KCET 2025 Question Paper with Solutions

A wooden block of mass $𝑀$ rests on a rough floor, while another block of the same mass is suspended from a point $𝑂$ by strings. To maintain equilibrium, we need to determine the coefficient of static friction between the block on the floor and the floor itself.

From the figure and given conditions, we can use the following equations for equilibrium:

The vertical component of tension:

$𝑇\sin 𝜃=𝑀𝑔\dots \left(𝑖\right)$

This equation accounts for the vertical force balance, where $𝑇$ is the tension in the string, and $𝜃$ is the angle made by the string with the vertical.

The horizontal component of tension:

$𝑇\cos 𝜃=𝜇𝑀𝑔\dots \left(𝑖𝑖\right)$

Here, $𝜇$ represents the coefficient of static friction.

To find $𝜇$, divide equation (ii) by equation (i):

$\frac{𝑇\cos 𝜃}{𝑇\sin 𝜃}=\frac{𝜇𝑀𝑔}{𝑀𝑔}$

$\cot 𝜃=𝜇$

Therefore, the coefficient of static friction is:

$𝜇=\cot 𝜃$

To determine the mass of the block, consider the following analysis using the given data:

Friction: Since the block is moving, we deal with kinetic friction in this scenario.

Acceleration Calculation:

$𝑎=\frac{𝑣-𝑢}{𝑡}=\frac{20-0}{3}=\frac{20}{3}{\text{m/s}}^2$

Here, $𝑣$ is the final velocity, $𝑢$ is the initial velocity, and $𝑡$ is the time.

Net Force Calculation:

The net force acting on the block can be expressed as:

$𝐹-{𝑓}_{𝑘}=𝑚\times 𝑎$

where $𝐹=20\text{N}$ is the applied force and ${𝑓}_{𝑘}$ is the force due to kinetic friction.

Kinetic Friction Force:

${𝑓}_{𝑘}=\text{coefficient of kinetic friction}\times \text{normal force}=0.25\times 𝑚\times 𝑔$

Using the Net Force Equation:

Substitute the values into the equation:

$20-0.25\times 𝑚\times 10=𝑚\times \frac{20}{3}$

Solving for $𝑚$:

$20-2.5𝑚=\frac{20}{3}𝑚$

Rearranging the terms gives:

$20=\frac{20}{3}𝑚+2.5𝑚$

Combine the terms:

$20=\left(\frac{20+7.5}{3}\right)𝑚$

$20=\frac{27.5}{3}𝑚$

Simplifying the equation:

$𝑚=\frac{20\times 3}{27.5}$

Therefore,

$𝑚\approx 2.2\text{kg}$

Thus, the mass of the block is approximately 2.2 kg.

To determine the work done by the net force during the displacement of the body, we need to evaluate the change in kinetic energy (ΔKE) as the object moves from $𝑥=0$ to $𝑥=2\text{ m}$.

The initial velocity $𝑢$ when $𝑥=0$ is zero because $𝑣=𝑘{𝑥}^{3/2}$ and substituting $𝑥=0$ gives $𝑣=0$.

The final kinetic energy at $𝑥=2\text{ m}$ can be found by calculating the velocity at this point:

$𝑣=𝑘\cdot (2{)}^{3/2}$

Given $𝑘=2\text{SI units}$, the velocity at $𝑥=2$ is:

$𝑣=2\cdot 2^{3/2}$

The change in kinetic energy is:

$\Delta \text{KE}=\frac{1}{2}𝑚{𝑣}^2-\frac{1}{2}𝑚{𝑢}^2$

Substituting $𝑚=0.25\text{kg}$ and $𝑢=0$:

$\Delta \text{KE}=\frac{1}{2}\times 0.25\times (2\times 2^{3/2}{)}^2-0$

Calculate $(2\times 2^{3/2}{)}^2$:

$=(2^1\times (2^{3/2}){)}^2=2^2\times (2^3)=2^5=32$

So the change in kinetic energy is:

$\Delta \text{KE}=\frac{1}{2}\times 0.25\times 32$

Finally, solving gives:

$=4\text{ J}$

Hence, the work done by the net force during the displacement is 4 Joules.

