Properties Of Matter Practice Questions

$\;\;\text{ Change \in pressure }\;=\;\frac{1}{2} \rho v^2$ $\;4.5 \times 10^{4-}2.0 \times 10^4=\;\frac{1}{2} \times 10^3 \times {\text{v}}^2$ $\;2.5 \times 10^4=\;\frac{1}{2} \times 10^3 \times {\text{v}}^2$ $\; {\text{v}}^2=50$ $\; {\text{v}}={\sqrt{50}}$ $\;\;\text{ Velocity of water }\;=\sqrt{ {\text{V}}}={\sqrt{50}}$ $\;= {\text{V}}=50$

$\; {\text{V}}_{ {\text{T}}}=\;{2 {\text{g}}}/{9} \;{ {\text{R}}^2 [\rho _{ {\text{B}}}-\rho _{ {\text{L}}}] }/{η}$ $\;\Rightarrow {\text{V}}_{ {\text{T}}}=\;\frac{2}{9} \times \;\frac{10 \times (10^{-4}) ^2}{9.8 \times 10^{-6}} [10^{5-}10^3]$ $\;\Rightarrow {\text{V}}_{ {\text{T}}}=224.5$when ball fall from height (h)$\; {\text{V}}=\sqrt{2 {\text{gh}}}$ $\; {\text{h}}= (\;{ {\text{V}}^2}/{2 {\text{g}}}) =2518 {\text{m}}$

$\; {\text{v}}_1=\;\text{ volume immersed \in water. }\;$ $\; {\text{v}}_2=\;\text{ volume immersed \in oil. }\;$ $\; {\text{v}}_1 \rho _{ {\text{w}}} {\text{g}}+ {\text{v}}_2 \rho _{ {\text{o}}} {\text{g}}= ( {\text{v}}_1+ {\text{v}}_2) \rho _{ {\text{c}}} {\text{g}}$ $\; {\text{v}}_1+\;{ {\text{v}}_2 \rho _{ {\text{o}}}}/{\rho _{ {\text{w}}}}= ( {\text{v}}_1+ {\text{v}}_2) \;{\rho _{ {\text{c}}}}/{\rho _{ {\text{w}}}}$ $\;= {\text{v}}_1+0.8 {\text{v}}_2=0.9 {\text{v}}_1+0.9 {\text{v}}_2$ $\;=0.1 {\text{v}}_1=0.1 {\text{v}}_2$ $\; {\text{v}}_1: {\text{v}}_2=1: 1$

$\; {\text{T}}= {\text{mg}}$ $\;\sigma =\;{ {\text{T}}}/{ {\text{A}}}=\;{ {\text{mg}}}/{ {\text{A}}}$ $\;\;{(\sigma {\text{A}} ℓ) {\text{g}}}/{ {\text{A}}}$ $\;\Rightarrow ℓ=\;{\sigma }/{\rho {\text{g}}}=\;\frac{1.2 \times 10^8 \times 3}{6 \times 10^4 \times 10}=600$

Surface tension will be less as temperature increases${\text{h}}=\;{2 {\text{T}} \cos \theta }/{\rho {\text{gr}}}$Height of capillary rise will be smaller in hot water and larger in cold water.

$\; {\text{B}}=-\;{\Delta {\text{P}}}/{ (\;{\Delta {\text{V}}}/{ {\text{V}}}) }$ $\;- (\;{\Delta {\text{V}}}/{ {\text{V}}}) =\;{\rho {\text{gh}}}/{ {\text{B}}}$ $=\;\frac{1000 \times 10 \times 4000}{2 \times 10^9}$ $\;=2 \times 10^{-2}[- {\text{ve}} \;\text{ sign represent compression }\;]$

