JEE Main 2024 9th April Morning Shift Question Paper with Solutions

Let's denote the total distance covered by the particle as $2d$, where $d$ is the distance for each half. To calculate the average speed, we need to find the total distance traveled and divide it by the total time taken.

For the first half of the journey, the particle covers the distance $d$ at a speed of $6 \, \text{m/s}$. The time taken for this part of the journey can be calculated using the formula $\text{time} = \frac{\text{distance}}{\text{speed}}$. So,

$\text{time}_1 = \frac{d}{6}$

For the second half of the journey, the distance $d$ is further divided into two parts, each covered in equal time intervals. Given the speeds are $9 \, \text{m/s}$ and $15 \, \text{m/s}$ respectively, let's call the equal time intervals $t$. The distances covered in these intervals can be found by $\text{distance} = \text{speed} \times \text{time}$.

For the part covered at $9 \, \text{m/s}$:

$d_1 = 9t$

For the part covered at $15 \, \text{m/s}$:

$d_2 = 15t$

Since these two parts together make up the second half of the journey,

$d_1 + d_2 = d$

$9t + 15t = d$

This gives us $24t = d$, and from this, we can find $t = \frac{d}{24}$.

The total time for the second half of the journey is the sum of the times for the two parts, which are equal ($t$ each), so the total time for the second half is $2t$. Since $t = \frac{d}{24}$,

$\text{time}_2 = 2 \times \frac{d}{24} = \frac{d}{12}$

The total time taken for the entire journey is the sum of the times for the first and second halves:

$\text{total time} = \text{time}_1 + \text{time}_2 = \frac{d}{6} + \frac{d}{12}$

$\text{total time} = \frac{2d}{12} + \frac{d}{12} = \frac{3d}{12} = \frac{d}{4}$

The total distance is $2d$, and the total time is $\frac{d}{4}$. Therefore, the average speed is calculated as:

$\text{average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{2d}{\frac{d}{4}} = \frac{2d}{1} \times \frac{4}{d} = 8 \, \text{m/s}$

Thus, the correct answer is Option D: 8 m/s.

At $C$

$$\begin{aligned} & \frac{v_1}{2}+\frac{v_1-5}{2}+\frac{v_1+10-v_2}{1}=0 \\ & v_2=2 v_1+\frac{15}{2} \quad \text{.... (i)} \end{aligned}$$

At $A$

$$\begin{aligned} & \frac{v_2-5}{4}+\frac{v_2}{4}+\frac{v_2-10-v_1}{1}=0 \\ & 6 v_2=4 v_1+45 \quad \text{.... (ii)} \end{aligned}$$

$$\begin{aligned} & \Rightarrow v_1=0 \\ & \text { and } v_2=\frac{15}{2} \\ & \therefore \quad i=\frac{5}{2} \mathrm{~A} \\ & \Rightarrow n=25 \end{aligned}$$

The least count of a vernier caliper is defined as the smallest distance that it can measure and is calculated by the difference in length between one main scale division and one vernier scale division. It can be represented as:

$\text{Least Count} = \text{Main scale division} - \text{Vernier scale division}$

Given that one main scale division is equal to $\mathrm{m}$ units and the $\mathrm{n}^{\text {th }}$ division of main scale coincides with the $(n+1)^{\text {th }}$ division of the vernier scale, this means that $n$ divisions on the main scale is equal to $(n+1)$ divisions on the vernier scale.

Since one main scale division is $\mathrm{m}$ units, $n$ divisions on the main scale would be $n \times \mathrm{m}$ units. If $n$ divisions on the main scale are equal to $(n+1)$ divisions on the vernier scale, we can determine the length of one vernier scale division as

$\text{Length of one vernier scale division} = \frac{n \times \mathrm{m}}{n+1}$

Thus, the least count, which is the difference between one main scale division and one vernier scale division, is:

$\text{Least Count} = \mathrm{m} - \frac{n \times \mathrm{m}}{n+1} = \mathrm{m} \left(1 - \frac{n}{n+1} \right) = \mathrm{m} \left( \frac{n+1-n}{n+1} \right) = \mathrm{m} \left( \frac{1}{n+1} \right)$

This simplifies to:

$\text{Least Count} = \frac{\mathrm{m}}{\mathrm{n + 1}}$

Therefore, the correct option is:

Option B: $\frac{\mathrm{m}}{\mathrm{n+1}}$

To find the angular velocity ($\omega$) of the wheel after $10$ seconds, we first need to understand the relationship between the force applied through the string, the torque produced by this force, and how this torque affects the wheel's angular acceleration ($\alpha$).

