NEET-UG 2024

V.C = MSD $-$ VSD ..... (1)

given : $(N+1)\mathrm{VSD}=N$ MSD

$\mathrm{VSD}=\left(\frac{N}{N+1}\right)\mathrm{MSD}$ ..... (2)

From (1) and (2)

$$\begin{array}{rl}& V.C=(MSD)-\frac{N}{N+1}(MSD)\\ & =MSD\left(1-\frac{N}{N+1}\right)=\frac{MSD}{N+1}\\ & =\frac{0.01}{N+1}=\frac{1}{100(N+1)}\end{array}$$

Solid angle $d\Omega =\frac{dA}{r^2}$ has dimensions $[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^0]$

Strain $=\frac{\Delta l}{l}$ has dimensions $[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^0]$

Angle measured in radians is also dimensionless $[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^0]$

$$\theta =\frac{I}{r}$$

The force acting on a particle at a given time $t$ is defined as $F=\alpha t^2+\beta t$, where $\alpha$ and $\beta$ are constants. We need to determine which factor is dimensionless if $\alpha$ and $\beta$ are constants.

Using the principle of homogeneity of dimensions:

$\begin{array}{rl}& [F]=[\alpha t^2]=[\beta t]\\ & [\alpha ]=\frac{[F]}{[t^2]}\text{and}[\beta ]=\frac{[F]}{[t]}\\ & \therefore [\alpha ][t]=[\beta ]\\ & \therefore \frac{\alpha t}{\beta }=\text{dimensionless}\end{array}$

Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially. After that, the slope of v-t curve is constant and positive.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of v-t curve is constant and negative.

$$\begin{array}{rl}& F=\left(M_1+M_2\right)a\\ & a=\frac{10}{2+3}=2{\mathrm{ms}}^{-2}\\ & F^{\prime }=M_2(2)=3\times 2\mathrm{N}=6\mathrm{N}\end{array}$$

When a particle moves along a circular path with a uniform speed, it is important to understand the motion characteristics in terms of both velocity and acceleration. Velocity is a vector quantity which means it has both magnitude and direction, while acceleration is the rate of change of velocity with respect to time.

First, let's analyze the velocity:

Although the speed (magnitude of the velocity vector) remains constant in uniform circular motion, the direction of the velocity vector continuously changes as the particle progresses along the circle. Since velocity includes both the magnitude and the direction, any change in either results in a change in velocity. Consequently, in uniform circular motion, the velocity of the particle is not constant but varies due to the continuous change in direction.

Next, consider the acceleration:

In circular motion, there is always an acceleration directed towards the center of the circle, known as centripetal acceleration. This acceleration is responsible for changing the direction of the velocity vector, thereby keeping the particle moving in a circle, despite the speed being constant. The formula for centripetal acceleration is:

$$a_c=\frac{v^2}{r}$$

where $v$ is the speed of the particle and $r$ is the radius of the circle. This acceleration is always directed towards the center of the circle and varies with the square of the speed and inversely with the radius of the circle.

Therefore, given that both the velocity and acceleration change, the correct choice is:

Option D - Varying velocity and varying acceleration

It states that in uniform circular motion, both the velocity and acceleration of the particle vary – velocity due to continuous changes in direction and acceleration due to its consistent inward (centripetal) direction towards the center of the circle.

To find the instantaneous power delivered by a force, we can use the formula:

$$P=\vec{F}\cdot \vec{v}$$

where $P$ is the power, $\vec{F}$ is the force vector, and $\vec{v}$ is the velocity vector of the particle.

In this question, the force exerted is given as a constant $5N$. Since no direction is specified, and the displacement is given in a scalar form, we assume the force acts along the direction of displacement. Therefore, we can treat the vectors as scalars for simplicity.

The displacement of the particle is given by:

$$x=2t-1$$

The velocity, $\vec{v}$, is the derivative of displacement with respect to time. Deriving the displacement equation with respect to time $t$ gives:

$$v=\frac{dx}{dt}=\frac{d}{dt}(2t-1)=2$$

The instantaneous power can now be calculated as:

$$P=F\cdot v=5\times 2=10\text{Watts}$$

Thus, the instantaneous power delivered by the force at any time $t$ is $10\text{Watts}$.

The correct answer is Option A: 10.

In a completely inelastic collision, the two bodies stick together and move with a common final velocity. Here, before the collision, body $A$ is moving with a velocity $v_1$ and body $B$ is at rest. The conservation of momentum must hold true because no external forces are acting on the system.

The momentum before the collision is only due to body $A$ since body $B$ is at rest. Therefore, the total initial momentum $p_{\text{initial}}$ of the system is given by:

$$p_{\text{initial}}=m_A{v}_1+m_B{v}_0=m{v}_1+0=m{v}_1$$

where:

• $m_A$ and $m_B$ are the masses of bodies $A$ and $B$ respectively,


• $m$ is the mass of each body,


• $v_1$ is the velocity of body $A$,


• $v_0=0$ is the velocity of body $B$ because it is initially at rest.

