Properties Of Matter – NEET UG PhysicsPractice Questions & PYQs

Bernoulli's principle dictates that as the speed of a fluid increases, the pressure within that fluid decreases. Because the wind blows rapidly across the exterior of the roof while the air inside the house remains stationary, the atmospheric pressure inside exerts a greater force than the lower pressure outside, creating a net lift. This pressure difference is determined by the relationship $\Delta P = \frac{1}{2} \rho v^2$ which results \in an upward force when applied across the surface area of the roof. Substituting the given density of 1.2 kg per cubic meter, velocity of 40 meters per second, and area of 250 square meters, the total upward force is $F = \frac{1}{2} \times 1.2 \times 40^2 \times 250 = 2.4 \times 10^5\,\text{N}$

The relationship between the weight $W$ suspended from a wire and the resulting elongation $\Delta l$ is described by Young's modulus, defined as:
$Y = \frac{W \cdot l}{A \cdot \Delta l}$
Rearranging this formula to solve for weight gives:
$W = \frac{Y \cdot A \cdot \Delta l}{l}$
This expression shows that for a fixed length $l$, cross-sectional area $A$, and extension $\Delta l$, the required weight is directly proportional to the Young's modulus of the material.

Because the steel and brass wires share identical physical dimensions and must stretch by the same amount to remain at the same level, the ratio of the weights applied to them is determined solely by the ratio of their Young's moduli. Given that the Young's modulus of steel is twice that of brass, the weight added to the steel wire must be twice the weight added to the brass wire, resulting in a ratio of $2:1$.

Thermal expansion causes a substance's volume to increase as temperature rises, which inversely results in a decrease in density. This relationship is captured by the equation $\rho_T = \rho_0(1 - \gamma \Delta T)$, where $\rho_T$ is the final density, $\rho_0$ is the initial density, $\gamma$ represents the coefficient of volume expansion, and $\Delta T$ is the change in temperature. The fractional change in density, expressed as $\frac{\rho_0 - \rho_T}{\rho_0}$, is mathematically equivalent to the product $\gamma \Delta T$. Substituting the provided values of $\gamma = 5 \times 10^{-4} \text{K}^{-1}$ and $\Delta T = 40 \text{ K}$ into this expression gives $(5 \times 10^{-4} \text{K}^{-1}) \times (40 \text{ K})$, which simplifies to $0.020$.

The principle of calorimetry states that the heat liberated by a hot object equals the heat absorbed by a cooler one in an isolated system. In this scenario, steam at 100°C releases heat by condensing into liquid water and then cooling down to 80°C. If $m$ represents the mass of the steam, the total heat given up is:

$540m + m(1)(100 - 80) = 560m$

Meanwhile, 20 g of water initially at 10°C absorbs heat to reach 80°C, which is calculated as:

$20(1)(80 - 10) = 1400$

By setting the heat released equal to the heat absorbed, we find $560m = 1400$, which yields $m = 2.5$. The total final mass of the water is the sum of the original 20 g and the 2.5 g that condensed from the steam, resulting in 22.5 g.

Newton's Law of Cooling states that the rate of change in an object's temperature is proportional to the difference between its average temperature and the constant temperature of its surroundings, represented as $\frac{\theta_1 - \theta_2}{t} = K\left(\frac{\theta_1 + \theta_2}{2} - \theta_0\right)$ where $\theta_0$ is the ambient temperature and $K$ is a constant.

For the first cooling interval, the water temperature drops from 70°C to 60°C in 5 minutes, yielding the equation $\frac{70 - 60}{5} = K(65 - \theta_0)$, which simplifies to $2 = K(65 - \theta_0)$. During the subsequent 5 minutes, the temperature drops from 60°C to 54°C, resulting in $\frac{60 - 54}{5} = K(57 - \theta_0)$, which is $\frac{6}{5} = K(57 - \theta_0)$. By dividing these two equations to eliminate $K$, we obtain $\frac{2}{6/5} = \frac{65 - \theta_0}{57 - \theta_0}$, which reduces to $\frac{5}{3} = \frac{65 - \theta_0}{57 - \theta_0}$. Cross-multiplying this equation leads to $285 - 5\theta_0 = 195 - 3\theta_0$, solving to a final ambient temperature of $\theta_0 = 45^\circ$C.

The surface area of a liquid stores potential energy proportional to the magnitude of the area and the surface tension of the material. When multiple small droplets coalesce into a single larger drop, the total volume $V$ remains constant, but the total surface area decreases, which causes the stored excess potential energy to be released. To find this energy, we express the initial and final surface areas in terms of $V$. The initial surface area, calculated by summing the areas of the small drops, simplifies to $\frac{3V}{r}$, while the final surface area of the single large drop is expressed as $\frac{3V}{R}$. By multiplying the reduction in total surface area by the surface tension $T$, we arrive at the total energy released:

$\text{Energy released} = \frac{3VT}{r} - \frac{3VT}{R}$

The relationship between the extension of a wire and its physical dimensions is governed by Young's modulus, expressed as $Y = \frac{Fl}{A \Delta l}$. Since the volume $V$ of the copper wire remains constant, its cross-sectional area can be substituted as $A = \frac{V}{l}$, leading to the equation:
$\Delta l = \frac{F l^2}{V Y}$
Because the applied force $F$, volume $V$, and Young's modulus $Y$ are all constants, the extension $\Delta l$ is directly proportional to the square of the length $l^2$, ensuring that plotting $\Delta l$ versus $l^2$ yields a straight line.