I. Final total kinetic energy equals initial

II. Linear momentum is always conserved

III. During collision kinetic energy gets parthy converted to potential energy

$\therefore$ All statements are correct

$\begin{array}{cc}& x_{\mathrm{cm}}=\frac{\sum \mathrm{mixi}}{\sum \mathrm{mi}}\\ & =\frac{1\times 0+2\times 1+3\times \frac{1}{2}}{6}=\frac{7}{12}\\ & y_{\mathrm{cm}}=\frac{\sum \mathrm{miyi}}{\sum \mathrm{mi}}=\frac{1\times 0+2\times 0+3\times \frac{\sqrt{\surd 3}}{2}}{6}=\frac{3\sqrt{\surd 3}}{12}\end{array}$

Explanation

The relationship between the torque, moment of inertia, and angular acceleration for each wheel can be established with the following equations:

For the smaller wheel:

${𝑇}_1={𝐼}_1{𝛼}_1$

where ${𝑇}_1$ is the torque applied to the smaller wheel, ${𝐼}_1=4\text{kg}\cdot {\text{m}}^2$ is its moment of inertia, and ${𝛼}_1$ is the angular acceleration.

The angular acceleration ratio between the two wheels is given by:

${𝛼}_2{𝑟}_2={𝛼}_1{𝑟}_1$

where ${𝛼}_2$ is the angular acceleration of the larger wheel, ${𝑟}_1=0.2\text{m}$, and ${𝑟}_2=0.3\text{m}$.

Given that a torque of 10 Nm is applied to the smaller wheel:

$10={𝛼}_1\times 4$

Solving for ${𝛼}_1$:

${𝛼}_1=\frac{10}{4}=\frac{5}{2}{\text{rad/s}}^2$

Using the relation for the angular accelerations:

${𝛼}_2\times 0.3=\left(\frac{5}{2}\right)\times 0.2$

${𝛼}_2=\frac{5}{3}{\text{rad/s}}^2$

Finally, for the larger wheel, the torque ${𝑇}_2$ is given by:

${𝑇}_2={𝐼}_2{𝛼}_2$

where ${𝐼}_2=20\text{kg}\cdot {\text{m}}^2$. Plugging in the values:

${𝑇}_2=20\times \frac{5}{3}=\frac{100}{3}\text{Nm}$

Thus, the angular accelerations and the torque can be summarized as follows:

Angular acceleration of the smaller wheel ${𝛼}_1=\frac{5}{2}{\text{rad/s}}^2$

Angular acceleration of the larger wheel ${𝛼}_2=\frac{5}{3}{\text{rad/s}}^2$

Torque on the larger wheel ${𝑇}_2=\frac{100}{3}\text{Nm}$

Angular momentum is conserved in an elliptical orbit, meaning:

${𝐿}_{𝑝}={𝐿}_{𝑎}$

where ${𝐿}_{𝑝}$ is the angular momentum at perihelion and ${𝐿}_{𝑎}$ is the angular momentum at aphelion.

The perihelion is the point in the orbit where the planet is closest to the Sun. Therefore, the radius at perihelion (${𝑟}_{𝑝}$) is less than the radius at aphelion (${𝑟}_{𝑎}$):

${𝑟}_{𝑝}<{𝑟}_{𝑎}$

Because angular momentum is conserved, we have:

$𝑚{𝑣}_{𝑝}{𝑟}_{𝑝}=𝑚{𝑣}_{𝑎}{𝑟}_{𝑎}$

From this, it follows that the velocity at perihelion (${𝑣}_{𝑝}$) is greater than the velocity at aphelion (${𝑣}_{𝑎}$):

${𝑣}_{𝑝}>{𝑣}_{𝑎}$

The total (mechanical) energy E of a satellite in a circular orbit of radius $𝑟=𝑅+ℎ$ is the sum of its kinetic and potential energies. You can show that

Gravitational potential energy:

