Properties of Matter

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

Given, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2: \gamma=2$

The process described is adiabatic. In an adiabatic process (a process during which no heat is gained or lost), the following relationship holds:

$P_1^{1-\gamma} \cdot T_1^\gamma = P_2^{1-\gamma} \cdot T_2^\gamma$

Where :

- $P_1$ and $P_2$ are the initial and final pressures,

- $T_1$ and $T_2$ are the initial and final temperatures,

- $\gamma$ is the ratio of specific heats (given as 2).

Given that $P_1 = 600 \, \text{Pa}$, $T_1 = 300 \, \text{K}$, $T_2 = 150 \, \text{K}$, and $\gamma = 2$, solve for $P_2$ :

$\frac{P_2}{P_1} = \left(\frac{T_1}{T_2}\right)^{\gamma/(1-\gamma)}$

$\Rightarrow P_2 = P_1 \left(\frac{T_1}{T_2}\right)^{\gamma/(1-\gamma)}$

$\Rightarrow P_2 = 600 \, \text{Pa} \times \left(\frac{300 \, \text{K}}{150 \, \text{K}}\right)^{2/(1-2)}$

$\Rightarrow P_2 = 150 \, \text{Pa}$

Using the ideal gas law in the form $\rho = \frac{PM}{RT}$ (where R is the specific gas constant), and rearranging to find the relationship between $\rho_1$ and $\rho_2$ :

$\frac{\rho_1}{\rho_2} = \frac{P_1}{P_2} \cdot \frac{T_2}{T_1}$

$\Rightarrow \rho_2 = \rho_1 \cdot \frac{P_2}{P_1} \cdot \frac{T_1}{T_2}$

$\Rightarrow \rho_2 = 0.2 \, \text{kg/m}^3 \cdot \frac{150 \, \text{Pa}}{600 \, \text{Pa}} \cdot \frac{300 \, \text{K}}{150 \, \text{K}}$

$\Rightarrow \rho_2 = 0.1 \, \text{kg/m} ^3$

Next, use the principle of conservation of mass flow rate to determine the velocity of the gas at the lower and upper ends.

Mass flow rate, $\frac{d m}{d t} = \rho_1 \cdot A_1 \cdot v_1 = \rho_2 \cdot A_2 \cdot v_2$,

The mass flow rate is conserved, and given by $\alpha = \rho \cdot A \cdot v$. Thus, you have:

$\alpha = \rho_1 \cdot A_1 \cdot v_1$

$\Rightarrow 0.8 \, \text{kg/s} = 0.2 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^2 \cdot v_1$

$\Rightarrow v_1 = \frac{0.8 \, \text{kg/s}}{0.2 \, \text{kg/m}^3 \cdot 0.1 \, \text{m}^2} = 40 \, \text{m/s}$

Similarly, for the upper end :

$\alpha = \rho_2 \cdot A_2 \cdot v_2$

$\Rightarrow v_2 = \frac{0.8 \, \text{kg/s}}{0.1 \, \text{kg/m}^3 \cdot 0.4 \, \text{m}^2} = 20 \, \text{m/s}$

Work done on the gas $=$ total energy $=\Delta \mathrm{K}+\Delta \mathrm{U}+$ internal energy

$ \begin{aligned} & \mathrm{P}_1 \mathrm{~A}_1 \Delta x_1-\mathrm{P}_2 \mathrm{~A}_2 \Delta x_2=\frac{1}{2} \Delta m v_2^2-\frac{1}{2} \Delta m v_1^2+\Delta m g h+\frac{f}{2} \\\\ &\left(\mathrm{P}_2 \Delta \mathrm{V}_2-\mathrm{P}_1 \Delta \mathrm{V}_1\right) 2 \mathrm{P}_1 \frac{\Delta \mathrm{v}_1}{\Delta m}-2 \mathrm{P}_2 \frac{\Delta \mathrm{v}_2}{\Delta m}=\frac{\mathrm{v}_2^2-\mathrm{v}_1^2}{2}+g h \\\\ & \frac{2 \times 600}{0.2}-\frac{2 \times 150}{0.1}=\frac{20^2-40^2}{2}+10 \mathrm{~h} \\\\ & 6000-3000=\frac{400-1600}{2}+10 \mathrm{~h} \\\\ & 3000=-600+10 \mathrm{~h} \\\\ & 10 \mathrm{~h}=3000+600 \\\\ & \mathrm{~h}=360 \mathrm{~m} \end{aligned} $