Since they undergo a completely inelastic collision, bodies $A$ and $B$ stick together after the collision and hence move with a common velocity $v_2$. The total mass of the combined system post-collision is $m_A+m_B=m+m=2m$. The momentum after the collision is given by:

$$p_{\text{final}}=(m_A+m_B)v_2=2m{v}_2$$

Applying the conservation of momentum (since no external force implies momentum is conserved), we equate $p_{\text{initial}}$ and $p_{\text{final}}$:

$$m{v}_1=2m{v}_2$$

Dividing through by $m$ yields:

$$v_1=2v_2$$

Thus, solving for the ratio $\frac{v_1}{v_2}$ gives:

$$\frac{v_1}{v_2}=2$$

Therefore, the ratio of $v_1$ to $v_2$ is $2:1$, making the correct answer:

Option B: $2:1$

$$\begin{array}{rl}& \text{ Moment of inertia of rod }=I=\frac{m{\ell }^2}{12}\\ & \Rightarrow 2400=400\frac{{\ell }^2}{12}\\ & \Rightarrow 72={\ell }^2\\ & \Rightarrow \ell =\sqrt{72}=8.48\mathrm{cm}\simeq 8.5\mathrm{cm}\end{array}$$

In the case of pure rolling,

The topmost point will have velocity $2v$ while point $Q$ i.e. lowest point will have zero velocity. Hence point $P$ moves faster than point $Q$.

The acceleration due to gravity (g) on a planet is given by the formula:

$$g=G\frac{M}{R^2}$$

where:

• $G$ is the universal gravitational constant,
• $M$ is the mass of the planet, and
• $R$ is the radius of the planet.

In this question, we know that:

• The mass of the planet $M_p$ is $\frac{1}{10}$ of the mass of the earth $M_e$, so $M_p=\frac{M_e}{10}$.
• The diameter of the planet is half that of earth. Since the diameter is twice the radius, a diameter half that of earth implies the radius $R_p$ is also half the radius of the earth $R_e$. Therefore, $R_p=\frac{R_e}{2}$.

Using the formula for acceleration due to gravity and substituting the above information:

$$g_p=G\frac{M_p}{R_p^2}=G\frac{\frac{M_e}{10}}{{\left(\frac{R_e}{2}\right)}^2}$$

Simplify the expression:

$$g_p=G\frac{M_e}{10}\cdot \frac{4}{R_e^2}=\frac{4}{10}G\frac{M_e}{R_e^2}$$

Knowing that $G\frac{M_e}{R_e^2}$ is the acceleration due to gravity on Earth $g_e\approx 9.8{m/s}^2$, we substitute this value into the equation:

$$g_p=\frac{4}{10}\cdot 9.8{m/s}^2=3.92{m/s}^2$$

Thus, the acceleration due to gravity on the planet is $3.92{m/s}^2$, which corresponds to:

Option D: $3.92{m/s}^2$

Apply energy conservation,

$$\begin{array}{rl}& U_i+K_i=U_f+K_f\\ & \Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}m{v}^2\\ & \Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}\times m\times \frac{GM}{3R}\\ & \Rightarrow K_i=-\frac{1}{6}\frac{GMm}{R}+\frac{GMm}{R}\\ & K_i=\frac{5}{6}\frac{GMm}{R}\end{array}$$

First, we need to find the maximum force that can be applied to the steel wire within its elastic limit. This force can be calculated using the given area under stress and the stress limit provided by the elastic limit of steel.

Let's assume the area of cross-section of the wire is $A$. The force exerted can be given by:

$$F=\sigma \times A$$

where:

$\sigma =8\times 10^8{N/m}^2$ (Elastic limit of steel)

To calculate the elongation ($\Delta L$) under this force, we use Hooke's Law, which relates force, elongation, cross-sectional area, original length, and Young's modulus as follows:

$$F=\frac{Y\times A\times \Delta L}{L}$$

where:

$Y=2\times 10^{11}{N/m}^2$ (Young's modulus of steel),

$L=1\mathrm{m}$ (original length of the wire).

Substituting for $F$ from the earlier expression and rearranging the formula, we get:

$$\sigma \times A=\frac{Y\times A\times \Delta L}{L}$$

$$\sigma =\frac{Y\times \Delta L}{L}$$

$$\Delta L=\frac{\sigma \times L}{Y}$$

$$\Delta L=\frac{8\times 10^8\times 1}{2\times 10^{11}}\mathrm{meters}$$

$$\Delta L=0.004\mathrm{meters}$$

$$\Delta L=4\mathrm{mm}$$

Thus, the maximum elongation of the wire within the elastic limit is $4\mathrm{mm}$.

