Properties Of Matter Practice Questions

${B}=- {\Delta{P}}/{ \frac{\Delta{V}}{ {V}}}$ $|\frac{\Delta{V}}{{V}} |= \frac{\Delta{P}}{ {B}}$ $=\frac{4 \\times 10^{9}}{8 \\times 10^{10}}=\ \frac{1}{20}$ $\frac{\Deltal}{l}= \frac{1}{3} \times \frac{{\DeltaV}}{ {V}}=\frac{1}{60}$ Percentage change $= \frac{\Deltal}{l}\times 100 \%$ $= \frac{100}{60} \%=1.67 \%$

$\ {r} \to$ radius of capillary $\ {R} \to$ Radius of meniscus. From figure, $\frac{{r}}{ {R}}=\cos 30°$ ${R}= \frac{2 {r}}{\\sqrt{3}}= \frac{2 \\times 0.15 \\times 10^{-3}}{\\sqrt{3}}$ $= \frac{0.3}{\\sqrt{3}} \\times 10^{-3} \ {m}$ Height of capillary ${h}= \frac{2 {T}}{\rho {gR}}=2 \\sqrt{3} {T}$ $h= {2 \\times 0.05}/{667 \\times 10 \times ( \frac{0.3 \\times 10^{-3}}{\\sqrt{3}} )}$ ${h}=0.087 \ {m}$

$\frac{50-40}{300}=\beta (\frac{50+40}{2}-20 )$ $\frac{40- {T}}{300}=\beta (\frac{40+{T}}{2}-20 )$ $\therefore {T}= \frac{100}{3}$

Capillary rise ${h}= {2 {S} \cos \theta }{\rho {gr}}$ ${S}= {\rho {grh}}{2 \cos \theta }$ $=\ \frac{(900)(10) (15 \\times 10^{-5} ) (15 \\times 10^{-2} )}{2}$ $\ {S}=1012.5 \\times 10^{-4}$ $\ {S}=101.25 \\times 10^{-3}=101.25 \ {mN} / {m}$ In question closest integer is asked so closest integer $=101.00$

$\frac{\text{Energy} \text{stored} }{\text{Volume}}=1/2 (\text{Stress})^2/Y$ $u_1/u_2=1/4\Rightarrow 4u_1=u_2$ 4 1/{2Y}\[\frac{W. 4}{πd_1^2}]^2=1/{2Y}$[\frac{W . 4}{\pi d_2^2}]^2$$4=(d_1/d_2)^4$$\Rightarrow d_1/d_2=\sqrt2:1$

Rate of flow of water $= A_AV_A = A_BV_B$ $(40) V_A = (20)V_B$ $V_B = 2V_A$ ......(1)Using Bernoulli's theorem$P_A+1/2 \rho V_A^2=P_B+1/2\rho V_B^2$ $P_A-P_B=1/2 \rho (V_B^{2-}V_A^2)$ $700=1/2\times 1000(4V_A^{2-}V_A^2)$ $V_A$ = 0.68 m/s = 68 cm/s Rate of flow $= A_AV_A$ = (40) (68) = 2720 $cm^3$/s

$F_1= {\rho gh}/2\times A$ $F_2(\rho gh+{2\rho gh}/2)A$ $F_1/F_2=1/4$

Applying pressure equation from A to B ${P}_{0}+\rho .\frac{ {R} \omega ^{2}}{2}.{R}-\rho {gh}=\ {P}_{0}$ $\frac{\rho {R}^{2} \omega ^{2}}{2}=\rho {gh}$ ${h}=\frac{{R}^{2} \omega ^{2}}{2 {g}}=(5)^{2} \frac{\omega ^{2}}{2 {g}}$ $= \frac{25}{2} \frac{\omega ^{2}}{ {g}}$

FBD of droplet B + F = mg$B=(2/3\pi R^3)\rho g$ $F=T(2\pi R)$ $m=d(4/3\pi R^3)$ $(2/3\pi R^3)\rho g+T(2\pi R)=d(4/3\pi R^3)g$ $T(2\pi R)=(2/3\pi R^3)g[2d-\rho ]$ $R=\sqrt{ \frac{3T}{ (2d-\rho )g}}$

