Properties of Matter

Oil will float on water so, $(2)$ or $(4)$ is the correct option, But density of ball is more than that of oil, hence it will sinkin oil.

As shown in the figure, the wires will have the same Young's modulus (same material) and the length of the wire of area of cross-section $3A$ will be $\ell /3$ (same volume as wire $1$).
For wire $1,$
$y = {{F/A} \over {\Delta x/\ell }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$
For wire $2.$
$Y = {{F'/3A} \over {\Delta x/\left( {\ell /3} \right)}}........(ii)$
From $(i)$ and $(ii),$ ${F \over A} \times {\ell \over {\Delta x}} = {{F'} \over {3A}} \times {\ell \over {3\Delta x}} \Rightarrow F' = 9F$

The forces acting on the ball -

(1) mg = $V{\rho _1}g$ downward direction

(2) Thrust upward direction ( By Archimedes principle )

(3) Force of friction ( Buoynat force) upward direction

The ball reaches to its terminal speed $\left( {{v_t}} \right)$ when acceleration = 0.

So, weight $=$ Buoyant force $+$ Viscous force

$\therefore$ $V\rho {}_1g = V{\rho _2}g + kv_t^2$

$\therefore$ ${v_t} = \sqrt {{{Vg\left( {{\rho _1} - {\rho _2}} \right)} \over k}} $

From the figure it is clear that liquid $1$ floats on liquid $2.$ The lighter liquid floats over heavier liquid. Therefore we can conclude that ${\rho _1} < {\rho _2}$

Also ${\rho _3} < {\rho _2}$ otherwise the ball would have sink to the bottom of the jar.

Also ${\rho _3} > {\rho _1}$ otherwise the ball would have floated in liquid $1.$ From the above discussion we conclude that ${\rho _1} < {\rho _2} < {\rho _3}.$

In case of water, the meniscus shape is concave upwards. Also according to ascent formula $h = {{2T\,\cos \,\theta } \over {r\rho g}}$

The surface tension $(I)$ of soap solution is less than water. Therefore rise of soap solution in the capillary tube is less as compared to water. As in the case of water. the meniscus shape of soap solution is also concave upwards.

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

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

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

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