Force on each column = ${{mg} \over 4}$

Strain = ${{mg} \over {4AY}}$

$ = {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$

= 2.6 $\times$ 10^$-$7

We know,

$\Delta l = {{WL} \over {2AY}}$

$\Delta l = {{10 \times 1} \over {2 \times 5}} \times 100 \times {10^{ - 4}} \times 2 \times {10^{11}}$

$\Delta l = {1 \over 2} \times {10^{ - 9}} = 5 \times {10^{ - 10}}$ m

Option (d)

Viscous force = Weight

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

= 3.9 $\times$ 10^$-$10

$T = {{2{m_1}{m_2}g} \over {{m_1} + {m_2}}} = {{2 \times 3 \times 5 \times 10} \over 8}$

$ = {{75} \over 2}$

Stress $ = {T \over A}$

${{24} \over \pi } \times {10^2} = {{75} \over {2 \times \pi {R^2}}}$

${R^2} = {{75} \over {2 \times 24 \times 100}} = {3 \over {8 \times 24}}$

$\Rightarrow$ R = 0.125 m

R = 12.5 cm

Image

We have, P_A = P_B. [Points A & B at same horizontal level]

$\therefore$ ${P_{atm}} - {{2T} \over {{r_1}}} + \rho g(x + \Delta h) = {P_{atm}} - {{2T} \over {{r_2}}} + \rho gx$

$\therefore$ $\rho g\Delta h = 2T\left[ {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right]$

$ = 2 \times 7.3 \times {10^{ - 2}}\left[ {{1 \over {2.5 \times {{10}^{ - 3}}}} - {1 \over {4 \times {{10}^{ - 3}}}}} \right]$

$\therefore$ $\Delta h = {{2 \times 7.3 \times {{10}^{ - 2}} \times {{10}^3}} \over {{{10}^3} \times 10}}\left[ {{1 \over {2.5}} - {1 \over 4}} \right]$

= 2.19 $\times$ 10^$-$3 m = 2.19 mm

Hence, option (b).

At terminal speed

a = 0

F_net = 0

mg = F_v = 6$\pi$ $\eta $Rv

$v = {{mg} \over {6\pi \eta Rv}}$

$v = {{{\rho _w}{{4\pi } \over 3}{R^3}g} \over {6\pi \eta R}}$

$ = {{2{\rho _w}{R^2}g} \over {9\eta }}$

$ = {{400} \over {81}}$ m/s

= 4.94 m/s

For no sliding

f $\ge$ $\rho$av²

$\mu$mg $\ge$ $\rho$av²

$\mu$$\rho$Ahg $\ge$ $\rho$a2gh

$\mu \ge {{2a} \over A}$

Option (3)

${{\Delta T} \over {\Delta t}} = K({T_t} - {T_s})$

T_t = average temp.

_T = surrounding temp.

${{61 - 59} \over 4} = K\left( {{{61 + 59} \over 2} - 30} \right)$ ..... (1)

${{51 - 49} \over t} = K\left( {{{51 + 49} \over 2} - 30} \right)$ ..... (2)

Divide (1) & (2)

${t \over 4} = {{60 - 30} \over {50 - 30}} = {{30} \over {20}}$

So, t = 6 minutes.

no. of moles is conserved

n₁ + n₂ = n₃

P₁V₁ + P₂V₂ = P₃V

${{4S} \over {{r_1}}}\left( {{4 \over 3}\pi r_1^3} \right) + {{4S} \over {{r_2}}}\left( {{4 \over 3}\pi r_2^3} \right) = {{4S} \over {{r_3}}}\left( {{4 \over 3}\pi r_3^3} \right)$

