Properties Of Matter Practice Questions

Given, breaking stress of wires 1 and 2 are $\sigma _{1}=\sigma _{2}=1.25 \times 10^{9} {N} / {m}^{2}$Cross-sectional area of wire 1 ,$A_{1}=8 \times 10^{-7} {m}^{2}$Cross-sectional area of wire 2 ,$A_{2}=4 \times 10^{-7} {m}^{2}$Mass hanging from first wire, $m_{1}=20 {kg}$ Mass hanging from second wire,$m_{2}=10 {kg}$Acceleration due to gravity, $g=10 {ms}^{-2}$Let $m$ be the maximum mass placed in pan without breaking the wire.According to free body diagram of given figure, Since, $stress(\sigma )=\;\frac{\;\text{ Tension }\;(T)}{Area(A)}$ $\therefore \;\; T_{1}=\sigma A_{1}$ $=1.25 \times 10^{9} \times 8 \times 10^{-7}$ $=10.00 \times 10^{2}=1000 {N}$and $T_{2}=\sigma {A}_{2}$ $=1.25 \times 10^{9} \times 4 \times 10^{-7}$ $=5.00 \times 10^{2}=500 {N}$By using concept of tension in string$T_{2}=(10+ {m}) {g}$ $\Rightarrow \;\; 500=(10+m) 10$ $\Rightarrow \;\; m=50-10=40 {kg}$

By newton's law of cooling (with approximation) $\frac{\Delta{T}}{\Deltat}=-{C}({T}_{\text{arg }}-{T}_{{s}})$ $1^{\text{st }} \frac{-10°{C}}{5 {min}}=-{C}(70°{C}-25°{C})$ $\Rightarrow {C}=\frac{2}{45} {min}^{-1}$ $2^{\text{nd }} \frac{{T}-65}{5 {min}}=-{C}(\frac{{T}+65}{2}-25)$ $=-(\frac{2}{45})(\frac{{T}+15}{2})$ $\Rightarrow 9({T}-65)=-({T}+15)$ ⇒ $10 {T}=570$ \Rightarrow ${T}=57°{C}$

Given, pressure on submarine at a certain depth, $p_{i}=3 \times 10^{5} {Pa}$ Since, we know that at a certain depth or below the surface of a liquid of density $\rho$, pressure is given as $p_{i}=p_{0}+\rho g h$ $\Rightarrow \;\; 3 \times 10^{5}=p_{0}+\rho g h$ $\Rightarrow \;\; 3 \times 10^{5}=1 \times 10^{5}+\rho g h$ $( \because p_{0}=1 \times 10^{5} {Pa})$ $\Rightarrow \;\; 3 \times 10^{5}-1 \times 10^{5}=\rho g h$ $\Rightarrow \;\; 2 \times 10^{5}=\rho g h$....(i) When the depth is doubled, then final pressure will be $p_{f}=p_{0}+\rho g 2 h=1 \times 10^{5}+2(\rho g h)$ $=1 \times 10^{5}+2(2 \times 10^{5})$ [using Eq. (i)] $=1 \times 10^{5}+4 \times 10^{5}$ $\Rightarrow \;\; p_{f}=5 \times 10^{5} {Pa}$ $\therefore$ Percentage increase in pressure $=\;\frac{p_{f}-p_{i}}{p_{i}} \times 100=\;\frac{(5-3) \times 10^{5}}{3 \times 10^{5}} \times 100=\;\frac{200}{3} \%$

The given situation is shown below $r_{\text{A}}=6 {\text{cm}}$ $r_{\text{B}}=3 {\text{cm}}$ Excess pressure inside the soap bubble is given by $(\Delta\text{p})_{{\text{excess} }}={4 \text{S}}/{r_{\text{A}}}+{4 \text{S}}/{r_{\text{B}}}$ According to the given question, ${4 \text{S}}/{r_{\text{A}}}+{4 \text{S}}/{r_{\text{B}}}={4 \text{S}}/{\text{R}_{{\text{eq}}}}$ $\Rightarrow {1}/{r_{\text{A}}}+{1}/{r_{\text{B}}}={1}/{\text{R}_{\text{eq}}}$ $\Rightarrow \frac{1}{6}+\frac{1}{3}={1}/{\text{R}_{\text{eq}}}$ $\Rightarrow \frac{1+2}{6}={1}/{\text{R}_{{\text{eq}}}}$ $\Rightarrow {1}/{\text{R}_{{\text{eq}}}}=\frac{1}{2}$ $\Rightarrow \text{R}_{{\text{eq}}}=2 {\text{cm}}$

