A liquid of density 'ρ' is filled in a cubical container OABCDEFG of side b. The point ‘O’ is assumed to be origin. A small ball of density ρ0 is kept at (b2,b2,b2) as shown in figure and an acceleration 2gi^+gj^ is given to container. Uniform gravity exists along negative y-axis. Now, choose the correct option(s).

a → 0 = ( 2 g i ^ + g j ^ ) F → B = ρ V ( a → 0 − g → ) = 2 ρ V g ( i ^ + j ^ ) F → n e t = F → B + m g → = 2 ρ V g ( i ^ + j ^ ) − ρ 0 V g j ^ a → n e t = 2 ρ ρ 0 g ( i ^ + j ^ ) − g j ^ a → r e l a t i v e = a → n e t − a → 0

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A liquid of density $' ρ'$ is filled in a cubical container OABCDEFG of side b. The point ‘O’ is assumed to be origin. A small ball of density $ρ_{0}$ is kept at $(\frac{b}{2}, \frac{b}{2}, \frac{b}{2})$ as shown in figure and an acceleration $2 g \hat{i} + g \hat{j}$ is given to container. Uniform gravity exists along negative y-axis. Now, choose the correct option(s).

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If $ρ = ρ_{0}$ then magnitude of acceleration of small ball is $\sqrt{5} g .$

If $ρ = 2 ρ_{0}$ then small ball will hit the midpoint of edge GF

If $ρ = 3 ρ_{0}$ then small ball will hit the midpoint of edge GF.

If $ρ = ρ_{0}$ then net force on the small ball is zero.

answer is A, B, C.

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Detailed Solution

${\vec{a}}_{0} = (2 g \hat{i} + g \hat{j}) {\vec{F}}_{B} = ρ V ({\vec{a}}_{0} - \vec{g}) = 2 ρ V g (\hat{i} + \hat{j}) {\vec{F}}_{n e t} = {\vec{F}}_{B} + m \vec{g} = 2 ρ V g (\hat{i} + \hat{j}) - ρ_{0} V g \hat{j} {\vec{a}}_{n e t} = \frac{2 ρ}{ρ_{0}} g (\hat{i} + \hat{j}) - g \hat{j} {\vec{a}}_{r e l a t i v e} = {\vec{a}}_{n e t} - {\vec{a}}_{0}$