$r_1^2 + r_2^2 = r_3^2$

${r_3} = \sqrt {r_1^2 + r_2^2} $

In series combination $\Delta$l = l₁ + l₂

$Y = {{F/A} \over {\Delta l/l}} \Rightarrow \Delta l = {{Fl} \over {AY}}$

$ \Rightarrow \Delta l \propto {l \over Y}$

Equivalent length of rod after joining is = 2l

As, lengths are same and force is also same in series

$\Delta l = \Delta {l_1} + \Delta {l_2}$

${{{l_{eq}}} \over {{Y_{eq}}}} = {l \over {{Y_1}}} + {l \over {{Y_2}}} \Rightarrow {{2l} \over Y} = {l \over {{Y_1}}} + {l \over {{Y_2}}}$

$\therefore$ $Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}$

${T_1} = k({l_1} - {l_0})$

${T_2} = k({l_2} - {l_0})$

${{{T_1}} \over {{T_2}}} = {{{l_1} - {l_0}} \over {{l_2} - {l_0}}}$

${{{T_1}{l_2} - {T_2}{l_1}} \over {{T_1} - {T_2}}} = {l_0}$

The volume of a sphere is given by $\frac{4}{3}\pi R^3$.

So the volume of the two small mercury drops each of radius $R$ is $2\times \frac{4}{3}\pi R^3$.

When they coalesce to form a larger drop, the volume is conserved. So, the volume of the larger drop is also $2\times \frac{4}{3}\pi R^3$.

Let's denote the radius of this large drop as $R'$.

Therefore, $2\times \frac{4}{3}\pi R^3 = \frac{4}{3}\pi {R'}^3$.

Solving for $R'$, we get $R' = 2^{1/3}R$.

Now, the surface energy of a sphere is proportional to its surface area, and the surface area of a sphere is given by $4\pi R^2$.

So, the ratio of total surface energy before and after the change is:

$\frac{{2\times 4\pi R^2}}{{4\pi (2^{1/3}R)^2}} = \frac{{2}}{{2^{2/3}}} = {2^{1/3}}:1$

Suppose, I₀ be the actual length of metal wire and Y be its Young's modulus.

From Hooke's law,

$Y = {{T{I_0}} \over {A\Delta I}}$

where, $\Delta I = I - {I_0}$

$ \Rightarrow Y = {{T{I_0}} \over {A(I - {I_0})}}$ or $I - I = {{T{I_0}} \over {AY}}$

$\therefore$ ${{{I_1} - {I_0}} \over {{I_2} - {I_0}}} = {{{T_1}{I_0}} \over {AY}} \times {{AY} \over {{T_2}{I_0}}} = {{{T_1}} \over {{T_2}}} $

$\Rightarrow {I_1}{T_2} - {I_0}{T_2} = {I_2}{T_1} - {I_0}{T_1}$

$ \Rightarrow {I_0} = {{{I_1}{T_2} - {I_2}{T_1}} \over {{T_2} - {T_1}}} = {{{T_1}{I_2} - {T_2}{I_1}} \over {({T_1} - {T_2})}}$

$\beta = {{\Delta p} \over {{{\Delta V} \over V}}}$

$ \Rightarrow $ $\beta = {{\Delta \rho gh} \over {{{\Delta V} \over V}}} = {{1000 \times 9.8 \times 2 \times {{10}^3}} \over {{{1.36} \over {100}}}}$

$ \Rightarrow $ $\beta$ = 1.44 $\times$ 10⁹ N/m²

${P_1} - {P_0} = {{4S} \over b}$ ....(1)

${P_2} - {P_0} - {{4S} \over a}$ .....(2)

${P_2} - {P_1} = {{4S} \over R}$ ......(3)

eq(2) - eq(1) = eq(3)

$ \Rightarrow $ ${1 \over a} - {1 \over b} = {1 \over R}$

$ \therefore $ $R = {{ab} \over {b - a}}$

The nature of flow is determined by reynolds no

$R = {{\rho VD} \over \eta }$

If R < 1000 $ \to $ flow is steady

1000 < R < 2000 $ \to $ flow becomes unsteady

R > 2000 $ \to $ flow is turbulent

${R_1} = {{4 \times {{10}^3} \times 0.18 \times {{10}^{ - 3}}} \over {60 \times \pi \times {{10}^{ - 2}} \times {{10}^{ - 3}}}} = {{4 \times {{10}^5} \times 0.18} \over {60\pi }}$

