Figure shows a closed container completely filled with an ideal liquid of density ρ. In the liquid there is a spherical body of volume V and density σ attached to a string whose other end is attached to the roof of the container. The container is accelerating with an acceleration ' a ' towards right. The force exerted by the liquid on the spherical body when it is in equilibrium with respect to the liquid, will be:

Question

Figure shows a closed container completely filled with an ideal liquid of density ρ. In the liquid there is a spherical body of volume V and density σ attached to a string whose other end is attached to the roof of the container. The container is accelerating with an acceleration ' a ' towards right. The force exerted by the liquid on the spherical body when it is in equilibrium with respect to the liquid, will be: (A) V ρ √a2+g2+V σ a (B) V ρ g+V σ a (C) V ρ(g+a) (D) V ρ √a2+g2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a29ea9a5dd3b33298b3db61b6a4224d9-.png" /> From the frame of reference of liquid, effective gravity (resultant of weight and pseudo force per unit mass) =√a2+g2 ∴ Effective force due to liquid =V ρ √a2+g2 Aliter: The force exerted on spherical body by surrounding liquid is equal to sum of contact forces on displaced liquid by the surrounding liquid as shown in. Hence, the force exerted on spherical body by surrounding liquid =V ρ √a2+g2.

$V ρ a^{2} + g^{2} + Vσa$

$V ρ g + Vσa$

$V ρ (g + a)$

$V ρ a^{2} + g^{2}$

Vρa2+g2​+Vσa

Vρg+Vσa

Vρ(g+a)

Vρa2+g2​

$V ρ a^{2} + g^{2} + Vσa$

$V ρ g + Vσa$

$V ρ (g + a)$

$V ρ a^{2} + g^{2}$