$𝑈=-\frac{𝐺𝑀𝑚}{𝑟}$

Kinetic energy (for a circular orbit):

$𝐾=\frac{1}{2}\frac{𝐺𝑀𝑚}{𝑟}$

Total energy:

$𝐸=𝐾+𝑈=\frac{1}{2}\frac{𝐺𝑀𝑚}{𝑟}-\frac{𝐺𝑀𝑚}{𝑟}=-\frac{1}{2}\frac{𝐺𝑀𝑚}{𝑟}$

Since $𝐺$, $𝑀$ and $𝑚$ are constants, $𝐸$ varies as

$𝐸\propto -\frac{1}{𝑟}=-\frac{1}{𝑅+ℎ}.$

Answer: Option A $-\frac{1}{𝑅+ℎ}$.

The problem involves two wires, A and B, made from the same material. The ratio of their diameters is $1:2$, and the ratio of their lengths is $1:3$. When the same force is applied to both wires, we need to determine the ratio of their length increases.

Here's how we can solve it:

The formula for the increase in length ($\Delta \ell$) when a wire is stretched by a force $𝐹$ is given by:

$\Delta \ell =\frac{𝐹\ell }{𝐴𝑌}=\frac{𝐹\ell }{𝜋{𝑟}^2𝑌}$

Where:

$\ell$ is the original length,

$𝐴$ is the cross-sectional area of the wire,

$𝑌$ is Young's modulus,

$𝑟$ is the radius of the wire.

To find the ratio of the increase in lengths of wires A and B, we calculate:

$\frac{\Delta {\ell }_1}{\Delta {\ell }_2}=\left(\frac{{\ell }_1}{{\ell }_2}\right)\times {\left(\frac{{𝑟}_2}{{𝑟}_1}\right)}^2$

For wires A and B:

The length ratio ${\ell }_1:{\ell }_2$ is $1:3$.

The diameter ratio ($\Rightarrow$ radius ratio) is $1:2$, so $\frac{{𝑟}_2}{{𝑟}_1}=\frac{2}{1}$.

Substitute these values into the equation:

$\frac{\Delta {\ell }_1}{\Delta {\ell }_2}=\frac{1}{3}\times {\left(\frac{2}{1}\right)}^2=\frac{1}{3}\times 4=\frac{4}{3}$

Therefore, the ratio of the increase in their lengths is $4:3$.

To solve this problem, we apply the principles of fluid dynamics, specifically the continuity equation and Bernoulli's equation.

Using the Continuity Equation:

The continuity equation for fluid flow states that the product of the cross-sectional area of the pipe and the fluid velocity must remain constant:

${𝐴}_1{𝑉}_1={𝐴}_2{𝑉}_2$

Given:

${𝐴}_1=10{\text{cm}}^2=10\times {10}^{-4}{\text{m}}^2$

${𝑉}_1=1\text{m/s}$

${𝐴}_2=5{\text{cm}}^2=5\times {10}^{-4}{\text{m}}^2$

Solving for ${𝑉}_2$:

${𝑉}_2=\frac{{𝐴}_1{𝑉}_1}{{𝐴}_2}=\frac{10\times {10}^{-4}\times 1}{5\times {10}^{-4}}=2\text{m/s}$

Applying Bernoulli's Equation:

Bernoulli's equation for steady, incompressible flow along a streamline is:

${𝑃}_1+\frac{1}{2}𝜌{𝑉}_2^1={𝑃}_2+\frac{1}{2}𝜌{𝑉}_2^2$

Where:

${𝑃}_1=2000\text{Pa}$

$𝜌=1000{\text{kg/m}}^3$ (density of water)

Rearranging to find ${𝑃}_2$:

${𝑃}_2={𝑃}_1+\frac{1}{2}𝜌({𝑉}_2^1-{𝑉}_2^2)$

Substituting the known values:

${𝑃}_2=2000+\frac{1}{2}\times 1000\times (1^2-2^2)$

${𝑃}_2=2000+\frac{1}{2}\times 1000\times (1-4)$

${𝑃}_2=2000+500\times (-3)=2000-1500=500\text{Pa}$

Thus, the pressure of the water at the second point in the pipe is $500\text{Pa}$.

