A sphere of volume V is made of a material with lower density than wat

A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f₁V (f₁ < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f₂V. Then: [amp_mcq option1="f₂=f₁" option2="f₂ = (a/g) * f₁" option3="f₂ > f₁” option4=”f₂ = (g/(g-a)) * f₁” correct=”option1″]

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UPSC NDA-1 – 2023

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The correct option is A.

The fraction of volume submerged when an object floats in a fluid is determined by the ratio of the object’s density to the fluid’s density ($f = \rho_{object} / \rho_{fluid}$). This relationship arises from the condition that the buoyant force equals the weight of the object. Both the buoyant force (which depends on the effective acceleration) and the weight (which also depends on the effective acceleration) are directly proportional to the effective acceleration. Therefore, the ratio of densities, and hence the fraction of submerged volume, is independent of the magnitude of the uniform effective acceleration (like gravity) acting on the system.

On Earth, the buoyant force is $F_{B1} = \rho_w (f_1V) g$ and the weight is $W_1 = \rho_s V g$. Since it floats, $F_{B1} = W_1 \implies \rho_w f_1 V g = \rho_s V g \implies f_1 = \rho_s / \rho_w$. On the spaceship with effective acceleration $a$, the buoyant force is $F_{B2} = \rho_w (f_2V) a$ and the weight is $W_2 = \rho_s V a$. Since it floats, $F_{B2} = W_2 \implies \rho_w f_2 V a = \rho_s V a \implies f_2 = \rho_s / \rho_w$. Thus, $f_1 = f_2$. The magnitude of the effective acceleration affects the magnitude of the forces but not their ratio when determining the submerged fraction.

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