Q281JEE Main 2008MCQ

A jar is filled with two non-mixing liquids 1 and 2 having densities $\rho _1$ and $\rho _2$ respectively. A solid ball, made of a material of density $\rho _3$ , is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for $\rho _1, \rho _1$ and $\rho _3$ ?

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📖 Explanation

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 $\rho _1< \rho _2< \rho _3$ .

Q282JEE Main 2008MCQ

Q283JEE Main 2008MCQ

A spherical solid ball of volume $V$ is made of a material of density $\rho _1$ . It is falling through a liquid of density $\rho _2 (\rho _2< \rho _1)$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$ , i.e., $F_{\;\text{viscous }\;}=-k v^2(k>0)$ . The terminal speed of the ball is

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📖 Explanation

The forces acting on the ball -(1) ${\text{mg}}=V \rho _1 g$ downward direction(2) Thrust upward direction ( By Archimedes principle )(3) Force of friction ( Buoynat force) upward directionThe ball reaches to its terminal speed $(v_t)$ when acceleration $=0$ .So, weight = Buoyant force + Viscous force $\therefore V \rho _1 g=V \rho _2 g+k v_t^2$ $\therefore v_t=\sqrt{\;\frac{V g (\rho _1-\rho _2) }{k}}$

Q284JEE Main 2006MCQ

If the terminal speed of a sphere of gold (density $=19.5 {\text{kg}} / {\text{m}}^3$ ) is $0.2 {\text{m}} / {\ s}$ in a viscous liquid (density $=1.5 {\text{kg}} / {\text{m}}^3$ , find the terminal speed of a sphere of silver (density $=10.5 {\text{kg}} / {\text{m}}^3$ ) of the same size in the same liquid

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📖 Explanation

Let Terminal velocity $=v_t$ Upward viscous force $=$ downward weight of sphere $\;\Rightarrow 6 \pi η r v_t= (\;\frac{4}{3} \pi r^3) (\rho -\sigma ) g$ $\;\Rightarrow v_t=\;\frac{2 r^2(\rho -\sigma ) g}{9 η} . . . . . . . .(1)$ where, $\rho =$ density of substance of a body $\sigma =$ density of liquidNow let the terminal velocity of gold $=v_g$ and silver $=v_S$ .From equation (1), we can write $\;\;\frac{v_g}{v_s}=\;\frac{\rho _g-\sigma }{\rho _s-\sigma }=\;\frac{19.5-1.5}{10.5-1.5}=\;\frac{18}{9}=\;\frac{2}{1}$ $\; \therefore v_s=\;\frac{v_g}{2}=\;\frac{0.2}{2}=0.1$

Q285JEE Main 2006MCQ

A wire elongates by $l m m$ when a LOAD $W$ is hanged from it. If the wire goes over a pulley and two weights $W$ each are hung at the two ends, the elongation of the wire will be (in ${\text{mm}}$ )

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📖 Explanation

Case $(i)$ At equilibrium, $T=W$ $Y=\;\frac{W / A}{ℓ / L} \;\; . . .(1)$ Case $(i i)$ At equilibrium $T=W$ $\therefore Y=\;\frac{W / A}{\frac{ℓ / 2}{L / 2}} \Rightarrow Y=\;\frac{W / A}{ℓ / L}$ $\Rightarrow$ Elongation is the same.

Q286JEE Main 2005MCQ

If $S$ is stress and $Y$ is young's modulus of material of a wire, the energy stored in the wire per unit volume is

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📖 Explanation

Energy stored per unit volume of wire, $\;E=\;\frac{1}{2} \times \;\text{ stress }\; \times \;\text{ strain }\;$ $\; \therefore E=\;\frac{1}{2} \times \;\text{ stress }\; \times \;\frac{\;\text{ stress }\;}{Y}=\;\frac{1}{2} \;\frac{S^2}{Y}$ [As Young's modulus $(Y)=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}$ $\; \therefore \;\text{ Strain }\;=\;\frac{\;\text{ Stress }\;}{Y}$ ]

Q287JEE Main 2005MCQ

A $20 {\text{cm}}$ long capillary tube is dipped in water. The water rises up to $8 {\text{cm}}$ . If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be

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📖 Explanation

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 $20 {\text{cm}}$ .

Q288JEE Main 2004MCQ

A wire fixed at the upper end stretches by length $l$ by applying a force $F$ . The work done in stretching is

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📖 Explanation

Work done by constant force in displacing the object by a distance $ℓ$ . = Potential energy stored $=\;\frac{1}{2} \times$ Stress $\times$ Strain $\times$ Volume $=\;\frac{1}{2} \times \;\frac{F}{A} \times \;\frac{l}{L} \times A L$ $=\;\frac{1}{2} F l$

Q289JEE Main 2004MCQ

Spherical balls of radius $R$ are falling in a viscous fluid of viscosity $η$ with a velocity $v$ . The retarding viscous force acting on the spherical ball is

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📖 Explanation

From Stoke's law, viscous force acting on the ball falling into a viscous fluid $F=6 \pi η R v$ $\therefore F \propto R$ and $F \propto v$ hence $F$ is directly proportional to radius & velocity.

Q290JEE Main 2004MCQ

If two soap bubbles of different radii are connected by a tube

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📖 Explanation

Pressure inside the bubble, ${\text{P}}=p_0+\;\frac{4 T}{R}$ So $P \propto \;\frac{1}{R}$ where ${\text{R}}$ 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.

Q291JEE Main 2003MCQ

According to Newton's law of cooling, the rate of cooling of a body is proportional to $(\Delta \theta )^n$ , where $\Delta \theta$ is the difference of the temperature of the body and the surrounding, and $n$ is equal to

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📖 Explanation

$-\;\frac{d Q}{d t} \propto (\Delta \theta )$

Q292JEE Main 2002MCQ

A cylinder of height $20 m$ is completely filled with water. The velocity of efflux of water (in ${\text{ms}}^{-1}$ ) through a small hole on the side wall of the cylinder near its bottom is

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📖 Explanation

The velocity of efflux of water is given $v=\sqrt{2 g h}$ Here $h$ is the height of the free surface of water from the hole $\therefore v=\sqrt{2 \times 10 \times 20}=20 {\text{m}} / {\text{s}}$