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JEE Main 202324 JanuaryEvening Shift
Exam
JEE Main
Year
2023
Date
24 January
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2023 · 24 January · Evening Shift · Physics · Mechanical Properties of Fluids

A spherical ball of radius 1mm and density 10 . 5g cc ^ - 1 is...

A spherical ball of radius 1mm1mm and density 10.5gcc−110 . 5g{cc}^{- 1} is dropped in glycerine of coefficient of viscosity 9.89 . 8 poise and density 1.5gcc−11 . 5g{cc}^{- 1}. Viscous force on the ball when it attains constant velocity is 3696×10−xN3696 \times {10}^{- x}N. The value of xx is

(Given, g=9.8ms−2andπ=227g = 9 . 8m{s}^{- 2}\text{and}\pi = \dfrac{22}{7})

Correct answer

0

Step-by-step solution

Viscous force on the ball when it attains constant velocity i.e., terminal velocity is,

Fv=(mg−FB)(As,a=0){F}_{v} = \left(mg - {F}_{B}\right)\left(\text{As}, a = 0\right)

Here, FB{F}_{B} is buoyant force.

⇒Fv=Vρbg−Vρgg=Vg(ρb−ρg)=43π(10−3)3×9.8[(10.5−1.5)×103]≈3696×10−7N⇒ {F}_{v} = V{ρ}_{b}g - V{ρ}_{g}g = Vg\left({ρ}_{b} - {ρ}_{g}\right) = \dfrac{4}{3}\pi{\left({10}^{- 3}\right)}^{3} \times 9 . 8\left[\left(10 . 5 - 1 . 5\right) \times {10}^{3}\right] \approx 3696 \times {10}^{- 7}N

Thus, x=7x = 7.

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