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Q.

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 

Question Image

Match the enteries in Column I with respect to entries in Column II.
 

Column-IColumn-II
a)Liquid surface makes an angle α  with horizontal. tanα=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. tanβ= 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 isR)12
d)Angle between the line AB and free surface of the liquid is ϕ.

tanϕ=

S)0

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a

aP, bQ, cR, dS

b

aR, bQ, cP, dS

c

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

d

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

answer is B.

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Detailed Solution

Effective acceleration of free fall in an RF  attached to the container is  g2+(2g)2=5g. geff
  makes angle θ   with vertical, whence  tanθ=12

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(A) Water surface is perpendicular to  geff.
    α=θ=tan1(12)

(B) Effective buoyancy is in a direction opposite to geff.  String will be along  geff.

    β=90°θtanβ=tanθ=2

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(D) As explained above,   ϕ=α=tan1(12)

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