A container contains a liquid of density $ρ .$ It is accelerated so as to have a horizontal component of acceleration equal to $g$ and a vertically upward component also equal to $g$
Match the enteries in Column I with respect to entries in Column II.

Column-I Column-II
a) Liquid surface makes an angle $α$ with horizontal. $\tan α =$ P) 4
b) A wooden block is held under water with the help of a string tied to the base. The string makes an angle $β$ with horizontal. $\tan β =$ Q) 2
c) A point A on the left wall is at a height 2cm above the base. B is another point on the base at a distance $x$ from the left wall. Pressure at A and B are equal. Value of $x$ in cm is R) $\frac{1}{2}$
d) Angle between the line AB and free surface of the liquid is $ϕ .$
$\tan ϕ =$
S) 0

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$a - P, b - Q, c - R, d - S$

$a - R, b - Q, c - P, d - S$

a-S, b-R, c-Q, d-P

a-Q, b-R, c-S, d-P

answer is B.

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Detailed Solution

Effective acceleration of free fall in an $R F$ attached to the container is $\sqrt{g^{2} + {(2 g)}^{2}} = \sqrt{5} g .$ $g_{e f f}$
makes angle $θ$ with vertical, whence $\tan θ = \frac{1}{2}$

Question Image

(A) Water surface is perpendicular to $g_{e f f} .$
$∴ α = θ = \tan^{- 1} (\frac{1}{2})$

(B) Effective buoyancy is in a direction opposite to $g_{e f f} .$ String will be along $g_{e f f} .$

$\begin{array}{l} ∴ β = 90 ° - θ \\ \tan β = \tan θ = 2 \end{array}$

(D) As explained above, $ϕ = α = \tan^{- 1} (\frac{1}{2})$

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Column-I		Column-II
a)	Liquid surface makes an angle $α$ with horizontal. $\tan α =$	P)	4
b)	A wooden block is held under water with the help of a string tied to the base. The string makes an angle $β$ with horizontal. $\tan β =$	Q)	2
c)	A point A on the left wall is at a height 2cm above the base. B is another point on the base at a distance $x$ from the left wall. Pressure at A and B are equal. Value of $x$ in cm is	R)	$\frac{1}{2}$
d)	Angle between the line AB and free surface of the liquid is $ϕ .$ $\tan ϕ =$	S)	0