Properties Of Matter – NEET UG PhysicsPractice Questions & PYQs

Forming a soap bubble requires work to be done against surface tension to expand the soap film. Because a soap bubble is a thin liquid film with both an inner and an outer side exposed to air, the effective surface area is twice the area of the sphere. The work done, which equals the energy stored, is found by multiplying the surface tension by this total created surface area.

The surface area of a sphere is given by $A = 4\pi r^2$. Converting the radius from $2 \text{ cm}$ to $0.02 \text{ m}$, the area becomes $A = 4\pi \times (0.02)^2 = 16\pi \times 10^{-4} \text{ m}^2$. Since the bubble has two surfaces, the total energy $E$ is calculated using $E = 2 \times T \times A$. By substituting the surface tension $T = 0.03 \text{ N m}^{-1}$ into this expression, we get $E = 2 \times 0.03 \times 16\pi \times 10^{-4} \text{ J}$. Evaluating this results in $E = 0.96\pi \times 10^{-4} \text{ J}$, which is approximately $3.015 \times 10^{-4} \text{ J}$.

Elastic potential energy per unit volume, commonly known as energy density, quantifies the energy stored within a material due to deformation. This quantity relates the material's intrinsic stiffness, represented by Young's modulus $Y$, to the strain it undergoes, defined as the ratio of the change in length $\Delta L$ to the original length $L$. The formula for this energy density is:

$u = \frac{1}{2} Y \left( \frac{\Delta L}{L} \right)^2$

Given a Young's modulus of $2.0 \times 10^{11} \text{ Nm}^{-2}$, an extension of 1 mm, and an original length of 100 cm, the strain is calculated by converting units to meters as $10^{-3} / 1 = 10^{-3}$. Applying these values to the expression provides $u = \frac{1}{2} \times 2.0 \times 10^{11} \times (10^{-3})^2$. Simplifying this calculation results in $1.0 \times 10^{11} \times 10^{-6}$, yielding an energy density of $10^5 \text{ J/m}^3$.

The operation of a venturi-meter is fundamentally governed by Bernoulli's principle, which dictates that the total energy of a flowing fluid remains constant along a streamline. As fluid travels through the constricted section of the device, its velocity increases, which necessitates a corresponding decrease in fluid pressure according to the energy conservation relationship $P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$. By measuring the resulting pressure difference between the inlet and the narrow throat, one can accurately calculate the discharge rate of the fluid.

Pressure, defined by the formula $P = \frac{F}{A}$, is a scalar quantity because it exerts magnitude equally in all directions at any point within a fluid rather than acting in a single, specific direction. Unlike vectors, which require both magnitude and a defined direction that follows the parallelogram law of addition, pressure depends only on magnitude. Because coefficient of viscosity, surface tension, and relative density are also scalar quantities, labeling pressure as a vector quantity is incorrect.

Stokes' Law provides the mathematical principle to calculate the drag force acting on a sphere as it falls through a viscous fluid, expressed as $F = 6 \pi \eta r v$. In this equation, $\eta$ represents the viscosity of the fluid, $r$ is the radius of the sphere, and $v$ is its velocity. Given the diameter of the sphere is 1 mm, the radius is 0.5 mm, which is $0.5 \times 10^{-3}$ m. Substituting the given values for viscosity, which is 0.8 Pa s, and velocity, which is 2 m/s, the calculation becomes $F = 6 \times 3.14 \times 0.8 \times 0.5 \times 10^{-3} \times 2$. Solving this expression results in a viscous drag force of $15 \times 10^{-3}$ N.

When a spherical object moves through a viscous fluid at its terminal velocity, the resistive force opposing its motion, known as viscous drag, can be determined using Stokes' Law. This law states that the force is defined by the relationship $F = 6 \pi \eta r v_t$, where $\eta$ represents the fluid's viscosity, $r$ is the radius of the sphere, and $v_t$ is its terminal velocity.

To find the result, first convert all given values into standard SI units so that the calculation remains consistent. The radius of 5 mm becomes 5 × 10^-3 m, and the velocity of 10 cm s^-1 converts to 10 × 10^-2 m s^-1. By substituting the viscosity of 0.9 kg m^-1 s^-1 and these dimensions into the expression $F = 6 \times \pi \times 0.9 \times 5 \times 10^-3 \times 10 \times 10^-2$, the arithmetic yields 84.78 × 10^-4 N. Expressing this in proper scientific notation results in a final drag force of 8.48 × 10^-3 N.

