Q165 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Physics
A cube floats in water with $\left( \dfrac{1}{3} \right)^{th}$ part outside the surface of water. The same cube floats in another liquid with $\left( \dfrac{3}{4} \right)^{th}$ part outside the liquid. If the density of liquid is $\rho$, then the value of $\rho$ in g/cc is ($\rho_w$ = density of water = $1$ g/cc)
- A$\dfrac{2}{3}$
- B$\dfrac{4}{3}$
- Correct answerC$\dfrac{8}{3}$
- D$\dfrac{10}{3}$
Solution
Let $V$ be the volume of the cube and $\rho_c$ be its density.
When the cube floats in water, the volume submerged is $V - \dfrac{1}{3}V = \dfrac{2}{3}V$.
By the principle of flotation, the weight of the cube equals the buoyant force:
$V \rho_c g = \left( \dfrac{2}{3}V \right) \rho_w g$
$\rho_c = \dfrac{2}{3} \rho_w$
When the cube floats in the liquid, the volume submerged is $V - \dfrac{3}{4}V = \dfrac{1}{4}V$.
Applying the principle of flotation again:
$V \rho_c g = \left( \dfrac{1}{4}V \right) \rho g$
$\rho = 4 \rho_c$
Substituting the value of $\rho_c$:
$\rho = 4 \left( \dfrac{2}{3} \rho_w \right) = \dfrac{8}{3} \rho_w$
Given $\rho_w = 1$ g/cc, we get:
$\rho = \dfrac{8}{3}$ g/cc
Answer: $\dfrac{8}{3}$
When the cube floats in water, the volume submerged is $V - \dfrac{1}{3}V = \dfrac{2}{3}V$.
By the principle of flotation, the weight of the cube equals the buoyant force:
$V \rho_c g = \left( \dfrac{2}{3}V \right) \rho_w g$
$\rho_c = \dfrac{2}{3} \rho_w$
When the cube floats in the liquid, the volume submerged is $V - \dfrac{3}{4}V = \dfrac{1}{4}V$.
Applying the principle of flotation again:
$V \rho_c g = \left( \dfrac{1}{4}V \right) \rho g$
$\rho = 4 \rho_c$
Substituting the value of $\rho_c$:
$\rho = 4 \left( \dfrac{2}{3} \rho_w \right) = \dfrac{8}{3} \rho_w$
Given $\rho_w = 1$ g/cc, we get:
$\rho = \dfrac{8}{3}$ g/cc
Answer: $\dfrac{8}{3}$
Hence, the correct answer is option: 'C'.