A Spherical ball of radius 1 mm and density 10.5 g / cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g / cc. Viscous force on the ball when it attains constant velocity is 3696 × 10-x N. The value of x is (Given, g=9.8 m / s 2 and π=(22/9) )

Question

A Spherical ball of radius 1 mm and density 10.5 g / cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g / cc. Viscous force on the ball when it attains constant velocity is 3696 × 10-x N. The value of x is (Given, g=9.8 m / s 2 and π=(22/9) ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

When the ball attain terminal velocity F v =( mg - F B )(∵ a =0) = operatornameV σ b g - V ρℓ g = operatornameVg(σ b -ρℓ) =(4/3) π(10-3)3 × 9.8(10.5-1.5) × 103 =3696 × 10-7 N So , x =7

Q. A Spherical ball of radius 1mm and density 10.5g/cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5g/cc. Viscous force on the ball when it attains constant velocity is 3696×10−xN. The value of x is_______ (Given, g=9.8m/s2 and π=922​ )

When the ball attain terminal velocity Fv​=(mg−FB​)(∵a=0) =Vσb​g−Vρℓ​g =Vg(σb​−ρℓ​) =34​π(10−3)3×9.8(10.5−1.5)×103 =3696×10−7N So , x=7

When the ball attain terminal velocity
$F_{v} = (m g - F_{B}) (∵ a = 0)$
$= V σ_{b} g - V ρ_{ℓ} g$
$= Vg (σ_{b} - ρ_{ℓ})$
$= \frac{4}{3} π (1 0^{- 3})^{3} \times 9.8 (10.5 - 1.5) \times 1 0^{3}$
$= 3696 \times 1 0^{- 7} N$
So , $x = 7$