The internal pressure of a soap bubble is determined by the balance between the external atmospheric pressure and the excess pressure created by the surface tension of the film. This total pressure is defined by the equation $P_{\text{in}} = P_{\text{out}} + \frac{4T}{R}$, where the excess pressure component is given by $\Delta P = \frac{4T}{R}$. Because this excess pressure is inversely proportional to the radius, expanding the bubble increases the value of $R$, which forces the overall internal pressure to decrease.

Hydrostatic pressure at the base of a container is determined by the formula $P = \rho g h$, where $\rho$ represents the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the vertical depth of the liquid column. Because both vessels are filled with water to the exact same vertical height $h$, the pressure exerted on the base area is identical for both, regardless of their specific shapes or the total volume of water they contain.

When two rods are joined end-to-end, the system reaches a steady state where the rate of heat flow through the copper rod is identical to the rate of heat flow through the steel rod. Because heat cannot accumulate at the junction, the thermal energy entering from the copper side must pass into the steel side at the same rate, maintaining continuity of the heat current. This flow is governed by the relation $H = \frac{KA \Delta T}{l}$, where $K$ represents thermal conductivity, $A$ is the cross-sectional area, and $l$ is the length of the rod.

Given that the lengths and cross-sectional areas are identical for both materials, they cancel out when equating the heat currents:
$\frac{K_{\text{Cu}} A (100 - \theta)}{l} = \frac{K_{\text{steel}} A (\theta - 0)}{l}$
Substituting the provided conductivity values into this balance allows us to simplify the expression to:
$385 (100 - \theta) = 50 \theta$
Distributing the values yields $38500 - 385 \theta = 50 \theta$, which rearranges to $435 \theta = 38500$. Solving for $\theta$ results in a junction temperature of approximately $88.5^{\circ} \text{C}$.

When a coil spring is stretched, the wire undergoes a twisting deformation that alters its shape rather than its length or volume, meaning the elasticity of the spring is governed by its shear modulus ($G$). Because this behavior is a direct consequence of how such springs function, the assertion accurately identifies the relevant material property. The reason provided is false, however, because steel possesses a significantly higher tensile strength than copper, directly contradicting the claim that a copper spring is stronger.

When a spherical ball is released into a highly viscous medium, it begins with an initial velocity of $v = 0$ and accelerates due to gravity. As the viscous drag force increases with rising speed, the net downward force diminishes, causing the rate of acceleration to decrease until the ball reaches a steady state. Consequently, the graph of $v$ versus $t$ starts at the origin and curves upward, eventually leveling off horizontally to represent the attainment of terminal velocity.

When a ball falls through a fluid at a constant velocity, it has reached terminal speed, meaning the downward pull of gravity is exactly balanced by the sum of the upward buoyant and viscous forces. The weight of the ball is $Mg$, and the buoyant force, which equals the weight of the displaced glycerine, is calculated as the product of the volume of the ball, the density of glycerine, and gravity. Given that the ball's mass $M = V \cdot d$ and the glycerine density is $\frac{d}{2}$, the buoyant force is $F_B = V \cdot (\frac{d}{2}) \cdot g = \frac{M}{d} \cdot \frac{d}{2} \cdot g$, which simplifies to $\frac{Mg}{2}$. Consequently, because the net force must be zero, the viscous force must account for the remaining weight of the ball, resulting in $F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}$.

Young's modulus characterizes the elasticity of a material by representing the ratio of longitudinal stress to longitudinal strain under an applied load. When mass $M$ is suspended from a wire of length $L$ and cross-sectional area $A$, the gravitational force creates a stress of $\frac{Mg}{A}$. Since the wire stretches to a new length $L_1$, the elongation is $L_1 - L$, which corresponds to a strain of $\frac{L_1 - L}{L}$. By dividing the applied stress by this strain, the resulting expression for Young's modulus is $\frac{MgL}{A(L_1 - L)}$.

When a capillary tube is placed into a liquid, the water rises until the upward surface tension force at the meniscus equals the downward gravitational force acting on the water column. This equilibrium is established by the balance between the weight of the water and the vertical component of the surface tension force:
$mg = 2\pi r T \cos \theta$
In this relationship, $m$ represents the mass of the water, $r$ is the radius of the tube, $T$ is the surface tension, and $\theta$ is the contact angle. Since the liquid type, temperature, and tube material remain constant, the surface tension, contact angle, and acceleration due to gravity are all fixed values.

Because these variables are constant, the mass of the water column scales directly with the radius of the tube, represented as $m \propto r$. Doubling the radius of the capillary tube from $r$ to $2r$ must therefore result in doubling the mass of the supported water. Given that the initial tube supports 5.0 g, the new mass is determined by the proportion $\frac{m_2}{5.0} = \frac{2r}{r}$, which confirms that the mass of water in the wider tube is 10.0 g.

