NEETPhysicsMotion in One Dimension
Match List I with List II for a freely falling body dropped from rest. (A) Total distances fallen in 1 s, 2 s, 3 s (I) 1:3:5 (B) Distances fallen in the 1^ st , 2^ nd , and 3^ rd seconds (II) 1:2:3 (C) Velocities at the end of 1 s, 2 s, 3 s (III) 1:( 2 -1):( 3 - 2 ) (D) Times taken to fall successive equal distances h (IV) 1:4:9 Choose the correct answer from the options given below:
Options
- A(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
- B(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
- C(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
- D(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Correct answer
B. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Step-by-step solution
For a freely falling body dropped from rest, the initial velocity u = 0 and acceleration a = g . (A) Total distance fallen in time t is S = 1 2 gt^2 . Thus, S t^2 . For t = 1, 2, 3 s, the ratio is 1^2 : 2^2 : 3^2 = 1:4:9 . (B) Distance fallen specifically in the n^ th second is S_n = g 2 (2n-1) . Thus, S_n (2n-1) . For n = 1, 2, 3 , the ratio is 1:3:5 . (C) Velocity at time t is v = gt . Thus, v t . For t = 1, 2, 3 s, the ratio is 1:2:3 . (D) Time taken to fall a distance h from rest is t = 2h g . The time to fall