To determine the temperature at the junction $𝑥$ of three identical metal rods, follow this analysis:

Assuming No Heat Loss: The rods are made of the same material and are identical, ensuring consistent heat flow throughout.

Thermal Resistance: Let the thermal resistance of each rod be $𝑅$. The temperature at the junction is ${𝑇}_{𝑥}$.

Heat Flow Equilibrium: Since the system reaches a steady state, the heat flowing into the junction from each side should equal the heat flowing out:

$𝐻={𝐻}_1+{𝐻}_2$

Where:

$𝐻$ is the heat flowing into the junction from both ends.

${𝐻}_1$ and ${𝐻}_2$ are the heat flows from the sides of the junction.

Application: Set up the heat balance equation:

$\frac{{𝑇}_{𝑥}-0}{𝑅}=\frac{90-{𝑇}_{𝑥}}{𝑅}+\frac{90-{𝑇}_{𝑥}}{𝑅}$

Simplifying: Substitute and solve:

${𝑇}_{𝑥}=\frac{180}{3}={60}^{\circ }\text{C}$

Therefore, the temperature at the junction $𝑥$ is ${60}^{\circ }\text{C}$. This calculation assumes perfect thermal conduction and no heat loss outside the system.

According to the first law of thermodynamics, the change in internal energy ($\Delta 𝑈$) of a system can be expressed as:

$\Delta 𝑈=𝑄-𝑊$

Where $𝑄$ is the heat absorbed by the system and $𝑊$ is the work done by the system.

For paths 1 and 2 from state A to state B, the change in internal energy is the same, regardless of the path taken:

$\Delta {𝑈}_1=\Delta {𝑈}_2$

Therefore, the heat absorbed and work done along the different paths follows the relationship:

${𝑄}_1-{𝑊}_1={𝑄}_2-{𝑊}_2$

This equation signifies that the net change in energy (considering both heat and work) is equal for both paths due to the same change in internal energy from state A to state B.

The mean kinetic energy of the atoms in an ideal gas is directly proportional to the temperature measured in Kelvin. At an initial temperature of 27°C, the mean kinetic energy is ${𝐸}_1$. When the temperature is increased to 327°C, we can determine the new mean kinetic energy ${𝐸}_2$ using the relationship:

$\frac{{𝐸}_1}{{𝐸}_2}=\frac{{𝑇}_1}{{𝑇}_2}$

To find ${𝐸}_2$, we can rearrange this equation:

${𝐸}_2={𝐸}_1\frac{{𝑇}_2}{{𝑇}_1}$

Converting the given temperatures from Celsius to Kelvin:

Initial temperature, ${𝑇}_1=27°𝐶+273=300𝐾$

Final temperature, ${𝑇}_2=327°𝐶+273=600𝐾$

Substituting these values into the equation:

${𝐸}_2={𝐸}_1\left(\frac{600}{300}\right)={𝐸}_1\times 2=2{𝐸}_1$

Therefore, when the temperature increases to 327°C, the mean kinetic energy of the atoms will be $2{𝐸}_1$.

To understand the displacement ${𝑥}_0$ where kinetic energy (KE) equals potential energy (PE) in simple harmonic motion (SHM), consider the following explanation:

At the position ${𝑥}_0$, the kinetic energy is equal to the potential energy. Mathematically, this is represented by:

$\frac{1}{2}𝑚{𝜔}^2({𝐴}^2-{𝑥}^2)=\frac{1}{2}𝑚{𝜔}^2{𝑥}^2$

Here, the left side of the equation represents the kinetic energy, and the right side represents the potential energy. Simplifying the equation:

${𝐴}^2-{𝑥}^2={𝑥}^2$

Rearranging terms, we get:

${𝐴}^2=2{𝑥}^2$

Solving for $𝑥$:

${𝑥}^2=\frac{{𝐴}^2}{2}$

$𝑥=\frac{𝐴}{\sqrt{\surd 2}}$

Thus, the value of $\left|{𝑥}_0\right|$ is $\frac{𝐴}{\sqrt{\surd 2}}$.

