Properties Of Matter Practice Questions

At terminal velocity, net force on the ball is Zero.$\;M g=f+B$ $\;\Rightarrow M g=f+V_{\;\text{ball }\;} \rho _o g \;\; . . .(i)$ $\;\;\text{ Volume of ball }\;=\;\frac{M}{\rho }$ $\;\;\text{ From eq (i), }\;$ $\;M g=f+\;\frac{M}{\rho } \rho _o g$ $\;\Rightarrow f=M g-\;\frac{M}{\rho } \rho _o g$ $\;\Rightarrow f=M g (1-\;\frac{\rho _o}{\rho })$

$\; {\text{A}}_1 {\text{V}}_1= {\text{A}}_2 {\text{V}}_2$ $\;750 \times 10^{-4} {\text{V}}_1=500 \times 10^{-6} \times 0.3$ $\; {\text{V}}_1=\;\frac{500 \times 3 \times 10^{-3}}{750} {\text{m}} / {\ s}$ $\;=2 \times 10^{-3} {\text{m}} / {\ s}$ $\;\;{ {\text{dh}}}/{ {\text{dt}}}=-2 \times 10^{-3} {\text{m}} / {\ s}$

$\;1000 \times \;\frac{4 \pi }{3}(1)^3=\;\frac{4 \pi }{3} {\text{R}}^3$ $\; {\text{R}}=10 {\text{mm}}$ $\; {\text{T}} \times 1000 \times 4 \pi (10^{-3}) ^{2-} {\text{T}} \times 4 \pi (10 \times 10^{-3}) ^2=\Delta {\text{E}}$ $\;\Delta {\text{E}}=4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}$ $\;\Delta {\text{E}}=7.92 \times 10^{-4} {\text{J}}$Option 2.

By equation of continuity$\; {\text{A}}_1 {\text{V}}_1= {\text{A}}_2 {\text{V}}_2$ $\; (2 {\text{cm}}^2) (4 {\text{cm}} / {\ s})= (10 \times 10^{-2} {\text{cm}}^2) ( {\text{v}})$ $\;\;{8 {\text{cm}}^3}/{ {\ s}}=10^{-1} {\text{cm}}^2( {\text{v}})$ $\; {\text{V}}=80 {\text{cm}} / {\ s}$

$\;\;{\Delta {\text{Q}}}/{\Delta {\text{t}}}=- {\text{K}} ( {\text{T}}- {\text{T}}_0)$ $\;\;{\Delta {\text{Q}}}/{\Delta {\text{t}}}=- {\text{K}} ( {\text{T}}_{ {\text{avg}}}- {\text{T}}_0)$(i) $\;{ {\text{ms}} \times 12}/{2}=- {\text{K}} (\;\frac{98+86}{2}-22)$ $\;6=-\;{ {\text{K}}}/{ {\text{ms}}} [\;\frac{98+86}{2}-22]$ $\;6=-\;{ {\text{K}}}/{ {\text{ms}}}[70] . . . . . . . . . . \;\text{ (i) }\;$(ii) $\;{ {\text{ms}} \times 6}/{\Delta {\text{t}}}=- {\text{K}} (\;\frac{75+69}{2}-22)$ $\;{6}/{\Delta {\text{t}}}=-\;{ {\text{K}}}/{ {\text{ms}}}(50)$ ............. (II)(ii) $÷$ (i)$\;\;{6}/{\Delta {\text{t}}(6)}=\;\frac{50}{70}$ $\;\Delta {\text{t}}=\;\frac{7}{5}=1.4 {\min}$

