Properties of Matter

Since the process is adiabatic,

$P{V^\gamma }$ = constant for gas inside bubble.

Thus, $P{V^{(1 - \gamma )}}.\,{T^\gamma } = $ constant

$ \Rightarrow {P^{1 - \gamma }}_{bottom}{T^\gamma }_{bottom} = P_y^{1 - \gamma }T_y^\gamma = $ constant

$ \Rightarrow {({P_0} + {\rho _1}gH)^{ - 2/3}}(T_0^{5/3}) = ({P_0} + {\rho _1}g{(H - y)^{ - 2/3}}){T^{5/3}}$

$ \Rightarrow {T_0} = {\left[ {{{{P_0} + {\rho _1}g(H - y)} \over {{P_0} + {\rho _1}gH}}} \right]^{2/5}}$

Buoyancy force = Weight of fluid displaced

= (mass of fluid displaced)g

$ = V{\rho _l}g$ .... (i)

Where V = Volume of fluid displaced

= Volume of the bubble

$PV = nRT$

$ \Rightarrow V = {{nRT} \over P} = {{nRT} \over {{P_0}(H - y){\rho _l}g}}$ .... (ii)

Where P is the pressure of the bubble at an arbitrary location at a distance 'y' from the bottom.

Put the value of temperature from eq. (i)

$V = {{nR} \over {[{P_0} + (H - y){\rho _l}g]}} \times {{{T_0}{{[{P_0} + (H - y){\rho _l}g]}^{2/5}}} \over {{{[{P_0} + H{\rho _l}g]}^{2/5}}}}$

$ = {{nR{T_0}} \over {{{[{P_0} + (H - y){\rho _l}g]}^{3/5}}{{[{P_0} + H{\rho _l}g]}^{2/5}}}}$ .... (iii)

From eq. (i) and (iii) Buoyance force

$ = {{nR{T_0}{\rho _1}g} \over {{{[{P_0} + (H - y){\rho _l}g]}^{3/5}}{{[{P_0} + H{\rho _l}g]}^{2/5}}}}$

Pressure is acting on area $A B C D$, we have

Pressure Force $=\int_{0}^{h}\left(\mathrm{P}_{0}+\rho g x\right) \times d x \times 2 \mathrm{R}$

Force due to pressure (push)

$\begin{aligned}& =\left[\left(\mathrm{P}_{0}+\rho g \frac{x^{2}}{2}\right) 2 \mathrm{R}\right]_{0}^{h} \\\\ & =\left(\mathrm{P}_{0} h+\rho g h^{2} / 2\right) 2 \mathrm{R} \\\\& =2 \mathrm{P}_{0} h \mathrm{R}+\rho g h^{2} \mathrm{R}\end{aligned}$

Surface tension force (push) $=T \times 2 R$

$\therefore \quad$ Net Force $=2 \mathrm{P}_{0} h \mathrm{R}+\rho g h^{2} \mathrm{R}-2 \mathrm{TR}$

(A) A bimetallic strip is used to convert a temperature change into mechanical displacement. The strip consists of two strips of different metals having different thermal coefficients.

(B) Steam engine is a device which converts heat energy into mechanical energy and heat is supplied into the engine through the medium of steam. Thus energy conversion is right answer here.

(C) The incandescent light bulb works on incandescence, which is the emission of light radiations caused by heating the filament.

(D) Fuse wire works on the principle of heat and melting, when excess current passes in the circuit, due to heat produced by the excess current fuse melts and prevent the damage.

Case $(i)$
At equilibrium, $T=W$
$Y = {{W/A} \over {\ell /L}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ...\left( 1 \right)$
Case $(ii)$ At equilibrium $T=W$
$\therefore$ $Y = {{W/A} \over {{{\ell /2} \over {L/2}}}} \Rightarrow Y = {{W/A} \over {\ell /L}}$
$ \Rightarrow $ Elongation is the same.

Let Terminal velocity = v_t

Upward viscous force = downward weight of sphere

$ \Rightarrow 6\pi \eta r{v_t} = \left( {{4 \over 3}\pi {r^3}} \right)\left( {\rho - \sigma } \right)g$

$ \Rightarrow {v_t} = {{2{r^2}\left( {\rho - \sigma } \right)g} \over {9\eta }}$ ........ (1)

where, $\rho $ = density of substance of a body

            $\sigma $ = density of liquid

Now let the terminal velocity of gold = v_g and silver = v_s.

