Question

A ball rises to the surface of a liquid with constant velocity. The density of the liquid is four times the density of the material of the ball. The frictional force of the liquid on the rising ball is greater than the weight of the ball by a factor of

• A. $2$
  
• B. $3$
  
• C. $4$
  
• D. $6$
  

Answer 1

The correct option is B $3$

Let $F_f=$ Force of friction

$F_b=$ Force of bouyancy

$F_w=$ Weight of ball

Archimedes' principle:

$F_b=F_w+F_f$

Given: $F_b=\left(4P_B\right)gV$

$F_w=V{P}_{B}g$

$F_f=F_b-F_w=3P_{B}gV$

$\Rightarrow \frac{F_f}{F_w}=\frac{3P_{B}gV}{P_{B}gV}=3:1$

$\Rightarrow F_f=3F_w$