The elastic potential energy stored within a stretched wire is equivalent to the work performed by the gravitational force pulling on the mass as the material elongates. Since the wire's tension starts at zero and increases linearly to $Mg$ as it extends by $l$, the average force acting throughout the process is simply $\frac{Mg}{2}$. Multiplying this average force by the total extension $l$ gives the stored energy:

$U = \frac{1}{2} Mgl$

The physical principle governing this problem is thermal expansion, where the change in length of a material is defined by the product of its original length, the coefficient of linear expansion, and the change in temperature. For the difference in length between the two rods to remain invariant despite temperature fluctuations, their individual increases in length, denoted as $\Delta L$, must be equal. Equating the thermal expansion products for both materials leads to the following condition:

$\alpha_{\mathrm{Cu}} L_{\mathrm{Cu}} = \alpha_{\mathrm{Al}} L_{\mathrm{Al}}$

By substituting the provided coefficient values of 1.7 × 10⁻⁵ K⁻¹ for copper and 2.2 × 10⁻⁵ K⁻¹ for aluminium, along with the 88 cm length of the copper rod, the relationship becomes:

$1.7 \times 10^{-5} \times 88 = 2.2 \times 10^{-5} \times L_{\mathrm{Al}}$

Solving this expression for the unknown length of the aluminium rod, which is expressed as $L_{\mathrm{Al}} = \frac{1.7 \times 88}{2.2}$, results in 68 cm.

The terminal velocity of a spherical body falling through a viscous medium is defined by the relationship $v_T = \frac{2r^2(\sigma - \rho)g}{9\eta}$, where $r$ is the radius of the sphere, $\sigma$ is the density of the sphere, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $\eta$ is the coefficient of viscosity. Since gravity and viscosity remain constant throughout the fall, the ratio of the terminal velocities for two such spheres is proportional to the square of their radii and the difference in density between each material and the medium.

Using the expression $\frac{v_1}{v_2} = \left(\frac{r_1}{r_2}\right)^2 \left(\frac{\sigma_1 - \rho}{\sigma_2 - \rho}\right)$, the ratio of the radii squared is $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$. By substituting the density values $\sigma_1 = 8\rho_2$, $\sigma_2 = \rho_2$, and $\rho = 0.1\rho_2$, the density ratio becomes $\frac{8\rho_2 - 0.1\rho_2}{\rho_2 - 0.1\rho_2} = \frac{7.9\rho_2}{0.9\rho_2} = \frac{79}{9}$. Multiplying the radius factor by the density factor yields $\frac{1}{4} \times \frac{79}{9} = \frac{79}{36}$.

The velocity at which fluid exits a small hole near the bottom of an open tank is determined by Torricelli's Law, which states that the exit velocity is $v = \sqrt{2gh}$ as potential energy converts into kinetic energy. The total flow rate is the product of the cross-sectional area of the orifice and this velocity, represented by $Q = A \sqrt{2gh}$.

The cross-sectional area of 2 mm$^2$ is equivalent to $2 \times 10^{-6}$ m$^2$. With the height at 2 m and gravity as 10 m/s$^2$, the exit velocity becomes $\sqrt{2 \times 10 \times 2}$, which is $\sqrt{40}$ m/s. Multiplying this velocity by the cross-sectional area provides a flow rate of approximately $12.6 \times 10^{-6}$ m$^3$/s.

The internal pressure of a soap bubble exceeds atmospheric pressure because of the surface tension acting on its two distinct interfaces, resulting in an excess pressure defined by $\Delta P = \frac{4T}{R}$. This internal pressure must balance the hydrostatic gauge pressure exerted by a fluid column of height $Z_0$, which is calculated using the expression $\rho g Z_0$. Equating these two values leads to the following relationship for the depth:

$Z_0 = \frac{4T}{\rho g R}$

By substituting the known values for surface tension, radius, density, and gravitational acceleration, the equation becomes $Z_0 = \frac{4 \times 2.5 \times 10^{-2}}{10^3 \times 10 \times 10^{-3}}$. This calculation yields $Z_0 = 0.01$ m, which is equivalent to 1 cm.

Ductility is defined by the ability of a material to undergo significant plastic deformation before fracture, whereas brittleness is characterized by failure with little to no plastic deformation. In a stress-strain curve, the separation between the ultimate strength point and the fracture point corresponds to the extent of plastic flow and necking; consequently, this distance is negligible for brittle materials like $X$ and substantial for ductile materials like $Y$. This confirms that material $X$ is brittle and material $Y$ is ductile.