The torque ($\tau$) produced by the force ($F$) is given by the product of the force and the radius ($r$) of the wheel through which the force is applied:

$\tau = F \cdot r$

Given that $F = 40 \, \mathrm{N}$ and $r = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m}$, the torque can be calculated as:

$\tau = 40 \cdot 0.1 = 4 \, \mathrm{Nm}$

The torque is related to the angular acceleration ($\alpha$) and the moment of inertia ($I$) of the wheel by the equation:

$\tau = I \cdot \alpha$

Given that $I = 0.40 \, \mathrm{kg \cdot m}^2$, we can rearrange the above formula to solve for $\alpha$:

$\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10 \, \mathrm{rad/s}^2$

With the angular acceleration ($\alpha$), we can calculate the angular velocity ($\omega$) after a given time ($t$) using the formula:

$\omega = \omega_0 + \alpha \cdot t$

Where $\omega_0$ is the initial angular velocity. Since the wheel starts from rest, $\omega_0 = 0$. Thus, for $t = 10 \, \mathrm{s}$:

$\omega = 0 + 10 \cdot 10 = 100 \, \mathrm{rad/s}$

Therefore, the angular velocity ($\omega$) of the wheel after $10$ seconds is $100 \, \mathrm{rad/s}$, so $x = 100$.

For an adiabatic process, the work done by the gas can be found using the formula:

$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

Given that the volume is doubled ($V_2 = 2V_1$) and the adiabatic constant $\gamma = \frac{3}{2}$, we can manipulate the ideal gas law and the adiabatic process relationship to find an expression for work done in terms of the initial conditions and $\gamma$.

Recall, for an adiabatic process, $PV^{\gamma} = \text{constant}$, so we can write:

$P_1V_1^{\gamma} = P_2V_2^{\gamma}$

Using the fact that $V_2 = 2V_1$, we can express $P_2$ in terms of $P_1$ and $V_1$ as follows:

$P_1V_1^{\gamma} = P_2(2V_1)^{\gamma}$

$P_2 = P_1 \left( \frac{V_1}{2V_1} \right)^{\gamma}$

$P_2 = P_1 \left( \frac{1}{2} \right)^{\gamma}$

The work done then becomes:

$W = \frac{P_1 V_1 - P_1 \left( \frac{1}{2} \right)^{\gamma} \cdot 2V_1}{\gamma - 1}$

$W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\gamma} \cdot 2}{\gamma - 1}$

Plugging in $\gamma = \frac{3}{2}$, we get:

$W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\frac{3}{2}} \cdot 2}{\frac{3}{2} - 1}$

$W = P_1V_1 \frac{1 - \sqrt{\frac{1}{2}} \cdot 2}{\frac{1}{2}}$

$W = 2P_1V_1 (1 - \sqrt{\frac{1}{2}})$

Since $P_1V_1 = nRT$ for 1 mole ($n = 1$) of gas at temperature $T$, we can further simplify:

$W = 2RT(1 - \sqrt{\frac{1}{2}})$

$W = 2RT(1 - \frac{1}{\sqrt{2}})$

$W = RT(2 - \sqrt{2})$

Therefore, the correct option is:

Option B $\mathrm{RT}[2-\sqrt{2}]$.

To derive the dimensional formula of latent heat, we need to understand what latent heat actually refers to. Latent heat is the amount of heat absorbed or released by a substance during a change in its physical state (phase) that occurs without changing its temperature.

