Properties Of Matter Practice Questions

Given, breaking stress of wire, $\sigma =\frac{24}{\pi } \times 10^{2} {\text{Nm}}^{-2}$ Free body diagram of 5kg block is given as where, a is common acceleration. Value of acceleration due to gravity, $g=10 {\text{ms}}^{-2}$ From free body diagram of block of mass 5 kg 5g − T = 5a \Rightarrow 5 × 10 − T = 5a ⇒ 50 − T = 5a ...(i) Free body diagram of 3 kg block is given as From free body diagram of block of mass 3 kg, T − 3g = 3a ⇒ T - 3 ×10 = 3a ⇒ T − 30 = 3a ...(ii) Add Eqs. (i) and (ii), we get 50 − T + T − 30 = 5a + 3a ⇒ 20 = 8a ⇒ $a=2.5 {\text{ms}}^{-2}$ Substituting the value of a in Eq. (i), we get 50 − T = 5 × 2.5 \Rightarrow T = 37.5 N Let us assume the minimum radius of wire is r. The breaking stress is expressed as $\sigma =\frac{\text{T}}{\pi r^{2}}$ $\frac{24}{\pi } \times 10^{2}=\frac{37.5}{\pi r^{2}}$ \Rightarrow $r^{2}=\frac{37.5}{24 \times 10^{2}}=\frac{1}{64}$ \Rightarrow $r=\frac{1}{8} {\text{m}}=\frac{100}{8} {\text{cm}}=12.5 {\text{cm}}$ Thus, the minimum radius of wire should be 12.5 cm.

By energy conservation $\frac{1}{2} \.\frac{{YA}}{{L}} \.{x}^{2}=\frac{1}{2} {mv}^{2}$ $\frac{0.5 \\times 10^{9} \\times 10^{-6} \\times (0.04)^{2}}{0.1}$ $=\frac{20}{1000} {v}^{2}$ $\therefore {v}^{2}=400$ ${v}=20 {m} / {s}$

In series combination $\Delta{l}=l_{1}+l_{2}$ ${Y}=\frac{{F} / {A}}{\Deltal/ l} \Rightarrow\Deltal=\frac{{F} l}{{AY}}$ $\Rightarrow\Deltal\propto\frac{l}{{Y}}$ Equivalent length of rod after joing is $=2 l$ As, lengths are same and force is also same in series $\Deltal=\Deltal_{1}+\Deltal_{2}$ {l_{{cq}}}/{{Y}_{{cq}}}=\frac{l}{{Y}_{1}}+\frac{l}{{Y}_{2}} \ $\Rightarrow \frac{2 l}{{Y}}=\frac{l}{{Y}_{1}}+\frac{l}{{Y}_{2}}$ $\therefore {Y}=\frac{2 {Y}_{1} {Y}_{2}}{{Y}_{1}+{Y}_{2}}$

Given, The volumetric strain is $\;\frac{\Delta V}{V}=1.36 \%$ The depth beneath the water surface, $h=2 {km}$ The pressure inside the water surface up to $2 {km}$, $p=\rho g h$ Substituting the values in the above equation, we get $p=1000 \times 9.81 \times 2000$ $p=19.62 \times 10^{6} {Pa}$ The bulk modulus of the object, $\beta =\;\frac{p}{\frac{\Delta V}{V}}$ Substituting the values in the above equation, we get $\beta =\;\frac{19.62 \times 10^{6}}{\frac{1.36}{100}}$ $\beta =1.44 \times 10^{9} {N} / {m}^{2}$ Hence, the ratio of the hydraulic stress to the corresponding hydraulic strain will be $1.44 \times 10^{9} {N} / {m}^{2}$.

Given, decrease in volume $(\Delta\text{V} / \text{V})=-\frac{0.5}{100}$ Bulk modulus of rubber, $\text{B}=9.8 \times 10^{8} {\text{Nm}}^{-2}$ Density of sea water, $\rho =10^{3} {\text{kgm}}^{-3}$ Acceleration due to gravity, $g=9.8 {\text{ms}}^{-2}$ Let h be the depth at which ball is dipped. Since, $\text{B}=\frac{\Delta\text{p}}{-\Delta\text{V} / \text{V}}$ where, $\Delta\text{p}$ is change in pressure $=\rho \text{gh}$ $\therefore \rho \text{gh}=-\text{B} \frac{\Delta\text{V}}{\text{V}}$ $\Rightarrow h=-\frac{1}{\rho g} \text{B} \frac{\Delta\text{V}}{\text{V}}$ $\Rightarrow h=-\frac{1}{10^{3} \times 9.8} \times 9.8 \times 10^{8} \times (-\frac{0.5}{100})$ $=0.5 \times 10^{8-3-2}=0.5 \times 10^{3}=500 {\text{m}}$

