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NEET PHYSICSEasy

Two bodies of different masses mam_a and mbm_b are dropped from two different heights aa and bb. The ratio of the time taken by the two to cover these distances is:

A

a : b

B

b : a

C

√a : √b

D

a² : b²

Step-by-Step Solution

  1. Identify the Physics Principle: For a body dropped from rest (u=0u=0) under gravity, the motion is uniformly accelerated with acceleration gg. The time of fall is independent of the mass of the body (Galileo's observation) .
  2. Apply Kinematic Equation: Use the second equation of motion, s=ut+12at2s = ut + \frac{1}{2}at^2.
  • For the first body: Height =a= a, Time =t1= t_1. a=0+12gt12  ⟹  t1=2aga = 0 + \frac{1}{2}gt_1^2 \implies t_1 = \sqrt{\frac{2a}{g}}
  • For the second body: Height =b= b, Time =t2= t_2. b=0+12gt22  ⟹  t2=2bgb = 0 + \frac{1}{2}gt_2^2 \implies t_2 = \sqrt{\frac{2b}{g}}
  1. Calculate Ratio: t1t2=2a/g2b/g=ab\frac{t_1}{t_2} = \frac{\sqrt{2a/g}}{\sqrt{2b/g}} = \frac{\sqrt{a}}{\sqrt{b}} Thus, the ratio is a:b\sqrt{a} : \sqrt{b} .
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