In a transverse wave, particles oscillate in a direction perpendicular to the wave's direction of propagation. Therefore, the angle between the particle velocity and wave velocity is $\frac{𝜋}{2}$ radians, except when the particle crosses the mean position.

$\text{ Since spheres are electrostatically shielded from each other, }𝜙=0\text{. }$

To determine the electric flux through sphere 'A', we can apply Gauss's Law, which states that the electric flux $𝜙$ through a closed surface is equal to the charge ${𝑄}_{\text{enc}}$ enclosed by that surface divided by the permittivity of free space ${𝜀}_0$.

For sphere 'A', which passes through the center of the dipole, the enclosed charge is only the negative charge $-𝑞$, since the positive charge $+𝑞$ lies outside of sphere 'A'.

Thus, the electric flux through sphere 'A' is calculated as:

${𝜙}_{\text{sphere }𝐴}=\frac{-𝑞}{{𝜀}_0}$

This result indicates that sphere 'A' encloses a net negative charge, contributing to a negative flux value.

To determine the work done in moving a charge from point B to point A, we use the formula:

${\text{W}}_{\text{ext}}=𝑞\Delta 𝑉$

Where:

$𝑞$ is the charge being moved, given as $5\text{mC}$ or $5\times {10}^{-3}\text{C}$.

$\Delta 𝑉$ is the change in electric potential between the two points: ${𝑉}_{𝑓}-{𝑉}_{𝑖}$.

Given:

Potential at point A, ${𝑉}_{𝐴}=-3\text{V}$

Potential at point B, ${𝑉}_{𝐵}=5\text{V}$

Substitute these values into the formula:

${\text{W}}_{\text{ext}}=5\times {10}^{-3}({𝑉}_{𝐴}-{𝑉}_{𝐵})$

$=5\times {10}^{-3}(-3-5)$

$=5\times {10}^{-3}\times (-8)$

$=-40\times {10}^{-3}\text{J}$

$=-0.04\text{J}$

Thus, the work done in carrying the charge from point B to point A is $-0.04\text{J}$.

For a conducting sphere with uniform surface charge distribution:

Inside the sphere ($𝑟<𝑅$): The electric field $𝐸=0$. This is because the electric field within a conductor in electrostatic equilibrium is zero.

Outside the sphere ($𝑟\ge 𝑅$): The electric field behaves like that of a point charge. It can be represented by the equation $𝐸=\frac{𝑘𝑄}{{𝑟}^2}$, where $𝑘$ is Coulomb's constant, $𝑄$ is the total charge on the sphere, and $𝑟$ is the distance from the center of the sphere.

Thus, the graph depicting the electric field as a function of distance $𝑟$ should show a flat line at $𝐸=0$ for $𝑟<𝑅$ and then a curve following the inverse square law $𝐸=\frac{𝑘𝑄}{{𝑟}^2}$ for $𝑟\ge 𝑅$.

The potential energy (PE) of an electric dipole in a uniform electric field is given by the formula:

$\mathrm{PE}=\overrightarrow{-𝑝}\cdot \overrightarrow{𝐸}=-𝑝𝐸\cos 𝜃$

Let's evaluate the potential energy for different angles:

(a) $𝜃={180}^{\circ }$:

$\mathrm{PE}=-𝑝𝐸\cos {180}^{\circ }=𝑝𝐸$

This corresponds to option (ii) in Column-II.

(b) $𝜃={120}^{\circ }$:

$\mathrm{PE}=-𝑝𝐸\cos {120}^{\circ }=-𝑝𝐸\left(\frac{-1}{2}\right)=\frac{𝑝𝐸}{2}$

This corresponds to option (iii) in Column-II.

(c) $𝜃={90}^{\circ }$:

$\mathrm{PE}=-𝑝𝐸\cos {90}^{\circ }=0$

This corresponds to option (iv) in Column-II.

