Physics And Measurement Question 266
A spherical body of mass m and radius r is allowed to fall in a medium of viscosity $ \eta $. The time in which the velocity of the body increases from zero to 0.63 times the terminal velocity (v) is called time constant $ (\tau ) $. Dimensionally $ \tau $ can be represented by: $ \tau = \frac{m}{6\pi\eta r} $
Options:
A) $ \frac{mr^{2}}{6\pi \eta } $
B) $ \sqrt{( \frac{6\pi mr\eta }{g^{2}} )} $
C) $ \frac{m}{6\pi \eta rv} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ F=6\pi \eta rv $
$ \Rightarrow mg=6\pi \eta rv $ (dimensionally) Check options dimensionally
BETA
