A coin dropped in the lift it takes time 0.5 s to reach the floor when lift is staionary it takes time t when lift is moving up

t₁ > t₂

Explanation:

A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂,  t₁ = t₂,  t₁ >> t₂ ,  t₂ > t₁

Solution:

Newton's law of motion is given by:

s = ut + (1/2)gt²;

where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.

u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.

hence:

When stationary:

Hence t₂ < t₁, this means that t₁ > t₂.

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the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this