Properties of Matter

Force of viscosity can be written as follows:

$f = \eta A{{dv} \over {dy}}$

Here $\eta $ is coefficient of viscosity, A is face area of plate and dv / dy is velocity gradient. We can assume that velocity of fluid at the bottom (h depth below) is zero; hence ${{dv} \over {dy}} = {{{u_0}} \over h}$; hence force of viscocity can be written as follows:

$f = \eta A{{{u_0}} \over h}$

We can see that force of viscosity is inversely proportional to h; hence option (a) is correct.

We can see that force of viscosity is proportional to area of plate; hence option (b) is wrong.

Tangential stress can be written as ${f \over A} = \eta {{{u_0}} \over h}$.

We can see that options (c) and (d) are also correct.

We know that,

$h = {{2\sigma \cos \theta } \over {r\,\rho g}}$..... (1)

where r is inner radius of capillary tube; $\sigma$ is surface tension; $\theta$ is angle between water and wall of capillary tube; h is height of water in capillary tube and g is acceleration due to gravity. From Eq. (1),

$h \propto {1 \over r}$

Therefore, for a given material of the capillary tube, h decreases with increase in r.

Hence, option (A) is correct.

h depends on $\sigma$, therefore option (B) is incorrect.

In a lift going up with a constant acceleration a, the force due to surface tension remains same but weight of the water increases i.e., W_lift = $\pi$r²h$\rho$(g + a). Thus, the height of the water rise in a capillary becomes

$h = {{2\sigma \cos \theta } \over {r\,\rho (g + a)}} = {{2\sigma \cos \theta } \over {r\, \rho{g_{eff}}}}$,

where g_eff is the effective g in the lift. Hence, h decrease if the experiment is performed in a lift going up with a constant acceleration.

From (1), h $\propto$ cos$\theta$ and not $\theta$, therefore, option (D) is incorrect.

We know,

Rate of flow of liquid through narrow tube,

${{dv} \over {dt}}$ = ${{\pi {{\Pr }^4}} \over {8\eta l}}$

Both tubes are connected in series so rate of flow of liquid is same.

$\therefore\,\,\,$ ${{\pi {P_1}r_1^4} \over {8\eta {l_1}}}$ = ${{\pi {P_2}r_2^4} \over {8\eta {l_2}}}$

$ \Rightarrow $$\,\,\,$ ${{{P_1}{r_1}^4} \over {{l_1}}}$ = ${{{P_2}{r_2}^4} \over {{l_2}}}$

Given,

P₂ = 4P₁   and   ${l_2}$ = ${{{l_1}} \over 4}$

$ \Rightarrow $$\,\,\,$ ${{{P_1}{r_1}^4} \over {{l_1}}}$ = ${{4{P_1}{r_2}^4} \over {{{{l_1}} \over 4}}}$

$ \Rightarrow $$\,\,\,$ ${r_2}^4$ = ${{{r_1}^4} \over {16}}$

$ \Rightarrow $$\,\,\,$ r₂ = ${{{r_1}} \over 2}$

To determine how the stress in the leg changes when a man's linear dimensions increase by a factor of 9, we can analyze the relationship between the dimensions and the stress on the legs.

First, let's establish some relationships:

1. Linear dimensions (length, width, height) increase by a factor of 9.


2. Density remains the same.

Stress is defined as force per unit area:

$ \text{Stress} = \dfrac{\text{Force}}{\text{Area}} $

Since density remains constant, the volume and thus the mass of the man will change according to the cube of the linear dimensions. Since the linear dimension changes by a factor of 9, the volume (and hence the mass) changes by a factor of:

$ 9^3 = 729 $

Therefore, the weight (force due to gravity) also increases by a factor of 729.

The cross-sectional area of the leg is proportional to the square of the linear dimensions. Thus, if the linear dimension increases by a factor of 9, the cross-sectional area increases by a factor of:

$ 9^2 = 81 $

Now, substituting these factors into the stress formula:

$ \text{New Stress} = \dfrac{\text{New Force}}{\text{New Area}} = \dfrac{729 \times \text{Original Force}}{81 \times \text{Original Area}} $

Simplifying, we get:

$ \text{New Stress} = 9 \times \dfrac{\text{Original Force}}{\text{Original Area}} = 9 \times \text{Original Stress} $

Thus, the stress in the leg increases by a factor of 9. Therefore, the correct answer is:

Option B: 9

The rate of radiation energy emitted by a body of surface area A, emissivity e, and kept at an absolute temperature T is given by Stefan's law

dQ_e/dt = $\sigma$eAT⁴,

where $\sigma$ is Stefan's constant. If this body is kept at a surrounding temperature T₀ then rate of radiation energy absorbed by the body is

dQ_a/dt = $\sigma$eAT$_0^4$.

Thus, the net rate of energy loss by the body is

$d{Q_i}/dt = d{Q_e}/dt - d{Q_a}/dt = \sigma eA({T^4} - T_0^4)$.