The formula for latent heat ($L$) is given by:

$Q = mL$

where:

$Q$ = Heat absorbed or released (with the dimension of energy $[\mathrm{ML}^2\mathrm{T}^{-2}]$),

$m$ = Mass of the substance ($[\mathrm{M}]$),

$L$ = Latent heat.

To find the dimensions of latent heat, we rearrange the formula to solve for $L$:

$L = \frac{Q}{m}$

Knowing the dimensions of $Q$ (energy, which is equivalent to work done, with dimensions $[\mathrm{ML}^2\mathrm{T}^{-2}]$) and $m$ (mass, with dimensions $[\mathrm{M}]$), we can substitute these into the equation to find $L$'s dimensions:

$L = \frac{[\mathrm{ML}^2\mathrm{T}^{-2}]}{[\mathrm{M}]}$

This simplifies to:

$L = [\mathrm{L}^2\mathrm{T}^{-2}]$

Therefore, the dimensional formula of latent heat is:

$L = [\mathrm{M}^0 \mathrm{L}^2 \mathrm{T}^{-2}]$

So, the correct option is:

Option C $\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

To solve this problem, we will use the concepts of vector addition and magnitudes involving the dot product. Given the angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}\right)$, we can use the properties of dot products and magnitudes to find the required integer value of $n$.

First, let's use the condition $|\vec{a} + \vec{b}| = \sqrt{2} |\vec{a} - \vec{b}|$. Square both sides to remove the square roots:

$|\vec{a} + \vec{b}|^2 = 2 |\vec{a} - \vec{b}|^2$

Now we will expand both sides using the formula for the magnitude of the sum and difference of vectors:

$|\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$

And

$|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}$

Substitute these back into the original equation:

$\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 2 (\vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b})$

Expand the right-hand side:

$\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 2 \vec{a} \cdot \vec{a} - 4 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{b}$

Rearrange all terms to one side to combine like terms:

$\vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{a} + 4 \vec{a} \cdot \vec{b} - 2 \vec{b} \cdot \vec{b} = 0$

Combine like terms:

$- \vec{a} \cdot \vec{a} + 6 \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{b} = 0$

We know that:

$\vec{a} \cdot \vec{a} = |\vec{a}|^2 \quad \text{and} \quad \vec{b} \cdot \vec{b} = |\vec{b}|^2 \quad \text{and} \quad \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)$

Since the angle between $\vec{a}$ and $\vec{b}$ is $\cos^{-1}\left(\frac{5}{9}\right)$, we have:

$\cos(\theta) = \frac{5}{9}$

Substitute these back into the equation:

$-|\vec{a}|^2 + 6 |\vec{a}| |\vec{b}| \left(\frac{5}{9}\right) - |\vec{b}|^2 = 0$

Simplify it further:

$-|\vec{a}|^2 + \frac{30}{9} |\vec{a}| |\vec{b}| - |\vec{b}|^2 = 0$

$-|\vec{a}|^2 + \frac{10}{3} |\vec{a}| |\vec{b}| - |\vec{b}|^2 = 0$

We know that $|\vec{a}| = n |\vec{b}|$. Substitute this into the equation:

$-(n |\vec{b}|)^2 + \frac{10}{3} (n |\vec{b}|) |\vec{b}| - |\vec{b}|^2 = 0$

Simplify it:

$-n^2 |\vec{b}|^2 + \frac{10}{3} n |\vec{b}|^2 - |\vec{b}|^2 = 0$

Factor out $|\vec{b}|^2$:

$|\vec{b}|^2 \left(-n^2 + \frac{10}{3} n - 1\right) = 0$

Since $|\vec{b}|^2 \neq 0$, we can solve:

$-n^2 + \frac{10}{3} n - 1 = 0$

Multiply through by 3 to clear the fraction:

$-3n^2 + 10n - 3 = 0$

Rearrange it to match standard quadratic form:

$3n^2 - 10n + 3 = 0$

Solve this quadratic equation using the quadratic formula:

