Properties Of Matter Practice Questions

We need to evaluate two statements about fluid mechanics. Statement I: "Pressure of a fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid." This is FALSE . By Pascal's law, pressure in a fluid at rest acts at every point within the fluid equally in all directions. Fluid pressure exists throughout the fluid volume, not just at solid surfaces. Statement II: "Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension." This is TRUE . Interior molecules are surrounded by neighbours on all sides (net force is zero), while surface molecules lack neighbours above, resulting in fewer attractive interactions and higher potential energy. The liquid minimizes this excess surface energy by minimizing its surface area - this tendency manifests as surface tension. The correct answer is Option (4) : Statement I is false but Statement II is true.

The terminal velocity $v_t$ of a sphere moving through a viscous fluid is given by Stoke's Law: $v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$.
Given $r = \frac{2 \, mm}{2} = 0.1 \, cm$, $\rho = 10.5 \, g/cm^3$, $\sigma = 1.5 \, g/cm^3$, $\eta = 10 \, Poise = 10 \, g/(cm \cdot s)$, and $g = 10 \, m/s^2 = 1000 \, cm/s^2$.
Substituting the values into the formula:
$v_t = \frac{2}{9} \times \frac{(0.1)^2 \times (10.5 - 1.5) \times 1000}{10}$
$v_t = \frac{2}{9} \times \frac{0.01 \times 9 \times 1000}{10}$
$v_t = \frac{2}{9} \times \frac{90}{10} = \frac{2}{9} \times 9 = 2.0 \, cm/s$
The terminal velocity is $2.0 \, cm/s$.

Equating dimensions on both sides of $t = A \rho^a r^b \eta^c \sigma^d$:
$[T] = \$[M L^{-3}]^a [L]^b \$[M L^{-1} T^{-1}]^c \$[M L^{-3}]^d = M^{a+c+d} L^{-3a+b-c-3d} T^{-c}$
Comparing the powers:
For $T$: $-c = 1 \implies c = -1$.
For $M$: $a+c+d = 0 \implies a+d = -c = 1$.
For $L$: $-3a + b - c - 3d = 0 \implies b = 3(a+d) + c = 3(1) - 1 = 2$.
The required ratio is $\frac{b+c}{a+d} = \frac{2-1}{1} = 1$.

The initial length of each rod is $L = \frac{120}{2} = 60 \text{ cm}$, and the temperature change is $\Delta T = 100^\circ C - 30^\circ C = 70^\circ C$.
The change in length for the aluminum rod is $\Delta L_{Al} = L \alpha_{Al} \Delta T = 60 \times (24 \times 10^{-6}) \times 70 = 0.1008 \text{ cm}$.
The change in length for the steel rod is $\Delta L_{St} = L \alpha_{St} \Delta T = 60 \times (1.2 \times 10^{-5}) \times 70 = 0.0504 \text{ cm}$.
The final length of the composite rod is $L_f = L_{total} + \Delta L_{Al} + \Delta L_{St} = 120 + 0.1008 + 0.0504$.
$L_f = 120.1512 \text{ cm} \approx 120.15 \text{ cm}$.

The capillary rise $h$ is determined by the formula $h = \frac{2T \cos \theta}{r \rho g}$, where $T$ is surface tension, $r$ is tube radius, $\rho$ is density, $\theta$ is the contact angle, and $g$ is gravity.
Since the same liquid is used in both cases, $T_1 = T_2 = T$ and the density $\rho$ is constant.
Assuming a consistent contact angle $\theta$, the relationship becomes $h \propto \frac{1}{r}$.
Given the radii $r_1 > r_2$, the inverse proportionality dictates that $h_1 < h_2$.
Therefore, both conditions $T_1 = T_2$ and $h_1 < h_2$ are satisfied.

For a cubical block floating in a liquid, the weight of the block must be balanced by the buoyant force: $\rho_b V_{total} g = \rho_{\ell} V_{submerged} g$.
Given the total volume $V_{total} = A H$ and submerged volume $V_{submerged} = A h$, where $A$ is the base area and $h$ is the submerged height, the equation simplifies to $\rho_b H = \rho_{\ell} h$.
Solving for the submerged height $h$, we get:
$h = \frac{\rho_b}{\rho_{\ell}} H$
Substituting the given values $\rho_b = 600 \, \text{kg/m}^3$, $\rho_{\ell} = 900 \, \text{kg/m}^3$, and $H = 8.0 \, \text{cm}$:
$h = \frac{600}{900} \times 8.0 \, \text{cm} = \frac{2}{3} \times 8.0 \, \text{cm} \approx 5.33 \, \text{cm}$
Rounding to one decimal place, the submerged height is $5.3 \, \text{cm}$.