If surrounding temperature is reduced by $\Delta$T₀ (<< T₀) then net rate of energy lost by the body becomes

${{d{Q_f}} \over {dt}} = \sigma eA\left[ {{T^4} - {{({T_0} - \Delta {T_0})}^4}} \right]$

$ = \sigma eA\left[ {{T^4} - T_0^4{{\left( {1 - {{\Delta {T_0}} \over {{T_0}}}} \right)}^4}} \right]$

$ \approx \sigma eA\left[ {{T^4} - T_0^4\left( {1 - 4{{\Delta {T_0}} \over {{T_0}}}} \right)} \right]$

$ = {{d{Q_i}} \over {dt}} + 4\sigma eAT_0^3\Delta {T_0}$.

Thus, decrease in the surrounding temperature increases the rate of energy loss by $4\sigma eAT_0^3\Delta {T_0}$. To maintain the body temperature, the human being needs to increase the rate of energy generation (through chemical reactions in the body) by $4\sigma eAT_0^3\Delta {T_0}$ to balance the increase in rate of energy loss.

The human beings can also maintain the body temperature (without increasing the rate of energy generation) by reducing the surface area (e.g., by curling up) of their body. If surrounding temperature is reduced by $\Delta$T₀ and surface area of the body is reduced by $\Delta$A (<< A) then net rate of energy loss by the body becomes

${{d{Q_f}} \over {dt}} = 4\sigma e(A - \Delta A)\left[ {{T^4} - {{({T_0} - \Delta {T_0})}^4}} \right]$

$ \approx {{d{Q_i}} \over {dt}} + 4\sigma eAT_0^3\Delta {T_0} - \sigma e\Delta A({T^4} - \Delta T_0^4)$.

Thus, the rate of energy loss by radiation remains same $(d{Q_f}/dt = d{Q_i}/dt)$ if reduction in surface area is

$\Delta A = {{4AT_0^3\Delta {T_0}} \over {{T^4} - T_0^4}} \approx A{{\Delta {T_0}} \over {\Delta T}}$,

where, the body temperature is $T = {T_0} + \Delta T$.

We assume body to be black i.e., e = 1. The amount of energy emitted by the body in one second is

$d{Q_e}/dt = \sigma eA{T^4} = \sigma eA{({T_0} + \Delta T)^4}$

$ \approx \sigma eAT_0^4(1 + 4\Delta T/{T_0})$

$ = (1)(1)(460)(1 + 4(10)/300) = 521$ J.

The amount of energy lost by the body in one second is

$d{Q_i}/dt = \sigma eA({T^4} - T_0^4) = \sigma eA{({T_0} + \Delta T)^4} - T_0^4)$

$ \approx eA\sigma T_0^4(4\Delta T/{T_0}) = 61$ J.

Bulk modulus, B  =  ${{\Delta P} \over {{{\Delta V} \over V}}}$

V_i  =  $\pi $ (a + $\Delta $a)²b

V_f  =  $\pi $a²b

$ \therefore $   $\Delta $V = 2$\pi $ab$\Delta $a

$ \therefore $   ${{{\Delta V} \over V}}$ = ${{2\pi ab\Delta a} \over {\pi {a^2}b}}$ = ${{2\Delta a} \over a}$

$ \therefore $   $\Delta $P = B $ \times $ ${{{\Delta V} \over V}}$

      = B $ \times $ ${{2\Delta a} \over a}$

Normal Force (N) = $\Delta $P $ \times $ A

       = B $ \times $ ${{2\Delta a} \over a}$ $ \times $ (2$\pi $a)b

       =   4$\pi $B$\Delta $ab

$ \therefore $   Frictional force = $\mu $N = (4$\pi $$\mu $Bb)$\Delta $a

Young's Modulus of the material of the rod is

$Y = {{Stress} \over {Strain}} = {{(F/A)} \over {(\Delta l/l)}}$

Here, Y remains maximum, when $\Delta$l is of least count.

That is,

${Y_{\max }} = \left[ {{{50\pi \times {{10}^3}N} \over {\pi {{(5 \times {{10}^{ - 3}})}^2}{m^2}}}} \right]\left[ {{{1m} \over {0.01 \times {{10}^{ - 3}}m}}} \right]$

$ = 2 \times {10^4}N/{m^2}$

v₁ = velocity of water when it leak from hole

v₂ = velocity of water when it reach the ground.

From Bernoulli's principle,

${1 \over 2}\rho {v_1}^2 + \rho gh$ = ${1 \over 2}\rho {v_2}^2$

$ \Rightarrow $   ${v_1}^2$ + 2gh = ${v_2}^2$

From Torricelli's theorem,

v₁ = $\sqrt {2gH} $

$ \therefore $   ${v_2}^2$ = 2gh + 22gH

From continuity equation,

${A_1}{v_1}$ = ${A_2}{v_2}$

$ \Rightarrow $   $\pi {r^2} \times \sqrt {2gH} $ = $\pi {x^2}\sqrt {2g\left( {h + H} \right)} $

