Answer:
Volume of the ball, \[V=\frac{m}{\rho }\] Mass of the liquid displaced, \[m'=V{{\rho }_{0}}=\frac{m}{\rho }.{{\rho }_{0}}\] When the body falls with a constant velocity, Viscous force \[=\] Effective weight of the ball \[F=\]Weight of the ball ? Up thrust \[=mg-m'g\] or \[F=mg-\frac{m{{\rho }_{0}}}{\rho }.g=mg\left( 1-\frac{{{\rho }_{0}}}{\rho } \right)\].
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