The answer is: Option A - 4 mm.

$$\begin{array}{rl}& \text{ Excess force }=T\times 2\pi R\\ & =\frac{7}{100}\times 2\times 3.14\times \frac{4.5}{100}\\ & =197.82\times 10^{-4}\\ & =19.8\times 10^{-3}\mathrm{N}\\ & =19.8\mathrm{mN}\end{array}$$

Given the properties and conditions of the metallic bar, we are required to calculate the compressive force developed due to heating. Key inputs include the Young's modulus (E), coefficient of linear thermal expansion (α), change in temperature ($\Delta T$), and the original dimensions of the bar.

First, compute the linear expansion of the bar if it were free to expand. The change in length ($\Delta L$) due to thermal expansion can be computed through the formula:

$$\Delta L=\alpha L_0\Delta T$$

Given:

• $$\alpha =10^{-5}{}^{\circ }{C}^{-1}$$
• $$L_0=1\text{ m}$$
• $$\Delta T=100^{\circ }C$$

Thus:

$$\Delta L=10^{-5}\times 1\times 100=0.001\text{ m}$$

This is the change in length that the bar would undergo if not constrained.

However, in this scenario, the bar is constrained and does not actually expand. This constraint induces a compressive stress (constrained thermal stress) in the bar, which can be calculated using the formula relating stress, Young's modulus, and strain:

$$\sigma =Eϵ$$

Where the strain ($ϵ$) under constrained conditions due to thermal expansion is:

$$ϵ=\frac{\Delta L}{L_0}=\frac{0.001}{1}=0.001$$

Therefore:

$$\sigma =0.5\times 10^{11}\times 0.001=5\times 10^7{N/m}^2$$

This stress is the force per unit area. To find the compressive force, we need to multiply this stress by the cross-sectional area of the bar:

$$F=\sigma A=5\times 10^7{N/m}^2\times 10^{-3}{\mathrm{m}}^2=5\times 10^4\mathrm{N}$$

Thus, the compressive force developed in the bar is $50\times 10^3\mathrm{N}$.

Hence, the correct answer is Option B:

$$50\times 10^3\mathrm{N}$$

Path $bc$ is an isochoric process.

$\therefore$ Work done by gas along path $bc$ is zero.

At same temperature, curve with higher volume corresponds to lower pressure.

$$\begin{array}{rl}& V_3>V_2>V_1\\ & \Rightarrow P_1>P_2>P_3\end{array}$$

(We draw a straight line parallel to volume axis to get this)

In the equation for simple harmonic motion (SHM), $x=5\sin \left(\pi t+\frac{\pi }{3}\right)\mathrm{m},$ the general form $x=A\sin (\omega t+\varphi )$ can be used to identify the parameters of SHM, where:

• A is the amplitude.
• ω (omega) is the angular frequency.
• φ (phi) is the phase constant.
• t is the time.

Comparing the given equation with the standard form:

• The amplitude A is 5 m, as that is the coefficient of sine in the equation.
• The angular frequency ω is $\pi$ rad/s.

The angular frequency $\omega$ is related to the time period $T$ of the motion through the formula:

$\omega =\frac{2\pi }{T}$.

Given that $\omega =\pi$, we can substitute and solve for $T$:

$$\begin{array}{rl}\pi =\frac{2\pi }{T}& ⟹T=\frac{2\pi }{\pi }=2\text{ seconds}.\end{array}$$

Hence, the amplitude of the motion is 5 m and the time period is 2 s. Thus, the correct answer is:

Option B: 5 m, 2 s

The period of oscillation, $T$, of a simple pendulum is determined by the formula:

$$T=2\pi \sqrt{\frac{L}{g}}$$

where:

• $L$ is the length of the pendulum
• $g$ is the acceleration due to gravity

The mass of the bob does not factor into the equation for the period.

Let's first denote the original length of the pendulum as $L$ and the original period of oscillation as $T_1$. Hence,

$$T_1=2\pi \sqrt{\frac{L}{g}}$$

When the length of the pendulum is halved, the new length $L^{\prime }$ would be $\frac{L}{2}$. Thus, the new period $T_2$ can be calculated as:

$$T_2=2\pi \sqrt{\frac{\frac{L}{2}}{g}}=2\pi \sqrt{\frac{L}{2g}}=2\pi \left(\frac{1}{\sqrt{2}}\right)\sqrt{\frac{L}{g}}=\frac{1}{\sqrt{2}}\cdot 2\pi \sqrt{\frac{L}{g}}=\frac{T_1}{\sqrt{2}}$$

We are given that the new period $T_2$ is $\frac{x}{2}{T}_1$. Therefore, we can set up the equation:

$$\frac{T_1}{\sqrt{2}}=\frac{x}{2}{T}_1$$

To find the value of $x$, we solve for $x$:

$$\frac{1}{\sqrt{2}}=\frac{x}{2}$$

Multiplying both sides by 2:

$$\frac{2}{\sqrt{2}}=x$$

Simplify to:

$$x=\sqrt{2}$$

Hence, the correct answer is:

Option B: $\sqrt{2}$

For uniformly charged spherical shell,

$$\begin{array}{rl}& V=\frac{kq}{R}(\text{ For }r\le R)\\ & \therefore V_C=V_P\\ & V_C-V_P=\text{ Zero }\end{array}$$