According to Pascal's law, $\;\frac{F_{1}}{A_{1}}=\;\frac{F_{2}}{A_{2}}$ Initially, $\;\; \;\frac{100 {g}}{A_{1}}=\;\frac{m g}{A_{2}}$ . . . (i) Finally, $\;\; \;\frac{M g}{16 A_{1}}=\;{m g}/{(\;\frac{A_{2}}{16})}$ . . . (ii) On dividing Eqs. (i) by (ii), we get $\;\frac{100 \times 16}{M}=\;\frac{1}{16}$ $M=25600 {kg}$

At terminal speed ${a}=0$ ${F}_{{net}}=0$ ${mg}={F}_{{v}}=6 \pi \eta {Rv}$ ${v}=\frac{{mg}}{6 \pi \eta {Rv}}$ ${v}={\rho _{{w}} \frac{4 \pi }{3} {R}^{3} {g}}/{6 \pi \eta {R}}$ $=\frac{2 \rho _{w} R^{2} g}{9 \eta }$ $=\frac{400}{81} {m} / {s}$ $=4.94 {m} / {s}$

no. of moles is conserved ${n}_{1}+{n}_{2}={n}_{3}$ ${P}_{1} {V}_{1}+{P}_{2} {V}_{2}={P}_{3} {V}$ $\frac{4 {S}}{{r}_{1}}(\frac{4}{3} \pi {r}_{1}^{3})+\frac{4 {S}}{{r}_{2}}(\frac{4}{3} \pi {r}_{2}^{3})$ $=\frac{4 {S}}{{r}_{3}}(\frac{4}{3} \pi {r}_{3}^{3})$ ${r}_{1}^{2}+{r}_{2}^{2}={r}_{3}^{2}$ ${r}_{3}=\\sqrt{{r}_{1}^{2}+{r}_{2}^{2}}$

Given, radius of oil drop, $r=2.0 \times 10^{-5} {\text{m}}$ Density of oil drop, $\rho =1.2 \times 10^{3} {\text{kgm}}^{-3}$ Viscosity of liquid, $\eta =1.8 \times 10^{-5} {\text{Nsm}}^{-2}$ To neglect buoyancy, consider the density of air, ρa = 0 Viscous force acting on drop can be given as $F=6 \pi \eta \text{rV}$ ...(i) Here, v is terminal velocity. The terminal velocity of drop can be given as $v=\frac{2 r^{2}(\rho -\rho _{a}) g}{9 \eta }$ $\Rightarrow v=\frac{2 \times (2.0 \times 10^{-5})^{2}(1.2 \times 10^{3}-0) \times 9.8}{9 \times 1.8 \times 10^{-5}}$ $=5.807 \times 10^{-2} {\text{ms}}^{-1}$ Substituting all values in Eq. (i), we get $\text{F}=6 \pi \times 1.8 \times 10^{-5} \times 2.0 \times 10^{-5} \times 5.807 \times 10^{-2}$ $=3.94 \times 10^{-10} {\text{N}}$ $\simeq 3.9 \times 10^{-10} {\text{N}}$

We know that, $Y=3 K(1-2 \sigma )$ $\Rightarrow \;\; \sigma =\;\frac{1}{2}(1-\;\frac{Y}{3 K})$ Also, $\;\; Y=2 η(1+\sigma )$ $\Rightarrow \;\; \sigma =\;\frac{Y}{2 η}-1$ ... (ii) On comparing Eqs. (i) and (ii), we get $(1-\;\frac{Y}{3 K}) \;\frac{1}{2}=\;\frac{Y}{2 η}-1$ On solving, we get $K=\;\frac{η Y}{9 η-3 Y} {N} / {m}^{2}$