$\DeltaP_{1}=0.01=4 {T} / {R}_{1}$ .....(1) $\Delta{P}_{2}=0.02=4 {T} / {R}_{2}$ ......(2) Equation $(1)÷(2)$ $\frac{1}{2}= \frac{R_{2}}{R_{1}}$ $R_{1}=2 R_{2}$ $\frac{V_{1}}{V_{2}}= \frac{R_{1}^{3}}{R_{2}^{3}}= \frac{8 R_{2}^{3}}{R_{2}^{3}}=8$

Volume ${V}= \frac{4 \pi }{3} \ {r}^{3}$ $= \frac{4 \pi }{3} \\times (1)^{3}=4.19 \ {cm}^{3}$ $\ {a}=9.8 \ {cm} / {s}^{2}$ $\ {B}-\ {mg}=\ {ma}$ ${m}=\frac{{B}}{ {g}+ {a}}$ ${m}=\frac{({V} \rho _{\omega } {g} )}{ {g}+ {a}}={{V} \rho _{\omega }}/{1+ \frac{ {a}}{ {g}}}$ $=\ \frac{(4.19) \\times 1}{1+\ \frac{9.8}{980}}$ $=\ \frac{4.19}{1.01}=4.15 \ {gm}$

Elastic unit $=\ \frac{F}{A}$ $379\times 10^{6}= \frac{400}{\pi r^{2}}$ $r=0.57 \times 10^{-3} \ {m}$Diameter $=2 \times 0.57\times 10^{-3} {m}$ $=1.16 \ {mm}$

$V^{2}-U^{2}=2 g h$ $\Rightarrow V^{2}=3$ $\Rightarrow V=\\sqrt{3} \ {m} / {s}$From equation of continuity$\ {A} \ {V}=a v$ $A \\sqrt{3}=10^{-4} \\times (1)$ $\Rightarrow A=5.77\times 10^{-5} \ {m}^{2}$

Since height of water column is constanttherefore, water inflow rate $Q_{\in}$ = water outflow rate $Q_{\in}$ = $10^{-4} m^3 s^{-1}$ $Q_{\text{out}}$ = Au = $10^{-4} \times \sqrt{2gh}$ $10^{-4}$ = $10^{-4} \sqrt{20\times h}$ h = $1/{20}$ m h = 5 cm ∴ correct answer is (4)

$\int w o d=\ \frac{4}{5} \int$water$\ \frac{v}{2} \int _{\ { oil }} g+\ \frac{v}{2} \int _{\ {water}} g=\int _{\ {wood}} g$ $\Rightarrow∫$ oil $+\int$ water $=2\ (\int$ wood $)$ $\Rightarrow∫$ oil $+\int$ water $=\ \frac{8}{5} \int$ water$\Rightarrow\ \frac{\int { oil }}{\int {water}}=0.6$

For a soap bubble, the volume $V = \frac{4}{3} \pi r^3$ increases at a constant rate $\frac{dV}{dt} = k$, which implies $V \propto t$, and consequently $r \propto t^{1/3}$. The pressure inside the bubble is given by the excess pressure formula $P_{\in} = P_{atm} + \frac{4S}{r}$, where $S$ is the surface tension. Substituting $r \propto t^{1/3}$ into the pressure equation, we obtain $P_{\in} = P_{atm} + \frac{C}{t^{1/3}}$. This mathematical relationship shows that as time $t$ increases, the pressure $P_{\in}$ decreases and asymptotically approaches the atmospheric pressure $P_{atm}$. This corresponds to a curve that starts at a high value and decays towards a horizontal asymptote.

${Re}= \frac{\rho D V}{\eta }= \frac{10^{3} \times 10^{-1} \times 100\times 10^{-3}}{60\times \pi \times (10^{-2} ) \times 1\times 10^{-3}}\times 4$ $\Rightarrow \frac{4}{6 \pi } \times 10^{5}=2.1 \times 10^{4}$ Order of 4

$V_{T}=\ \frac{2}{9} \ \frac{r^{2}(p-\sigma ) g}{\eta }$ $V_{T} \propto r^{2}$ $R_{b i g}=(27)^{1 / 3} r_{\ {small}}$ $\Rightarrow R_{b i g}=3 r_{\ {small}}$ $\frac{v_{1}}{v_{2}}= (\ \frac{r_{1}}{r_{2}}\ )^{2}= \frac{R^{2}}{(R / 3)^{2}}=9$