Given, cross-sectional area of the tank, $A=0.4 {m}^{2}$ Cross-sectional area of the opening at $B$, $A=4 {cm}^{2}=1 \times 10^{-4} {m}^{2}$ Load applied on the top of the tank, $m=24 {kg}$ Height of the water level, $h=40 {cm}$ From the figure,Using the Bernoulli's principle equation, $p_{0}+\;\frac{m g}{A}+\rho g h+\;\frac{\rho v_{A}^{2}}{2}=p_{0}+\;\frac{\rho v_{B}^{2}}{2}$ $\Rightarrow \;\; 0+\;\frac{(24) 10}{0.4}+1000 \times 10 \times 0.4+\;\frac{\rho (0)^{2}}{2}=0+\;\frac{1000 \times v_{B}^{2}}{2}$ $v_{B}=3 \frac{m}{s}$ Hence, the value of $v$ to the nearest integer is 3 .

Given, the radius of the spherical drop $r=(\;\frac{3}{40 \pi })^{1 {/} 3} \times 10^{-3} {cm}$ The volume of 100 spherical drops, $V=100 \times \;\frac{4 \pi r^{3}}{3}$ Substituting the value of the radius in the above equation, we get $V=100 \times \;\frac{4 \pi }{3} \times \;\frac{3}{40 \pi } \times 10^{-9}$ $V=10^{-8} {cm}^{3}$ Now, Volume of the film $=$ Area of the film $\times$ Thickness of the film $\Rightarrow$ Volume of the film $=4 \times t$ The film is made up of 100 spherical drops. $\therefore$ The volume of the film $=$ Volume of 100 spherical drops $4 t=10^{-8} \Rightarrow t=25 \times 10^{-12} {m}$ The thickness of the oleic layer, $t_{0}=0.01 \times t=25 \times 10^{-14} {m}$ Comparing with $x \times 10^{-14}$, we get $x=25$

${T}_{1}={k}(l_{1}-l_{0})$ ${T}_{2}={k}(l_{2}-l_{0})$ $\frac{{T}_{1}}{{T}_{2}}=\frac{l_{1}-l_{0}}{l_{2}-l_{0}}$ $\frac{{T}_{1} l_{2}-{T}_{2} l_{1}}{{T}_{1}-{T}_{2}}=l_{0}$

Given, the Young's modulus of the steel rod, Y = 2 ×$10^11$ Pa Thermal coefficient of the steel rod, α = $10^{-5}$ºC The length of the steel rod, I = 4 m The area of the cross-section, A = 10 $\text{cm}^2$ The temperature difference, ΔT = 400°C As we know that, Thermal strain = α ΔT Using the Hooke's law Young's modulus $(\text{Y})={{ \text{Thermal} \text{stress} }}/{{ \text{Thermal} \text{strain} }}=\frac{\text{F} / \text{A}}{\alpha \Delta\text{T}}$ Thermal stress, F = YA α Δ T Substitute the values in the above equation, we get $\text{F}=2 \times 10^{11} \times 10 \times 10^{-4} \times 10^{-5} \times (400)$ $=8 \times 10^{5} {\text{N}}$ Comparing with, $\text{F}=x \times 10^{5} {\text{N}}$ The value of the x = 8.

Let inner and outer radii of hollow cylindrical column are r and R, respectively. Given that, r = 50 cm, R = 100 cm Mass supported on four columns, $\text{M}=50 \times 10^{3} {\text{kg}}$ Mass supported on each column, $m=\frac{\text{M}}{4}$ $\Rightarrow m=\frac{50 \times 10^{3}}{4}=12.5\times 10^{3} {\text{kg}}$ Now, weight, $w=m g=12.5 \times 9.8 \times 10^{3} {\text{N}}=1225 \times 10^{5} {\text{N}}$ Area of cross-section of each column $\text{A}=\pi (\text{R}^{2}-r^{2})$ $=3.14\{(100)^{2}-(50)^{2}\} \times 10^{-4} {\text{m}}^{2}=2.35 {\text{m}}^{2}$ Young's modulus, $\text{Y}=2.0 \times 10^{11} {\text{Pa}}$ By using Hooke's law, Stress = Y × Strain ∴ Compressive strain $={{ \text{Stress} }}/{\text{Y}}=\frac{\text{W}}{\text{AY}}$ Substituting the values, we get Compressive strain $=\frac{1.225 \times 10^{5}}{2.35 \times 2.0 \times 10^{11}}=2.60 \times 10^{-7}$

$\frac{4}{3} \pi R^{3}+\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R^{3}$ $R'=2^{\frac{1}{3}} R$ .......(i) $A_{i}=2[4 \pi R^{2}]$ $A_{f}=4 \pi R'^{2}$ $\frac{U_{i}}{U_{f}}=\frac{A_{i}}{A_{f}}=\frac{2 R^{2}}{2^{2 / 3} R^{2}}=2^{1 / 3}$

