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Match the following
Column-I Column-II
A Cube of mass m and density $σ$ is pulled by a force ${\vec{F}}_{x} = m g \hat{i}$ in an accelerating liquid of density $ρ$ . The value of $\frac{a_{x}}{a_{y}}$ cube is    P $\frac{ρ}{σ}$
B
The sphere of mass m is pulled horizontally by force $F_{x}$ = mg in non – accelerating liquid. The value of $\frac{a_{x}}{a_{y}}$ of sphere is

Q 1:1
C
A closed cube of length x is placed in a rotating liquid. The value of $a_{x} / g,$ (where $a_{x}$ is the horizontal acceleration of the cube ) is

R $\frac{σ}{ρ - σ}$
D
The string of a pendulum makes an angle $θ$ with vertical when the cart moves with an acceleration a. If the cart is filled with liquid of density $ρ < σ$ (density of the bob), the value of the angle of inclination with vertical becomes $θ^{1}$ then    $\frac{\tan θ}{\tan θ^{1}}$
S $\frac{2 σ + ρ}{2 (σ - ρ)}$

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$a \to q, b \to p, c \to s, d \to r$

$a \to r, b \to q, c \to s, d \to p$

$a \to p, b \to s, c \to q, d \to r$

$a \to s, b \to r, c \to p, d \to q$

answer is C.

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Detailed Solution

A) $F_{B} = ρ (\frac{m}{σ}) g_{e f f} = m (\frac{ρ}{σ}) \sqrt{a^{2} + g^{2}}$

FBC of cube w.r.t container

So, $\cos θ = {\frac{g}{\sqrt{a^{2} + g^{2}}}}_{} a n d_{} \sin θ = \frac{a}{\sqrt{a^{2} + g^{2}}}$

Let $a_{x} = a c c$ of cube w.r.t container in x-direction $m g + F_{B} \sin θ - \frac{m g}{2} = m a_{x}$

$a_{x}^{1} = \frac{g}{2} + \frac{ρ}{σ} a \Rightarrow a_{x}^{1} = \frac{g}{2} + \frac{ρ}{σ} . \frac{g}{2} H e n c e, a_{x}^{1} = \frac{g}{2} + \frac{ρ}{σ} a \Rightarrow a_{x}^{1} = \frac{g}{2} + \frac{ρ}{σ} . \frac{g}{2}$

In Y-direction :

$F_{B} \cos θ - m g = m a_{y} \Rightarrow m (\frac{ρ}{σ}) \sqrt{a^{2} + g^{2}} \frac{g}{\sqrt{a^{2} + g^{2}}} - m g = m a_{y}$

FBD for sphere

$a_{y} = g [\frac{ρ}{σ} - 1] = g [\frac{ρ}{σ}] ∴ \frac{a_{x}}{a_{y}} = \frac{(2 σ + ρ)}{2 (ρ - σ)}$

B) $a_{x} = g$

$a_{y} = \frac{- m g + m (\frac{ρ}{σ}) g}{m} = (\frac{ρ - σ}{σ}) g ∴ \frac{a_{x}}{a_{y}} = \frac{σ}{ρ - σ}$

$C) (P_{1} - P_{2}) A = m_{} a_{x}$

FBD of cylinder

$\frac{ρ ω^{2} 2}{} [{(2 x)}^{2} - x^{2}] A = m_{} a_{x} \frac{ρ}{2} (\frac{2 g}{3 x}) 3 x^{2} A = (A - x) σ a_{x} ∴ \frac{a_{x}}{g} = \frac{ρ}{σ}$

D) When cart is not filled with liquid, $\tan θ = \frac{a}{g}$

FBD of pendulum

$\frac{W h e n_{} c a r t_{} i s_{} f i l l e d_{} w i t h_{} l i q u i d}{C a r t : (T + F_{B}) \sin θ^{1} = m a} \dots \dots (1) \tan θ^{1} = \frac{a}{g} \dots . (2) (F_{B} + T) \cos θ^{1} = m g \dots . . (3) ∴ \frac{\tan θ^{1}}{\tan θ} = \frac{1}{1} = \frac{θ^{1}}{θ} = \frac{1}{1}$

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