Properties Of Matter – NEET UG PhysicsPractice Questions & PYQs

The force required to stretch a wire depends on its Young's Modulus $Y$, cross-sectional area $A$, original length $l$, and extension $\Delta l$ through the relation $F = \frac{YA \Delta l}{l}$. Because both wires are made of the same material and possess equal volumes, increasing the area of the second wire to $3A$ requires its length to decrease to $l/3$ to conserve volume. We determine the force $F'$ for the second wire by substituting these parameters into the formula, giving $F' = \frac{Y(3A)\Delta l}{l/3}$. Simplifying this equation reveals $F' = 9 \frac{YA \Delta l}{l}$, which confirms that the force needed for the second wire is exactly $9F$.

The rate of heat production in a viscous liquid is equivalent to the power dissipated by the viscous drag force, which is the product of the drag force and the terminal velocity. The drag force is proportional to the product of the sphere's radius and velocity, implying that the power dissipated is proportional to $r \cdot V_T^2$. Given that the terminal velocity $V_T$ scales with the square of the radius as $V_T \propto r^2$, substituting this into the power relationship results in a proportionality of $r \cdot (r^2)^2$, which simplifies to $r^5$.

Wien's Displacement Law establishes that the product of the wavelength of maximum spectral radiance and the absolute temperature of a black body is constant, while the Stefan-Boltzmann law defines the total power radiated as proportional to the fourth power of that absolute temperature. By relating the initial state at wavelength $\lambda_0$ and temperature $T$ to the final state at wavelength $\frac{3}{4}\lambda_0$ and temperature $T'$, we use the constant nature of $\lambda_{\text{max}}T$ to determine the new temperature. Specifically, the relationship $\lambda_0 T = \frac{3}{4}\lambda_0 T'$ indicates that the new temperature is $T' = \frac{4}{3}T$.

Substituting this temperature relationship into the power proportionality $P \propto T^4$, the ratio of the new power $nP$ to the initial power $P$ is calculated by raising the temperature change factor to the fourth power:

$\frac{nP}{P} = \left(\frac{T'}{T}\right)^4 = \left(\frac{4}{3}\right)^4$

Evaluating this expression results in $n = \frac{256}{81}$.

Thermal resistance behaves like electrical resistance, where rods joined side-by-side create a parallel configuration, ensuring both rods experience the same temperature difference across their lengths. In this arrangement, the total thermal conductance is the sum of the conductances of the individual rods, which leads to the governing equation:
$\frac{K_{eq} A_{total}}{l} = \frac{K_1 A_1}{l} + \frac{K_2 A_2}{l}$
Since each rod possesses a cross-sectional area $A$ and length $l$, the total area equals $A_{total} = 2A$, while the individual areas are $A_1 = A_2 = A$. Substituting these dimensions into the relationship produces:
$\frac{K_{eq} (2A)}{l} = \frac{K_1 A}{l} + \frac{K_2 A}{l}$
By canceling the common term $A/l$ from every side, the expression simplifies to $2 K_{eq} = K_1 + K_2$, which ultimately determines the equivalent thermal conductivity:
$K_{eq} = \frac{K_1 + K_2}{2}$

The Stefan-Boltzmann law establishes that the power radiated by a black body sphere is directly proportional to its surface area and the fourth power of its absolute temperature, meaning $P \propto r^2 T^4$. Given an initial radius of 12 cm, power of 450 watt, and temperature of 500 K, we can calculate the final power by comparing the states through the ratio:
$\frac{P_2}{P_1} = \left( \frac{r_2}{r_1} \right)^2 \left( \frac{T_2}{T_1} \right)^4$
When the radius $r_2$ is half of $r_1$ and the temperature $T_2$ is double $T_1$, the equation becomes:
$\frac{P_2}{450} = \left( \frac{1}{2} \right)^2 (2)^4$
Solving this expression shows that the power increases by a factor of 4, resulting in a final power of 1800 watt.

Bulk modulus measures a material's resistance to compression by quantifying the relationship between applied pressure and the resulting volumetric strain. Since the volume of a sphere scales with the cube of its radius, a small relative change in radius creates a volumetric strain three times larger, expressed mathematically as $\frac{\Delta V}{V} = 3 \frac{\Delta R}{R}$. By substituting this relationship into the fundamental definition of bulk modulus, $B = \frac{P}{\Delta V/V}$, the expression becomes $B = \frac{P}{3(\Delta R/R)}$. Solving for the fractional decrease in radius simply requires rearranging these terms to isolate $\frac{\Delta R}{R}$, which yields $\frac{p}{3B}$.