From equation (1), we can write

${{{v_g}} \over {{v_s}}} = {{{\rho _g} - \sigma } \over {{\rho _s} - \sigma }}$ $ = {{19.5 - 1.5} \over {10.5 - 1.5}}$ = ${{18} \over 9}$ $ = {2 \over 1}$

$\therefore$ ${v_s} = {{{v_g}} \over 2}$ = ${{0.2} \over 2}$ = 0.1

$ \begin{aligned} & \pi r^2=\text { Area of base of cylinder in air } \\ & 3 \pi r^2=\text { Area of base of cylinder in water. } \\ & 4 \pi r^2=\text { Cross-sectional area of the cylinder. } \end{aligned} $

$ \begin{aligned} &\text { Equating the forces, we get }\\ &\begin{aligned} & {\left[\mathrm{P}_0+\rho g h_1\right] \pi\left(4 r^2\right)+\left(\frac{\rho}{3}\right) \pi r^2 .4 h g} \\ & =\left[\mathrm{P}_0+\rho g\left(h_1+h\right)\right] \pi\left(3 r^2\right)+\mathrm{P}_0 \pi r^2 \\ & \rho g h_1 \cdot 4 \pi r^2+\frac{\rho g h 4 \pi r^2}{3}=\rho g h_1 3 \pi r^2+\rho h g 3 \pi r^2 \\ & \rho g h_1 \pi r^2=\rho g h \pi r^2\left(3-\frac{4}{3}\right) \\ & h_1=h\left(\frac{5}{3}\right) \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Equation the forces, we get }\\ &\begin{gathered} \mathrm{P}_0 \pi\left(4 r^2\right)+(\rho / 3) 4 \pi r^2 h g=\left[\mathrm{P}_0+\rho g h_2\right] \pi\left(4 r^2\right) \\ (\rho / 3) 4 \pi r^2 h g=\rho g h_2 3 \pi r^2 \end{gathered} \end{aligned} $

$ h_2=\frac{4 h}{9} $

For $h_2<\frac{4 h}{9}$, Cylinder does not move up because further buoyant force decreases while the weight of block remains same.

In freely falling elevator $g$ = 0

Water fills the tube entirely in gravity less condition. Hence, length of water column in the capillary tube is 20 cm.

Energy stored per unit volume of wire,

$E = {1 \over 2} \times \,stress\, \times \,strain$

$\therefore$ $E = {1 \over 2} \times \,stress\, \times \,{{stress} \over Y} = {1 \over 2}{{{S^2}} \over Y}$

[ As Young's modulus(Y) = ${{Stress} \over {Strain}}$

$\therefore$ Strain = ${{Stress} \over Y}$ ]

Weight of liquid column of height $\mathrm{H}, \mathrm{W}=\rho \mathrm{V} g$

$=\frac{\rho\left(\pi d^{2}\right)}{4} \mathrm{Hg}$ ....... (i)

Let us consider a small element $d x$ of mass $d m$ at a distance $x$ (as shown in figure).

Centripetal force required, in rotating this mass $=d m \omega^{2}$

$\therefore$ Total force required to rotate mass of length $\mathrm{L}$,

$\begin{aligned} \mathrm{F} & =\int d m \omega^{2} \\ & =\int_{0}^{\mathrm{L}} \frac{\rho\left(\pi d^{2}\right)}{4} x d x \omega^{2} \\ & =\frac{\rho\left(\pi d^{2}\right)}{4} \frac{\mathrm{L}^{2}}{2} \omega^{2} ~~........ (\mathrm{ii})\end{aligned}$

This centripetal force is provided by weight of liquid column (of height H), Here on comparing equation (i) and (ii), we get

$\begin{aligned} W & =\mathrm{F} \\ \frac{\rho\left(\pi d^{2}\right)}{4} \mathrm{Hg} & =\frac{\rho\left(\pi d^{2}\right)}{4} \frac{\mathrm{L}^{2}}{2} \omega^{2} \\ \mathrm{H} & =\frac{\mathrm{L}^{2} \omega^{2}}{2 g} \end{aligned}$

Work done by constant force in displacing the object by a distance $\ell $.

= Potential energy stored

$ = {1 \over 2} \times $ Stress $ \times $ Strain $ \times $ Volume

$ = {1 \over 2} \times {F \over A} \times {l \over L} \times AL$

$ = {1 \over 2}Fl$

Pressure inside the bubble, P $ = {p_0} + {{4T} \over R}$

So $P \propto {1 \over R}$ where R is the radius of the bubble. It means pressure inside a smaller bubble is greater than the inside of a bigger bubble.

So when two bubbles are connected by a tube, air will flow from smaller bubble to bigger bubble and the size of bigger bubble will increase.

From Stoke's law,

viscous force acting on the ball falling into a viscous fluid

$F = 6\pi \eta Rv$

$\therefore$ $F \propto R$ and $F \propto v$

hence $F$ is directly proportional to radius & velocity.

Newton's law of cooling states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings. Mathematically, this law is often written as :

$ \frac{d\theta}{dt} \propto \Delta \theta $

In this formula, $ \frac{d\theta}{dt} $ is the rate of cooling (i.e., the rate of change of temperature with respect to time), and $ \Delta \theta $ is the difference in temperature between the body and the environment.

The law does not specify the temperature difference to an exponential power; it simply states a linear proportionality. Therefore, the correct answer is :

Option D : one.

The velocity of efflux of water is given $v = \sqrt {2gh} $

Here $h$ is the height of the free surface of water from the hole

$\therefore$ $v = \sqrt {2 \times 10 \times 20} = 20m/s$