$ \Rightarrow $   x²  =  r²$\sqrt {{H \over {H + g}}} $

$ \Rightarrow $   x = r ${\left( {{H \over {H + g}}} \right)^{{1 \over 4}}}$

Here r = 3R $-$ ${{2R} \over L}$ x

$ \therefore $   Extension in the wire of length dx,

dl = ${{Fdx} \over {AY}}$

= ${{Mg\,dx} \over {\pi {r^2}\,Y}}$

= ${{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}$

$ \therefore $   Change in wire length,

$\Delta $L = $\int\limits_0^L {dl} $

= $\int\limits_0^L {{{Mg\,dx} \over {\pi {{\left( {3R - {{2R} \over L}x} \right)}^2}Y}}} $

= ${{Mg} \over {\pi Y}}\int\limits_0^L {{{dx} \over {{{\left( {3R - {{2R} \over L}x} \right)}^2}}}} $

= ${{Mg} \over {\pi Y}}\left[ { - {1 \over {\left( {3R - {{2R} \over L}x} \right)}} \times \left( { - {L \over {2R}}} \right)} \right]_0^L$

= ${{Mg} \over {\pi Y}}$ $\left[ {\left( {{L \over {2{R^2}}} - {L \over {6{R^2}}}} \right)} \right]$

= ${{Mg} \over {\pi Y}}\left( {{{2L} \over {6{R^2}}}} \right)$

= ${{MgL} \over {3\pi {R^2}Y}}$

$ \therefore $   The equilibrium extended length of the wire,

= L + $\Delta $L

= L + ${{MgL} \over {3\pi {R^2}Y}}$

= L (1 + ${1 \over 3}$ ${{Mg} \over {\pi {R^2}Y}}$)

Equation of motion for the mass,

ma = mg $-$ kv

$ \Rightarrow $    ${{dv} \over {dt}} = {{mg - kv} \over m}$

$ \Rightarrow $   $\int\limits_0^v {{{dv} \over {mg - kv}}} = {1 \over m}\int\limits_0^t {dt} $

$ \Rightarrow $   $ - {1 \over k}\left[ {\ln \left( {mg - kv} \right)} \right]_0^v = {t \over m}$

$ \Rightarrow $   $\ln \left( {{{mg - kv} \over {mg}}} \right) = - {{kt} \over m}$

$ \Rightarrow $   $1 - {{kv} \over {mg}} = {e^{ - {{kt} \over m}}}$

$ \Rightarrow $   ${{kv} \over {mg}} = 1 - {e^{ - {{kt} \over m}}}$

$ \Rightarrow $   $v = {{mg} \over k}\left( {1 - {e^{ - {{kt} \over m}}}} \right)$

ma $=$ mg $-$ k $ \times $ ${{mg} \over k}$ (1 $-$ e$^{ - {{kt} \over m}}$)

$=$  mg $-$ mg + mge$^{ - {{kt} \over m}}$

a $=$ g e$^{ - {{kt} \over m}}$

The acceleration due to gravity at a radial distance r (r < R) from the centre of a sphere of constant mass density $\rho$ is given by

$g = {{G\left( {{4 \over 3}\pi {r^3}} \right)\rho } \over {{r^2}}} = {{4\pi \rho G} \over 3}r$.

Consider a spherical shell of radius r and thickness dr.

Let P and P + dP be the fluid pressures inside and outside the shell. Consider a small element of area A and mass $dm = \rho Adr$ on the shell. The forces on this element are gravitational force (dm g) direted inwards, the force due to pressure of the fluid outside the shell $((P + dP)A)$ directed inwards, and the force due to pressure of the fluid inside the shell (PA) directed outwards. In equilibrium, net force on the element is zero, i.e.,

$(\rho Adr){4 \over 3}\pi \rho Gr + (P + dP)A = PA$, or

$dP = - {4 \over 3}\pi G{\rho ^2}r\,dr$.

Note that the pressure decreases with increase in r. The pressure is zero at the surface of the sphere i.e., P = 0 at r = R. Integrate to get

$\int_0^{P(r)} {dP = P(r) = - {4 \over 3}\pi G{\rho ^2}\int_R^r {rdr} } $

$ = {2 \over 3}\pi G{\rho ^2}({R^2} - {r^2})$.

Now, $P(r = 0) = {{2\pi } \over 3}{\rho ^2}G{R^2} \ne 0$

${{P(r = 3R/4)} \over {P(r = 2R/3)}} = {{{R^2} - {{(3R/4)}^2}} \over {{R^2} - {{(2R/3)}^2}}} = {7 \over {16}} \times {9 \over 5} = {{63} \over {80}}$

${{P(r = 3R/5)} \over {P(r = 2R/5)}} = {{{R^2} - {{(3R/5)}^2}} \over {{R^2} - {{(2R/5)}^2}}} = {{16} \over {25}} \times {{25} \over {21}} = {{16} \over {21}}$

${{P(r = R/2)} \over {P(r = R/3)}} = {{{R^2} - {{(R/2)}^2}} \over {{R^2} - {{(R/3)}^2}}} = {3 \over 4} \times {9 \over 8} = {{27} \over {32}} \ne {{20} \over {27}}$

We know, $Y = {{Stress} \over {Strain}}$

According to graph,

Slope of curve $ = {{Change\,in\,strain} \over {Change\,in\,stress}} = {1 \over Y}$

(Slope)_P > (Slope)_Q $\therefore$ Y_P < Y_Q

P has more tensile strength than Q as it sustains more stress after elastic limit.

