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Question: A solid spherical ball is attached to a rigid rod radially and this arrangement is kept inside a liq...

A solid spherical ball is attached to a rigid rod radially and this arrangement is kept inside a liquid. The density of the liquid is twice that of the ball and the rod is fixed to the container at the bottom.

Now, the container is given possible acceleration (a→)(\overrightarrow{a}) as given in List-I. List-II gives the ratio of tension in the rod to the weight of the ball. Match the two lists.

List-IList-II
(P) a→=0\overrightarrow{a}=0(1) 0
(Q) a→=g2↓\overrightarrow{a}=\frac{g}{2}\downarrow(2) 3
(R) a→=g↓\overrightarrow{a}=g\downarrow(3) 2
(S) a→=g↑\overrightarrow{a}=g\uparrow(4) 1
(5) 12\frac{1}{2}
A

(P)-4, (Q)-5, (R)-1, (S)-3

B

(P)-1, (Q)-2, (R)-3, (S)-4

C

(P)-2, (Q)-3, (R)-4, (S)-5

D

(P)-3, (Q)-4, (R)-5, (S)-1

Answer

(P)-4, (Q)-5, (R)-1, (S)-3

Explanation

Solution

Let WbW_b be the weight of the ball, VV its volume, and ρb\rho_b its density. The liquid density is ρl=2ρb\rho_l = 2\rho_b. The weight of the ball is Wb=ρbVgW_b = \rho_b V g. The buoyant force in an accelerating frame is F→B=ρlV(a→−g→)\overrightarrow{F}_B = \rho_l V (\overrightarrow{a} - \overrightarrow{g}). The net force on the ball is F→net=F→B+W→b+T→=mba→\overrightarrow{F}_{net} = \overrightarrow{F}_B + \overrightarrow{W}_b + \overrightarrow{T} = m_b \overrightarrow{a}. Taking upward as positive, g→=−gj^\overrightarrow{g} = -g\hat{j}. The tension T→\overrightarrow{T} is downwards, so T→=−Tj^\overrightarrow{T} = -T\hat{j}.

(P) a→=0\overrightarrow{a}=0: 2ρbV(0−(−gj^))−Wbj^−Tj^=0  ⟹  2Wb−Wb−T=0  ⟹  T=Wb2\rho_b V (0 - (-g\hat{j})) - W_b\hat{j} - T\hat{j} = 0 \implies 2W_b - W_b - T = 0 \implies T=W_b. Ratio = 1. (Q) a→=−g2j^\overrightarrow{a}=-\frac{g}{2}\hat{j}: 2ρbV(−g2j^−(−gj^))−Wbj^−Tj^=ρbV(−g2j^)  ⟹  2ρbV(g2j^)−Wbj^−Tj^=−Wb2j^  ⟹  Wb−Wb−T=−Wb2  ⟹  T=Wb22\rho_b V (-\frac{g}{2}\hat{j} - (-g\hat{j})) - W_b\hat{j} - T\hat{j} = \rho_b V (-\frac{g}{2}\hat{j}) \implies 2\rho_b V (\frac{g}{2}\hat{j}) - W_b\hat{j} - T\hat{j} = -\frac{W_b}{2}\hat{j} \implies W_b - W_b - T = -\frac{W_b}{2} \implies T=\frac{W_b}{2}. Ratio = 1/2. (R) a→=−gj^\overrightarrow{a}=-g\hat{j}: 2ρbV(−gj^−(−gj^))−Wbj^−Tj^=ρbV(−gj^)  ⟹  2ρbV(0)−Wbj^−Tj^=−Wbj^  ⟹  −Wb−T=−Wb  ⟹  T=02\rho_b V (-g\hat{j} - (-g\hat{j})) - W_b\hat{j} - T\hat{j} = \rho_b V (-g\hat{j}) \implies 2\rho_b V (0) - W_b\hat{j} - T\hat{j} = -W_b\hat{j} \implies -W_b - T = -W_b \implies T=0. Ratio = 0. (S) a→=gj^\overrightarrow{a}=g\hat{j}: 2ρbV(gj^−(−gj^))−Wbj^−Tj^=ρbV(gj^)  ⟹  2ρbV(2gj^)−Wbj^−Tj^=Wbj^  ⟹  4Wb−Wb−T=Wb  ⟹  T=2Wb2\rho_b V (g\hat{j} - (-g\hat{j})) - W_b\hat{j} - T\hat{j} = \rho_b V (g\hat{j}) \implies 2\rho_b V (2g\hat{j}) - W_b\hat{j} - T\hat{j} = W_b\hat{j} \implies 4W_b - W_b - T = W_b \implies T=2W_b. Ratio = 2.

Matches: (P)-4, (Q)-5, (R)-1, (S)-3.