The potential $V$ at any point, at distance $r$ from centre of dipole $=\frac{KP\cos \theta }{r^2}$

At axial point where $\theta =0^{\circ },V=\frac{KP}{r^2}=\frac{9\times 10^9\times 4\times 10^{-6}}{2^2}=9\times 10^3\mathrm{V}$

At axial point where $\theta =180^{\circ },V=\frac{-KP}{r^2}=-9\times 10^3\mathrm{V}$

$$\begin{array}{rl}\text{ Current in circuit }i& =\frac{10}{4+1}=2\mathrm{A}\\ \text{ Terminal voltage }& =E-iR\\ & =10-2\times 1=8\mathrm{V}\end{array}$$

To solve this problem, we first need to understand how the resistance changes when we cut the wire into equal parts and how it behaves when connected in different configurations (series and parallel).

Given:

• Original length of the wire, $l$.
• Total resistance of the wire $R=100\Omega$.
• The wire is divided into 10 equal parts.

Resistance of each part:

Since the wire is divided into 10 equal parts, the length of each part is $\frac{l}{10}$. Resistance is proportional to length (as long as the cross-sectional area and material of the wire remain constant). Therefore, the resistance of each part, denoted as $r$, is $\frac{1}{10}$th of the total resistance:

$$r=\frac{R}{10}=\frac{100\Omega }{10}=10\Omega$$

First 5 parts in series:

When resistors are connected in series, the total resistance is the sum of the individual resistances:

$$R_{\text{series}}=5\times 10\Omega =50\Omega$$

Next 5 parts in parallel:

When resistors are connected in parallel, the total resistance $R_{\text{parallel}}$ can be calculated using the reciprocal formula:

$$\frac{1}{R_{\text{parallel}}}=\frac{1}{10\Omega }+\frac{1}{10\Omega }+\frac{1}{10\Omega }+\frac{1}{10\Omega }+\frac{1}{10\Omega }=5\times \frac{1}{10\Omega }=\frac{5}{10\Omega }=\frac{1}{2\Omega }$$

Thus,

$$R_{\text{parallel}}=2\Omega$$

Final combination in series:

The total resistance of the combination, where the series and parallel groups are again connected in series, will be:

$$R_{\text{total}}=R_{\text{series}}+R_{\text{parallel}}=50\Omega +2\Omega =52\Omega$$

Thus, the correct answer to the resistance of the final combination is:

Option B $52\Omega$

To find the ratio of power outputs when two heaters with different power ratings are connected first in series and then in parallel, we need to understand how the total power output varies based on the type of connection.

Heater Specifications:

• Power of heater A $(P_A)=1\text{kW}=1000\text{W}$
• Power of heater B $(P_B)=2\text{kW}=2000\text{W}$

Scenario 1: Series Connection

When resistors (or heaters in this case) are connected in series, the total resistance ($R_{\text{series}}$) is the sum of the individual resistances ($R_A$ and $R_B$).

Using the formula for electrical power: $P=\frac{V^2}{R}$,

where $P$ is power, $V$ is voltage, and $R$ is resistance, we can express the resistance of each heater as:

$$R_A=\frac{V^2}{P_A}$$

$$R_B=\frac{V^2}{P_B}$$

Substitute the given power values:

$$R_A=\frac{V^2}{1000}$$

$$R_B=\frac{V^2}{2000}$$

Then the total resistance for the series connection is:

$$R_{\text{series}}=R_A+R_B=\frac{V^2}{1000}+\frac{V^2}{2000}=\frac{3V^2}{2000}$$

The total power output in series ($P_{\text{series}}$) is:

$$P_{\text{series}}=\frac{V^2}{R_{\text{series}}}=\frac{V^2}{\frac{3V^2}{2000}}=\frac{2000}{3}\text{W}$$

Scenario 2: Parallel Connection

For parallel connections, the total resistance ($R_{\text{parallel}}$) is given by:

$$\frac{1}{R_{\text{parallel}}}=\frac{1}{R_A}+\frac{1}{R_B}=\frac{1}{\frac{V^2}{1000}}+\frac{1}{\frac{V^2}{2000}}=\frac{3}{2V^2}$$

Reformulate to find $R_{\text{parallel}}$:

$$R_{\text{parallel}}=\frac{2V^2}{3}$$

And the total power output in parallel ($P_{\text{parallel}}$) is:

$$P_{\text{parallel}}=\frac{V^2}{R_{\text{parallel}}}=\frac{V^2}{\frac{2V^2}{3}}=\frac{3V^2}{2}$$

However, simplifying,

$$P_{\text{parallel}}=\frac{3}{2}{V}^2$$

The ratio of powers is then:

$$\frac{P_{\text{series}}}{P_{\text{parallel}}}=\frac{\frac{2000}{3}}{\frac{3V^2}{2}}$$

Solving and simplifying,

$$\text{Ratio}=\frac{\frac{2000}{3}}{\frac{3\times V^2}{2}}=\frac{2000\times 2}{3\times 3\times V^2}=\frac{4000}{9V^2}$$

Given that we know one ratio of the actual power values, we simplify further. For calculating power in simple terms, consider voltage to be normalized (taken out of the fraction):

$$\frac{P_{\text{series}}}{P_{\text{parallel}}}=\frac{\frac{2000}{3}}{\frac{6000}{2}}=\frac{2000\times 2}{3\times 6000}=\frac{4000}{18000}=\frac{2}{9}$$

Therefore, the ratio of the power outputs when the heaters are connected first in series and then in parallel is 2:9, which corresponds to Option B.

