A) 80 cm/s
B) 40 m/s
C) 10 cm/s
D) 5 cm/s
Correct Answer: D
Solution :
Let the radius of ball is r and density is \[\rho ,\] falling freely in a liquid whose density is \[\sigma \] and coefficient of viscosity t|. When it attains a terminal velocity v it is subjected to two forces. (i) Effective force acting downwards \[=V(\rho -\sigma )g=\frac{4}{3}\pi {{r}^{3}}(\rho -\sigma )g\] (ii) Viscous force acting upward = 6nr}rv Since, ball is moving with a constant speed v, there is no acceleration in it, the net force acting on it, must be zero. That is, \[6\,\pi \eta \,rv=\frac{4}{3}\pi {{r}^{3}}(\rho -\sigma )g\] \[\Rightarrow \] \[v=\frac{2}{9}\frac{{{r}^{2}}(\rho -\sigma )g}{\eta }\] \[\Rightarrow \] \[v\propto {{r}^{2}}\] Here \[{{v}_{1}}=20\,cm/s,\,{{r}_{1}}=2\,cm,\,{{v}_{2}}=?,\,\,{{r}_{2}}=\,1\,cm\] \[\therefore \] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\frac{2}{{{v}_{2}}}=\frac{{{(2)}^{2}}}{{{(1)}^{2}}}\] \[\Rightarrow \] \[{{v}_{2}}=20/4\,=5\,cm/s\]You need to login to perform this action.
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