There is large deformation between the elastic limit and the fracture point for material P as compared to material Q. Hence, P is more ductile than Q.

After the elastic limit, Q breaks soon as compared to P. So, Q is more brittle than P.

Let $V=\frac{4}{3} \pi r^3$ be the volume of the spheres $P$ and $Q$ of equal radii $r$. The forces acting on the sphere $P$ are its weight $\rho_1 V g$, tension from the string $T$, and the buoyancy force $\sigma_1 V g$.



Similarly, forces on the sphere $Q$ are $\rho_2 V g, T$, and $\sigma_1 V g$. In equilibrium, the net force on the spheres $P$ and $Q$ are separately zero i.e.,

$ \begin{aligned} & T+\rho_1 V g=\sigma_1 V g .........(1) \\\\ & T+\sigma_2 V g=\rho_2 V g ..........(2) \end{aligned} $

The tension $T > 0$ because the string is taut. Thus, equation (1) gives $\rho_1<\sigma_1$ and equation (2) gives $\rho_2>$ $\sigma_2$. Eliminate $T$ from equations (1) and (2) to get

$ \sigma_1-\rho_1=\rho_2-\sigma_2 $ ..........(3)



Now, consider the situation when the sphere $P$ moves in liquid $L_2$ and the sphere $Q$ moves in liquid $L_1$. These spheres will attain the terminal velocities $\vec{v}_{\mathrm{P}}$ and $\vec{v}_{\mathrm{Q}}$ after some time. The direction of the velocity (upwards or downwards) will depend on the density of the sphere in comparison to the density of the liquid. Let us consider the case when $\rho_1>\sigma_2$. In this case, the velocity of the sphere $P$ is downwards. From equation (3), if $\rho_1>\sigma_2$ then $\rho_2<\sigma_1$. If the density of a sphere is less than the density of the liquid in which it is immersed, it will move up. Thus, the velocity of the sphere $Q$ is upwards i.e., the directions of $\vec{v}_{\mathrm{P}}$ and $\vec{v}_{\mathrm{Q}}$ are opposite.

Hence, $\vec{v}_{\mathrm{P}} \cdot \vec{v}_{\mathrm{Q}}<0$.

The forces on the sphere $P$ are its weight $\rho_1 V g$, buoyancy force $\sigma_2 V g$, and viscous drag $6 \pi \eta_2 r v_{\mathrm{P}}$ (see figure). Similarly, the forces on the sphere $Q$ are $\rho_2 V g, \sigma_1 V g$ and $6 \pi \eta_1 r v_{\mathrm{Q}}$. Net forces on the spheres are zero when they move with terminal velocities i.e.,

$ \begin{aligned} & 6 \pi \eta_2 r v_{\mathrm{P}}+\rho_1 V g=\sigma_2 V g ...........(4) \\\\ & 6 \pi \eta_1 r v_{\mathrm{Q}}+\sigma_2 V g=\rho_1 V g ..........(5) \end{aligned} $

Solving equations (4) and (5), we get

$ \begin{aligned} & v_{\mathrm{P}}=\frac{2 r^2\left(\rho_1-\sigma_2\right)}{9 \eta_2} .........(6) \\\\ & v_{\mathrm{Q}}=\frac{2 r^2\left(\sigma_1-\rho_2\right)}{9 \eta_1} .........(7) \end{aligned} $

By dividing equation (6) by (7), we get

$ \frac{\left|\vec{v}_{\mathrm{P}}\right|}{\left|\vec{v}_{\mathrm{Q}}\right|}=\frac{\eta_1}{\eta_2} \frac{\rho_1-\sigma_2}{\sigma_2-\rho_2}=\frac{\eta_1}{\eta_2} . $

Pressure at interface A must be same from both the sides to be in equilibrium.

$\therefore$ $\left( {R\cos \alpha + R\sin \alpha } \right){d_2}g = \left( {R\cos \alpha - R\sin \alpha } \right){d_1}g$
$ \Rightarrow {{{d_1}} \over {{d_2}}} = {{\cos \alpha + \sin \alpha } \over {\cos \alpha - \sin \alpha }} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}$

Young's modulus $Y = {{stress} \over {strain}}$
$stress = Y \times strain$
$Stress$ in steel wire $=$ Applied $pressure$
$Pressure$ $=$ $stress$ $=$ $Y \times \,strain$
$Strain = {{\Delta L} \over L} = \alpha \Delta T$ (As length is constant)
$ = 2 \times {10^{11}} \times 1.1 \times {10^{ - 5}} \times 100$
$ = 2.2 \times {10^8}Pa$

