NEET Practice Questions (MCQs) with Answers & Solutions

(c) Since, upthrust (F) $=\mathrm{V\sigma g}$   i.e. F∝V

Submarine has a large volume ,so buoyant force  is more than the weight. Hence it floats on water surface. When it has to go inside water, it is filled with water so that weight is more than the buoyant force acting on submarine. So it sinks.

(b) For streamline flow, Reynold's number ${\mathrm{N}}_{\mathrm{R}}\propto \frac{\mathrm{r\rho }}{\mathrm{\eta }}$ should be less. For less value of ${\mathrm{N}}_{\mathrm{R}}$, radius and density should be small and viscosity should be high.

(a) ${\mathrm{d}}_{\mathrm{A}}=2\mathrm{cm}$and ${\mathrm{d}}_B=4 \mathrm{cm}$               $\therefore {\mathrm{r}}_{\mathrm{A}}=1\mathrm{cm}$ and ${\mathrm{r}}_{\mathrm{B}}=2 \mathrm{cm}$

From equation of continuity , av= constant

$\therefore \frac{{\mathrm{v}}_{\mathrm{A}}}{{\mathrm{v}}_{\mathrm{B}}}=\frac{{\mathrm{a}}_{\mathrm{B}}}{{\mathrm{a}}_{\mathrm{A}}}=\frac{\mathrm{\pi }({\mathrm{r}}_{\mathrm{B}}{)}^2}{\mathrm{\pi }({\mathrm{r}}_{\mathrm{A}}{)}^2}={\left(\frac{2}{1}\right)}^2\Rightarrow {\mathrm{v}}_{\mathrm{A}}=4{\mathrm{v}}_{\mathrm{B}}$

(a) Velocity head $\mathrm{h}=\frac{{\mathrm{v}}^2}{2\mathrm{g}}=\frac{(5{)}^2}{2\times 10}=1.25\mathrm{m}$

(b) Bernoulli's theorem for unit mass of liquid

$\frac{\mathrm{P}}{\mathrm{\rho }}+\frac{1}{2}{\mathrm{v}}^2= \mathrm{constant}$

As the liquid starts flowing, it pressure energy decreases 

$$\begin{gathered} \frac{1}{2}{\mathrm{v}}^2=\frac{{\mathrm{P}}_1-{\mathrm{P}}_2}{\mathrm{\rho }}\Rightarrow \frac{1}{2}{\mathrm{v}}^2=\frac{3.5\times {10}^{-5}-3\times {10}^5}{{10}^3} \\ =\frac{2\times 0.5\times {10}^5}{{10}^3}\Rightarrow {\mathrm{v}}^2=100\Rightarrow \mathrm{v}=10 \mathrm{m}/\mathrm{s} \end{gathered}$$

(c)

$$\begin{gathered} When the \tan k is full upto height h, velocity at the hole, v=\sqrt{2gh} \\ Let the rate of decrease height of water=-\frac{dh}{dt} \\ So, u\sin g continuity equation, \\ -A\frac{dh}{dt}=A_{0}v=A_0\sqrt{2gh} \\ {\int }_0^{t}dt=-{\int }_{H_1}^{H_2}\frac{A}{A_0}\frac{dh}{\sqrt{2gh}} \\ t=-\frac{A}{A_0}\times \frac{2}{\sqrt{2g}}{\left(\sqrt{h}\right)}_{H_1}^{H_2}=\frac{A}{A_0}\times \sqrt{\frac{2}{g}}\left(\sqrt{H_1}-\sqrt{H_2}\right) \end{gathered}$$

Time taken for the level to fall from H to H'

${\mathrm{t}}_1=\frac{\mathrm{A}}{{\mathrm{A}}_0}\sqrt{\frac{2}{\mathrm{g}}}\left[\sqrt{H}-\sqrt{H\text{'}}\right]$

According to problem-the time taken for the level to fall from h to $\frac{\mathit{h}}{\mathit{2}}$

$\mathrm{t}=\frac{\mathrm{A}}{{\mathrm{A}}_0}\sqrt{\frac{2}{\mathrm{g}}}\left[\sqrt{\mathrm{h}}-\sqrt{\frac{\mathrm{h}}{2}}\right]$

and similarly time taken for the level to fall from $\frac{\mathrm{h}}{2}$ to zero

${\mathrm{t}}_2=\frac{\mathrm{A}}{{\mathrm{A}}_0}\sqrt{\frac{2}{\mathrm{g}}}\left[\sqrt{\frac{\mathrm{h}}{2}}-0\right]$

$\therefore \frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}=\frac{1-{\displaystyle \frac{1}{\sqrt{2}}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1$

(b) $\mathrm{v}=\sqrt{2\mathrm{gh}}=\sqrt{2\times 10\times 20}=20 \mathrm{m}/\mathrm{s}$

(a) Pressure at the bottom of tank $\mathrm{P}=\mathrm{h\rho g}=3\times {10}^5\frac{\mathrm{N}}{{\mathrm{m}}^2}$ 

Pressure due to lipid column ${\mathrm{P}}_{\mathrm{I}}=3\times {10}^5-1\times {10}^5=2\times {10}^5$

and velocity of water $\mathrm{v}=\sqrt{2\mathrm{gh}}$

$\therefore \mathrm{v}=\sqrt{\frac{2{\mathrm{P}}_{\mathrm{I}}}{\mathrm{\rho }}}=\sqrt{\frac{2\times 2\times {10}^5}{{10}^3}}=\sqrt{400} \mathrm{m}/\mathrm{s}$

(a) The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second.
Volume of water flowing out per second

$=\mathrm{Av}=\mathrm{A}\sqrt{2\mathrm{gh}}$                             ....(i) 

Volume of water flowing in per second

$=70 {\mathrm{cm}}^3/\sec$                             .....(ii)

From (i) and (ii) we get

$$\begin{gathered} \mathrm{A}\sqrt{2\mathrm{gh}}=70\Rightarrow 1\times \sqrt{2\mathrm{gh}}=70 \\ \Rightarrow 1\times \sqrt{{}_{}2\times 980\times \mathrm{h}}=70 \\ \therefore \mathrm{h}=\frac{4900}{1960}=2.5 \mathrm{cm} \end{gathered}$$