Let $I_{0}$ be the original length, $A$ be the area of cross-section, $\alpha$ be the coefficient of linear expansion, $\Delta l$ be the change in length and $Y$ be the Young's modulus of elasticity. As, $\;\; I_{1}=I_{0}(1+\alpha \Delta T)$ $\Rightarrow \;\; I_{1}-I_{0}=I_{0} \alpha \Delta T \Rightarrow \Delta I=I_{0} \alpha \Delta T$ Initially, $Y=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}=\;\frac{T / A}{\Delta l / I_{0}}$ $\Rightarrow \;\; Y=\;\frac{T_{1} / A}{(I_{1}-I_{0}) / I_{0}}$ . . . (i) Finally, $Y=\;\frac{T_{2} / A}{(I_{2}-I_{0}) / I_{0}}$ . . . (ii) Now, from Eqs. (i) and (ii), we get $\;\;\frac{T_{1} / A}{(I_{1}-I_{0}) / I_{0}} \;\; =\;\frac{T_{2} / A}{(I_{2}-I_{0}) / I_{0}}$ $\Rightarrow \;\;\frac{T_{1}}{I_{1}-I_{0}}= \;\; \;\frac{T_{2}}{I_{2}-I_{0}}$ $\Rightarrow \;T_{1} I_{2}-T_{1} I_{0} \;\; =T_{2} I_{1}-T_{2} I_{0}$ $\Rightarrow \;T_{1} I_{2}-T_{2} I_{1} \;\; =(T_{1}-T_{2}) I_{0}$ $\Rightarrow \;I_{0}=\;\frac{T_{1} I_{2}-T_{2} I_{1}}{T_{1}-T_{2}}$ $\;\text{ or }\;\;I_{0}=\;\frac{T_{2} I_{1}-T_{1} I_{2}}{T_{2}-T_{1}}$

When two soap bubbles of radii a and $b(b>a)$ coalesce, it can be shown as followsPressure inside the bubble $1, p_{1}=p_{0}+\;\frac{4 T}{a}$ where, $p_{0}$ is the initial pressure, $T$ is the surface tension and $a$ is the radius of the bubble $1 .$ Similarly, pressure inside the bubble $2, p_{2}=p_{0}+\;\frac{4 T}{b}$ where, $b$ is the radius of the bubble 2 . Excess pressure at common surface is given by $p_{\;\text{ex }\;}=p_{1}-p_{2}$ Let, $r$ be the radius of common surface, then $\Rightarrow \;\; \;\frac{4 T}{r}=(p_{0}+\;\frac{4 T}{a})-(p_{0}+\;\frac{4 T}{b})$ $\Rightarrow \;\; \;\frac{4 T}{r}=p_{0}+\;\frac{4 T}{a}-p_{0}-\;\frac{4 T}{b}$ $\Rightarrow \;\; \;\frac{4 T}{r}=\;\frac{4 T}{a}-\;\frac{4 T}{b} \Rightarrow \;\frac{1}{r}=\;\frac{1}{a}-\;\frac{1}{b} \Rightarrow \;\frac{1}{r}=\;\frac{b-a}{a b}$ $\Rightarrow r=\;\frac{a b}{b-a}$, which is the required expression for radius of curvature at common surface.

Given, radius of small drop $=r$ Radius of big drop $=R$ Surface tension $=T$ and mechanical equivalent of heat $=J$ As, small drops combine to form big drop. $\therefore$ Volume of big drop $(V_{B})=n \times$ Volume of small drop $(V_{S})$ $\Rightarrow \;\; \;\frac{4}{3} \pi R^{3}=n \cdot \;\frac{4}{3} \pi r^{3}$ $\Rightarrow \;\; n r^{3}=R^{3}$ $\Rightarrow \;\; r=\;\frac{R}{n^{1 / 3}}$. . . (i) Surface energy of small drop $(E_{S})=$ Surface tension $(T) \times$ Area $(A)$ $\Rightarrow \;\; E_{S}=n \times 4 \pi r^{2} T$ and $E_{B}=4 \pi R^{2} T$ Now, change in energy will be $\Delta E=E_{B}-E_{S}=4 \pi T(n r^{2}-R^{2})$ $\therefore$ Heat energy per unit volume $=\;\frac{\Delta E}{V}=\;\frac{4 \pi T(n r^{2}-R^{2})}{J \times \;\frac{4}{3} \pi R^{3}}$ $=\;\frac{3 T}{J}(\;\frac{n r^{2}}{R^{3}}-\;\frac{1}{R})=\;\frac{3 T}{J}(n \;\frac{R^{2}}{n^{2 / 3} R^{3}}-\;\frac{1}{R})$ =\;\frac{3 T}{J}\[;\frac{n^{1 / 3}}{R}-;\frac{1}{R}]$;;$[from Eq. (i)]$=;\frac{3 T}{J}[;\frac{1}{r}-;\frac{1}{R}]$

