Properties of Matter

$\begin{aligned} & \text { (A) } w_{\text {all }}=k_f-k_i=0 \\ & w_g+w_B+w_v+w_{\text {ext }}=0 \\ & \mathrm{mgd}-\rho_w \cdot v \cdot g d-6 \pi \eta r v d+w_{\text {ext }}=0 \\ & \quad(\text { slowly } v=0) \end{aligned}$

$\begin{aligned} & \mathrm{w}_{\text {ext }}=\rho_{\mathrm{w}} \operatorname{vgd}-\mathrm{mgd}=\left(1000 \times \frac{4}{3} \times \frac{22}{7} \times\left(\frac{3}{2} \times 10^{-2}\right)^3-\frac{22}{7} \times 10^{-3}\right) \mathrm{gd} \\ & \mathrm{w}_{\text {ext }}=\frac{22}{7} \times 10^{-3}\left[\frac{9}{2}-1\right] \times 10 \times 0.7=\frac{22}{7} \times 10^{-3} \times \frac{7}{2} \times 7 \\ & \mathrm{~W}_{\text {ext }}=77 \times 10^{-3} \mathrm{~J}=0.077 \mathrm{~J} \end{aligned}$

$\begin{aligned} & \text { (B) } \mathrm{w}_{\mathrm{g}}+\mathrm{w}_{\mathrm{B}}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}} \quad\left(\mathrm{k}_{\mathrm{i}}=0\right) \\ & \frac{1}{2} \times \frac{22}{7} \times 10^{-3} \mathrm{v}^2=77 \times 10^{-3} \\ & \mathrm{v}^2=\frac{77 \times 7 \times 2}{22} \\ & \mathrm{v}=7 \mathrm{~m} / \mathrm{s} \end{aligned}$

(C) $\mathrm{H}=\frac{\mathrm{v}^2}{2 \mathrm{~g}}=\frac{49}{20}=2.45 \mathrm{~m}$

For adiabatic process:

$\frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^\gamma $

$\Rightarrow \mathrm{P}_1 \mathrm{~V}_1^\gamma=\mathrm{P}_2 \mathrm{~V}_2^\gamma $

$ \Rightarrow $ $\left(\mathrm{P}_{a 1}+\frac{4 \mathrm{~S}}{\gamma_1}\right)\left(\frac{4}{3} \pi r_1^3\right)^{5 / 3} $

$=\left(\mathrm{P}_{a 2}+\frac{4 \mathrm{~S}}{r_2}\right)\left(\frac{4}{3} \pi r_2^3\right)^{5 / 3}$

$ \Rightarrow $ $ \left(\frac{r_1}{r_2}\right)^5=\frac{P_{a 2}+4 S / r_2}{P_{a 1}+4 S / r_1} $

Also, $\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1}$

$ \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\left(\mathrm{P}_{a 2}+4 \mathrm{~S} / r_2\right)^{2 / 5}}{\left(\mathrm{P}_{a 2}+\frac{4 \mathrm{~S}}{r_1}\right)^{2 / 5}} $

At same temperature,

$ \mathrm{P}_1 \mathrm{~V}_1=\mathrm{P}_2 \mathrm{~V}_2 $

$ \Rightarrow $ $\left(\mathrm{P}_{a 1}+\frac{4 \mathrm{~S}}{r_1}\right)\left(\frac{4}{3} \pi r_1^3\right)=\left(\mathrm{P}_{a 2}+\frac{4 \mathrm{~S}}{r_2}\right)\left(\frac{4}{3} \pi r_2^3\right) $

$ \Rightarrow $ $\left(\frac{r_1}{r_2}\right)^3=\frac{\mathrm{P}_{a_2}+4 \mathrm{~S} / r_2}{\mathrm{P}_{a_1}+4 \mathrm{~S} / r_1}$

$({p_1} - {p_2})ds = \rho dsd\sqrt 2 (g\sin 45^\circ - a)$

$({p_1} - {p_2})ds = \rho d(g - a\sqrt 2 )$

$\beta = {{({p_1} - {p_2})} \over {\rho gd}} = \left( {1 - {{a\sqrt 2 } \over g}} \right)$

When $a = {g \over {\sqrt 2 }},\beta = 0$

When $a = {g \over 2},\beta = \left( {{{\sqrt 2 - 1} \over {\sqrt 2 }}} \right)$

${n_1} > > ({n_1} - {n_2}) = \Delta n$

$ \because $ ${p_1} = {{{n_1}RT} \over {{N_A}}}$ and ${p_2} = {{{n_2}RT} \over {{N_A}}}$

$F = ({n_1} - {n_2}){k_B}TS = \Delta n{k_B}TS$

$V = {{\Delta n{k_B}TS} \over \beta }$

$\Delta n{k_B}TS = {\ln _1}S\beta v$

$ \Rightarrow {n_1}\beta vl = \Delta n{k_B}T$

Total number of molecules/second

$ = {{({n_1}vdt)S} \over {dt}}$

$ = {n_1}vS = {{\Delta n{k_B}TvS} \over {\beta vl}} = \left( {{{\Delta n} \over l}} \right)\left( {{{{k_B}T} \over \beta }} \right)S$

As $\Delta $n will decrease with time, therefore rate of molecules getting transfer decreases with time.

Heights if only single material tubes are used of sufficient length,

${h_1} = {{2R\cos \theta } \over {\rho rg}}$

$ = {{2 \times 0.075 \times \cos 0^\circ } \over {1000 \times 2 \times {{10}^{ - 4}} \times 10}} = 7.5\,cm$

${h_2} = {{2T\cos \theta '} \over {\rho rg}}$

$ = {{2 \times 0.075 \times \cos 60^\circ } \over {1000 \times 2 \times {{10}^{ - 4}} \times 10}} = 3.75\,cm$

Case - I

Case - II

2. Liquid will rise only upto height of 5 cm and meniscus will adjust by changing is radius of curvature. If the liquid goes up in tube 2 then it will not be able to support the weight of the liquid.

3. Weight of water in meniscus will be different in two cases because angle of contact is different.

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.

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} . $

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.

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)

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.

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).