Physics - Fluid Question with Solution | TestHub

Work done in pushing the ball $$W=(v\rho g)d-(v\sigma g)d$$ Where, $\rho \to$ Density of water $\sigma \to$ Density of ball $$\begin{array}{rl}& \Rightarrow W=\frac{4}{3}\pi R^3\times 10\times 0.7\left[1000-\frac{3}{4}\times \frac{10^{-3}}{R^3}\right]\\ & W=0.077\mathrm{J}\end{array}$$ [1 is correct] $\Rightarrow$ When ball is released at bottom same work (i.e. $0.077\mathrm{J}$ ) is done on ball. $$\begin{array}{rl}& \therefore \frac{1}{2}m{v}^2=0.077\\ & v=\sqrt{\frac{0.077\times 2}{\frac{22}{7}\times 10^{-3}}}\\ & =7\mathrm{m}/\mathrm{s}\end{array}$$ [2 is correct] $$\Rightarrow \text{ also, }H=\frac{v^2}{2g}=\frac{7\times 7}{2\times 10}=2.45\mathrm{m}$$ [3 is incorrect] $\Rightarrow$ Net force $F_{\text{net }}=v\sigma g-v\sigma g=0.11\mathrm{N}$ Also, viscous force is maximum when $v=7\mathrm{m}/\mathrm{s}$ $$\begin{array}{rl}& \therefore {\left(F_v\right)}_{\max }=6\pi \eta rv\\ & =6\times \frac{22}{7}\times 10^{-3}\left(\frac{3}{2}\times 10^{-2}\right)\times 7\\ & =18\times 11\times 10^{-5}\mathrm{N}\end{array}$$ Now, $$\frac{F_{\text{net }}}{{\left(F_v\right)}_{\max }}=\frac{500}{9}$$ [4 is correct]