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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2024 · 9 April · Morning Shift · Physics · Mechanical Properties of Fluids

A sphere of relative density σ and diameter D has concentric cavity...

A sphere of relative density σ\sigma and diameter DD has concentric cavity of diameter dd. The ratio of Dd\frac{D}{d}, if it just floats on water in a tank is :

Options

  1. A(σ−2σ+2)1/3\left(\frac{\sigma-2}{\sigma+2}\right)^{1 / 3}
  2. B(σσ−1)1/3\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}
  3. C(σ−1σ)1/3\left(\frac{\sigma-1}{\sigma}\right)^{1 / 3}
  4. D(σ+1σ−1)1/3\left(\frac{\sigma+1}{\sigma-1}\right)^{1 / 3}

Correct answer

B. (σσ−1)1/3\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}

Step-by-step solution

weight (w)=43π(D3−d38)σg(w)=\frac{4}{3} \pi\left(\frac{\mathrm{D}^3-\mathrm{d}^3}{8}\right) \sigma \mathrm{g} Buoyant force (Fb)=1×43π(D38)⋅g\left(\mathrm{F}_{\mathrm{b}}\right)=1 \times \frac{4}{3} \pi\left(\frac{\mathrm{D}^3}{8}\right) \cdot \mathrm{g} For Just Float ⇒\Rightarrow\mathrm{w}=\mathrm{F}b_{\mathrm{b}}⇒(D3−d3)σ=D3⇒1−d3D3=1σ⇒1−1σ=(dD)3⇒(σσ−1)13=(Dd)\begin{aligned} & \Rightarrow\left(D^3-d^3\right) \sigma=D^3 \\ & \Rightarrow 1-\frac{d^3}{D^3}=\frac{1}{\sigma} \\ & \Rightarrow 1-\frac{1}{\sigma}=\left(\frac{d}{D}\right)^3 \\ & \Rightarrow\left(\frac{\sigma}{\sigma-1}\right)^{\frac{1}{3}}=\left(\frac{D}{d}\right)\end{aligned}$$
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