Question

A body falls freely under the action of gravity from height h above the ground.

Column – i	Column – ii
(a) P.E. = 2(K.E.)	(P) constant at every point
(b) P.E. = K.E.	(Q) at height (h/3)
(c) P.E. = (1/2) (K.E.)	(R) at height (2h/3)
(d) P.E.+ K.E.	(S) at height (h/2)

(A) a – P, b – Q, c – R, d – S 

(B) a – Q, b – P, c – S, d – R 

(C) a – S, b – R, c – Q, d – P 

(D) a – R, b – S, c – Q, d – P

Answer 1

Correct option: (D) a – R, b – S, c – Q, d – P

Explanation:

(1) PE + PK for a body is always constant.

(2) initial PE of body at height h = mgh

When body falls freely from height, then at height

(h/2), PE = mg(h/2)

i.e. at height (h/2) remaining PE of body is converted into KE hence

KE = mgh – mg(h/2) = mg(h/2) at (h/2), PE = PK

(3) at height (h/3), PE = mg(h/3)

remaining PE is converted into KE hence

KE = mgh – mg(h/3) = (2/3) mgh

(PE / PK) = [{mg(h/3)} / {(2/3)mgh}] = (1/2)

KE = 2PK i.e. PE = (1/2)KE

(4) at height (2h / 3), PE = mg(2h / 3) = {(2mgh) / 3}

KE = mgh – (2/3)mgh = (1/3)mgh

(PE / KE) = (2/1)

PE = 2KE