A)
B)
C)
D)
Correct Answer: C
Solution :
Acceleration of ball \[a=\frac{mg-6\pi \eta rv}{m}\] Terminal velocity of ball \[v=\frac{2}{9}\frac{(\sigma -\rho ){{r}^{2}}g}{\eta }\] As temperature increases Viscosity of gas \[\uparrow \] Viscosity of liquid \[\downarrow \]
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