As we know that the nature of flow of water is determined by Reynold's number, $R_{ {e}}=\;\frac{\rho v D}{η}$ Here, $\rho =10^{3} {kgm}^{-3}, r=0.5 {cm}=0.5 \times 10^{-2} {m}$ $D=2 r=1 \times 10^{-2} {m}$ $η=10^{-3} {Pa}- {s}$ $Q_{1}=0.18 \frac{L}{min}=\;\frac{0.18 \times 10^{-3}}{60} {m}^\frac{3}{s}$ $Q_{2}=0.48 {L} {/} \min =\;\frac{0.48 \times 10^{-3}}{60} {m}^\frac{3}{s}$ Also, $v=\;\frac{Q}{\pi r^{2}} \therefore v_{1}=\;\frac{Q_{1}}{\pi r^{2}}$ Also, $v=\;\frac{Q}{\pi r^{2}} \therefore v_{1}=\;\frac{Q_{1}}{\pi r^{2}}$ $\therefore R_{ {e}_{\;\text{initial }\;}}=\;\frac{\rho v_{1} D}{η}=\;\frac{\rho Q_{1} D}{\pi r^{2} η}$ $=\;\frac{10^{3} \times 0.18 \times 10^{-3}}{\pi \times (0.5 \times 10^{-2})^{2} \times 60} \times \;\frac{1 \times 10^{-2}}{10^{-3}}$ $=381.97 ∼eq 382$ and $R_{ {e}_{\;\text{final }\;}}=\;\frac{\rho v_{2} D}{η}=\;\frac{\rho Q_{2} D}{\pi r^{2} η}$ $=\;\frac{10^{3} \times 0.48 \times 10^{-3} \times 1 \times 10^{-2}}{\pi \times (0.5 \times 10^{-2})^{2} \times 60 \times 10^{-3}}$ $=1019.09$ We know that, When $R_{ {e}}< 1000$, the flow of water is considered to be steady. When $1000< R_{ {e}}< 2000$, the flow of water becomes unsteady. When $R_{ {e}}>2000$, the flow of water becomes turbulent. As the $R_{e_{\;\text{final }\;}}$ value lies in between 1000 and 2000 and $R_{ {e}_{\;\text{initial }\;}}$ value is less than 1000 , so the nature of flow of water will be steady flow to the unsteady one.

$\frac{\Delta{T}}{\Delta{t}}={K}({T}_{{t}}-{T}_{{s}})$ ${T}_{{t}}=$ average temp ${T}_{{S}}=$ surrounding temp. $\frac{61-59}{4}={K}(\frac{61+59}{2}-30)$ .........(1) $\frac{51-49}{{t}}={K}(\frac{51+49}{2}-30)$ .........(2) Divide (1) & (2) $\frac{{t}}{4}=\frac{60-30}{50-30}=\frac{30}{20}$ so, ${t}=6$ minutes

Given, The change in the length of the wire $A, \Delta L_{A}=2 {mm}=0.002 {m}$ The change in the length of the wire $B, \Delta L_{B}=4 {mm}=0.004 {m}$ The force subjected to the wire, $F=2 {N}$ The radius of the wire $B$ is 4 times the radius of the wire $A$, i.e., $\;\frac{r_{B}}{r_{A}}=\;\frac{4}{1}$ Since, the wire is made of the same material, so the Young's modulus of the elasticity of the wire is same. $\Rightarrow \;\; Y_{A}=Y_{B}$ Using Hooke's law, Stress $=Y$ (Strain) $\;\frac{F}{A}=Y(\;\frac{\Delta L}{L})$ $\Rightarrow \;\; L=\;\frac{Y \Delta L A}{F}$ $\Rightarrow \;\; \;\frac{L_{A}}{L_{B}}=\;\frac{Y_{A}}{Y_{B}} \times \;\frac{\Delta L_{A}}{\Delta L_{B}} \times \;\frac{A_{A}}{A_{B}} \times \;\frac{F_{B}}{F_{A}}$ $\Rightarrow \;\; \;\frac{L_{A}}{L_{B}}=\;\frac{Y_{A}}{Y_{B}} \times \;\frac{0.002}{0.004} \times \;\frac{\pi r_{A}^{2}}{\pi r_{B}^{2}} \times \;\frac{2}{2}$ $\Rightarrow \;\; \;\frac{L_{A}}{L_{B}}=\;\frac{0.002}{0.004} \times \;\frac{r_{A}^{2}}{16 r_{A}^{2}}$ $\Rightarrow \;\; \;\frac{L_{A}}{L_{B}}=\;\frac{a}{b}=\;\frac{1}{32}$ Comparing this equation with $1 / x$, we get the value of the $x$ is 32 .