In a U-tube filled with immiscible liquids, hydrostatic equilibrium dictates that the pressure at the horizontal interface between the oil and water must be equal on both sides. When the water level rises by 65 mm on the side open to the atmosphere, the water column on that side effectively reaches a height of 130 mm above the interface, accounting for the 65 mm rise and the 65 mm drop on the side where the oil is poured. The total height of the oil column, which balances this pressure, extends from the interface to 10 mm above the water level on the other side, resulting in a total height of 140 mm. Equating the pressures exerted by these two columns leads to the equation $\rho_{oil} g \times 140 = \rho_{water} g \times 130$. Using the standard density of water, $\rho_{oil} = \frac{130}{140} \times 10^3$, which confirms the density of the oil is approximately 928 kg/m^3.

When a rod experiences a change in temperature $\Delta\theta$, its new length is given by $l' = l(1 + \alpha \Delta\theta)$, where $l$ is the initial length and $\alpha$ is the coefficient of linear expansion. For two rods with initial lengths $l_1$ and $l_2$ and expansion coefficients $\alpha_1$ and $\alpha_2$, the new lengths at a different temperature are $l_1' = l_1(1 + \alpha_1 \Delta\theta)$ and $l_2' = l_2(1 + \alpha_2 \Delta\theta)$. The difference in their lengths at any temperature becomes $l_2' - l_1' = (l_2 - l_1) + (\alpha_2 l_2 - \alpha_1 l_1) \Delta\theta$. For this difference to remain constant regardless of the temperature change, the term involving $\Delta\theta$ must equate to zero, which leads to the relationship:

$\alpha_1 l_1 = \alpha_2 l_2$

According to Wien's displacement law, the wavelength $\lambda_m$ at which a black body emits the maximum amount of radiation is inversely proportional to its absolute temperature $T$, defined by the relationship $\lambda_m = \frac{b}{T}$ where $b$ is Wien's constant. Using the provided values of $b = 2.88 \times 10^6$ nmK and $T = 5760$ K, the peak wavelength is calculated as $\lambda_m = \frac{2.88 \times 10^6}{5760}$ nm, which results in $500$ nm.

Since the energy emission reaches its peak value at $500$ nm, the radiation intensity $U_2$ at this wavelength must be higher than the intensity $U_1$ at $250$ nm. Because the black body curve is peaked at $500$ nm and decreases as the wavelength deviates from this value, the energy emitted at the peak wavelength is necessarily greater than the energy emitted at any other wavelength, confirming that $U_2 > U_1$.

The principle of energy conservation states that gravitational potential energy transforms into heat as the ice descends, though only a portion of this energy actually contributes to melting. Since the ice absorbs one-quarter of the total gravitational potential energy $Mgh$, we equate this fraction to the thermal energy required to melt the mass $m$, which is given by $mL$.

Equating these two expressions leads to the formula $\frac{Mgh}{4} = mL$. By canceling the mass $m$ and rearranging the terms to isolate the height, we find $h = \frac{4L}{g}$. Using the latent heat of 3.4 × 10^5 J/kg and gravity at 10 N/kg, substituting these values yields 136,000 meters, which simplifies to 136 km.

The equilibrium of the cylinder is governed by the principle of flotation, which dictates that its total weight must be exactly balanced by the total buoyant force exerted by the two distinct liquids. Given the cylinder has a cross-sectional area $A$ and a total length $L$, its weight can be expressed as $W = (A \cdot L \cdot d) \cdot g$, where $d$ is the density of the cylinder and $g$ is the acceleration due to gravity. This weight is countered by the combined upthrust from the two liquids, which depends on the volume of the cylinder submerged in each layer.

The cylinder is partially submerged in both liquids: a length of $pL$ is immersed in the denser liquid of density $n\rho$, and the remaining length of $(L - pL)$ is immersed in the lighter liquid of density $\rho$. Expressing the total upthrust as the sum of the buoyant forces from both layers, we have $\rho \cdot A \cdot (L - pL) \cdot g + n\rho \cdot A \cdot pL \cdot g$. Equating the weight to this total upthrust and cancelling the common terms $A$ and $g$ from both sides simplifies the relationship to $d \cdot L = \rho (L - pL) + n\rho \cdot pL$. Dividing the entire equation by the total length $L$ yields $d = \rho (1 - p) + n\rho p$, which algebraically rearranges to the final expression $d = \{1 + (n - 1)p\} \rho$.

The work done in stretching a liquid film is stored as potential energy, which is determined by the surface tension of the liquid and the total change in surface area, expressed as $W = T \cdot \Delta A_{total}$. Because a liquid film possesses two surfaces in contact with the air, the change in area must be calculated for both sides. The initial area is 8 cm², and the final area is 20 cm², which yields an increase of 12 cm² per side.

Since there are two surfaces, the total increase in surface area is 24 cm², or 24 x 10⁻⁴ m² when converted to standard SI units. Substituting the provided work of 3 x 10⁻⁴ J into the equation $T = \frac{W}{\Delta A_{total}}$, we arrive at $T = \frac{3 \times 10^{-4}}{24 \times 10^{-4}}$. Solving this expression results in a surface tension of 0.125 Nm⁻¹.