When the bubble gets detached, Buoyant force $=$ force due to surface tension

Force due to excess pressure $=$ upthrust
Access pressure in air bubble $ = {{2T} \over R}$
${{2T} \over R}\left( {\pi {r^2}} \right) = {{4\pi {R^3}} \over {3T}}{\rho _w}g$
$ \Rightarrow {r^2} = {{2{R^4}{\rho _w}g} \over {3T}}$
$ \Rightarrow r = {R^2}\sqrt {{{2{\rho _w}g} \over {3T}}} $

Length of the air column above mercury in the tube is,
$P + x = {P_0}$
$ \Rightarrow P = \left( {76 - x} \right)$
$ \Rightarrow 8 \times A \times 76 = \left( {76 - x} \right) \times A \times \left( {54 - x} \right)$
$\therefore$ $x=38$
Thus, length of air column $=54-38=16cm.$

Resistance of initially given kettle

$R = \rho {l \over A}$

$ = \rho {L \over {\pi {{(d/2)}^2}}} = {{4\rho L} \over {\pi {d^2}}}$

Resistance of two replaced kettles

${R_1} = {{\rho L} \over {\pi {d^2}}}$ and

${R_2} = {{\rho L} \over {\pi {d^2}}}$

So, ${R_1} = {R_2} = {R \over 4}$

If wires are in parallel then equivalent resistance

${R_P} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {R \over 8}$ ....... (1)

If wires are in series then equivalent resistance

${R_S} = {R_1} + {R_2} = {R \over 2}$ ..... (2)

Let V be the applied voltage, m = 0.5 kg be the mass of the water, and S be the specific heat of the water. Initially, the heat produced by a resistance R₁ in time t₁ = 4 min is V²t₁/R₁. This heat is used to raise the temperature of the water by $\Delta$T = 40 K. Thus,

${V^2}{t_1}/{R_1} = mS\Delta T$. ....... (3)

Let t_s and t_p be the time taken to raise the temperature of same amount of water by $\Delta$T when the resistances are connected in series and parallel. Thus,

${V^2}{t_s}/{R_s} = mS\Delta T$ ....... (4)

${V^2}{t_p}/{R_p} = mS\Delta T$ ....... (5)

Divide equation (4) by (3) and use the equation (1) to get ${t_s} = {{{R_s}} \over {{R_1}}}{t_1} = {1 \over 2} \times 4 = 2$ min. Similarly, divide equation (5) by (3) and use the equation (2) to get ${t_p} = {{{R_p}} \over {{R_1}}}{t_1} = {1 \over 8} \times 4 = 0.5$ min.

The velocity of the efflux depends on the hydrostatic pressure difference and its given by

$v = \sqrt {2{g_{eff}}H} $.

After comin gout the jar, water follows the projectile motion. Time taken in traversing a vertical distance h is

$t = \sqrt {2h/{g_{eff}}} $.

The horizontal distance covered in this time is

$d = vt = \sqrt {2{g_{eff}}H} \sqrt {2h/{g_{eff}}} = 2\sqrt {Hh} $,

which is independent of g_eff. In the case of free fall, hydrostatic pressure difference is zero (g_eff = 0) and hence the water does not come out.

Let R be the radius of the meniscus formed with a contact angle $\theta$.

By geometry, this radius makes an angle $\theta + {\alpha \over 2}$ with the horizontal where

$\cos \left( {\theta + {\alpha \over 2}} \right) = b/R$. ...... (1)

Let P₀ be the atmospheric pressure and P₁ be the pressure just below the meniscus. Excess pressure on the concave side of meniscus of radius R is given by

${P_0} - {P_1} = 2S/R$ ..... (2)

The hydrostatic pressure gives

${P_0} - {P_1} = h\rho g$ ..... (3)

From (2) and (3) we get,

${{2S} \over R} = h\rho g$

or, $h\rho g = {{2s} \over b}\cos \left( {\theta + {\alpha \over 2}} \right)$ (By putting value of R)

$h = {{2S} \over {b\rho g}}\cos \left( {\theta + {\alpha \over 2}} \right)$

The dynamics of the spray gun follow the principle of conservation of mass or the continuity equation from fluid dynamics. This principle states that the mass flow rate of a fluid entering a system must equal the mass flow rate leaving a system. For incompressible fluids like liquids, or compressed air where changes in density are negligible, this is often simplified to the rate of volume flow or the continuity equation:

$ A_1v_1 = A_2v_2 $

where

$A_1$ is the cross-sectional area of the piston,

$v_1$ is the speed of the piston,

$A_2$ is the cross-sectional area of the nozzle, and

$v_2$ is the velocity of the fluid (air) leaving the nozzle.

We can use this equation to find the speed of air leaving the nozzle. Given that the piston's speed is 5 mm/s and its radius is 20 mm, we have

$ A_1 = \pi (20\, \text{mm})^2 = 400\pi\, \text{mm}^2 $

and

$ v_1 = 5\, \text{mm/s} $

The nozzle's radius is 1 mm, thus

$ A_2 = \pi (1\, \text{mm})^2 = \pi\, \text{mm}^2 $

Substituting the values into the continuity equation, we can solve for $v_2$:

$ A_1v_1 = A_2v_2 \rightarrow 400\pi\, \text{mm}^2 \times 5\, \text{mm/s} = \pi\, \text{mm}^2 \times v_2 \rightarrow v_2 = \frac{400\pi\, \text{mm}^2 \times 5\, \text{mm/s}}{\pi\, \text{mm}^2} = 2000\, \text{mm/s} $

Converting this to meters per second, we get

$ v_2 = 2000\, \text{mm/s} \times \frac{1\, \text{m}}{1000\, \text{mm}} = 2\, \text{m/s} $

Therefore, the air comes out of the nozzle with a speed of 2 m/s, which corresponds to Option C.

