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Match the List I with the List-II and select the correct answer using the code given below the lists.
List – I List – II
(P)
Three identical uniform rods each of mass ‘m’ and length ‘L’ are joined as shown in figure. The system is hinged about one end in a vertical plane. At equilibrium position, the value of $' \tan θ'$ is
(1) $\frac{1}{3}$
(Q)
Two identical uniform rods each of mass ‘m’ and length ‘L’ are joined as shown in figure. The system is hinged about one end in a vertical plane. At equilibrium position, the value of ‘tan $θ$ ’ is
(2) $\frac{3}{4}$
(R) A ball collides with a smooth and fixed inclined plane (of inclination $θ$ and coefficient of restitution $e = \frac{1}{4}$ ) after falling vertically through a distance h. If it moves horizontally just after the impact, the value of tan $θ$ is (3) 1
(S)
A ball is projected with velocity u at an angle ‘ $θ$ ’ with the horizontal. The value of “tan $θ$ ” for which radius at the highest point is half the range is
(4) $\frac{1}{2}$

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Q-1

answer is A, B, C, D.

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Detailed Solution

In equilibrium, the COM of system should lie vertically under the hinged point.

$\begin{array}{l} {P) x}_{c m} = \frac{M \frac{l}{2} + M l + M \frac{l}{2}}{3 M} = \frac{2 l}{3} \\ y_{c m} = \frac{M .0 + M \frac{l}{2} + M l}{3 M} = \frac{l}{2} \\ \tan θ = \frac{y_{c m}}{x_{c m}} = \frac{3}{4} (u n d e r e q u i l i b r i u m) \end{array}$

$\begin{array}{l} Q) x_{c m} = \frac{M \frac{l}{2} + M l}{2 M} = \frac{3 l}{4} \\ y_{c m} = \frac{M \cdot 0 + M \frac{l}{2}}{2 M} = \frac{l}{4} \\ \tan θ = y_{c m} / x_{c m} = 1 / 3 \end{array}$

$\begin{array}{l} R) P e r p e n d i c u l a r t o p l a n e, e = \frac{v \sin θ}{u \cos θ}, \\ P a r a l l e l t o p l a n e, u \sin θ = v \cos θ \\ \Rightarrow v = u \tan θ \\ ∴ e = \tan^{2} θ \\ \Rightarrow 1 / 4 = \tan^{2} θ \\ \Rightarrow \tan θ = \frac{1}{2} \end{array}$

$\begin{array}{l} S) \frac{u^{2} \sin 2 θ}{2 g} = \frac{u^{2} \cos^{2} θ}{g} \\ \Rightarrow \tan θ = 1 \end{array}$

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