Thus, the correct matching is:

$𝑎\to$ (ii)

$𝑏\to$ (iii)

$𝑐\to$ (iv)

The false statement is Option A.

Reasoning:

By definition, on an equipotential surface the potential $𝑉$ is the same everywhere, so any two points have

$\Delta 𝑉=0$.

The work done in moving a charge $𝑞$ through a potential difference $\Delta 𝑉$ is

$𝑊=𝑞\Delta 𝑉=𝑞\times 0=0.$

Hence it’s incorrect to say the work is “not zero.”

The other options are all true:

B: Equipotentials are surfaces of constant $𝑉$.

C: In a uniform field, equipotentials are equally spaced parallel planes.

D: $𝐄$ is always perpendicular to equipotential surfaces.

The only correct statement is Option D.

Reasoning:

Gauss’s law in integral form reads

${\Phi }_{𝐸}={\oint }_{𝑆}𝐄\cdot 𝑑𝐀=\frac{{𝑄}_{enclosed}}{{𝜀}_0},$

where the surface integral is taken over a closed surface $𝑆$.

It holds for any closed surface, regardless of its shape or how the enclosed charge is distributed.

Open surfaces (Option A) aren’t allowed because the “enclosed” concept only makes sense for closed surfaces.

Symmetry is not required for the validity of Gauss’s law—it only makes field calculations easier (so Option B is false).

Charges outside the surface contribute zero net flux, but Gauss’s law itself still applies (so Option C is false).

Hence:

Option D is correct.

To determine the terminal voltage across the cell in the given circuit, we start by calculating the current ($𝑖$) flowing through the circuit. The formula for the current is:

$𝑖=\frac{\text{Voltage}}{\text{Resistance}}=\frac{2}{4}=\frac{1}{2}\text{ A}$

Next, we calculate the terminal voltage using the formula for terminal voltage, which accounts for the internal resistance of the cell:

$\text{Terminal voltage}=\text{emf}-𝑖\times 𝑟$

where:

$\text{emf}$ is the electromotive force of the cell (2 V in this case),

$𝑖$ is the current calculated above ($\frac{1}{2}$ A),

$𝑟$ is the internal resistance of the cell (0.1 Ω).

Substituting these values into the formula:

$\text{Terminal voltage}=2-(\frac{1}{2}\times 0.1)=2-0.05=1.95\text{ V}$

Therefore, the terminal voltage across the cell is 1.95 V.

To find the equivalent electromotive force (emf) of the two cells connected in parallel, we use the formula for the equivalent emf, ${𝐸}_{\text{eq}}$, which is given by:

${𝐸}_{\text{eq}}=\frac{\frac{{𝐸}_1}{{𝑟}_1}+\frac{{𝐸}_2}{{𝑟}_2}}{\frac{1}{{𝑟}_1}+\frac{1}{{𝑟}_2}}=\frac{{𝐸}_1{𝑟}_2+{𝐸}_2{𝑟}_1}{{𝑟}_1+{𝑟}_2}$

This result represents the combined effect of the two cells' emfs and their internal resistances. The equivalent emf ${𝐸}_{\text{eq}}$ will lie between ${𝐸}_1$ and ${𝐸}_2$. Given that ${𝐸}_2>{𝐸}_1$ and ${𝑟}_2>{𝑟}_1$, the equivalent emf ${𝐸}_{\text{eq}}$ will be closer to ${𝐸}_2$ rather than ${𝐸}_1$ because the cell with the larger emf and larger internal resistance will have a more significant influence on the total emf value. Therefore, it can be concluded that ${𝐸}_1<{𝐸}_{\text{eq}}<{𝐸}_2$, and ${𝐸}_{\text{eq}}$ is nearer to ${𝐸}_2$.

