A sphere of brass released in a long liquid column attains a terminal speed v0. If the terminal speed attained by sphere of marble of the same radius and released in the same liquid is n v0, then the value of n will be (Given: The specific gravities of brass, marble and the liquid are 8.5 , 2.5 and 0.8 respectively.)

Question

A sphere of brass released in a long liquid column attains a terminal speed v0. If the terminal speed attained by sphere of marble of the same radius and released in the same liquid is n v0, then the value of n will be (Given: The specific gravities of brass, marble and the liquid are 8.5 , 2.5 and 0.8 respectively.) (A) (5/17) (B) (17/77) (C) (11/31) (D) (17/5)

Tardigrade Admin · Accepted Answer

vC=(2/9) r2 ((ρ-ρ prime) g/η) Here, (n v0/v0)=(ρ text marble -ρ text liquid /ρ text brass -ρ text liquid )=(2.5-0.8/8.5-0.8)=(1.7/7.7) or n=(17/77)

$\frac{5}{17}$

$\frac{17}{77}$

$\frac{11}{31}$

$\frac{17}{5}$

$v_{C} = \frac{2}{9} r^{2} \frac{( ρ - ρ ^{'} ) g}{η}$
Here, $\frac{n v _{0}}{v _{0}} = \frac{ρ _{marble} - ρ _{liquid}}{ρ _{brass} - ρ _{liquid}} = \frac{2.5 - 0.8}{8.5 - 0.8} = \frac{1.7}{7.7}$
or $n = \frac{17}{77}$

175​

7717​

3111​

517​

vC​=92​r2η(ρ−ρ′)g​ Here, v0​nv0​​=ρbrass ​−ρliquid ​ρmarble ​−ρliquid ​​=8.5−0.82.5−0.8​=7.71.7​ or n=7717​

$\frac{5}{17}$

$\frac{17}{77}$

$\frac{11}{31}$

$\frac{17}{5}$

$v_{C} = \frac{2}{9} r^{2} \frac{( ρ - ρ ^{'} ) g}{η}$
Here, $\frac{n v _{0}}{v _{0}} = \frac{ρ _{marble} - ρ _{liquid}}{ρ _{brass} - ρ _{liquid}} = \frac{2.5 - 0.8}{8.5 - 0.8} = \frac{1.7}{7.7}$
or $n = \frac{17}{77}$