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

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

A spherical ball of radius 1 × 10^ -4 ~m and density 10^5 ~kg / m^3...

A spherical ball of radius 1×10−4 m1 \times 10^{-4} \mathrm{~m} and density 105 kg/m310^5 \mathrm{~kg} / \mathrm{m}^3 falls freely under gravity through a distance hh before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of hh is approximately: (The coefficient of viscosity of water is 9.8×10−6 N s/m29.8 \times 10^{-6} \mathrm{~N} \mathrm{~s} / \mathrm{m}^2 )

Options

  1. A2296 m2296 \mathrm{~m}
  2. B2518 m2518 \mathrm{~m}
  3. C2249 m2249 \mathrm{~m}
  4. D2396 m2396 \mathrm{~m}

Correct answer

B. 2518 m2518 \mathrm{~m}

Step-by-step solution

VT=2 g9R2[ρB−ρL]η&⇒VT=29×10×(10−4)29.8×10−6[105−103]&⇒VT=224.5\begin{aligned}& \mathrm{V}_{\mathrm{T}}=\frac{2 \mathrm{~g}}{9} \frac{\mathrm{R}^2\left[\rho_{\mathrm{B}}-\rho_{\mathrm{L}}\right]}{\eta} \& \Rightarrow \mathrm{V}_{\mathrm{T}}=\frac{2}{9} \times \frac{10 \times\left(10^{-4}\right)^2}{9.8 \times 10^{-6}}\left[10^5-10^3\right] \& \Rightarrow \mathrm{V}_{\mathrm{T}}=224.5\end{aligned} when ball fall from height (h) V=2ghh=(V22 g)=2518 m\begin{aligned}& \mathrm{V}=\sqrt{2 \mathrm{gh}} \\ & \mathrm{h}=\left(\frac{\mathrm{V}^2}{2 \mathrm{~g}}\right)=2518 \mathrm{~m}\end{aligned}
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