- 1(current)
Q. No.If the mass of a bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is \( \dfrac{x}{2}\) times its original time period. The value of \(x\) is:
| 1. | \(\sqrt2\) | 2. | \(2\sqrt3\) |
| 3. | \(4\) | 4. | \(\sqrt3\) |
Subtopic: Â Angular SHM |
 68%
Level 2: 60%+
NEET - 2024
Hints
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly
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%
Level 2: 60%+
Hints
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 |
 75%
Level 2: 60%+
AIPMT - 2000
Hints
A simple pendulum oscillating in air has a period of \(\sqrt3\) s. If it is completely immersed in non-viscous liquid, having density \(\left(\dfrac14\right)^{\text{th}}\) of the material of the bob, the new period will be:
1. \(2\sqrt3\) s
2. \(\dfrac{2}{\sqrt3}\) s
3. \(2\) s
4. \(\dfrac{\sqrt 3}{2}\) s
1. \(2\sqrt3\) s
2. \(\dfrac{2}{\sqrt3}\) s
3. \(2\) s
4. \(\dfrac{\sqrt 3}{2}\) s
Subtopic: Â Angular SHM |
 54%
Level 3: 35%-60%
NEET - 2023
Hints
The time period of a simple pendulum in a stationary lift is \(T.\) If the lift accelerates upward with an acceleration of \(\dfrac g 6\) (where \(g\) is the acceleration due to gravity), then the time period of the pendulum would be:
| 1. | \(\sqrt{\dfrac{6}{5}} ~T \) | 2. | \(\sqrt{\dfrac{5}{6}} ~T\) |
| 3. | \(\sqrt{\dfrac{6}{7}}~T\) | 4. | \(\sqrt{\dfrac{7}{6}} ~T\) |
Subtopic: Â Angular SHM |
 80%
Level 1: 80%+
Please attempt this question first.
Hints
Please attempt this question first.
If a simple pendulum is brought deep inside a mine from the earth's surface, its time period of oscillation will:
| 1. | increase |
| 2. | decrease |
| 3. | remain the same |
| 4. | be any of the above, depending on the length of the pendulum |
Subtopic: Â Angular SHM |
 68%
Level 2: 60%+
Hints
If a simple pendulum be suspended in an elevator which is moving upward, its time period is found to decrease by \(2\%.\) The acceleration of the elevator is (in magnitude):
1. \(2\%\) of \(g\)
2. \(1\%\) of \(g\)
3. \(4\%\) of \(g\)
4. \(102\%\) of \(g\)
1. \(2\%\) of \(g\)
2. \(1\%\) of \(g\)
3. \(4\%\) of \(g\)
4. \(102\%\) of \(g\)
Subtopic: Â Angular SHM |
 77%
Level 2: 60%+
Please attempt this question first.
Hints
Please attempt this question first.
- 1(current)
Q. No.
© 2025 GoodEd Technologies Pvt. Ltd.