Physics M.C.Q (1 Marks)

Acceleration (slope of $v / t$ curve) of ball first decreases and after some time it becomes zero.

$\quad =6 \times 3.14 \times 0.9 \times 5 \times 10^{-3} \times 10 \times 10^{-2}$

$\quad=847.8 \times 10^{-5}\,N$

$\quad=8.48 \times 10^{-3}\,N$

$F_{B}=M^{\prime} g=v \frac{d}{2} y=\frac{M g}{2}$

$F_{v}+F_{B}=F_{g}$

$F_{v}=m g-\frac{m g}{2}=\frac{m g}{2}$

Reading of mercury barometer, $h=76 cm$

Reading of liquid barometer $= h ^{\prime}$

$h ^{\prime} d _{i} g = h d _{ hg } g$

$h ^{\prime}=\frac{76 \times 13600}{760}=1360 cm$

$=13.6 m$

$\Rightarrow \rho_{oil} =\frac{1000(15)}{20}=750 \mathrm{kg} \mathrm{m}^{-3}$

volume flow rate $=\mathrm{Av}=\mathrm{A} \sqrt{2 \mathrm{gh}}$

${=\left(2 \times 10^{-6}\right)(2 \times 10 \times 2)^{1 / 2}}$

${=4 \sqrt{10} \times 10^{-6} \mathrm{m}^{3} / \mathrm{s}}$

${\cong 12.6 \times 10^{-6} \mathrm{m}^{3} / \mathrm{s}}$

$\frac{v_{1}}{v_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2} \frac{\left(\sigma_{1}-\rho\right)}{\left(\sigma_{2}-\rho\right)}$$=\left(\frac{1}{2}\right)^{2}\left(\frac{8 \rho_{2}-0.1 \rho_{2}}{\rho_{2}-0.1 \rho_{2}}\right)$$=\frac{79}{36}$

$where\,v = terminal\,velocity$

$\therefore \,The\,rate\,of\,production\,of\,heat = power$

$ = force \times terminal\,velocity$

$ \Rightarrow power = 6\pi \eta rv \cdot v = 6\pi \eta r{v^2}\,\,\,\,\,\,\,\,...\left( i \right)$

$Terminal\,velocity\,v = \frac{{2{r^2}\left( {\rho  - \sigma } \right)}}{{9\eta }};\,\,\,\therefore v \propto {r^2}$

$Now,\,power = 6\pi \eta r\left[ {\frac{{4{r^2}{{\left( {\rho  - \sigma } \right)}^2}}}{{81{\eta ^2}}}{g^2}} \right]or\,power \propto {r^5}.$

Shear stress on the fluid $=\frac{F}{A}=\frac{0.1\; N }{0.1 \;m ^{2}}$

Strain rate $=\frac{v}{l}=\frac{0.085 \;m {s}^{-1}}{0.3 \times 10^{-3}\; m }$

Coefficient of viscosity, $\eta=\frac{\text { Shear stress }}{\text { Strain rate }}$

$=\left(\frac{0.1\; N }{0.1 \;m ^{2}}\right) \times \frac{\left(0.3 \times 10^{-3}\; m \right)}{0.085\; ms ^{-1}}$

$=3.5 \times 10^{-3} Pa\cdot s$

${p_C} = {p_a} + {\rho _{water}}g{h_{water'}}$

$where\,{h_{water}} = CE = \left( {65 + 65} \right)mm = 130\,mm$

$pressure\,at\,point\,B,{P_B} = {P_u} + {\rho _{oil}}g{h_{oil}}$

$where\,{h_{oil}} = AB = \left( {65 + 65 + 10} \right)\,mm = 140\,mm$

In liquid, Pressure is same at same liquid level,

${p_B} = {p_c} \Rightarrow {\rho _{oil}}g{h_{oil}} = {\rho _{water}}g\,{h_{water}}$

${\rho _{oil}} = \frac{{130 \times {{10}^3}}}{{140}} = \frac{{13}}{{14}} \times {10^3} = 928.57\,kg\,{m^{ - 3}}$

$A = area\,of\,cross - section\,of\,cylinder$

Using law of floatation,

$Weight\,of\,cylinder = Upthrust\,by\,two\,liquids$

$L \times A \times d \times g = n\rho  \times \left( {pL \times A} \right)g + \rho \left( {L - pL} \right)Ag$

$d = np\rho  + \rho \left( {1 - p} \right) = \left( {np + 1 - p} \right)\rho $

$d = \left\{ {1 + \left( {n - 1} \right)p} \right\}\rho $

$ = \frac{{2 \times {{10}^{ - 3}} \times {{10}^4} \times {{10}^2}}}{{{{10}^{ - 4}}}}$

$ = 2 \times {10^7}N/{m^2}$

 

$ = [250 \times 1 + 250 \times 0.85] = 250\;[1.85]\frac{g}{{c{m^2}}}$

$ = 462.5\frac{g}{{c{m^2}}}$

When substances are mixed in equal masses then density $ = \frac{{2{\rho _1}{\rho _2}}}{{{\rho _1} + {\rho _2}}} = 3$

==> $2{\rho _1}{\rho _2} = 3({\rho _1} + {\rho _2})$ .......$(ii)$

By solving $(i)$ and $(ii)$ we get ${\rho _1} = 6$ and ${\rho _2} = 2$.

