A particle A of diameter 10 microns settles in an oil of specific gravity 0.9 and viscosity 10 poise under Stoke's law. A particle B with diameter 20 microns settling in the same oil will have a settling velocity
A. Same as that of A
B. One fourth as that of A
C. Twice as that of A
D. Four times as that of A
Answer: Option B
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A. Thermal conductivity
B. Electrical conductivity
C. Specific gravity
D. Electrical resistivity
A. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {\frac{{\text{x}}}{{\text{r}}}} \right)^{\frac{1}{7}}}$$
B. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {\frac{{\text{r}}}{{\text{x}}}} \right)^{\frac{1}{7}}}$$
C. $$\frac{{\text{V}}}{{{{\text{V}}_{\max }}}} = {\left( {{\text{x}} \times {\text{r}}} \right)^{\frac{1}{7}}}$$
D. None of these
A. d
B. $$\frac{1}{{\text{d}}}$$
C. $$\sigma $$
D. $$\frac{l}{\sigma }$$
A. $$\frac{{4\pi {\text{g}}}}{3}$$
B. $$\frac{{0.01\pi {\text{gH}}}}{4}$$
C. $$\frac{{0.01\pi {\text{gH}}}}{8}$$
D. $$\frac{{0.04\pi {\text{gH}}}}{3}$$
From stokes law,
Ut=(g*Dp^2*(Rho p-Rho))/18mue
“Ut” directly proportional to “Dp” then
=> (Ut)a / (Ut)b = (Dp^2)a / (Dp^2)b
=> (Ut)a / (Ut)b = (10^2) / (20^2)
Therefore; (Ut)a / (Ut)b = 1/4