- 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:
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 |
58%
From NCERT
Please attempt this question first.
Hints
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Two pendulums of length \(121\) cm and \(100\) 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 |
68%
From NCERT
NEET - 2022
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Hints
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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 |
68%
From NCERT
AIPMT - 1999
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
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