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Properties of Matter question

2024 · 8 Apr · Shift 2 · Q73

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Properties of Matter
2024 · 8 Apr · Shift 2 · Q73

Properties of Matter question

2024 · 8 Apr · Shift 2 · Q73

JEE MainPhysicsProperties of MatterMCQ+4 / −1

A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil = 0.8, specific gravity of ice = 0.9): JEE Main 2024 (Online) 8th April Evening Shift Physics - Properties of Matter Question 44 English

JEE Main 2024 (Online) 8th April Evening Shift Physics - Properties of Matter Question 44 English

A
5 : 4
B
9 : 10
C
8 : 9
D
1 : 1

View written solutionFree

Correct answer: D

Let the cube have total volume $V$ .

Suppose:
- volume immersed in water $= V_w$
- volume immersed in kerosene $= V_k$
Since the cube floats partly in both liquids, $V_w + V_k = V$
Use the condition of floating equilibrium.
Weight of ice $=$ total buoyant force due to water and kerosene.

Density relation using specific gravity:
- density of ice $= 0.9\rho_w$
- density of kerosene $= 0.8\rho_w$
- density of water $= \rho_w$
Therefore, $\rho_{ice} V g = \rho_w V_w g + \rho_k V_k g$

Substituting, $0.9\rho_w V g = \rho_w V_w g + 0.8\rho_w V_k g$

Cancel $\rho_w g$ : $0.9V = V_w + 0.8V_k$
Use $V = V_w + V_k$ .

Substitute into the equation: $0.9(V_w + V_k) = V_w + 0.8V_k$

Expanding, $0.9V_w + 0.9V_k = V_w + 0.8V_k$

Rearranging, $0.9V_k - 0.8V_k = V_w - 0.9V_w$ $0.1V_k = 0.1V_w$ $V_k = V_w$
Hence, the required ratio is $V_w : V_k = 1 : 1$
Check options:
- A: $5:4$ ❌
- B: $9:10$ ❌
- C: $8:9$ ❌
- D: $1:1$ ✅

Therefore, the correct answer is D.

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