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Step-by-Step Solution
Step 1: Write down the formula for the time period of a simple pendulum
The time period $T$ of a simple pendulum of length $L$ is given by:
$ T = 2\pi \sqrt{\frac{L}{g}} $
where $g$ is the acceleration due to gravity.
Step 2: Observe the relationship between $T$ and $L$
The quantity $g$ is constant, so $T$ is directly proportional to $\sqrt{L}$. Therefore, if the length $L$ changes, $T$ changes in proportion to the square root of that change.
Step 3: Increase the length by 21%
If the original length is $L$, increasing it by 21% means the new length $L'$ is:
$ L' = L + 21\%\times L = 1.21\,L $
Step 4: Write down the new time period $T'$
Using the standard formula for the time period with the new length $L'$:
$ T' = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{1.21\,L}{g}} $
The ratio of the new time period to the old time period is:
$ \frac{T'}{T} = \frac{2\pi \sqrt{\frac{1.21\,L}{g}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{1.21} \,.$
Step 5: Calculate the percentage increase in the time period
Since $ \sqrt{1.21} \approx 1.1 $, the new time period is about 1.1 times the original time period. The percentage increase is:
$ \left( \frac{T' - T}{T} \right) \times 100\%
= (\sqrt{1.21} - 1) \times 100\%
\approx (1.1 - 1)\times 100\%
= 0.1 \times 100\%
= 10\% \,.
$
Hence, the time period increases by approximately 10% when the length is increased by 21%.