Wien's displacement Law explains this behavior by establishing that the wavelength at which a body emits maximum radiation intensity, $\lambda_m$, is inversely proportional to its absolute temperature, described by $\lambda_m = \frac{\text{constant}}{T}$. As the iron piece heats up, its temperature increases, causing the peak emission wavelength to shift toward the shorter end of the visible spectrum. This migration moves the perceived color from longer-wavelength red to shorter-wavelength hues like reddish yellow, eventually reaching a white-hot state once the temperature is high enough to emit a balanced mix of all visible wavelengths.

The rate at which thermal energy transfers through a metal rod is defined by the formula $H = \frac{KA\Delta T}{L}$. When comparing two rods of the same length and temperature difference, the specific value of the heat flow rate becomes directly dependent on the product of the material's thermal conductivity and its cross-sectional area, represented by the term $KA$.

Since the problem states that the heat conduction rate in the first rod is four times greater than that in the second, we can equate the two scenarios using the relation $H_1 = 4H_2$. Plugging the standard heat conduction formula into this equality results in:

$\frac{K_1 A_1 \Delta T}{L} = 4 \frac{K_2 A_2 \Delta T}{L}$

After dividing both sides by the common factors of $\Delta T$ and $L$, the relationship simplifies directly to $K_1 A_1 = 4 K_2 A_2$.

To determine which wire stretches the most, one must examine the relationship derived from Young's modulus, $\Delta L = \frac{4FL}{\pi D^2 Y}$. Given that the force and material properties remain constant for all wires, the extension $\Delta L$ depends solely on the ratio $\frac{L}{D^2}$. Calculating this factor for each wire allows for a direct comparison of their relative extensions.

Evaluating this ratio for each option yields $\frac{200 \text{ cm}}{(0.2 \text{ cm})^2} = 5 \times 10^3 \text{ cm}^{-1}$ for the first wire, $\frac{300 \text{ cm}}{(0.3 \text{ cm})^2} = 3.3 \times 10^3 \text{ cm}^{-1}$ for the second, $\frac{50 \text{ cm}}{(0.05 \text{ cm})^2} = 20 \times 10^3 \text{ cm}^{-1}$ for the third, and $\frac{100 \text{ cm}}{(0.1 \text{ cm})^2} = 10 \times 10^3 \text{ cm}^{-1}$ for the fourth. Since the third wire (option C) results in the highest value of $20 \times 10^3 \text{ cm}^{-1}$, it experiences the largest extension under the same tension.

When a substance heats up, its density decreases because its volume expands. The coefficient of volume expansion, denoted by the Greek letter $\gamma$, quantifies this thermal response and is related to density and temperature changes through the following equation:

$\rho_{T_2} = \frac{\rho_{T_1}}{1 + \gamma \Delta T}$

Starting with an initial density of 998 kg/m³ at 20 °C and a final density of 992 kg/m³ at 40 °C, the temperature change $\Delta T$ is 20 °C. Substituting these values into the equation yields $992 = 998 / (1 + 20 \gamma)$. Rearranging the terms to isolate $\gamma$ results in $1 + 20 \gamma = 998 / 992$. Subtracting 1 from both sides leads to $20 \gamma = (998 - 992) / 992$, which simplifies to $20 \gamma = 6 / 992$. Dividing both sides by 20 results in $\gamma = 3 \times 10^{-4}$ /°C.

The principle of continuity dictates that for an incompressible fluid flowing through a pipe, the velocity increases as the cross-sectional area decreases, which makes the fluid speed greatest at the narrowest point. When we apply Bernoulli's equation for horizontal flow, expressed as $P + \frac{1}{2}\rho v^2 = \text{constant}$, it becomes clear that an increase in kinetic energy caused by higher velocity requires a corresponding decrease in static pressure to maintain energy balance. Therefore, the fluid reaches its maximum velocity and minimum pressure at the narrowest part, while at the widest part of the pipe, the velocity is at its minimum and the pressure is at its maximum.

The ability of a liquid to wet a solid surface is determined by the interplay between the adhesive forces pulling the liquid toward the solid and the cohesive forces holding the liquid together. This physical interaction is explicitly quantified by the angle of contact, denoted as $\theta$, which forms at the junction where the liquid, solid, and gas phases meet. A small or acute angle of contact signifies that adhesive forces are strong enough to cause the liquid to spread and wet the surface, whereas an obtuse angle indicates that cohesive forces dominate, preventing the liquid from wetting the material.

