Question

A body of density D1 and mass M is moving

downwards in glycerine of density D2. What is the

viscous force acting on it?

(a) Mg(1-(D2/D1) b) Mg(1-D1/D2)

(c) MgD1 (d) MgD2​

Answer 1

Answer:

(a) $F = Mg(1-\frac{D_{2}}{D_{1}})$

Explanation:

We know that,

$F_{viscous}=6\pi\eta rv_{t}$

And,

$v_{t}=\frac{2}{9}\frac{r^{2}g}{\eta}(D_{1}-D{2})$

Substituting the value of $v_{t}$, we get

$F_{viscous}=6\pi\eta r*\frac{2}{9}\frac{r^{2}g}{\eta}(D_{1}-D{2})\\\\F_{viscous}=\frac{4}{3}\pi r^{3}*g*(D_{1}-D{2})\\\\F_{viscous}=V*g*(D_{1}-D{2})\\\\F_{viscous}=\frac{M}{D_{1}}*g*(D_{1}-D{2})\\\\F_{viscous}=Mg(\frac{D_{1}}{D_{1}}-\frac{D_{2}}{D_{1}})\\\\F_{viscous}=Mg(1-\frac{D_{2}}{D_{1}})$

Answer 2

Answer:

mg (1 - $\frac{D2}{D1}$)

Explanation:

Mg is acting downward and the buoyant force is opposing it. (Newton III rd Law)

buoyant force = -$rho$ . g . Vol

Viscous force = mg - buoyant force

F = mg - rho.g.V

  = D1.V.g - D2.g.V

  = Vg (D1 - D2)

  = m/D1.g (D1 - D2)

  = mg ( 1 - D2/D1)