Radius, $r=\frac{d}{2}=0.25 \times 10^{-3} m$

Compressional force, $F=50000 N$

Pressure at the tip of the anvil

$p=\frac{\text { Force }}{\text { Arear }}=\frac{F}{\pi r^{2}}$

$=\frac{50000}{\pi\left(0.25 \times 10^{-3}\right)^{2}}$

$=2.55 \times 10^{11} Pa$

Therefore, the pressure at the tip of the anvil is $2.55 \times 10^{11} \;Pa$

Diameter of the heel, $d=1 cm =0.01 m$ Radius of the heel, $r=d / 2=0.005 m$

Area of the heel $=\pi r^{2}=\pi(0.005)^{2}=7.85 \times 10^{-5} m ^{2}$

Force exerted by the heel on the floor:

$F=m g=50 \times 9.8=490 N$

Pressure exerted by the heel on the floor:

$P=\frac{\text { Force }}{\text { Area }}=\frac{490}{7.85 \times 10^{-5}}$

$=6.24 \times 10^{6} \;N m ^{-2}$

Therefore, the pressure exerted by the heel on the horizontal floor is $6.24 \times 10^{6} \,Nm ^{-2}$

Area of the hinged door, $a=20 cm ^{2}=20 \times 10^{-4} m ^{2}$

Density of water, $\rho_{1}=10^{3} kg / m ^{3}$

Density of acid, $\rho_{2}=1.7 \times 10^{3} kg / m ^{3}$

Height of the water column, $h_{1}=4 m$

Height of the acid column, $h_{2}=4 m$

Acceleration due to gravity, $g=9.8$

Pressure due to water is given as:

$P_{1}=h_{1} \rho_{1} g$

$=4 \times 10^{3} \times 9.8$

$=3.92 \times 10^{4} Pa$

Pressure due to acid is given as:

$P_{2}=h_{2} \rho_{2} g$

$=4 \times 1.7 \times 10^{3} \times 9.8$

$=6.664 \times 10^{4} Pa$

Pressure difference between the water and acid columns:

$\Delta P=P_{2}-P_{1}$

$=6.664 \times 10^{4}-3.92 \times 10^{4}$

$=2.744 \times 10^{4} Pa$

Hence, the force exerted on the door $=\Delta P \times a$ $=2.744 \times 10^{4} \times 20 \times 10^{-4}$

$=54.88 N$

Therefore, the force necessary to keep the door closed is $54.88 \;N .$

Let $m$ be mass ,

$m = Dv ; D =$ density of mass, $v =$ vol.

$d =$ density of water.

$m \times g = Dv \times g =60$

fully submerged in water

$( D - d ) v \times g =40$

dividing both equations

$\frac{ D - d }{ D }=\frac{2}{3}$

solving this :

$\frac{ D }{ d }=3$

$\therefore {A_2} = 2 \times {10^4}c{m^2}$$(g = 980 \approx {10^3}cm/{s^2})$

As there is no atmospheric pressure and liquid is stationary with respect to flask

so force exerted by the liquid on flask $=$ force exerted by flask on liquid $=10 N$

$\frac{\mathrm{F}_{1}}{\mathrm{A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{A}_{2}} \Rightarrow \mathrm{F}_{1} \mathrm{A}_{2}=\mathrm{F}_{2} \mathrm{A}_{1}$

as $\mathrm{A}_{1}<\mathrm{A}_{2}$

So $\mathrm{F}_{1}<\mathrm{F}_{2}$

$=\frac{\text { Area of bigger piston }}{\text { Area of smaller piston }}$

$=\frac{\pi\left(\frac{\mathrm{d}_{1}}{2}\right)^{2}}{\pi\left(\frac{\mathrm{d}_{2}}{2}\right)^{2}}=\frac{\mathrm{d}_{1}^{2}}{\mathrm{d}_{2}^{2}}=\frac{0.36}{0.01}=36$

 which means they have displaced the equal weights of liquids.

Weight of the oil displaced by larger block $=2 V \times 0.9 \times g$  $...(I)$

Weight of the unknown liquid displaced by smaller block $=V \times \rho \times g$ $...(II)$

Equating $(I)$ and $(I I),$ we have

$\rho=1.8 \mathrm{gm} / \mathrm{cm}^{3}$

$=\frac{2 d m V \cos 60^{\circ}}{d t}$

$=\frac{2(\rho A d x) V \cos 60^{\circ}}{d t}$

$=2 \rho A V^{2} \cos 60^{\circ}$

$=10^{3} \times 6 \times 10^{-4} \times 12^{2}$

$=86.4 \;N$

$\rho_{w} g \times \frac{10}{100}=\rho_{B} g \times \frac{12}{100} \Rightarrow \rho_{B}=\frac{10}{12} \rho_{w}$

$8 \times \rho_{y} \times g+2 \times \rho_{H g} \times g=10 \times \rho_{x} \times g$

$\therefore 8 \rho_{y}+2 \times 13.6=10 \times 3.36$

or $\rho_{y}=\frac{33.6-27.2}{8}=0.8 \mathrm{g} / \mathrm{cc}$

pressure due Weight of mercury in column is equal to atmospheric pressure, if g will vary, height should adjust so as to keep $W$ same.

 Leakage of air inside the tube will disturb the vacuum, due to which pressure inside the tube will become equal to atmospheric pressure and height of mercury will become $0.$

Evaporation of mercury does not cause the height to vary, because it doesn't not change the pressure at surface of reservoir.

$\mathrm{h}=\frac{\mathrm{P}_{1}-\mathrm{P}_{0}}{\rho \mathrm{g}}$

$=\frac{12 \times 10^{5}}{4 \times 10^{3} \times 10}=30 \mathrm{m}$