$\;\text{ (A) Surface Tension }\; =\;\frac{F}{ℓ}\;=\;{ {\text{MLT}}^{-2}}/{ {\text{L}}}= {\text{ML}}^{-1} {\text{T}}^{-2}$ $\;= {\text{Kgs}}^{-2}( {\text{IV}})$ $\;\;\text{ (B) Pressure }\;=\;\frac{F}{A}=\;{ {\text{MLT}}^{-2}}/{ {\text{L}}^2}$ $\;= {\text{kg}} {\text{m}}^{-1} {\ s}^{-2} \;\text{ (III) }\;$(C) $\;\text{ Viscosity }\;\;==\;{ {\text{F}}}/{ {\text{A}} (\;{ {\text{dV}}}/{ {\text{dz}}}) }=\;{ {\text{MLT}}^{-2}}/{ {\text{L}}^2 (\;{ {\text{LT}}^{-1}}/{ {\text{L}}}) }$ $\;= {\text{ML}}^{-1} {\text{T}}^{-1}= {\text{kg}} {\text{m}}^{-1} {\ s}^{-1}( {\text{I}})$ $\;\text{ (D) }\; \;\text{ Impulse }\;\;=∫ {\text{Fdt}}= {\text{MLT}}^{-2} \times {\text{T}}$ $\;= {\text{MLT}}^{-1}= {\text{Kgms}}^{-1} \;\text{ (II) }\;$ So A-(IV), B-(III), C-(I), D- (II)

Pressure at surface $= {\text{P}}_{ {\text{atm}}}=1 \times 10^5 {\text{Pa}}$ ${\text{v}}_{\;\text{surface }\;}=$ ?Pressure at ${\text{h}}=40 {\text{m}}$ depth$\; {\text{P}}= {\text{P}}_{ {\text{atm}}}+\rho {\text{gh}}$ $\; {\text{P}}=10^{5+}10^3 \times 10 \times 40$ $\; {\text{P}}=5 \times 10^5 {\text{Pa}}$ $\; {\text{V}}=1 {\text{cm}}^3$Temp. is constant$\; {\text{P}}_1 {\text{V}}_1= {\text{P}}_2 {\text{V}}_2$ $\;10^5 \times v=5 \times 10^5 \times 1$ $\;v=5 {\text{cm}}^3$

$\; {\text{Y}}=\;{ {\text{F}} / {\text{A}}}/{\frac{\Delta ℓ}{ℓ}}$ $\;\Rightarrow {\text{F}}=\;{ {\text{YA}}}/{ℓ} \Delta ℓ$ $\; (\;{ {\text{A}} \Delta ℓ}/{ℓ}) _1= (\;{ {\text{A}} \Delta ℓ}/{ℓ}) _2$ $\;\Rightarrow \;\frac{\Delta ℓ_2}{\Delta ℓ_1}=\;{ {\text{A}}_1}/{ {\text{A}}_2} \times \;\frac{ℓ_2}{ℓ_1}$ $\;\Rightarrow \;\frac{\Delta ℓ_2}{0.2}=\;\frac{1}{2.4 \times 2.4} \times \;\frac{2}{1}$ $\;\Rightarrow \Delta ℓ_2=6.9 \times 10^{-2} {\text{mm}}$

Tension $(F)=m g$ $\; \;\; =4 \times \;\frac{10}{4}=10 {\text{N}}$ $\;\Delta {\text{L}}=\;{ {\text{FL}}}/{ {\text{AY}}}$ $\;=\;\frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}}$ $\;=10^{-4} {\text{m}}=0.1 {\text{mm}}$

From graph:$\; {\text{F}}=\Delta {\text{L}}$ $\; {\text{Y}}=\;{ {\text{FL}}}/{ {\text{A}} \Delta {\text{L}}}$ $\; {\text{Y}}=\;{ {\text{L}}}/{ {\text{A}}}$ $\; {\text{Y}}=\;\frac{62.8 \times 10^{-2}}{\pi (2 \times 10^{-3}) ^2}$ $\;Y=5 \times 10^4 {\text{N}} / {\text{m}}^2$

