To find the spring constant (k) of a spring experimentally, a student commits 2% positive error in the measurement of time and 1% negative error in measurement of mass. The percentage error in determining value of k is :
- 0.03
- 0.01
- 0.04
- 0.05
The Correct Option is D
Solution and Explanation
The time period \( T \) of a spring is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}}. \]
Squaring both sides:
\[ T^2 \propto \frac{m}{k}. \]
Taking percentage errors:
\[ \frac{\Delta T^2}{T^2} = \frac{\Delta m}{m} - \frac{\Delta k}{k}. \]
Substituting the relationship \( \frac{\Delta T^2}{T^2} = 2 \frac{\Delta T}{T} \), we get:
\[ 2 \frac{\Delta T}{T} = \frac{\Delta m}{m} - \frac{\Delta k}{k}. \]
Rewriting for \( \frac{\Delta k}{k} \):
\[ \frac{\Delta k}{k} = \frac{\Delta m}{m} - 2 \frac{\Delta T}{T}. \]
Given:
\( \frac{\Delta T}{T} = 2\% \) (positive error),
\( \frac{\Delta m}{m} = -1\% \) (negative error).
Substitute the values:
\[ \frac{\Delta k}{k} = (-1\%) - 2(2\%) = -1\% - 4\% = -5\%. \]
Hence, the magnitude of the percentage error in \( k \) is:
\[ \left| \frac{\Delta k}{k} \right| = 5\%. \]
Final Answer: \( 5\% \) (Option 4)
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