In option (1),

$$\frac{10}{15}=\frac{10}{5+R_D}$$

The diode can conduct and have resistance $R_D=10\Omega$ because diode have dynamic resistance. In that case bridge will be balanced.

Given circuit is balanced Wheatstone bridge

$$\begin{array}{rl}{C}_{AB}& =1+1\\ & =2\mu F\end{array}$$

Given $V^{\prime }=V=$ Constant

$$\begin{array}{rl}\text{(i)}& C^{\prime }=\frac{{\epsilon }_{0}A}{d^{\prime }},C=\frac{{\epsilon }_{0}A}{d}\\ & d^{\prime }<d\\ & C^{\prime }>C\end{array}$$

Hence, final capacitance greater than initial capacitance,

$$\begin{array}{rl}\text{ (ii) }{U}^{\prime }& =\frac{1}{2}{C}^{\prime }{V}^2\\ U& =\frac{1}{2}C{V}^2\\ U^{\prime }& >U\end{array}$$

Hence final energy is greater than initial energy

$$\begin{array}{rl}\text{ (iii) }\frac{Q^{\prime }}{V^{\prime }}& =C^{\prime }\text{ and }\frac{Q}{V}=C\\ \frac{Q^{\prime }}{V^{\prime }}& \ne \frac{Q}{V}\end{array}$$

(iv) Product of charge and voltage

$$\begin{array}{rl}& X^{\prime }=Q^{\prime }V=C^{\prime }{V}^2\\ & X=QV=C{V}^2\\ & X^{\prime }>X\end{array}$$

The magnitude of magnetic field due to circular coil of N turns is given by

$$\begin{array}{rl}& B_C=\frac{{\mu }_{0}iN}{2R}\\ & =\frac{4\pi \times 10^{-7}\times 7\times 100}{2\times 0.1}\\ & =4.4\times 10^{-3}\mathrm{T}\\ & =4.4\mathrm{mT}\end{array}$$

According to modified Ampere's law

$$\oint B\cdot dI={\mu }_0\left(I_C+I_D\right)$$

For Loop $L_1{I}_C\ne 0$ and $I_D=0$

For Loop $L_2{I}_C=0$ and $I_D\ne 0$

Due to $\mathrm{KCL}{I}_C=I_D$

Time period of Oscillation, $T=2\pi \sqrt{\frac{I}{MB}}$

$$\begin{array}{rl}& \Rightarrow \frac{1}{4}=2\pi \sqrt{\frac{9.8\times 10^{-6}}{M\times 0.049}}\\ & \Rightarrow \frac{1}{16}=4{\pi }^2\times \frac{9.8\times 10^{-6}}{M\times 49\times 10^{-3}}\\ & \Rightarrow M=\frac{4{\pi }^2\times 9.8\times 10^{-6}}{49\times 10^{-3}}\times 16\\ & =\frac{4{\pi }^2\times 9.8\times 16\times 10^{-3}}{49}\\ & =12.8{\pi }^2\times 10^{-3}\times 10^{-2}\times 10^2\\ & =1280{\pi }^2\times 10^{-5}{\mathrm{Am}}^2\end{array}$$

North of magnet is moving away from solenoid 1 so end $B$ of solenoid 1 is South and as south of magnet is approaching solenoid 2 so end $C$ of solenoid 2 is South.

To match the magnetic materials with their respective magnetic susceptibilities, we must understand the characteristics of each type of material regarding its magnetic behavior. Here are the implications of each type:

Diamagnetic Materials: These materials are repelled by magnetic fields. The susceptibility ($\chi$) of diamagnetic materials is negative but very small, near zero. Therefore, $\chi >-1$ and closer to zero, reflects this property.

Ferromagnetic Materials: They have a very high, positive susceptibility because these materials can be permanently magnetized. Their susceptibility ($\chi$) is much greater than one ($\chi >>1$).

Paramagnetic Materials: These materials are weakly attracted by a magnetic field, and their susceptibility is positive. However, the value of $\chi$ for paramagnetic materials is small but greater than zero ($0<\chi <\epsilon$), where $\epsilon$ represents a small positive value.

Non-magnetic Materials: For materials that are considered non-magnetic, the susceptibility ($\chi$) is zero ($\chi =0$), as they are neither attracted nor repelled by magnetic fields.