For horizontal flow, Bernoulli's theorem becomes

${P \over \rho } + gh + {1 \over 2}{v^2} = k'$

$P + {1 \over 2}\rho {v^2} = k$ (h = 0)

Between points 1 and 2 :

${P_1} - {P_2} = {1 \over 2}{\rho _a}v_a^2$

Between points 2 and 3 :

${P_3} - {P_2} = {1 \over 2}{\rho _l}v_l^2$

Since, P₁ = P₃, we have

${1 \over 2}{\rho _l}v_l^2 = {1 \over 2}{\rho _a}v_a^2$

${{{v_l}} \over {{v_a}}} =\sqrt {{{{\rho _a}} \over {{\rho _l}}}} $

From figure, $k{x_0} + {F_B} = Mg$
$k{x_0} + \sigma {L \over 2}Ag = Mg$
[ as mass $=$ density $ \times $ volume ]
$ \Rightarrow k{x_0} = Mg - \sigma {L \over 2}Ag$
$ \Rightarrow {x_0} = {{Mg - {{\sigma LAg} \over 2}} \over k}$
$ = {{Mg} \over k}\left( {1 - {{LA\sigma } \over {2M}}} \right)$

According to Newton's law of cooling, the temperature goes on decreasing with time non-linearly.

Considering FBDs of spheres A and B, we get

Here, $x$ is the extension produced in the spring

At equilibrium $\Sigma \mathrm{F}_{\mathrm{A}}=0$ and $\Sigma \mathrm{F}_{\mathrm{B}}=0$

$ \begin{aligned} \frac{4}{3} \pi R^3(2 \rho) g & =k x+\frac{4}{3} \pi R^3 \rho g ...........(i)\\\\ k x+\frac{4}{3} \pi R^3(2 \rho) g & =\frac{4}{3} \pi R^3(3 \rho) g .........(ii)\end{aligned} $

Subtract (ii) from (i), we get

$ \begin{aligned} -k x & =k x-\frac{4}{3} \pi \mathrm{R}^3(2 \rho) g \\\\ x & =\frac{4 \pi \mathrm{R}^3 \rho g}{3 k} \end{aligned} $

(Option A is correct)

If we consider both the spheres as a system, then at equilibrium,

$ \begin{aligned} \text { total weight } & =\text { buoyant force } \\ \text { Total weight }(\mathrm{A}+\mathrm{B}) & =\frac{4}{3} \pi \mathrm{R}^3(4 \rho) g \\ \text { Total buoyant force } & =\frac{4}{3} \pi \mathrm{R}^3(4 \rho) g \end{aligned} $

Hence, sphere A (light sphere) is completely submerged (option D is correct)

To solve this problem, we need to use Hooke's Law, which states that the elongation (or stretch) of a material is directly proportional to the force applied and the length of the material, and inversely proportional to the cross-sectional area and the Young's modulus of the material. Mathematically, this relation can be expressed as:

$ \Delta L = \frac{F \cdot L}{A \cdot Y} $

where:

• $ \Delta L $ is the elongation
• $ F $ is the force applied
• $ L $ is the original length of the material
• $ A $ is the cross-sectional area of the material
• $ Y $ is the Young's modulus of the material

In this problem, the Young's modulus ($ Y $) and the force ($ F $) are the same for both wires because they are made of the same material (copper) and the forces applied are equal. Therefore, the elongation is dependent on the length and cross-sectional area of each wire.

For the thick wire with length $ 2L $ and radius $ 2R $, the cross-sectional area $ A_1 $ is given by:

$ A_1 = \pi (2R)^2 = 4\pi R^2 $

For the thin wire with length $ L $ and radius $ R $, the cross-sectional area $ A_2 $ is given by:

$ A_2 = \pi R^2 $

Using the formula for elongation for both wires, we get:

Elongation of the thick wire ($ \Delta L_\text{thick} $):

$ \Delta L_\text{thick} = \frac{F \cdot 2L}{4\pi R^2 \cdot Y} = \frac{F \cdot 2L}{4\pi R^2 Y} = \frac{FL}{2\pi R^2 Y} $

Elongation of the thin wire ($ \Delta L_\text{thin} $):

$ \Delta L_\text{thin} = \frac{F \cdot L}{\pi R^2 \cdot Y} = \frac{FL}{\pi R^2 Y} $

To find the ratio of the elongation in the thin wire to that in the thick wire, we divide $ \Delta L_\text{thin} $ by $ \Delta L_\text{thick} $:

