Properties Of Matter Practice Questions

$\;Y=\;\frac{F / A}{\Delta L / L}=\;\frac{F L}{A N}=\;{(500 {\text{N}})(3)}/{\pi \times 9 \times 10^{-6} \times 1 \times 10^{-4}}$ $\;=5 \times 10^{11} {\text{N}} / {\text{m}}^2$

$\;v=\;\frac{2}{9} \;\frac{r^2 g(\sigma -\rho )}{η}$ $\;η=\;\frac{2}{9} \times \;\frac{1.8 \times 1.8 \times 10^{-6} \times 10}{2.45 \times 10^{-2}}\{7825-925\}$ $\;\; \approx 2$

Using Bernoulli's theorem$\;P_1-P_2=\;\frac{1}{2} \rho v^2$ $\;v \propto {\sqrt{P_1-P_2}}$

$\;\;\text{ We know }\; \;\frac{\Delta P}{\Delta V / V}=-\beta$ $\;\Rightarrow \Delta P=-\beta \cdot \;\frac{\Delta V}{V}$ $\;\Rightarrow |\Delta P|=2.15 \times 10^9 \times \;\frac{2}{1000}=4.3 \times 10^6 N / m^2$ $\;\Rightarrow \Delta P=43 \times 10^5 N / m^2$

$\; {\text{Y}}=\;\frac{\;\text{ stress }\;}{\;\text{ strain }\;}$ $\; {\text{Y}}=\;{\;{ {\text{F}}}/{{\pi {\text{r}}^2}/{ℓ}}}/{ {\text{L}}}$ $\; {\text{F}}= {\text{Y}} \pi {\text{r}}^2 \times \;{ℓ}/{ {\text{L}}}$ .............(i) $\; {\text{Y}}=\;{\;{ {\text{F}} / 2}/{\frac{\pi }{2} / 4}}/{{\Delta ℓ}/{ {\text{L}}}}$ $\; {\text{F}}= {\text{Y}} \;{\Delta ℓ}/{ {\text{L}}} \times 2 \times \;{\pi {\text{r}}^2}/{4}$ From (i) $\; {\text{Y}} \pi {\text{r}}^2 \;{ℓ}/{ {\text{L}}}= {\text{Y}} \;{\Delta ℓ}/{ {\text{L}}} \;{\pi {\text{r}}^2}/{2}$ $\;\Delta ℓ=2 ℓ$

Young's modulus depends on the material not length and cross sectional area. So young's modulus remains same.

$\;\rho {\text{gh}}=\;{4 {\text{S}}}/{ {\text{R}}}$ $\;\Rightarrow {\text{R}}=\;\frac{4 \times 0.28}{8 \times 10^3 \times 10 \times 4 \times 10^{-4}}$ $\;\Rightarrow \;\frac{0.28}{8} {\text{m}}=\;\frac{28}{8} {\text{cm}}$ $\;\Rightarrow {\text{R}}=3.5 {\text{cm}}$ $\;\;\text{ Diameter }\;=7 {\text{cm}}$

$\; {\text{mg}}- {\text{F}}_{ {\text{B}}}- {\text{F}}_{ {\text{v}}}= {\text{ma}}$ $\; (\rho \;\frac{4}{3} \pi {\text{r}}^3) {\text{g}}- (\rho _{ {\text{L}}} \;\frac{4}{3} \pi {\text{r}}^3) {\text{g}}$ $-6 \pi η {\text{rv}}= {\text{m}} \;{ {\text{dv}}}/{ {\text{dt}}}$ $\;\;\text{ Let }\; \;{4}/{3 {\text{m}}} \pi {\text{R}}^3 {\text{g}} (\rho -\rho _{ {\text{L}}}) = {\text{K}}_1$ $\;\text{ and }\; \;{6 \pi η {\text{r}}}/{ {\text{m}}}= {\text{K}}_2$ $\;\;{ {\text{dv}}}/{ {\text{dt}}}= {\text{K}}_1- {\text{K}}_2 {\text{v}}$ $\;∫_{0}^{ {\text{v}}} \;{ {\text{dv}}}/{ {\text{K}}_1- {\text{K}}_2 {\text{v}}}=∫_{0}^{}{ {\text{t}}} {\text{dt}}$ $\;-\;{1}/{ {\text{K}}_2} \ln [ {\text{K}}_1- {\text{K}}_2 {\text{v}}] _0^{ {\text{v}}}= {\text{t}}$ $\;ℓ n (\;{ {\text{K}}_1- {\text{K}}_2 {\text{v}}}/{ {\text{K}}_1}) =- {\text{K}}_2 {\text{t}}$ $\; {\text{K}}_1- {\text{K}}_2 {\text{v}}= {\text{K}}_1 {\text{e}}^{- {\text{K}}_2 {\text{t}}}$ ${\text{v}}=\;{ {\text{K}}_1}/{ {\text{K}}_2} [1- {\text{e}}^{- {\text{K}}_2 {\text{t}}}]$