The Young's modulus is given by $Y = \frac{FL}{A \Delta L}$. Given that the force $F$ and the extension $\Delta L$ are the same for both wires, the ratio of Young's moduli is $\frac{Y_A}{Y_B} = \frac{L_A A_B}{L_B A_A}$.

Substituting the given values ($L_A = 6.0 \text{ cm}$, $A_A = 3.0 \times 10^{-5} \text{ m}^2$, $L_B = 5.4 \text{ cm}$, $A_B = 4.5 \times 10^{-5} \text{ m}^2$):
$\frac{Y_A}{Y_B} = \frac{6.0 \times 10^{-2} \times 4.5 \times 10^{-5}}{5.4 \times 10^{-2} \times 3.0 \times 10^{-5}}$
Simplifying the terms, we get:
$\frac{Y_A}{Y_B} = \frac{6.0 \times 4.5}{5.4 \times 3.0} = \frac{27}{16.2} = \frac{5}{3}$
Since the ratio is given as $x:3$, comparing $\frac{Y_A}{Y_B} = \frac{x}{3}$ with $\frac{5}{3}$ gives $x = 5$.

Using Bernoulli's principle and the continuity equation for a horizontal tube, the pressure difference is related to the flow velocity as:
$P_A - P_B = \frac{1}{2} \rho (v_B^2 - v_A^2) = \rho g h \implies v_B^2 - v_A^2 = 2gh$
Using the continuity equation $Q = A_A v_A = A_B v_B$, we can express velocities as $v_A = \frac{Q}{A_A}$ and $v_B = \frac{Q}{A_B}$. Substituting these into the Bernoulli result:
$Q^2 \left( \frac{1}{A_B^2} - \frac{1}{A_A^2} \right) = 2gh$
Given $A_A = 6 \text{ cm}^2$, $A_B = 3 \text{ cm}^2$, $h = 5 \text{ cm}$, and $g = 1000 \text{ cm/s}^2$:
$Q^2 \left( \frac{1}{3^2} - \frac{1}{6^2} \right) = 2 \times 1000 \times 5 \implies Q^2 \left( \frac{1}{9} - \frac{1}{36} \right) = 10000$
$Q^2 \left( \frac{4-1}{36} \right) = 10000 \implies Q^2 \left( \frac{3}{36} \right) = 10000 \implies Q^2 = 120000$
$Q = \sqrt{120000} = 200 \sqrt{3} \text{ cm}^3/\text{s}$

The tension $T$ generated in a wire fixed between two rigid supports due to a temperature change $\Delta \theta$ is given by the formula $T = Y A \alpha \Delta \theta$, where $Y$ is Young's modulus, $A$ is the cross-sectional area, and $\alpha$ is the coefficient of thermal expansion. Since $Y, A,$ and $\alpha$ are constant, we have $T \propto \Delta \theta$.

Initially, cooling from $27^\circ C$ to $-43^\circ C$ gives $\Delta \theta_1 = 27 - (-43) = 70^\circ C$, resulting in tension $T$. For a tension of $1.4 T$, the required temperature difference $\Delta \theta_2$ is:
$\Delta \theta_2 = 1.4 \times \Delta \theta_1 = 1.4 \times 70^\circ C = 98^\circ C$
Let the final temperature be $T_f$. The change is $\Delta \theta_2 = 27^\circ C - T_f$:
$27 - T_f = 98 \implies T_f = 27 - 98 = -71^\circ C$

The terminal velocity $v_t$ of a spherical drop is proportional to the square of its radius: $v_t \propto r^2$.
By conservation of volume, $64 \cdot (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$, which gives the radius $R$ of the larger drop as $R = (64)^{1/3}r = 4r$.
The ratio of terminal velocities is $\frac{v_2}{v_1} = \frac{R^2}{r^2} = \left(\frac{4r}{r}\right)^2 = 16$.
Therefore, the terminal velocity of the bigger drop is $v_2 = 16 \cdot v_1 = 16 \cdot 10 \text{ cm/s} = 160 \text{ cm/s}$.

Young's modulus $Y$ is defined as the ratio of stress to strain, $Y = \frac{\text{stress}}{\text{strain}}$.
In the given plot, strain is on the y-axis and stress is on the x-axis, so the slope $m$ of the curve is $m = \frac{\text{strain}}{\text{stress}} = \frac{1}{Y}$.
Rearranging for Young's modulus, we get $Y = \frac{1}{m}$, which indicates that $Y$ is inversely proportional to the slope of the line.
To maximize the Young's modulus $Y$, the slope $m$ must be minimized.
Visually inspecting the plot, the line for material $C$ is the closest to the horizontal axis (stress axis), meaning it has the smallest slope.
Therefore, material $C$ has the largest Young's modulus.