$ \text{Ratio} = \frac{\Delta L_\text{thin}}{\Delta L_\text{thick}} = \frac{\frac{FL}{\pi R^2 Y}}{\frac{FL}{2\pi R^2 Y}} = \frac{FL}{\pi R^2 Y} \cdot \frac{2\pi R^2 Y}{FL} = 2 $

Therefore, the ratio of the elongation in the thin wire to that in the thick wire is 2. The correct answer is:

Option C: 2.00

Newton's law of cooling

${{d\theta } \over {dt}} = - k\left( {\theta - {\theta _0}} \right)$

$ \Rightarrow {{d\theta } \over {\left( {\theta - {\theta _0}} \right)}} = - kdt$

Intergrating

$ \Rightarrow \log \left( {\theta - {\theta _0}} \right) = - kt + c$

Which represents an equation of straight line.

Thus the option $(a)$ is correct.

At equilibrium,
$2Tl = mg$
$T = {{mg} \over {2l}} = {{1.5 \times {{10}^{ - 2}}} \over {2 \times 30 \times {{10}^{ - 2}}}} = {{1.5} \over {60}}$
$ = 0.025\,N/m = 0.025Nm$

The high pressure pulse is a compression pulse. When this gets reflected from a rigid boundary (closed pipe), the reflected pressure wave has the same phase as the incident wave i.e., a compression pulse is reflected as a compression pulse. In case of open pipe, there is a phase change of 180$^\circ$ when it is reflected by an open end i.e., a compression pulse is reflected as a rarefaction pulse.

Consider a thin, uniform cylindrical shell that is closed at both ends and partially filled with water. This cylindrical shell is floating vertically in water, with exactly half of its height submerged. The relative density of the shell's material compared to water is denoted by ${\rho_c}$.

The inner radius of the cylindrical shell is $r_1$, the outer radius is $r_2$, the height is $h$, and the material density is $\rho$. The shell is filled with water, which has a density of $\rho_w$, up to a height of $x$. Given ${\rho_c = \frac{\rho}{\rho_w}}$, we know that the cylinder is in a half-submerged state, meaning $h/2$ of the cylinder's height is submerged in water.

In equilibrium, the weight of the cylindrical shell, including the water inside it, must equal the buoyant force (upthrust) acting on it. This relationship can be expressed as follows:

$ \left(\pi r_2^2-\pi r_1^2\right) h \rho g + \pi r_1^2 x \rho_w g = \pi r_2^2\left(\frac{h}{2}\right)\rho_w g $

By simplifying this equation, we get:

$ x = h\left[\frac{r_2^2}{r_1^2}\left(0.5-\rho_c\right)+\rho_c\right] = h\left[0.5+\frac{r_2^2-r_1^2}{r_1^2}\left(0.5-\rho_c\right)\right] ...........(i) $

From equation (i), we can determine the condition under which $x$ is greater than $0.5h$. This occurs if $\rho_c < 0.5\left(\because r_2 > r_1\right)$.

$W = T \times \,\,$ change in surface area
$W = 2T4\pi \left[ {{{\left( 5 \right)}^2} - {{\left( 3 \right)}^2}} \right] \times {10^{ - 4}}$
$ = 2 \times 0.03 \times 4\pi \left[ {25 - 9} \right] \times {10^{ - 4}}\,J$
$ = 0.4\pi \times {10^{ - 3}}\,J$
$ = 0.4\pi mJ$

From Bernoulli's theorem,
${P_0} + {1 \over 2}\rho v_1^2\rho gh = {P_0} + {1 \over 2}\rho v_2^2 + 0$
${v_2} = \sqrt {v_1^2 + 2gh} $
$ = \sqrt {0.16 + 2 \times 10 \times 0.2} $
$ = 2.03\,m/s$
From equation of continuity
${A_2}{v_2} = {A_1}{v_1}$
$\pi {{D_2^2} \over 4} \times {v_2} = \pi {{D_1^2} \over 4}{v_1}$
$ \Rightarrow \,\,{D_1} = {D_2}\sqrt {{{{v_1}} \over {{v_2}}}} = 3.55 \times {10^{ - 3}}m$

Let T₁, T₂, T₃ and T₄ be the temperatures at slab interfaces as shown in the figure.

The rate of heat flow across a slab, with thermal conductivity $\kappa $, width w, thickness l, length x, and temperature difference $\Delta$T, is given by

$dQ/dt = \kappa lw(\Delta T/x)$.

Thus, the heat flow rates through the given slabs are

${{d{Q_A}} \over {dt}} = {{(2K)(4Lw)({T_1} - {T_2})} \over L} = 8Kw({T_1} - {T_2})$ ...... (1)

${{d{Q_B}} \over {dt}} = {{(3K)(Lw)({T_2} - {T_3})} \over {4L}} = {3 \over 4}Kw({T_2} - {T_3})$ ...... (2)

${{d{Q_C}} \over {dt}} = {{(4K)(2Lw)({T_2} - {T_3})} \over {4L}} = 2Kw({T_2} - {T_3})$ ...... (3)

${{d{Q_D}} \over {dt}} = {{(5K)(Lw)({T_2} - {T_3})} \over {4L}} = {5 \over 4}Kw({T_2} - {T_3})$ ..... (4)

${{d{Q_E}} \over {dt}} = {{(6K)(4Lw)({T_3} - {T_4})} \over L} = 24Kw({T_3} - {T_4})$ ..... (5)

In the steady state, heat flow into the system is equal to the heat flow out of the system i.e., $d{Q_A}/dt = d{Q_E}/dt$.