$\;\Delta {\text{L}}=\;{ {\text{FL}}}/{ {\text{AY}}}$ $\;\;{\Delta {\text{L}}}/{ {\text{L}}}=\;{ {\text{F}}}/{ {\text{AY}}}$ $\;\;{\;{\Delta {\text{L}}_1}/{ {\text{L}}_1}}/{{\Delta {\text{L}}_2}/{ {\text{L}}_2}}=\;{ {\text{F}}_1}/{ {\text{F}}_2}=\;\frac{30}{10}=3$

Gases have less viscosity. Due to insoluble impurities like detergent surface tension decreases

$\;\;\frac{F_1}{A_1}=\;\frac{F_2}{A_2}$ $\;\;\frac{F_1}{\pi (7)^2}=\;\frac{10}{\pi \times (0.7)^2}$ $\;F_1=1000 {\text{N}}$

${\text{P}}+\rho {\text{gh}}+\;\frac{1}{2} \rho {\text{V}}^2=$ constant

Applying Bernoulli's equation${\text{P}}_1+\rho {\text{gh}}_1+\;\frac{1}{2} \rho {\text{v}}_1^2= {\text{P}}_2+\rho {\text{gh}}_2+\;\frac{1}{2} \rho v_2^2$ [h_1 & h_2. are height of point from any reference level]Given ${\text{V}}_1= {\text{V}}_2=0$ (for statement- 1 )$\therefore {\text{P}}_1- {\text{P}}_2=\rho {\text{g}} ( {\text{h}}_2- {\text{h}}_2)$For statement-2${\text{P}}_1+\;\frac{1}{2} \rho v_1^2= {\text{P}}_2+\;\frac{1}{2} \rho v_2^2$ $\; {\text{P}}_1- {\text{P}}_2=\rho {\text{gh}}$ $\; {\text{P}}_1- {\text{P}}_2=\;\frac{1}{2} \rho {\text{v}}_2^{2-}\;\frac{1}{2} \rho {\text{v}}_1^2$ $\;\rho {\text{gh}}=\;\frac{1}{2} \rho {\text{v}}_2^{2-}\;\frac{1}{2} \rho {\text{v}}_1^2$ $\;2 {\text{gh}}= {\text{v}}_2^{2-} {\text{v}}_1^2$Hence answer (4)

Capillary rise${\text{h}}=\;{2 {\text{T}} \cos \theta }/{\rho {\text{gr}}}$If $\theta =0^{\circ}$ then rise is non-zero$\therefore$ Statement-1 is incorrect.Option(1) is correct

Conceptual questions

${\text{F}}=\;\frac{1}{2} \rho ( {\text{v}}_1^{2-} {\text{v}}_2^2) {\text{A}}$ ${\text{F}}=\;\frac{1}{2} \times 1.2 \times (70^{2-}65^2) \times 2$ $=810 {\text{N}}$

$\;\;{ {\text{F}}}/{ {\text{A}}}= {\text{Y}} \;\frac{\Delta ℓ}{ℓ} \Rightarrow \Delta ℓ=\;{ {\text{F}} ℓ}/{ {\text{AY}}}$ $\;\Delta ℓ=\;\frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}}=2 \times 10^{-5}=20 \mu {\text{m}}$

$\; {\text{Y}}=\;\frac{\;\text{ Stress }\;}{\;\text{ Strain }\;}=\;{ {\text{F}} / {\text{A}}}/{\Delta ℓ / ℓ}=\;{ {\text{F}} ℓ}/{ {\text{A}} \Delta ℓ}$ $\;\Delta ℓ=\;{ {\text{F}} ℓ}/{ {\text{AY}}}$ $\; {\text{V}}= {\text{A}} ℓ \Rightarrow ℓ=\;{ {\text{V}}}/{ {\text{A}}}$ $\;\Delta ℓ=\;{ {\text{FV}}}/{ {\text{A}}^2 {\text{Y}}}$ {\text{Y}} & {\text{V}} is same for both the wires $\;\Delta ℓ \propto \;{ {\text{F}}}/{ {\text{A}}^2}$ $\;\;\frac{\Delta ℓ_1}{\Delta ℓ_2}=\;{ {\text{F}}_1}/{ {\text{A}}_1^2} \times \;{ {\text{A}}_2^2}/{ {\text{F}}_2}$ $\;\Delta ℓ_1=\Delta ℓ_2$ $\; {\text{F}}_1 {\text{A}}_2^2= {\text{F}}_2 {\text{A}}_1^2$ $\;\;{ {\text{F}}_1}/{ {\text{F}}_2}=\;{ {\text{A}}_1^2}/{ {\text{A}}_2^2}= (\;\frac{4}{1}) ^2=16$