$ = 0.0038 \times {10^5} = 380$

${R_2} = {{0.48} \over {0.18}} \times 380 = 1018$

P = P₀ + h$\rho$g = 3 $\times$ 10⁵ Pa

$ \Rightarrow $ h$\rho$g = 3 $\times$ 10⁵ $-$ 1 $\times$ 10⁵

$ \Rightarrow $ h$\rho$g = 2 $\times$ 10⁵

$ \therefore $ 2h$\rho$g = 4 $\times$ 10⁵

$ \therefore $ P' = P₀ + 4 $\times$ 10⁵

$ \therefore $ P' = 5 $\times$ 10⁵ Pa

$ \therefore $ % increase in pressure = ${{P' - P} \over P} \times 100$

$ = {{(5 - 3) \times {{10}^5}} \over {3 \times {{10}^5}}} \times 100$

$ = {{200} \over 3}$%

Assuming Hooke's law to be valid.

$T \propto (\Delta l)$

$T = k(\Delta l)$

Let, l₀ = natural length (original length)

$ \Rightarrow T = k(l - {l_0})$

so, ${T_1} = k({l_1} - {l_0})$ & ${T_2} = k({l_2} - {l_0})$

$ \Rightarrow {{{T_1}} \over {{T_2}}} = {{{l_1} - {l_0}} \over {{l_2} - {l_0}}}$

$ \Rightarrow {l_0} = {{{T_2}{l_1} - {T_1}{l_2}} \over {{T_2} - {T_1}}}$

R is the radius of bigger drop.

r is the radius of n water drops.

Water drops are combined to make bigger drop.

So,

Volume of n drops = volume of bigger drop

$n\left( {{4 \over 3}\pi {r^3}} \right) = {4 \over 3}\pi {R^3}$

$ \Rightarrow $ $R = r{n^{1/3}} \Rightarrow n = {\left( {{R \over r}} \right)^3}$

Loss in surface energy, $\Delta$U = T $ \times $ (Change in surface area)

$\Delta$U = T (n4$\pi$r² $-$ 4$\pi$R²)

$\Delta U = 4\pi T\left[ {{{\left( {{R \over r}} \right)}^3}{r^2} - {R^2}} \right] = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over J}$

$ \therefore $ ${{\Delta U} \over V} = {{4\pi T\left( {{{{R^3}} \over r} - {R^2}} \right)} \over {J \times {4 \over 3}\pi {R^3}}} = {{3T} \over J}\left[ {{1 \over r} - {1 \over R}} \right]$

Bulk modulus $K = {{ - \Delta P} \over {{{\Delta v} \over v}}} = {{ - \Delta Pv} \over {\Delta v}}$

We know, $\rho = {M \over V}$

So, ${{ - \Delta \rho } \over \rho } = {{\Delta v} \over v}$

$K = {{ - \Delta P} \over {\left( { - {{\Delta \rho } \over \rho }} \right)}} = {{\rho \Delta P} \over {\Delta \rho }}$

$\Delta \rho = {{\rho \Delta P} \over K}$

$\Delta \rho = {{\rho P} \over K}$

We know that,

$Y = 3K(1 - 2\sigma )$

$ \Rightarrow \sigma = {1 \over 2}\left( {1 - {Y \over {3K}}} \right)$ ..... (i)

Also, $Y = 2\eta (1 + \sigma )$

$ \Rightarrow \sigma = {Y \over {2\eta }} - 1$ .... (ii)

On comparing Eqs. (i) and (ii), we get

$\left( {1 - {Y \over {3K}}} \right){1 \over 2} = {Y \over {2\eta }} - 1$

On solving, we get

$K = {{\eta Y} \over {9\eta - 3Y}}$ N/m²

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

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

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

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

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

From Bernoulli's equation,

P + ${1 \over 2}\rho {v^2}$ = ${P \over 2} + {1 \over 2}\rho {V^2}$

$ \Rightarrow $ V = $\sqrt {{P \over \rho } + {v^2}} $

An elastic wire can be treated as a spring with

k = ${{YA} \over l}$

T = $2\pi \sqrt {{m \over k}} $

$ \Rightarrow $ f = ${1 \over {2\pi }}\sqrt {{k \over m}} $ = ${1 \over {2\pi }}\sqrt {{{YA} \over {ml}}} $