The height of the capillary rise is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$.
Assuming the contact angle $\theta$ and gravity $g$ are constant, the height is proportional to $h \propto \frac{T}{r \rho}$.
Using the relative error formula, the fractional change in $h$ is given by $\frac{\Delta h}{h} = \frac{\Delta T}{T} - \frac{\Delta r}{r} - \frac{\Delta \rho}{\rho}$.
Given that $T, r,$ and $\rho$ each decrease by $1 \%$, we have $\frac{\Delta T}{T} = -0.01$, $\frac{\Delta r}{r} = -0.01$, and $\frac{\Delta \rho}{\rho} = -0.01$.
Substituting these values: $\frac{\Delta h}{h} = (-0.01) - (-0.01) - (-0.01) = -0.01 + 0.01 + 0.01 = 0.01$.
Therefore, the height of the liquid in the tube increases by $1 \%$.

The work $W$ done in expanding a soap bubble (having two surfaces) is given by $W = T \times \Delta A = T \times 8\pi(r_2^2 - r_1^2)$.
Substituting the given values $T = 0.04 \, N/m$, $r_1 = 0.035 \, m$, and $r_2 = 0.07 \, m$:
$W = 0.04 \times 8 \times \frac{22}{7} \times ((0.07)^2 - (0.035)^2)$
$W = 0.32 \times \frac{22}{7} \times (0.0049 - 0.001225) = 0.32 \times \frac{22}{7} \times 0.003675$
$W = 0.003696 \, J$
Since $1 \, J = 10^6 \, \mu J$, $W = 3696 \, \mu J$.
Given $W = (15000 - x) \, \mu J$, we have $15000 - x = 3696$, so $x = 11304$.

[CORRECT_OPTION: 11304]

The shear stress $\tau$ is defined as $\frac{F}{A}$, where $F = 10^5 \, \text{N}$ and the cross-sectional area $A = \pi r^2 = \pi (0.04 \, \text{m})^2 = 0.0016 \pi \, \text{m}^2 = \frac{\pi}{625} \, \text{m}^2$.
The relationship between shear stress $\tau$, shear modulus $G$, and angular displacement $\theta$ (for small displacements, $\theta \approx \tan \theta = \gamma$) is given by $\tau = G \theta$.
Rearranging for $\theta$, we get $\theta = \frac{\tau}{G} = \frac{F}{AG}$.
Substituting the given values:
$\theta = \frac{10^5 \, \text{N}}{(\frac{\pi}{625} \, \text{m}^2) \times 10^{10} \, \text{N/m}^2} = \frac{6.25 \times 10^7}{\pi \times 10^{10}} = \frac{6.25 \times 10^{-3}}{\pi} = \frac{1}{160 \pi}$
Thus, the angular displacement $\theta$ is $\frac{1}{160 \pi}$.

$\;P_1=P_0+\rho g h=P_0+1000 \times 10 \times \;\frac{20}{100}$ $\;\Rightarrow \;\; P_1=P_0+2000$ $\;\;\text{ So, }\; P_2-P_1=\;\frac{2 S}{R}= (\;\frac{2 S}{1 \times 10^{-3}})$ $\;\Rightarrow \;\; P_2=P_0+2100 \;\; \;\text{ (given) }\;$ $\;\;\text{ So, }\; P_0+2100-P_0-2000=2 S \times 10^3$ $\;\Rightarrow 100=2 S \times 10^3$ $\;\Rightarrow s= (\;\frac{1}{20}) =0.05$

$\;\Delta P=\;\frac{1}{2} \rho v^2$ $\;\;\frac{m g}{A}+\rho g \Delta h=\;\frac{1}{2} \rho v^2$ $\;\;\frac{500}{0.5}+10^3 \times 10 \times 0.7=\;\frac{1}{2} \times 10^3 v^2$ $\;1+7=\;\frac{1}{2} v^2$ $\;v=4 {\text{m}} / {\ s}$

$\;\;\frac{2 s}{r_1}=\;\frac{1}{2} \;\frac{2 s}{r_2}$ $\;r_1=2 r_2$ $\;\;\text{ Ratio }\;=\;\frac{\;\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3}=8$

Surface tension arises due to extra energy of the molecules at the molecules at the surface as compared at the interior of a liquid. The coefficient of viscosity for a liquid decreases with rise in temperature where as it increases for gases with increase in temperature. The flow is turbulent for a Reynold's number greater than 2000 . Stream lines never intersect in a steady flow.

The terminal velocity of ball of radius $R$ inside a liquid of viscosity $η$ can be written as $V_T=\;\frac{2 R^2 g}{n}(\sigma -\rho )$, where $\sigma$ is the density of ball and $\delta$ is the density of the liquid.Hence,$A$ is correct since $V_T \propto R^2$ gives a parabola on a graph$C$ is correct since $V_T \alpha \;\frac{1}{η}$ and $η$ varies with temperature$D$ is correct since $V_T \alpha (\sigma -\rho )$ i.e., varies with density of liquid.