Let's match these descriptions with the given List II in your question:

• Diamagnetic (A) matches with $0>\chi \ge -1$ (II).
• Ferromagnetic (B) corresponds to $\chi >>1$ (III).
• Paramagnetic (C) should be aligned with $0<\chi <\epsilon$ (IV).
• Non-magnetic (D) clearly fits $\chi =0$ (I).

Therefore, the correct answer matches:

• A-II, B-III, C-IV, D-I.

This aligns with Option A.

To understand which statement(s) is/are correct, we must explore the interactions between the magnetic field and different types of materials (magnetic, non-magnetic, conducting, non-conducting).

Statement A: If a sheet is magnetic and placed near a strong magnetic pole, it will experience a magnetic force due to the magnetic field. This force could either attract or repel the sheet depending on the polarity of the magnetic pole and the induced or inherent poles in the sheet. A force is needed to hold the sheet stationary against this magnetic force. Hence, statement A is true.

Statement B: If the sheet is non-magnetic, it will not experience any magnetic force because non-magnetic materials do not respond to magnetic fields in a manner where a force would be exerted on them. Thus, no external force is needed to hold a non-magnetic sheet in place near a magnetic pole. Statement B is false.

Statement C: If the sheet is conducting and you move it away from the magnetic pole, it will experience a change in the magnetic flux through it. According to Faraday's Law of Electromagnetic Induction, a change in magnetic flux induces an electromotive force (EMF) and consequently, currents known as eddy currents in the conductor. These induced currents produce their own magnetic fields, which interact with the original magnetic field, leading to a force (Lenz's Law). To move the sheet away with uniform velocity, a force must counteract this magnetic interaction. Therefore, statement C is true.

Statement D: If the sheet is non-conducting and non-polar, it will neither induce currents (since it's non-conducting) nor experience magnetic forces (since it's non-magnetic). Thus, moving such a sheet away from a magnetic pole with uniform velocity does not require overcoming any electromagnetic forces. Statement D is false.

Given these explanations, the correct statements are A and C only.

The correct choice is: Option B: A and C only.

$$\begin{array}{rl}& \Delta l=2\frac{I}{2}\sin 30^{\circ }\\ & =\frac{I}{2}\\ & M^{\prime }=mI/2\\ & =M/2\end{array}$$

An ideal transformer is a device that transfers electrical energy from one circuit to another through inductively coupled conductors—the transformer's coils. The primary coil (with $N_P$ turns) is connected to the input voltage ($V_P$), and the secondary coil (with $N_S$ turns) delivers the output voltage ($V_S$).

The relationship between the input (primary) voltage and output (secondary) voltage in a transformer is determined by the ratio of the number of turns in the primary coil to the number of turns in the secondary coil, expressed by the formula:

In this case, the given turns ratio is $\frac{N_P}{N_S}=\frac{1}{2}$. To find the correct expression for $\frac{V_S}{V_P}$, we need to invert the given ratio to reflect the relationship with respect to the voltages. This means:

Therefore, substituting into the voltage ratio equation:

This indicates that the secondary voltage ($V_S$) is twice the primary voltage ($V_P$). If we were to express the ratio $V_S:V_P$ based on this, it would be:

Option B $2:1$

This is the correct answer as it represents that, in accordance with the transformer's turns ratio, the voltage in the secondary coil is twice that in the primary coil.

Capacitive Reactance

$$\begin{array}{r}{X}_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\times 3.14\times 50\times 10\times 10^{-6}}\\ =\frac{1000}{3.14}\end{array}$$

$$\begin{array}{rl}& V_{\mathrm{rms}}=210\mathrm{V}\\ & i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C}\\ & \text{ Peak current }=\sqrt{2}{i}_{\mathrm{ms}}=\sqrt{2}\times \frac{210}{1000}\times 3.14=0.932\\ & =0.93\mathrm{A}\end{array}$$

Electromagnetic waves have several defining characteristics when they propagate through free space. Let's evaluate each of the options provided:

Option A: They are transverse in nature.

This is true. Electromagnetic waves are transverse waves, meaning the directions of the electric field and magnetic field oscillations are perpendicular to the direction of wave propagation. The electric field (E) and magnetic field (B) vectors are also perpendicular to each other and to the direction of propagation.

Option B: The energy density in the electric field is equal to the energy density in the magnetic field.

This is also true. In electromagnetic waves, the energy density stored in the electric field is equal to the energy density stored in the magnetic field. This is because the magnitudes of the electric and magnetic fields are related by $c=\frac{E}{B}$ where c is the speed of light in vacuum. The energy density for each is given by $\frac{1}{2}{ϵ}_0{E}^2$ for the electric field and $\frac{1}{2}\frac{B^2}{{\mu }_0}$ for the magnetic field. Given the relationship between E and B in a wave, these two energy densities are equal.

Option C: They travel with a speed equal to $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$.