$\;Y=3 η(1+\sigma )$ $\;Y=3 K(1-\sigma )$ $\;\Rightarrow 2 η(1+\sigma )=3 K(1-2 \sigma )$ $\;\Rightarrow \sigma = (\;\frac{3 K-2 η}{6 K+2 η})$

For uniform velocity net force $=0$ ${\text{B}}=6 \pi η r {\text{r}}$ $\;\rho \;\frac{4}{3} \pi r^3 g=6 \pi η r v$ $\;η=\;{2 \rho r^2 g}/{9 {\text{v}}}$ $\;η=\;\frac{2 \times 1750 \times (3 \times 10^{-3}) ^2 \times 10}{9 \times 0.35 \times 10^{-2}}$ $\;η=10 {\text{Pa}}- {\ s}$

Concept based

Elongation in wire $\delta =\;{ {\text{F}} ℓ}/{ {\text{AY}}}$ $\;\delta =\;\frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}}$ $\;\delta =4 \times 10^{-3} {\text{m}}$

Taking volume constant : $\;\frac{P_1}{T_1}=\;\frac{P_2}{T_2}$ $\Rightarrow P_2=\;{P_1}/{ {\text{T}}_1} \times {\text{T}}_2=\;\frac{270 \times (309)}{300}$ $=278 {\text{kPa}}$

${\text{A}}_1=10 {\text{cm}}^2$ $\Delta {\text{P}}= {\text{P}}_1- {\text{P}}_2=3 {\text{N}} / {\text{m}}^2 \;\text{ (given) }\;$By continuity equation$\; {\text{A}}_1 {\text{v}}_1= {\text{A}}_2 {\text{v}}_2$ $\; \therefore {\text{v}}_1=\;{ {\text{A}}_2}/{ {\text{A}}_1} {\text{v}}_2$ . . . . . (1) By bernoulli's equation$\;P_1+\;\frac{1}{2} \rho v_1{ }^2=P_2+\;\frac{1}{2} \rho v_2{ }^2$ $\;P_1-P_2=\;\frac{1}{2} \rho (v_2{ }^{2-}v_1{ }^2)$ $\;\Delta P=\;\frac{1}{2} \rho (v_2{ }^{2-}\;\frac{A_2^2}{A_1^2} v_2{ }^2)$ \;\Delta P=\;\frac{1}{2} \rho \[1- (;\frac{A_2}{A_1}) ^2]$ v_2^2$$;3=;\frac{1}{2} \times 1.25 \times 10^3 (1- (;\frac{5}{10}) ^2) v_2^2$$;3=;\frac{1}{2} \times 1.25 \times 10^3 [1-;\frac{1}{4}] v_2^2$$;3=;\frac{1}{2} \times 1.25 \times 10^3 \times ;\frac{3}{4} v_2^2$$\therefore {\text{v}}_2=8 \times 10^{-2} {\text{m}} / {\ s}$So discharge rate$= {\text{A}}_2 {\text{V}}_2$$;=5 \times 10^{-4} \times 8 \times 10^{-2}$$;=4 \times 10^{-5} {\text{m}}^3 / {\ s}$

Terminal velocity of a spherical body in liquid$\;V_t \propto r^2$ $\;\;\frac{\Delta V_t}{V_t}=2 \;\frac{\Delta r}{r}$ $\;\;\frac{\Delta V_t}{V_t} \times 100 \%=2 \times \;\frac{0.1}{5} \times 100=4 \%$ $\;\;\text{ Also, }\; V_t \propto r^2$ $\;\;\text{ Reason (R) is false }\;$

$\;B_{\;\text{water }\;}=\;\frac{-\Delta P}{ (\;\frac{\Delta V}{V}) }=\;\frac{-\Delta P}{\frac{0.01}{100}}$ $\;B_{\;\text{liquid }\;}=\;\frac{-\Delta P}{\frac{0.03}{100}}$ $\;\;{B_{\;\text{water }\;}}/{B_{\;\text{liquid }\;}}=3$ $\;x=1$