Applying Bernoulli's Equation ${P}_{1}+\frac{1}{2} \rho {v}_{1}^{2}+\rho {g} {y}_{1}$ $={P}_{2}+\frac{1}{2} \rho {v}_{2}^{2}+\rho {g} {y}_{2}$ ${P}+\frac{1}{2} \rho {v}^{2}=\frac{{P}}{2}+\frac{1}{2} \rho {\text{V}}^{2}$ $\frac{2 {P}}{2 \rho }+\frac{1}{2} \frac{\rho {v}^{2}}{\rho } \times 2={\text{V}}^{2}$ $\sqrt{\frac{{P}}{{\rho }}+{v}^{2}}=\text{V}$

$A_1v_1=A_2v_2$ ${v_{\min}/v_{\max}}=A_{\min}/A_{\max}$ $=(408/6.4)^2=9/16$

${U}_{ {i}}= (\rho {Sx}_{1} ) {g}.\frac{{x}_{1}}{2}$ $+ (\rho {Sx}_{2} ) {g} .\frac{{x}_{2}}{2}$ ${U}_{ {f}}= (\rho {Sx}_{ {f}} ) {g}.{{x}_{ {f}}}/{2} \\times 2$ By volume conservation ${Sx}_{1}+ {Sx}_{2}= {S} (2 {x}_{ {f}} )$ ${x}_{ {f}}=\frac{{x}_{1}+ {x}_{2}}{2}$ $\Delta{U}=\rho {Sg}[(\frac{{x}_{1}^{2}}{2}+\frac{{x}_{2}^{2}}{2} ) - {x}_{ {f}}^{2} ]$ $=\rho {Sg}[\frac{x_{1}^{2}}{2}+ \frac{x_{2}^{2}}{2}-(\frac{x_{1}+x_{2}}{2} )^{2} ]$ $= \frac{\rho {Sg}}{2}[\frac{x_{1}^{2}}{2}+ \frac{x_{2}^{2}}{2}-x_{1} x_{2} ]$ $= \frac{\rho {Sg}}{4}\ (x_{1}-x_{2} )^{2}$

$\frac{4}{3} \pi ({R}^{3}- {r}^{3} ) \rho _{ {m}} {g}= \frac{4}{3} \pi \ {R}^{3} \rho _{ {w}} {g}$ $1-(\frac{{r}}{ {R}} )^{3}= \frac{8}{27}$ $\Rightarrow \frac{{r}}{ {R}}=(\frac{19}{27} )^{1 / 3}= \frac{19^{1 / 3}}{3}$ $=0.88≃\frac{8}{9}$

$\rho _{ {\text{wire} }}=9 \\times 10^{-3} \frac{{kg}}{\ {cm}^{3}}$ $= \frac{9 \\times 10^{-3}}{10^{-6}} \ {kg} / {m}^{3}$ $=9000 {kg} / {m}^{2}$ (A = CSA of wire) (Y = $9 \\times 10^{10} \ {Nm}^{2}$ ) (Strain $=4.9 \\times 10^{-4}$ ) $\Rightarrow {L}=1 \ {m}= \frac{\lambda}{2} \Rightarrow \lambda=2 \ {m}$ $\Rightarrow {v}=\ {f} \lambda \Rightarrow \\sqrt{\frac{{T}}{\mu}}=\ {f} \lambda$ Where $Y=\frac{{T} / {A}}{\text{strain }} \Rightarrow {T}= {Y} .$ A. strain $\Rightarrow \sqrt{ \frac{\text{ Y.A. strain}}{ {m} / {L}}}= {f} \\times 2$ $\Rightarrow \sqrt{ \frac{\text{ Y.A.L. strain }}{{M}}}= {f} \\times 2$ $\Rightarrow \sqrt{\frac{{Y} \times {V}\times \text{ strain }}{ {M}}}={f}\times 2$ $\Rightarrow \sqrt{\frac{{Y} \times \text{ strain }}{\rho }}= {f} \\times 2$ ${f}= \frac{1}{2}.\sqrt{{{Y}\times {\text{strain}}}/{\rho }}$ $= \frac{1}{2} \\sqrt{ \frac{9 \\times 10^{10} \\times 4.9 \\times 10^{-4}}{9000}}$ ${f}= \frac{1}{2}.\sqrt{ \frac{9 \\times 10^{3}}{9} \\times 4.9}$ $=\ \frac{1}{2} \\sqrt{4900}=\ \frac{70}{2}=35 \ {Hz}$

After falling through h, the velocity be equal to terminal velocity $\sqrt{2 {gh}}= \frac{2}{9} \frac{{r}^{2} { g}}{\eta } (\rho _{l}-\rho )$ $\Rightarrow {h}= \frac{2}{81} \frac{{r}^{4} { g} (\rho _{l}-\rho )^{2}}{\eta ^{2}}$ $\Rightarrow {h} \propto{r}^{4}$

$m = \rho _0A (80)$...(ii)$m = \rho _0A (79)$ ....(ii)