The extension of a wire under tension is governed by the relation $\Delta L = \frac{FL}{AY}$, which links the change in length to the applied force, original length, cross-sectional area, and Young's modulus. Substituting the area of a wire as $A = \frac{\pi D^2}{4}$ transforms this into $\Delta L = \frac{4FL}{\pi D^2 Y}$, showing that the extension is directly proportional to the force and length, and inversely proportional to the square of the diameter and the Young's modulus.

Applying this principle to the system where the steel wire supports the combined weight of $(5m + 2m)g = 7mg$ and the copper wire supports $5mg$, the ratio of their extensions is determined by the product of individual ratios. Taking the ratio of extensions for steel and copper, $\frac{\Delta L_S}{\Delta L_C}$, involves multiplying the force ratio of $\frac{7}{5}$ by the length ratio $q$, the inverse square of the diameter ratio $(\frac{1}{p})^2$, and the inverse of the Young's modulus ratio $\frac{1}{s}$. Multiplying these factors together yields $\left(\frac{7}{5}\right) \cdot q \cdot \left(\frac{1}{p}\right)^2 \cdot \left(\frac{1}{s}\right)$, which simplifies to $\frac{7q}{5sp^2}$.

The thermal energy transferred through the stone slab provides the latent heat necessary to melt the ice, acting as a steady heat conduction process. According to the law of thermal conductivity, the total heat flow is proportional to the material's thermal conductivity, the slab's surface area, the temperature gradient, and the duration of heat flow, while being inversely proportional to the thickness of the slab.

Calculating the total heat absorbed by the ice involves multiplying the mass of melted ice by its latent heat of fusion, defined by the equation $Q = m L_{f}$. Simultaneously, the heat conducted through the slab in one hour is determined by the formula $Q = \frac{K A (T_{1}-T_{2}) t}{L}$

Substituting the given parameters into this conduction equation yields $Q = \frac{K \times 0.36 \times (100 - 0) \times 3600}{0.1}$. Because the heat required to melt the ice is equivalent to the heat conducted, we can write the equation as $4.8 \times 3.36 \times 10^5 = \frac{K \times 0.36 \times 100 \times 3600}{0.1}$. Solving for the thermal conductivity $K$, we find $K = \frac{4.8 \times 3.36 \times 10^5 \times 0.1}{0.36 \times 100 \times 3600}$. Performing these arithmetic operations results in a value of 1.24 J/m/s°C.

Heating a substance at a constant rate causes its temperature to rise proportionally to the absorbed energy during single-phase states, while energy input during a phase change is entirely consumed as latent heat without changing the temperature. Consequently, liquid oxygen starting at 50 K exhibits an initial upward slope on the temperature-time graph until it reaches the boiling point at 1 atm. During the subsequent phase transition to gas, the graph plateaus horizontally, after which the temperature resumes climbing as the gaseous oxygen heats toward 300 K.

The Stefan-Boltzmann law dictates that the energy radiated by a black body is given by the formula $Q = \sigma A T^4$, where $A$ denotes the surface area. For a star treated as a spherical body with radius $R$, the surface area is $A = 4 \pi R^2$, so substituting this into the equation yields $Q = \sigma (4 \pi R^2) T^4$. Isolating the temperature $T$ involves rearranging the terms to solve for $T^4$ and taking the fourth root to arrive at the final expression:

$T = \left( \frac{Q}{4 \pi R^2 \sigma} \right)^{1/4}$

A star radiating as a black body emits total power defined by the Stefan-Boltzmann law, where power depends on the surface area of the star and the fourth power of its temperature. With a stellar radius $r$ and temperature $T$, the total power output is $P = \sigma 4\pi r^2 T^4$. This radiant energy spreads evenly over a spherical shell at distance $R$ from the center, which has an surface area of $4\pi R^2$. Determining the radiant energy per unit area requires dividing the total power by the area of this larger sphere:

$I = \frac{P}{4\pi R^2} = \frac{\sigma 4\pi r^2 T^4}{4\pi R^2}$

Canceling the common factor $4\pi$ from both the numerator and the denominator yields the final intensity:

$I = \frac{\sigma r^2 T^4}{R^2}$

The rate of heat transfer through a metallic rod is directly proportional to its cross-sectional area and inversely proportional to its length, as described by the equation
$Q = \frac{K A (T_1 - T_2) t}{L}$
When the material is melted and reshaped, its total volume remains invariant. Reducing the radius to half its original value decreases the cross-sectional area to one-fourth, since the area $A = \pi R^2$ scales with the square of the radius. Because volume is conserved, this fourfold reduction in cross-sectional area necessitates that the rod must stretch to four times its original length. Substituting the new dimensions, $A' = A/4$ and $L' = 4L$, into the heat conduction formula results in a new heat flow of $Q' = Q/16$.

In a steady state, the rate of heat transfer through a material depends directly on the temperature difference between its ends and the cross-sectional area of the rod, while being inversely proportional to the rod's length. This flow is characterized by the material's thermal conductivity, denoted by $k$, which acts as the proportionality constant in the relationship and is conceptually analogous to electrical conductivity in the context of Ohm's law. Combining these factors yields the final expression for the rate of heat transfer:

$\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}$