Two identical simple pendulums are compared, one \((A)\) located on the surface of the earth and the other \((B)\) – at a height \((h)\) above the earth's surface: \(h=\dfrac{R}{1000}.\)
Their time periods are related as:
1. \(T_A\Big(1+\dfrac{1}{1000}\Big)=T_B\)
2. \(T_B\Big(1+\dfrac{1}{1000}\Big)=T_A\)
3. \(T_A\Big(1+\dfrac{1}{2000}\Big)=T_B\)
4. \(T_B\Big(1+\dfrac{1}{2000}\Big)=T_A\)
Add Note
Subtopic: Angular SHM |
58%
From NCERT
Other Reason
Highlight in NCERT
Please attempt this question first.
Hints
Please attempt this question first.
Upgrade Your Plan
Upgrade now and unlock your full question bank experience with daily practice.
Two pendulums of length \(121~\text{cm}\) and \(100~\text{cm}\) start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:
1.
\(8\)
2.
\(11\)
3.
\(9\)
4.
\(10\)
Add Note
Subtopic: Angular SHM |
68%
From NCERT
NEET - 2022
Other Reason
Highlight in NCERT
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Upgrade Your Plan
Upgrade now and unlock your full question bank experience with daily practice.