The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is 6.5 cms–1. Calculate the viscosity of the oil at 20°C. (Density of oil is \[1.7\times 10^{3}kgm^{-3}\] and density of copper is \[8.9\times 10^{3}kgm^{-3}\]) :
A lead sphere of mass m falls in viscous liquid with terminal velocity v0. Another lead sphere of mass M falls through the same viscous liquid with terminal velocity 4v0. the ratio M/m is :
A long capillary is dipped in a beaker containing water. Water rises in capillary upto some height \(h\). Match the statements in list-I with most appropriate effects on water level mentioned in list-II:
**List-I**
(A) Soap solution is added to water
(B) Arrangement taken in a freely falling lift
(C) In a lift accelerating uniformly upward
(D) Arrangement is taken in a lift accelerating uniformly downward
**List-II**
(p) \(h\) decreases
(q) \(h\) increases
(r) \(h\) remains same
(s) water will rise upto complete height of capillary
(t) water level in capillary goes below the outside level
Soap reduces surface tension, so \(h\) decreases (A-p). In a free fall, effective gravity \(g_{eff} = 0\), so water rises to full height (B-s). Upward acceleration increases \(g_{eff}\) hence \(h\) decreases (C-p). Downward acceleration decreases \(g_{eff}\) hence \(h\) increases (D-q).
The velocity of a small ball of mass \(M\) and density \(d\), when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(\frac{d}{2}\), then the viscous force acting on the ball will be
When the ball reaches terminal velocity, net force is zero: \(F_v + F_B = Mg\). The buoyant force is \(F_B = V \rho_{\text{glycerine}} g = V \left(\frac{d}{2}\right) g = \frac{Mg}{2}\). Thus, the viscous force is \(F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}\).