The given situation can be shown as below. Points A and B are at same horizontal level. So, the pressure at A and B will be equal i.e. $p_{\text{A}}=p_{\text{B}} ....$ (i) If $p_{0}$ is the atmospheric pressure, then from Bernoulli's theorem, $p_{\text{A}}=p_{0}-{2 \text{T}}/{r_{\text{A}}}+(h+y) \rho g$ $p_{\text{B}}=p_{0}-{2 \text{T}}/{r_{\text{B}}}+\rho g y$ Substituting the value of $p_{\text{A}}$ and $p_{\text{B}}$ in Eq. (i), we get $p_{0}-{2 \text{T}}/{r_{\text{A}}}+(h+y) \rho g=p_{0}-{2 \text{T}}/{r_{\text{B}}}+\rho g y$ $\Rightarrow 2 \text{T}({1}/{r_{\text{B}}}-{1}/{r_{\text{A}}})=\rho g y-(h+y) \rho g$ $\Rightarrow 2 \text{T}({1}/{r_{\text{B}}}-{1}/{r_{\text{A}}})=-\rho g h$ or$2 \text{T}({1}/{r_{\text{A}}}-{1}/{r_{\text{B}}})=\text{cg} h$ ...(ii) Here, $r_{\text{A}}=2.5 {\text{mm}}=2.5 \times 10^{-3} {\text{m}}$, and $\text{T}=7.3 \times 10^{-2} {\text{N}} / {\text{m}}$, $r_{\text{B}}=4 {\text{mm}}=4 \times 10^{-3} {\text{m}}, {g}=10 {\text{m}} / {\text{s}}^{2}$ and $\rho =1 \times 10^{3} {\text{kgm}}^{-3}$ Substituting the given value in Eq. (ii), we get $h=\frac{2 \times 7.3 \times 10^{-2}}{1 \times 10^{3} \times 10}(\frac{1}{2.5 \times 10^{-3}}-\frac{1}{4 \times 10^{-3}})$ $h=\frac{2 \times 7.3 \times 10^{-2}}{10^{-3} \times 10^{4}}(\frac{1}{2.5}-\frac{1}{4})$ $h=2.19 \times 10^{-3} {\text{m}}=2.19 {\text{mm}}$

${R}=\\sqrt{2 {gh}} \\times \\sqrt{\frac{(12-{h}) \\times 2}{{g}}}$ $\\sqrt{4 h(12-h)}=R$ For maximum ${R}$ $\frac{{d} {R}}{{dh}}=0$ $\Rightarrow{h}=6 {m}$

Let initial length and diameter be $l_{1}$ and $d_{1}$, whereas final length and diameter be $I_{2}$ and $d_{2}$. Given, $I_{2}=2 I_{1}, d_{2}=2 d_{1}, \Delta I_{1}=0.04 {m}$ By using formula of Young's modulus of elasticity, $Y \;\; =\;\frac{F \cdot l}{A \Delta l}\;$ $\because \;Y_{1} \;\; =Y_{2}$ $\Rightarrow \;\;\frac{F I_{1}}{A_{1} \times \Delta I_{1}}=\;\frac{F I_{2}}{A_{2} \times \Delta I_{2}}$ $\Rightarrow \;\;\frac{F I_{1}}{\pi (d_{1} / 2)^{2} \times 0.04}=\;\frac{F 2 I_{1}}{\pi (2 d_{1} / 2)^{2} \Delta I_{2}}$ $\Rightarrow \;\;\frac{1}{1 / 4 \times 0.04}=\;\frac{2}{\Delta I_{2}}$ $\Rightarrow \;\Delta I_{2}=0.02 {m}=2 {cm}$