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Physics · Fluid Mechanics

JEE Main 2025 — 28 January, Evening Shift — Question 61

a 400 g id cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water? (Given : density of water =1000 kg m−3=1000 \mathrm{~kg} \mathrm{~m}^{-3} )

  1. Option A:

    1400 cm31400 \mathrm{~cm}^{3}

  2. Option B:

    4000 cm34000 \mathrm{~cm}^{3}

  3. Option C:

    400 cm3400 \mathrm{~cm}^{3}

  4. Option D:

    600 cm3600 \mathrm{~cm}^{3}

    Correct

Answer: D

Step-by-step solution

Mg=FB⇒(400×10−3)=103×Vd\quad \mathrm{Mg}=\mathrm{F}_{\mathrm{B}} \Rightarrow\left(400 \times 10^{-3}\right)=10^{3} \times \mathrm{V}_{\mathrm{d}}

Vd=400×10−6 m3\mathrm{V}_{\mathrm{d}}=400 \times 10^{-6} \mathrm{~m}^{3}

(Vol.) outiside =(10×10−2)3−400×10−6=\left(10 \times 10^{-2}\right)^{3}-400 \times 10^{-6}

=600×10−6 m2=600 cm3=600 \times 10^{-6} \mathrm{~m}^{2}=600 \mathrm{~cm}^{3}

4

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Evening Shift
Subject
Physics
Chapter
Fluid Mechanics
Topic
Buoyancy and Archimedes' Principle

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