In the column-I, a system is described in each option and corresponding time period is given in the column-II. Suitably match them.
Column-I | Column-II | ||
(a) | A simple pendulum of length 'λ' oscillating with small amplitude in a lift moving down with retardation g/2. | (p) | T = 2π |
(b) | A block attached to an end of a vertical spring, whose other end is fixed to the ceiling of a stationary lift, stretches the spring by length 'λ' in equilibrium. It's time period when lift moves up with an acceleration g/2 is | (q) | T = 2π |
(c) | The time period of small oscillation of a uniform rod of length 'λ' smoothly hinged at one end. The rod oscillates in vertical plane. | (r) | T = 2π |
(d) | A cubical block of edge 'λ' and specific gravity 1/2 is in equilibrium with some volume inside water filled in a large fixed container. Neglect viscous forces and surface tension. The time period of small oscillations of the block in vertical direction is | (s) | T = 2π |
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
p, q, p, s
Sol. In frame of lift effective acceleration due to gravity is
downwards
∴ T = 2 π 
K λ = mg
∴ 
constant acceleration of lift has no effect in time period of oscillation.
∴ T = 2 π
= 2 π 
T = 2 π
= 2 π 
T = 2 π
= 2 π
= 2 π 
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