NEET Practice Questions (MCQs) with Answers & Solutions

The liquid flow is incompressible and steady. By the principle of conservation of mass, the volume flow rate must be constant throughout the tube. Therefore, the speed v in the tube is related to the speed v' through the holes by the equation πR^2v = nπr^2v', which gives v' = vR^2/nr^2.

Given pressure=150mm of Hg

Pumping rate of heart of a man=$\frac{dV}{dt}$=$\frac{{\displaystyle 5x{10}^3}}{{\displaystyle 60}}$ m³/s

Power of heart=P.$\frac{dV}{dt}$=ρgh.$\frac{dV}{dt}$ [P=ρgh]

=>(13.6x10³kg/m³) (10x0.15x5x10^-3)/60

=1.70W

In case of insufficient length, angle of contact adjusts itself so that water does not spill

$$\begin{gathered} If the surface area changes, it will change the surface energy as well. As the surface area is decrea\sin g, energy will be released. \\ Change in surface energy=T\times ∆A \\ If there is n no. if small drops; \\ Volume of a large drop=volume of n small drops \\ \frac{4}{3}{\mathrm{\pi R}}^3=\mathrm{n}\times \frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{Change} \mathrm{in} \mathrm{area}; \\ ∆\mathrm{A}=4{\mathrm{\pi R}}^2-\mathrm{n}\times 4{\mathrm{\pi r}}^2=3\left(\frac{4{\mathrm{\pi R}}^3}{3\mathrm{R}}-\mathrm{n}\times \frac{4{\mathrm{\pi r}}^3}{3\mathrm{r}}\right)=3\left(\frac{4{\mathrm{\pi R}}^3}{3\mathrm{R}}-\frac{4{\mathrm{\pi R}}^3}{3\mathrm{r}}\right)=3\left(\frac{\mathrm{V}}{\mathrm{R}}-\frac{\mathrm{V}}{\mathrm{r}}\right)=3\mathrm{V}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right) \\ \mathrm{E}=\mathrm{T}\times ∆\mathrm{A}=3\mathrm{VT}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right), \mathrm{released}. \end{gathered}$$

(d) The value of angle of contact determines whether a liquid will spread on the surface

Let m is the mass per unit length then rate of mass per sec=$\frac{mx}{t}=mv$

Rate of KE=$\frac{1}{2}\left(mv\right)v^2=\frac{1}{2}m{v}^3$

(c) Apparent weight = $\mathrm{V}(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}=\frac{\mathrm{m}}{\mathrm{\rho }}(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}$
where m = mass of the body,
$\mathrm{\rho }$ = density of the body
$\mathrm{\sigma }$ = density of water
If two bodies are in equilibrium then their apparent weight must be equal.

$$\begin{gathered} \therefore \frac{{\mathrm{m}}_1}{{\mathrm{\rho }}_1}({\mathrm{\rho }}_1-\mathrm{\sigma })=\frac{{\mathrm{m}}_2}{{\mathrm{\rho }}_2}({\mathrm{\rho }}_2-\mathrm{\sigma }) \\ \Rightarrow \frac{36}{9}(9-1)=\frac{48}{{\mathrm{\rho }}_2}({\mathrm{\rho }}_2-1) \end{gathered}$$

By solving we get ${\mathrm{\rho }}_2=3.$

(b) Difference of pressure between sea level and the top of hill
$∆\mathrm{P}=({\mathrm{h}}_1-{\mathrm{h}}_2)\times {\mathrm{\rho }}_{\mathrm{Hg}}\times \mathrm{g}=(75-50)\times {10}^{-2}\times {\mathrm{\rho }}_{\mathrm{Hg}}\times \mathrm{g}$         …(i)
and pressure difference due to h meter of air
$∆\mathrm{P}=\mathrm{h}\times {\mathrm{\rho }}_{\mathrm{air}}\times \mathrm{g}$                                                          …(ii)
By equating (i) and (ii) we get

$\mathrm{h}\times {\mathrm{\rho }}_{\mathrm{air}}\times \mathrm{g}=(75-50)\times {10}^{-2}\times {\mathrm{\rho }}_{\mathrm{Hg}}\times \mathrm{g}$

$\therefore \mathrm{h}=25\times {10}^{-2}\left(\frac{{\mathrm{\rho }}_{\mathrm{Hg}}}{{\mathrm{\rho }}_{\mathrm{air}}}\right)=25\times {10}^{-2}\times {10}^4=2500 \mathrm{m}$

Height of the hill = 2.5 km.

(b) If two liquid of equal masses and different densities are mixed together then density of mixture

$\mathrm{\rho }=\frac{2{\mathrm{\rho }}_1{\mathrm{\rho }}_2}{{\mathrm{\rho }}_1+{\mathrm{\rho }}_2}=\frac{2\times 1\times 2}{1+2}=\frac{4}{3}$