$\;{\Delta {\text{Y}}}/{ {\text{Y}}}\;=\;{\Delta {\text{m}}}/{ {\text{m}}}+\;\frac{\Delta ℓ}{ℓ}$ $\;=\;\frac{5}{500}+\;\frac{0.02}{2}=0.01+0.01$ $\;{\Delta {\text{Y}}}/{ {\text{Y}}}\;=0.02 \Rightarrow \% \;{\Delta {\text{Y}}}/{ {\text{Y}}}=2 \%$

Since density is negligible hence Buoyancy force will be negligibleAt terminal velocity.${\text{Mg}}\;=6 \pi η {\text{rv}}$ ${\text{V}}\;\propto \;{1}/{ {\text{r}}} \;\; \;\text{ (as mass is constant) }\;$Now, $\;{ {\text{v}}}/{ {\text{v}}^{'}}=\;{ {\text{r}}^{'}}/{ {\text{r}}}$ ${\text{r}}^{'}=2 {\text{r}}$So, ${\text{v}}^{'}=\;{ {\text{v}}}/{2}$

Volume constant$\;\;\frac{4}{3} \pi {\text{R}}^3=27 \times \;\frac{4}{3} \times \pi {\text{r}}^3$ $\; {\text{R}}^3=27 {\text{r}}^3$ $\; {\text{R}}=3 {\text{r}}$ $\; {\text{r}}=\;{ {\text{R}}}/{3}$ $\; {\text{r}}^2=\;{ {\text{R}}^2}/{9}$Work done $= {\text{T}} . \Delta {\text{A}}$ $\;=27 {\text{T}} (4 \pi {\text{r}}^2) - {\text{T}} 4 \pi {\text{R}}^2$ $\;=27 {\text{T}} 4 \pi \;{ {\text{R}}^2}/{9}-4 \pi {\text{R}}^2 {\text{T}}$ $\;=8 \pi {\text{R}}^2 {\text{T}}$

$\; {\text{P}}_1- {\text{P}}_0=3 ( {\text{P}}_2- {\text{P}}_0)$ $\;\;{4 {\text{T}}}/{ {\text{r}}_1}=3 \;{4 {\text{T}}}/{ {\text{r}}_2}$ $\; {\text{r}}_2=3 {\text{r}}_1$ $\;\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;{\;\frac{4}{3} \pi {\text{r}}_1^3}/{\frac{4}{3} \pi {\text{r}}_2^3}=\;\frac{1}{27}$

In vertical derection$2 {\text{T}} sin \;\theta =20$ using small angle approximation $sin \;\theta =\theta$ $\;\theta =\;\frac{1}{100}$ $\; \therefore \;\; {\text{T}}=\;\frac{10}{\theta }$ $\; {\text{T}}=1000 {\text{N}}$ $\;\text{ Change \in length }\; \Delta {\text{L}}\;=2 \sqrt{ {\text{x}}^{2+} {\text{L}}^2}-2 {\text{L}}$ $\;=2 {\text{L}} [1+\;{ {\text{x}}^2}/{2 {\text{L}}^2}-1]$ $\Delta {\text{L}}\;=\;{ {\text{x}}^2}/{ {\text{L}}}$ $\therefore \;\text{ Modulus of elasticity }\;=\;\frac{\;\text{ stress }\;}{\;\text{ strain }\;}$ $\;2 \times 10^{11}=\;{10^3}/{ {\text{A}} \times \;{ {\text{x}}^2}/{ {\text{L}}}} \times 2 {\text{L}}$ $\therefore \;\; {\text{A}}=1 \times 10^{-4} {\text{m}}^2$

$\; {\text{m}}=\;\text{ mass of small drop }\;$ $\; {\text{M}}=\;\text{ mass of bigger drop }\;$ $\; {\text{V}}_{ {\text{t}}}=\;\frac{2}{9} \;{ {\text{R}}^2(\rho -\sigma ) {\text{g}}}/{η}$ $\;8 \propto {\text{m}}= {\text{M}}$ $\;8 {\text{r}}^3= {\text{R}}^3 \Rightarrow {\text{R}}=2 {\text{R}}$ $\;\;\text{ as }\; {\text{V}}_{ {\text{t}}} \times {\text{R}}^2 \;\; \because \;\text{ Radius double so }\; {\text{V}}_{ {\text{t}}} \;\text{ becomes }\; 4 \;\text{ time }\;$ $\; \therefore 4 \times 10=40 {\text{cm}} / {\ s}$