After falling through h, the velocity be equal to terminal velocity

$\sqrt {2gh} $ = ${2 \over 9}{{{r^2}g} \over \eta }\left( {{\rho _l} - \rho } \right)$

$ \Rightarrow $ h = ${2 \over {81}}{{{r^4}g{{\left( {{\rho _l} - \rho } \right)}^2}} \over {{\eta ^2}}}$

$ \Rightarrow $ h $ \propto $ r⁴

${4 \over 3}\pi \left( {{R^3} - {r^3}} \right){p_m}\,g = {4 \over 3}\pi {R^3}{p_w}\,g$

$1 - {\left( {{r \over R}} \right)^3} = {8 \over {27}}$

$ \Rightarrow {r \over R} = {\left( {{{19} \over {27}}} \right)^{1/3}} = {{{{19}^{1/3}}} \over 3}$

$ = 0.88 \simeq {8 \over 9}$

Bulk Modulus, B = $\left( - \right){{\Delta P} \over {\Delta V/V}} $

$\Delta P = -\left( {{{\Delta V} \over V}} \right).B$

$ = -{{3\Delta L} \over L} \times B$

$ \therefore $ $|{{\Delta L} \over L}| = {{\Delta P} \over {3B}}$

$ \therefore $ % change, ${{\Delta L} \over L} \times 100\% $

= ${1 \over 3}{{\Delta P} \over B} \times 100$

= ${{4 \times {{10}^9}} \over {8 \times {{10}^{10}}}} \times 100$

= ${1 \over {60}} \times 100$ = 1.67

${u_i} = \left[ {dS{x_1}.{{{x_1}} \over 2} + dS{x_2}.{{{x_2}} \over 2}} \right]g\left\{ {dS{x_1} \to m,\,{{{x_1}} \over 2} \to h(C.O.M)} \right\}$

Here total volume remains same.

$ \therefore $ V_i = V_f

$ \Rightarrow $ S(x₁ + x₂) = S(h + h)

$ \Rightarrow $ h = ${{{x_1} + {x_2}} \over 2}$

u_f = (dSh)g${h \over 2} \times 2$

$ \Rightarrow $ ${u_f} = \left[ {dS\left( {{{{x_1} + {x_2}} \over 2}} \right) \times \left( {{{{x_1} + {x_2}} \over 4}} \right) \times 2} \right]g$

$ \therefore $ ${u_i} - {u_f} = dsg\left[ {{{x_1^2} \over 2} + {{x_2^2} \over 2} - {{{{\left( {{x_1} + {x_2}} \right)}^2}} \over 4}} \right]$

$ = dsg{{{{\left( {{x_1} - {x_2}} \right)}^2}} \over 4}$

B - mg = ma

$ \Rightarrow $ m = ${B \over {g + a}}$

= ${{V{\rho _w}g} \over {g + a}}$

= ${{V{\rho _w}} \over {1 + {a \over g}}}$

= ${{{4 \over 3}\pi {{\left( 1 \right)}^3} \times 1} \over {1 + {{9.8} \over {980}}}}$

= 4.15 gm

${{\Delta T} \over {\Delta t}} = k\left( {{{{T_f} + {T_i}} \over 2} - {T_0}} \right)$

$ \Rightarrow $ ${{50 - 40} \over {300}} = k\left( {{{90} \over 2} - 20} \right)$

$ \Rightarrow $ ${{40 - T} \over {300}} = k\left( {{{40 + T} \over 2} - 20} \right)$

$ \Rightarrow $ ${{10} \over {40 - T}} = {{25 \times 2} \over {40 + T - 40}}$

$ \Rightarrow $ ${1 \over {40 - T}} = {5 \over T}$

$ \Rightarrow $ $T = 200 - 5T$

$ \Rightarrow $ $6T = 200$

$ \Rightarrow $ $T = 33^\circ C$