Question:
lf the length of a seconds pendulum is increased by 2%, then in a day the pendulum
lf the length of a seconds pendulum is increased by 2%, then in a day the pendulum
Updated On: Mar 4, 2024
- loses 764 s
- loses 924 s
- loses 236 s
- loses 864 s
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The Correct Option is D
Solution and Explanation
Time period, T = $ 2 \pi \sqrt{ \frac{ l }{ g }} $
or $ \frac{ \triangle T }{ T } = \frac{ 1}{ 2} \frac{ \triangle l }{ l } $
For ls, $ \triangle T = \frac{ 1}{2} \bigg( \frac{ \triangle l }{ l } \bigg) T = \frac{ 1}{ 2} \times 0.02 \times T = 0.01 T = 0.01 s $
For a day, $ \triangle T = 24 \times 60 \times 60 \times 0.01 = 864 $s
or $ \frac{ \triangle T }{ T } = \frac{ 1}{ 2} \frac{ \triangle l }{ l } $
For ls, $ \triangle T = \frac{ 1}{2} \bigg( \frac{ \triangle l }{ l } \bigg) T = \frac{ 1}{ 2} \times 0.02 \times T = 0.01 T = 0.01 s $
For a day, $ \triangle T = 24 \times 60 \times 60 \times 0.01 = 864 $s
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