Viscous force on the ball when it attains constant velocity i.e., terminal velocity is,
Fv=(mg−FB)(As,a=0)
Here, FB is buoyant force.
⇒Fv=Vρbg−Vρgg=Vg(ρb−ρg)=34π(10−3)3×9.8[(10.5−1.5)×103]≈3696×10−7N
Thus, x=7.
JEE Main 2023 — Physics Mechanics
A spherical ball of radius 1mm and density 10.5gcc−1 is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5gcc−1. Viscous force on the ball when it attains constant velocity is 3696×10−xN. The value of x is
(Given, g=9.8ms−2andπ=722)
Held on 24 Jan 2023 · Verified 6 Jul 2026.
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