Physics MCQs for NEET — Practice Questions with Answers

(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

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

(d) 

$$\begin{gathered} \mathrm{A}=(0.1{)}^2=0.01 {\mathrm{m}}^2 \\ \mathrm{\eta }=0.01 \mathrm{Poise}=0.001 \mathrm{decapoise} (\mathrm{M}.\mathrm{K}.\mathrm{S}. \mathrm{unit}) \\ \mathrm{dv}=0.1 \mathrm{m}/\mathrm{s} \mathrm{and} \mathrm{F}=0.002 \mathrm{N} \\ \mathrm{F}=\mathrm{\eta A}\frac{\mathrm{dv}}{\mathrm{dx}} \\ \therefore \mathrm{dx}=\frac{\mathrm{\eta Adv}}{\mathrm{F}}=\frac{0.001\times 0.01\times 0.1}{0.002}=0.0005 \mathrm{m} \end{gathered}$$                                    

(b) $\mathrm{F}=6\mathrm{\pi \eta rv}$

(b) In the first 100 m body starts from rest and its velocity goes on increasing and after 100 m it acquire maximum velocity (terminal velocity). Further, air friction i.e. viscous force which is proportional to velocity is low in the beginning and maximum at $\mathrm{v}={\mathrm{v}}_{\mathrm{T}}$.
Hence work done against air friction in the first 100 m is less than the work done in next 100 m.

(c) If two drops of same radius r coalesce then radius of new drop is given by R

$\frac{4}{3}{\mathrm{\pi R}}^3=\frac{4}{3}{\mathrm{\pi r}}^3+\frac{4}{3}{\mathrm{\pi r}}^3\Rightarrow {\mathrm{R}}^3=2{\mathrm{r}}^3\Rightarrow \mathrm{R}=2^{1/3}\mathrm{r}$

If drop of radius r is falling in viscous medium then it acquire a critical velocity v and $\mathrm{v}\propto {\mathrm{r}}^2$

$$\begin{gathered} \frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}={\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^2={\left(\frac{{\displaystyle 2^{1/3}}}{\mathrm{r}}\right)}^2 \\ \Rightarrow {\mathrm{v}}_2=2^{2/3}\times {\mathrm{v}}_1=2^{2/3}\times 5=5\times 4^{1/3} \mathrm{cm}/\mathrm{s} \end{gathered}$$

(b) Rate of flow of liquid $\mathrm{V}=\frac{\mathrm{P}}{\mathrm{R}}$

where liquid resistance $\mathrm{R}=\frac{8\mathrm{\eta l}}{{\mathrm{\pi r}}^4}$

For another liquid resistance

$\mathrm{R}\text{'}=\frac{8\mathrm{\eta l}}{\mathrm{\pi }{\left({\displaystyle \frac{\mathrm{r}}{2}}\right)}^4}=\frac{8\mathrm{\eta l}}{{\mathrm{\pi r}}^4}.16=16\mathrm{R}$

For the series combination

${\mathrm{V}}_{\mathrm{New}}=\frac{\mathrm{P}}{\mathrm{R}+\mathrm{R}\text{'}}=\frac{\mathrm{P}}{\mathrm{R}+16\mathrm{R}}=\frac{\mathrm{P}}{17\mathrm{R}}=\frac{\mathrm{V}}{17}$