendulum.

Step 1: Calculate the time period for one oscillation

The mean time for 20 oscillations is given as 30 s. Therefore, the time period \(T\) for one oscillation is:

T = \frac{30}{20} = 1.5\, \text{s}

Step 2: Determine the expression for \(g\) and the error formula

Rearranging the formula for \(g\):

g = \frac{4\pi^2 L}{T^2}

The error in \(g\) can be calculated using the differential formula:

\left( \frac{\Delta g}{g} \right) = \left( \frac{\Delta L}{L} \right) + 2 \left( \frac{\Delta T}{T} \right)

Step 3: Calculate the percentage errors

• Given \(L = 55.0 \, \text{cm}\) with a least count of 1 mm, the error in \(L\):

\Delta L = 0.1 \, \text{cm}

\left( \frac{\Delta L}{L} \right) = \frac{0.1}{55.0} \approx 0.001818 \, \text{or} \, 0.1818\%

• The time \(T = 1.5 \, \text{s}\), with a least count of 1 s for 20 oscillations, resulting in:

\Delta T = \frac{1}{20} = 0.05 \, \text{s}

\left( \frac{\Delta T}{T} \right) = \frac{0.05}{1.5} \approx 0.0333 \, \text{or} \, 3.33\%

Step 4: Calculate total percentage error in g

Substitute these errors into the error formula for \(g\):

\left( \frac{\Delta g}{g} \right) = 0.001818 + 2 \times 0.0333 \approx 0.06844 \, \text{or} \, 6.844\%

Therefore, the percentage error in the determination of \(g\) is approximately \(6.80\%\). Thus, the correct answer is:

6.80%