- 1(current)
Q. No.Two spherical bobs of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\) and \(l_B\) respectively. If they are executing SHM with frequency as per the relation \(f_A=2f_B,\) Then:
1. \(l_A = \frac{l_B}{4}\)
2. \(l_A= 4l_B\)
3. \(l_A= 2l_B~\&~M_A=2M_B\)
4. \(l_A= \frac{l_B}{2}~\&~M_A=\frac{M_B}{2}\)
Subtopic: Â Angular SHM |
 72%
From NCERT
AIPMT - 2000
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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: Â Angular SHM |
 67%
From NCERT
AIPMT - 1999
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The period of oscillation of a simple pendulum of length \(L\) suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination \(\theta\), is given by:
1. \(2\pi\sqrt{\frac{L}{g\cos\theta}}\)
2. \(2\pi\sqrt{\frac{L}{g\sin\theta}}\)
3. \(2\pi\sqrt{\frac{L}{g}}\)
4. \(2\pi\sqrt{\frac{L}{g\tan\theta}}\)
1. \(2\pi\sqrt{\frac{L}{g\cos\theta}}\)
2. \(2\pi\sqrt{\frac{L}{g\sin\theta}}\)
3. \(2\pi\sqrt{\frac{L}{g}}\)
4. \(2\pi\sqrt{\frac{L}{g\tan\theta}}\)
Subtopic: Â Angular SHM |
 60%
From NCERT
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A simple pendulum hanging from the ceiling of a stationary lift has a time period \(T_1\). When the lift moves downward with constant velocity, then the time period becomes \(T_2\). It can be concluded that:
| 1. | \(T_2 ~\text{is infinity} \) | 2. | \(T_2>T_1 \) |
| 3. | \(T_2<T_1 \) | 4. | \(T_2=T_1\) |
Subtopic: Â Angular SHM |
 61%
From NCERT
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Q. No.
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