Match the difference in level and angle ( θ ) of line connecting O to A in List-I with the respective options in List-II. (Liquid is ideal in all cases and g = 10 m / s 2 ) List-I List-II (P) The diff

Question

Match the difference in level and angle ( $θ$ ) of line connecting O to A in List-I with the respective options in List-II. (Liquid is ideal in all cases and $g = 10 m / s^{2}$ )

List-I	List-II
	(P) The difference in the vertical height between O and A is 1.25 m
	(Q) The difference in the vertical height between O and A is 2.5 m
	(R) $θ = tan^{- 1} (\frac{1}{2})$
	(S) $θ = tan^{- 1} (\frac{1}{4})$
	(T) $θ = tan^{- 1} (\frac{5}{4})$

TestHub · Accepted Answer

(I) Rotating frame, parabolic free surface: $y = \frac{ω ^{2} x ^{2}}{2 g} = \frac{25}{20} = 1.25 m \to P$ and $tan θ = \frac{y}{x} = \frac{5}{4} \to T$

(II) y = 1.25 m → P and $tan θ = \frac{1.25}{5} = \frac{1}{4} \to S$

(III) Accelerating container: $tan θ = \frac{a}{g} = \frac{5}{10} = \frac{1}{2} \to R$ and y = x tanθ = 5 × 1/2 = 2.5 m → Q

(IV) $tan θ = \frac{p _{2} - p _{1}}{p _{2} + p _{1}} = \frac{1}{2} \to R$ and $sin θ = \frac{1}{5}, y = 2 R sin θ = 2 \times \frac{5 5}{4} \times \frac{1}{5} = 2.5 m \to Q$

I → P,T; II → P,S; III → Q,R; IV → Q,R, so option (A) is correct.

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