$X\text{ is copper, }Y\text{ is nichrome, }Z\text{ is semiconductor }$

In a steady‐state circuit the charge per second crossing any section must be the same, so the current $𝐼$ is constant. However, for a non-uniform wire:

• Drift speed

${𝑣}_{𝑑}=\frac{𝐼}{𝑛𝑞𝐴}$

depends on the local cross-sectional area $𝐴$, so ${𝑣}_{𝑑}$ varies.

• Electric field

$𝐸=𝜌𝐽=𝜌\frac{𝐼}{𝐴}$

also varies with $𝐴$.

Therefore only the current remains the same throughout.

Answer: Option C (Current only).

To understand how the resistance of a metal wire varies with its diameter, while keeping other factors like length and temperature constant, we use the formula for resistance:

$𝑅=\frac{𝜌\ell }{𝐴}=\frac{𝜌\ell }{𝜋{𝑟}^2}$

Where:

$𝑅$ is the resistance.

$𝜌$ is the resistivity of the material.

$\ell$ is the length of the wire.

$𝐴$ is the cross-sectional area of the wire.

$𝑟$ is the radius of the wire.

Since $𝐴=𝜋{𝑟}^2$, the resistance $𝑅$ is inversely proportional to the square of the radius:

$𝑅\propto \frac{1}{{𝑟}^2}$

When considering the diameter $𝑑$ of the wire, which is twice the radius ($𝑑=2𝑟$), the relation can be rewritten in terms of the diameter:

$𝑅\propto \frac{1}{(𝑑/2{)}^2}=\frac{4}{{𝑑}^2}$

This indicates that as the diameter of the wire increases, the resistance decreases quadratically. The relationship between resistance $𝑅$ and diameter $𝑑$ is a hyperbolic function, as resistance is inversely proportional to the square of the diameter.

$\begin{array}{cc}& \frac{f}{\ell }=\frac{{𝜇}_{0}i,i_2}{t𝜋r}\\ & \text{ If they carry opposite currently they repeat each other }\end{array}$

$\ell =1m;r=2\mathrm{cm};N=5000;t=7A;\sin 𝜃\approx 1$

$B=\frac{{𝜇}_0\mathrm{Ni}}{\ell }\sin 𝜃$

$\begin{array}{cc}& 𝐵=\frac{4𝜋\times {10}^{-7}\times 5000\times 7\times 1}{1}\\ & 𝐵=4.396\times {10}^{-2}T\end{array}$

$\begin{array}{cc}& \text{ is }=\mathrm{ig}(h+s)\\ & i=\mathrm{ig}(\frac{g}{s}+1)\end{array}$

More 'S', less i

So milliammeter will have more shunt resistance.

The correct choice is Option D.

• Diamagnetic materials develop an induced magnetic moment opposite to an applied field, so their volume susceptibility is small and negative (χ < 0).

• They’re actually repelled (not attracted) by a magnet.

• Their permeability μ is just below 1 (not as high as 1000).

• And unlike paramagnets, their susceptibility is essentially independent of temperature.

The correct statement is Option B:

• Option B

A straight current-carrying conductor has circular magnetic field lines around it.

Why the others are incorrect:

Option A

A neutral, current-carrying wire does not produce an electric field around it (externally it remains electrically neutral).

Option C

Fleming’s Left-Hand Rule gives the direction of force on a current in an external magnetic field, not the field created by the current itself. The field around a straight conductor follows the Right-Hand Thumb Rule.

Option D

Inside a long, closely wound solenoid the magnetic field is essentially uniform along its length.

The magnetic field $𝐵$ around an infinitely long, straight conductor carrying a steady current can be described using the equation:

$𝐵=\frac{{𝜇}_0𝑖}{2𝜋𝑟}$

where:

$𝐵$ is the magnetic field,

${𝜇}_0$ is the permeability of free space,

$𝑖$ is the current through the conductor, and

$𝑟$ is the perpendicular distance from the conductor.

From this equation, we can see that $𝐵$ is inversely proportional to $𝑟$, which implies:

$𝐵\propto \frac{1}{𝑟}$

This relationship is characteristic of a rectangular hyperbola, meaning the graph of magnetic field $𝐵$ versus distance $𝑟$ will form a hyperbolic shape.