This statement is true. The speed of electromagnetic waves in vacuum is given by $c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$, where ${\mu }_0$ is the magnetic permeability of free space and ${\epsilon }_0$ is the electric permittivity of free space. This relationship derives from Maxwell’s equations in a vacuum.

Option D: They originate from charges moving with uniform speed.

This statement is false. Electromagnetic waves are not generally produced by charges moving with a uniform speed; rather, they are produced by charges that are accelerating. Uniform motion (where velocity is constant and acceleration is zero) does not result in radiation of electromagnetic waves. If a charge is accelerating – changing either the speed or direction of its motion – it emits electromagnetic radiation.

Therefore, the correct answer is Option D, as it is the property that is not true for electromagnetic waves traveling in free space.

In prism, $r_1+c=A$

$$\begin{array}{rl}& r_1=90^{\circ }-c\text{.... (1)}\\ & \sin c=\frac{1}{\mu }\Rightarrow \cos c=\frac{\sqrt{{\mu }^2-1}}{\mu }\end{array}$$

$\Rightarrow$ Apply Snell's law, on incidence surface

$$\begin{array}{r}1\cdot \sin 30^{\circ }=\mu \sin \left(r_1\right)\Rightarrow 1\times \frac{1}{2}=\mu \times \sin \left(90^{\circ }-c\right)\\ \frac{1}{2}=\mu \times \frac{\sqrt{{\mu }^2-1}}{\mu }\end{array}$$

On squaring $\frac{1}{4}={\mu }^2-1$

$$\Rightarrow {\mu }^2=\frac{5}{4}\Rightarrow \mu =\frac{\sqrt{5}}{2}$$

The magnifying power of a telescope when viewing distant objects can be calculated using the formula:

$$\text{Magnifying Power}(M)=\frac{\text{Focal Length of Objective Lens}(f_o)}{\text{Focal Length of Eyepiece Lens}(f_e)}$$

Given in the problem:

• Focal Length of Objective Lens, $f_o=140\mathrm{cm}$
• Focal Length of Eyepiece Lens, $f_e=5.0\mathrm{cm}$

Substituting these values into the formula gives:

$$M=\frac{140\mathrm{cm}}{5.0\mathrm{cm}}=28$$

Thus, the magnifying power of the telescope is 28. The correct answer is Option B.

In Young’s double-slit experiment, if a monochromatic source is replaced by white light, which contains multiple wavelengths, the patterns observed on the screen will differ significantly.

First, remember that the pattern formed in the double-slit experiment consists of both bright and dark fringes due to constructive and destructive interference, respectively. The position and intensity of these fringes depend on the wavelength of the light used. For monochromatic light (light of a single wavelength), the interference pattern is stable, with bright and dark fringes evenly spaced. Each fringe is uniformly bright or dark.

When using white light, which is a combination of various wavelengths of light, each color, or each wavelength, forms its own interference pattern with slightly different fringe spacing. This occurs because the separation between fringes $\Delta y$ is given by the formula:

$$\Delta y=\frac{\lambda L}{d}$$

Where:

• $\lambda$ is the wavelength of light
• $L$ is the distance from the slits to the screen
• $d$ is the separation between the slits

Since different wavelengths have different values for $\lambda$, each color’s fringes will be at slightly different positions. The result is that near the center of the pattern, where there is the least path difference, all wavelengths constructively interfere to form a bright white central fringe. However, moving away from the center, the fringes start to show different colors as the path difference between the light from the two slits increases. This causes a dispersion of colors with different orders of fringes dominated by different colors. The constructive and destructive interference patterns of different wavelengths slightly offset one another.

Thus:

• Option A - "Interference pattern will disappear" is incorrect because the interference pattern does not disappear but changes due to the dispersion of colors.
• Option B - "There will be a central dark fringe surrounded by a few coloured fringes" is incorrect as the central fringe in the presence of white light is bright, not dark.
• Option C - "There will be a central bright white fringe surrounded by a few coloured fringes" is correct. This is because all the wavelengths interfere constructively at the center, creating a bright white fringe, succeeded by colored fringes due to the varying interference conditions for each wavelength.
• Option D - "All bright fringes will be of equal width" is incorrect. The width and spacing of the fringes vary by wavelength, so the interference pattern will not have uniform fringe widths.

Therefore, the correct answer is Option C: "There will be a central bright white fringe surrounded by a few coloured fringes."

According to Brewster's law, reflected rays are completely polarized and refracted rays are partially polarized.

Let's analyze each statement one by one to determine the correctness:

Statement I: "Atoms are electrically neutral as they contain equal number of positive and negative charges."

This statement is correct. Atoms are composed of protons, neutrons, and electrons. Protons have a positive charge, electrons have a negative charge, and neutrons have no charge. In a neutral atom, the number of protons (positive charges) equals the number of electrons (negative charges), thus making the atom electrically neutral. The positive and negative charges balance each other, resulting in no overall charge.

Statement II: "Atoms of each element are stable and emit their characteristic spectrum."