Use equations (1) and (5) to get

${T_1} - {T_2} = 3({T_3} - {T_4})$ ..... (6)

Similarly, the steady state condition for the interface at temperature T₂ is

$d{Q_A}/dt = d{Q_B}/dt + d{Q_C}/dt + d{Q_D}/dt$,

which gives (by using equations (1)-(4))

$2({T_1} - {T_2}) = {T_2} - {T_3}$ ..... (7)

Use equations (6) and (7) to get

${T_3} - {T_4} = (1/3)({T_1} - {T_2}) = (1/6)({T_2}/{T_3})$.

Also, equations (2)-(4) give $d{Q_C}/dt = d{Q_B}/dt + d{Q_D}/dt$.

Let V be the volume of each sphere and let T be the tension in the string.

Buoyant force on sphere A is U_A = d_FVg

Buoyant force on sphere B is U_B = d_FVg

Weight of A is W_A = d_AVg

Weight of B is W_B = d_BVg

The free body diagrams of A and B are as follows. (see Fig. 11.34)

For equilibrium,

U_A = T + W_A and U_B + T = W_B

i.e. d_FVg = T + d_AVg ..... (i)

and d_FVg + T = d_BVg .... (ii)

From Eq. (i) ${d_F} = {T \over {Vg}} + {d_A}$. Hence, d_F > d_A. So choice (a) is correct.

From Eq. (ii) ${d_B} = {T \over {Vg}} + {d_F}$. Hence d_B > d_F. So choice (b) is also correct.

Eliminating T from Eqs. (i) and (ii) we get $2{d_F} = {d_A} + {d_B}$, which is choice (d).

So, the correct choices are (a), (b) and (d).

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.

We have

Surface energy = Tension $\times$ Area

That is, E_surface = T $\times$ 4$\pi$R²

where T = 0.11 N/m and R = 1.4 $\times$ 10^$-$4 m. Therefore, the surface energy is

${E_{surface}} = T \times 4\pi {R^2} = 0.11 \times 4 \times 3.14 \times {(1.4 \times {10^{ - 3}})^2} = 2.7 \times {10^{ - 6}}J$

Consider a small element of length dl on the drop-dropper interface. The force on this element is $dF = T\,dl$, and it makes an angle $\theta$ from the horizontal. Resolve $dF$ in the horizontal and the vertical directions to get, $d{F_h} = T\,dl\cos \theta $ and $d{F_v} = T\,dl\sin \theta $. By symmetry, the force $d{F_h}$ on two diametrically opposite elements on circular interface is equal and opposite. Thus, total force in the horizontal direction, ${F_h} = \int {d{F_h} = 0} $. Integrate the vertical component over the circular interface to get ${F_v} = T(2\pi r)\sin \theta = 2\pi rT(r/R) = 2\pi {r^2}T/R$.

The drop falls down when the weight of the drop becomes larger than the surface tension. Hence,

$mg = {{2\pi {r^2}T} \over R}$

That is,

${4 \over 3}\pi {R^3}\rho g = {{2\pi {r^2}T} \over R}$

where

r = 5 $\times$ 10^$-$4 m ; $\rho$ = 10³ kg m^$-$3 ; g = 10 m/s² ; T = 0.11 Nm^$-$1

Therefore,

${R^4} = {{3{r^2}t} \over {2\rho g}} = {{3 \times {{(5 \times {{10}^{ - 4}})}^2} \times 0.11} \over {2 \times {{10}^3} \times 10}} = 4.125 \times {10^{ - 12}}$ m⁴

$ \Rightarrow R = 1.425 \times {10^{ - 3}}$ m

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.

Given that,

${r_2} > {r_1}$

Just inside of soap bubble, the pressure

${P_1} = {P_0} + {{4T} \over {{r_1}}}$ ..... (i)

${P_2} = {P_0} + {{4T} \over {{r_2}}}$ .... (ii)

Hence, ${r_2} > {r_1}$ then ${P_1} > {P_2}$

Therefore, air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases.

Volume flow rate (V) of an incompressible fluid in the steady flow remains constant.

$V=a\times v$

Where $a$ = area of cross-section and $v$ = velocity

If '$v$' decreases, '$a$' increases and vice-versa.

When steam of water moves up, its speed (v) decreases and therefore 'a' increases, i.e., the water spreads out like a fountain. When steam of water from hose pipe moves down, its speed increases and as a result, area of cross-section decreases.

Therefore, statement - 1 is true.

Statement - 2 is the correct explanation of statement - 1.

As the bubble moves upwards, besides the buoyancy force (the cause of which is pressure difference), only the force of gravity and the force of viscosity will act.