$\; {\text{mg}}- {\text{F}}_{ {\text{B}}}- {\text{F}}_{ {\text{v}}}= {\text{ma}}$ $\; {\text{a}}=0 \;\text{ for constant velocity }\;$ $\; {\text{mg}}- {\text{F}}_{ {\text{B}}}= {\text{F}}_{ {\text{v}}}$ $\; {\text{F}}_{ {\text{v}}}= {\text{mg}}- {\text{v}} \rho _0 {\text{g}}= {\text{mg}}-\;{ {\text{m}}}/{\rho } \rho _0 {\text{g}}= {\text{mg}} (1-\;\frac{\rho _0}{\rho })$

$\;\beta =\;{-\Delta {\text{P}}}/{{\Delta {\text{V}}}/{ {\text{V}}}}$ $\;\Delta {\text{P}}=-\beta \;{\Delta {\text{V}}}/{ {\text{V}}}$ $\;\rho {\text{gh}}=-\beta \;{\Delta {\text{V}}}/{ {\text{V}}}$ $\;10^3 \times 10 \times {\text{h}}=-9 \times 10^8 \times (-\;\frac{0.02}{100})$ $\;\Rightarrow {\text{h}}=18 {\text{m}}$

Lets say radius of small droplets is ${\text{r}}$ and that of big drop is ${\text{R}}$ $\;\;\frac{4}{3} \pi {\text{R}}^3=1000 \;\frac{4}{3} \pi {\text{r}}^3$ $\; {\text{R}}=10 {\text{r}}$ $\; {\text{U}}_{ {\text{i}}}=1000 (4 \pi {\text{r}}^2 {\text{S}})$ $\; {\text{U}}_{ {\text{f}}}=4 \pi {\text{R}}^2 {\text{S}}$ $\;=100 (4 \pi {\text{r}}^2 {\text{S}})$ $\; {\text{U}}_{ {\text{f}}}=\;\frac{1}{10} {\text{U}}_{ {\text{i}}}$

${\text{A}}=80 {\text{m}}^2$Using Bernonlli equation$\; {\text{A}} ( {\text{P}}_2- {\text{P}}_1) =\;\frac{1}{2} \rho ( {\text{V}}_1^{2-} {\text{V}}_2^2) {\text{A}}$ $\; {\text{mg}}=\;\frac{1}{2} \times 1 (70^{2-}50^2) \times 80$ $\; {\text{mg}}=40 \times 2400$ $\; {\text{m}}=9600 {\text{kg}}$

$\;\rho ({ }_3^4 \pi r^3) 1000={ }_3^4 \pi R^3 \rho$ $\;R=10 r$ $\;E_1=1000 \times 4 \pi r^2 \times S$ $\;E_2=4 \pi (10 r)^2 S$ $\;\;\frac{E_1}{E_2}=\;\frac{10}{1}, x=10$

$1000 \;\frac{4}{3} \pi {\text{r}}^3\;=\;\frac{4}{3} \pi {\text{R}}^3$ $10 {\text{r}}\;= {\text{R}}$ ${\text{R}}\;=10 {\text{r}}$ $\;\frac{\;\text{ S.E. of }\; 1000 \;\text{ drops }\;}{\;\text{ S.E. of Big drop }\;}\;=\;{1000 (4 \pi {\text{r}}^2) {\text{T}}}/{4 \pi {\text{R}}^2 {\text{T}}}$ $\;=\;{1000 \times {\text{r}}^2}/{(10 {\text{r}})^2}=10=\;{10}/{ {\text{x}}}$ $\therefore \;\; x\;=1$