This statement requires modification for accuracy. Atoms emit their characteristic spectrum when they are excited and then return to a lower energy state. The emission of a spectrum is not linked directly to the stability of an atom; it is linked to the electronic transitions within the atom. The term "stable" in relation to atoms typically refers to the overall energy state of an atom being at its lowest or in a ground state. However, atoms can emit radiation whether they are in a stable ground state or in an excited state. Moreover, not all atoms are inherently stable; some may be radioactive and unstable. The part that atoms emit their characteristic spectrum under certain conditions (like excitation) is correct, but linking this property strictly to stability is misleading.

Based on the analysis:

• Statement I is correct.
• Statement II is partially correct, as it is accurate that each element can emit a characteristic spectrum when excited, but incorrect in generalizing that all such atoms are stable.

Therefore, the correct option is:

Option C: Statement I is correct but Statement II is incorrect.

$$\begin{array}{rl}& \text{ Energy difference }\Delta E=\frac{hc}{\lambda }\\ & \therefore \lambda \propto \frac{1}{\Delta E}\\ & (\Delta E{)}_{6-2}>(\Delta E{)}_{5-2}>(\Delta E{)}_{4-2}>(\Delta E{)}_{3-2}\\ & {\lambda }_{6-2}<{\lambda }_{5-2}<{\lambda }_{4-2}<{\lambda }_{3-2}\end{array}$$

A-III, B-IV, C-II, D-I

To determine the mass number and atomic number of the final product $Q$ in the provided nuclear reactions, we need to analyze the impact of each type of decay (alpha, positron emission, beta-minus decay, and electron capture) on the mass number and atomic number of the initial element $X={}_{82}^{290}X$:

1. Alpha decay ($\alpha$ decay): In alpha decay, an alpha particle (which is a ${}_2^{4}He$ nucleus) is emitted. This reduces the mass number by 4 units and the atomic number by 2 units.

$${}_{82}^{290}X\stackrel{\alpha }{\to }{}_{80}^{286}Y$$

2. Beta plus decay (positron emission, $e^{+}$): During positron emission, a proton in the nucleus is transformed into a neutron, and a positron is emitted. This decreases the atomic number by 1 but does not change the mass number.

$${}_{80}^{286}Y\stackrel{e^{+}}{\to }{}_{79}^{286}Z$$

3. Beta-minus decay (${\beta }^{-}$ decay): In a beta-minus decay, a neutron in the nucleus converts into a proton and an electron (beta particle) and an antineutrino are emitted. This results in an increase in the atomic number by 1, while the mass number remains unchanged.

$${}_{79}^{286}Z\stackrel{{\beta }^{-}}{\to }{}_{80}^{286}P$$

4. Electron capture ($e^{-}$ capture): During electron capture, an atomic electron is absorbed by the nucleus, causing a proton to convert into a neutron. This process decreases the atomic number by 1 without altering the mass number.

$${}_{80}^{286}P\stackrel{e^{-}}{\to }{}_{79}^{286}Q$$

From the calculations above, the mass number of $Q$ is 286, and its atomic number is 79. Comparing these values with the multiple choices:

Option A: Mass number = 280, Atomic number = 81
Option B: Mass number = 286, Atomic number = 80
Option C: Mass number = 288, Atomic number = 82
Option D: Mass number = 286, Atomic number = 81

None of these match exactly, implying an error in the problem or the answer options. Based on the correct atomic number calculations and assuming option D is meant to be atomic number 79, then the corrected choice would have been:

D: Mass number = 286, Atomic number = 79

de-Broglie wavelength $\lambda =\frac{h}{P}=\frac{h}{mv}=\frac{h}{\sqrt{2mE}}$ where $E=\frac{1}{2}m{v}^2$

Squaring both sides,

$$\begin{array}{rl}& {\lambda }^2=\frac{h^2}{4m^{2}E}\\ & \Rightarrow \frac{1}{{\lambda }^2}=\text{ (constant) }E\end{array}$$

Graph passes through origin with constant slope.

A. The energy of a photon is $E=hv$.

This statement is correct. The energy of a photon is given by the equation:

$$E=hv$$

where $E$ is the energy, $h$ is Planck's constant, and $v$ (also written as $\nu$) is the frequency of the photon.

B. The velocity of a photon is $c$.

This statement is also correct. In free space (vacuum), the velocity of light, and hence the velocity of a photon, is constant and is denoted by $c$.

C. The momentum of a photon, $p=\frac{hv}{c}$.

This statement is correct as well. The momentum $p$ of a photon can be expressed as:

$$p=\frac{E}{c}=\frac{hv}{c}$$

D. In a photon-electron collision, both total energy and total momentum are conserved.

This statement is correct. During collisions involving photons and electrons, such as Compton scattering, both energy and momentum are conserved.

E. Photon possesses positive charge.

This statement is incorrect. Photons are electrically neutral particles and do not possess any charge.

Based on the analysis, the correct statements are A, B, C, and D. Therefore, the correct answer is:

Option B
A, B, C, and D only