- 1(current)
Q. No.A uniform rod of length \(l\) is suspended by an end and is made to undergo small oscillations. The time period of small oscillation is \(T\). Then, the acceleration due to gravity at this place is:
| 1. | \(4\pi^2\dfrac{l}{T^2}\) | 2. | \(\dfrac{4\pi^2}{3}\dfrac{l}{T^2}\) |
| 3. | \(\dfrac{8\pi^2}{3}\dfrac{l}{T^2}\) | 4. | \(\dfrac{12\pi^2l}{T^2}\) |
Subtopic: Angular SHM |
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A physical pendulum consists of a uniform rod \(AB\) of mass \(m\) and length \(L,\) suspended from one end \(A\) – so as to rotate freely under gravity. If it is displaced slightly from its mean position, it executes SHM. Let the maximum kinetic energy of the rod be \(E_0.\)
If the time period of a simple pendulum of the same length is \(T_0,\) then the time period of this pendulum is:
If the time period of a simple pendulum of the same length is \(T_0,\) then the time period of this pendulum is:
| 1. | \({\Large\sqrt\frac23}T_0\) | 2. | \({\Large\sqrt\frac{1}{12}}T_0\) |
| 3. | \({\Large\sqrt\frac32}T_0\) | 4. | \(T_0\) |
Subtopic: Angular SHM |
From NCERT
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An elastic ball is projected vertically upward with a speed \(u,\) and it returns to the ground and rebounds, the motion is periodic with a period \(T.\) A simple pendulum, having a length equal to maximum altitude attained by this ball, would have a time period of:
| 1. | \(T\) | 2. | \(\pi T\) |
| 3. | \(\pi\sqrt2T\) | 4. | \(\dfrac{\pi}{\sqrt 2}T\) |
Subtopic: Angular SHM |
52%
From NCERT
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Hints
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