Question

7If a ball of steel (density p = 7.8 g cm ) attains a terminal velocity of 10 cm s 'when falling in a water (Coefficient of
Viscosity nwater = 8.5 x 10+ Pa. s) then its terminal velocity in glycerin (p = 1.2 g cm, n=132 Pa. s.) would be, nearly
\(A*) 6.25 x 104 cm s' (B) 6.45 x 10-4 cm s-1 (C) 1.5 x10 cm s-1 (D) 1.6 10-5 cm
cms-
(AIEEE-2011)

PLZZ if you the solution then only reply otherwise I will report spammers.​

Answer 1

Answer:

Terminal velocity of ball when falling in liquid is given by

(ρb−ρl)Vg=6πηrv...eq(1)

1. case

when ball falling in tank of water of density ρwater=1gcm−3, ρball=7.8gcm−3ηwater=8.5×10−4Pa.s and terminal velocity v=10cms−1

Put all the above value in eq(1)

⇒(7.8−1)Vg=6π×8.5×10−4r×10...eq(2)

2. case

when ball falling in tank of glycerin

ρball=7.8gcm−3ηglycerin=13.2Pa.s and terminal velocity v1

put all the value in eq(1)

⇒(7.8−1.2)Vg=6π×13.2r×v1...eq(3)

divide eq(3) b eq(1)

7.8−17.8−1.2=8.5×10−4×1013.2×v1

⇒v1=6.8×13.26

hope it will help you

Answer 2

Explanation:

I think the A is the correct answer