Newton's Law of Cooling dictates that the excess temperature of a body above its surroundings decreases exponentially over time. We calculate this excess temperature by subtracting the room temperature $T$ from the body's initial temperature of $3T$, giving us an initial difference of $2T$. After 10 minutes, the body reaches $2T$, which means the excess temperature has dropped from $2T$ to $T$. Since the decay is exponential, this indicates the excess temperature is halved every 10 minutes. Extending this logic to the next 10 minutes, the excess temperature will halve once more, changing from $T$ to $T/2$. The final temperature of the body is found by adding this excess back to the room temperature, yielding $T + T/2 = \frac{3}{2} T$.

When two identical bodies are brought into thermal contact within an isolated system, the principle of conservation of energy dictates that the heat lost by the hotter body equals the heat gained by the colder body. If the heat capacity of the material were constant, the final equilibrium temperature would be the simple arithmetic mean of the two initial temperatures, which is 50°C. However, since the material's heat capacity increases as the temperature rises, the body starting at 100°C possesses a greater capacity to store thermal energy than the body starting at 0°C.

Because the warmer body has a higher heat capacity, it effectively exerts a stronger influence on the final equilibrium state than the colder body. As energy is exchanged, the colder body absorbs heat, but the hotter body releases it while maintaining a relatively higher capacity to retain energy at its given temperature range. This asymmetry causes the system to settle at a temperature that is weighted toward the hotter starting point. Consequently, the final common temperature $T_f$ must be greater than 50°C.

The phenomenon of capillary rise is defined by the equation $h = \frac{2T \cos \theta}{r \rho g}$. Because the height $h$, surface tension $T$, and capillary radius $r$ are held constant across the three scenarios, the product of density and the cosine of the contact angle must be constant, which implies $\cos \theta \propto \rho$. Given that the densities are ordered as $\rho_1 > \rho_2 > \rho_3$, it necessarily follows that $\cos \theta_1 > \cos \theta_2 > \cos \theta_3$. Since the cosine function is monotonically decreasing within the range $[0, \pi/2)$, larger values of cosine correspond to smaller angles of contact, resulting in the relationship $0 \leq \theta_1 < \theta_2 < \theta_3 < \pi/2$.

The rate of heat flow through a metal rod in steady state is directly proportional to the temperature difference between its two ends, a relationship expressed by the equation $\frac{dQ}{dt} \propto \Delta T$. In the first instance, the temperature difference is 110°C - 100°C = 10°C, and in the second instance, the difference is 210°C - 200°C = 10°C. Because the temperature difference remains the same in both scenarios, the rate of heat flow remains constant at 4.0 J/s.

When a capillary tube is placed in water, the liquid rises to an equilibrium height $h$ where the upward force from surface tension perfectly balances the weight of the liquid column. If the physical length of the tube is shorter than $h$, the water does not overflow because the meniscus radius of curvature adjusts to a larger value to compensate for the reduced height. Consequently, the liquid rises exactly to the brim of the capillary tube and remains stable there without spilling over.

Wien's displacement law establishes that the wavelength of maximum spectral intensity is inversely proportional to the absolute temperature, defined by the relation $\lambda_m \propto \frac{1}{T}$. Since the wavelength of light increases across the visible spectrum from violet to red, the peak wavelengths for the three stars follow the sequence $\lambda_P < \lambda_R < \lambda_Q$. Given this inverse relationship, a shorter peak wavelength must correspond to a higher temperature, leading to the conclusion that $T_P > T_R > T_Q$.

The principle of continuity for incompressible fluids dictates that the volume flow rate into a system must equal the volume flow rate out of it, ensuring mass conservation. In the cylindrical spray pump, the liquid enters with a cross-sectional area $\pi R^2$ at speed $V$, creating an inlet flow rate of $\pi R^2 V$. This same volume of liquid must pass through $n$ fine holes at the exit, where each hole has a cross-sectional area $\pi r^2$ and the liquid reaches an ejection speed $v$, resulting in a total exit flow rate of $n \pi r^2 v$. Equating the inlet and exit flow rates leads to the balance equation $\pi R^2 V = n \pi r^2 v$ and isolating the unknown ejection speed $v$ gives the final expression $v = \frac{V R^2}{n r^2}$

Compressibility quantifies the extent to which a substance decreases in volume when subjected to pressure, defined mathematically as the inverse of the bulk modulus. In an ocean, the immense weight of the water column exerts hydrostatic pressure at the bottom, which induces a fractional reduction in volume. This fractional compression is directly proportional to the applied pressure and the material's specific compressibility coefficient.

The hydrostatic pressure is calculated using the formula $P = \rho g h$ where the density $\rho$ is 1000 kg/m^3, acceleration due to gravity $g$ is 9.8 m/s^2, and depth $h$ is 2700 m. Multiplying these values provides the total pressure acting at the base of the ocean. The fractional compression is then determined using the relation $\frac{\Delta V}{V} = P \times K$ where $K$ is the compressibility 45.4 x 10^-11 Pa^-1. Combining these values leads to a fractional compression of approximately 1.2 x 10^-2.