They’ll have the same momentum.

Reasoning (from de Broglie’s relation):

$𝜆=\frac{ℎ}{𝑝}⟹𝑝=\frac{ℎ}{𝜆}.$

If an electron and a photon share the same λ, they share the same p.

According to Lenz's Law, a changing magnetic flux through a coil induces an electromotive force (EMF) that opposes the change in flux. When a bar magnet is pushed towards the coil, the change in magnetic flux through the coil causes the galvanometer pointer to deflect in one direction (towards X, in this case). Conversely, when the bar magnet is pulled away from the coil, the change in magnetic flux is in the opposite direction. Thus, the galvanometer pointer will deflect in the opposite direction, which is towards ${𝑋}^1$. This occurs as the system attempts to oppose the change in magnetic flux by inducing a current that produces a magnetic field opposing the original direction.

To find the magnetic flux through a square loop of side 2 meters that lies in the Y-Z plane, given the magnetic field $\overrightarrow{B}=(5\hat{i}+3\hat{j}-4\hat{k})T$, we proceed as follows:

Determine the area vector:

Since the loop is in the Y-Z plane, its area vector $\bar{A}$ is perpendicular to this plane and points along the X-axis. The magnitude of the area vector is given by the area of the square, which is $2\times 2=4$ square meters. The area vector is thus $\bar{A}=4\hat{i}$.

Calculate the magnetic flux:

The magnetic flux $𝜙$ through the loop is given by the dot product of the magnetic field $\bar{B}$ and the area vector $\bar{A}$:

$𝜙=\bar{B}\cdot \bar{A}=(5\hat{i}+3\hat{j}-4\hat{k})\cdot 4\hat{i}$

Performing the dot product, we have:

$𝜙=(5\times 4)+(3\times 0)+(-4\times 0)=20\mathrm{Wb}$

Thus, the magnitude of the magnetic flux through the square loop is $20\mathrm{Wb}$.

Domestic mains use alternating voltage and thus also alternating current. That matches:

Option D: AC voltage and AC current.

The sinusoidal voltage from an AC generator at any instant is represented by the equation $𝑉=311\sin 314𝑡$.

To determine the root mean square (rms) value of the voltage, we use the formula for the rms value of a sinusoidal voltage:

${𝑉}_{\text{rms}}=\frac{{𝑉}_0}{\sqrt{\surd 2}}=\frac{311}{\sqrt{\surd 2}}=220\text{V}$

Here, ${𝑉}_0=311\text{V}$ is the peak voltage.

To find the frequency, we use the relationship between angular frequency ($𝜔$) and frequency ($𝑓$):

$𝜔=2𝜋𝑓$

Given $𝜔=314$, we set:

$314=2𝜋𝑓$

Solving for $𝑓$, we find:

$𝑓=\frac{314}{2𝜋}\approx 50\text{Hz}$

Thus, the rms value of the voltage is 220 V, and the frequency is 50 Hz.

$\begin{array}{cc}& {𝑋}_{𝐶}=24\Omega \\ & {𝑋}_{𝐿}=16\Omega \\ & 𝑅=6\Omega \\ & 𝑍=\sqrt{\surd {𝑅}^2+{({𝑋}_{𝐶}-{𝑋}_{𝐿})}^2}\\ & =\sqrt{\surd 6^2+8^2}\\ & =\sqrt{\surd 100}=10\Omega \end{array}$

$\begin{array}{cc}& i=\frac{V}{Z}=\frac{100}{10}=10A\\ & \text{ Net voltage across capacitor and inductor }\\ & \begin{array}{cc}& =i(X_C-X_L)\\ & =10(24-16)\\ & =10\left(8\right)\\ & =80V\end{array}\end{array}$

The correct matching is

i. Microwave → c. 0.1 m to 1 mm

ii. Visible light → a. 700 nm to 400 nm

iii. Ultraviolet → d. 400 nm to 1 nm

iv. X‐rays → b. 1 nm to 10⁻³ nm

That corresponds to Option A.