- 1(current)
Q. No.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. \(T_B\Big(1+\dfrac{1}{2000}\Big)=T_A\)
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. \(T_B\Big(1+\dfrac{1}{2000}\Big)=T_A\)
Subtopic: Angular SHM |
60%
Level 2: 60%+
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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. | \(10\) |
Subtopic: Angular SHM |
70%
Level 2: 60%+
NEET - 2022
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NEET 2021 - Achiever Batch - Aryan Raj Singh
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NEET 2021 - Achiever Batch - Aryan Raj Singh
The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite
Subtopic: Angular SHM |
70%
Level 2: 60%+
AIPMT - 1999
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NEET 2021 - Achiever Batch - Aryan Raj Singh
Hints
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NEET 2021 - Achiever Batch - Aryan Raj Singh
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Q. No.
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