$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 3$, $b = -10$, and $c = 3$:

$n = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6}$

This results in two possible solutions for $n$:

$n = \frac{18}{6} = 3 \quad \text{and} \quad n = \frac{2}{6} = \frac{1}{3}$

However, since $n$ is given to be an integer, we take:

$\boxed{3}$

Let's first determine the resistance of the coil when connected to a DC supply. The current drawn by the coil in a DC circuit can be used to calculate its resistance using Ohm's law:

$R = \frac{V}{I}$

Given the DC supply voltage $V_{DC} = 20 \, \mathrm{V}$ and the current $I_{DC} = 5 \, \mathrm{A}$, the resistance $R$ is:

$R = \frac{20 \, \mathrm{V}}{5 \, \mathrm{A}} = 4 \, \Omega$

Next, let's use the information given for the AC supply. When connected to an AC supply, the coil's impedance $Z$ can be determined using the given current. The total voltage and current in an AC circuit are related to the impedance by the formula:

$Z = \frac{V}{I}$

Given the AC supply voltage $V_{AC} = 20 \, \mathrm{V}$ and the current $I_{AC} = 4 \, \mathrm{A}$, the impedance $Z$ is:

$Z = \frac{20 \, \mathrm{V}}{4 \, \mathrm{A}} = 5 \, \Omega$

The impedance $Z$ of the coil in an AC circuit is composed of both the resistance $R$ and the inductive reactance $X_L$, related by:

$Z = \sqrt{R^2 + X_L^2}$

We already know that $R = 4 \, \Omega$. We can now solve for the inductive reactance $X_L$:

$5 = \sqrt{4^2 + X_L^2}$

Squaring both sides of the equation:

$25 = 16 + X_L^2$

Solving for $X_L$:

$X_L^2 = 25 - 16$

$X_L^2 = 9$

$X_L = \sqrt{9}$

$X_L = 3 \, \Omega$

The inductive reactance $X_L$ is also related to the inductance $L$ and the angular frequency $\omega$ by the formula:

$X_L = \omega L$

where $\omega = 2 \pi f$. Given the frequency $f = 50 \, \mathrm{Hz}$ and using $\pi = 3$, we find:

$\omega = 2 \times 3 \times 50$

$\omega = 300 \, \mathrm{rad/s}$

Now we can solve for the inductance $L$:

$X_L = 300 L$

$3 = 300 L$

$L = \frac{3}{300}$

$L = 0.01 \, \mathrm{H}$

Since $1 \, \mathrm{H} = 1000 \, \mathrm{mH}$, the self inductance of the coil is:

$L = 0.01 \, \mathrm{H} \times 1000 \, \mathrm{mH/H} = 10 \, \mathrm{mH}$

Therefore, the self inductance of the coil is $10 \, \mathrm{mH}$.

Let's analyze each statement carefully:

Statement (I): When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.

To understand this statement, let's recall the nature of images formed by concave lenses. A concave lens (diverging lens) always forms images that are virtual, erect, and smaller than the object, regardless of the object's position. The term "centre of curvature" is typically associated with mirrors rather than lenses. Therefore, this statement is incorrect because a concave lens does not form a real image that could be said to be at the centre of curvature on the other side.

Statement (II): Concave lens always forms a virtual and erect image.

Now, considering the second statement, a concave lens (diverging lens) indeed always forms virtual, erect, and diminished images. This is a fundamental property of concave lenses.

Given this analysis:

The correct answer is Option A: Statement I is false but Statement II is true.

Due to constant component of magnetic field $F = 0$

Due to variable component

$$\begin{aligned} & F_1=0 \\ & \text { and, } F_2+F_3=0 \\ & \begin{aligned} \text { and, } F_4 & =\left(\mathrm{B}_0 4_x\right) i \mathrm{~L} \\ & =5 \times 4 \times 2 \times 2 \times 2 \\ & =160 \mathrm{~N} \end{aligned} \end{aligned}$$

To determine the energy equivalent of a mass, we use Einstein's mass-energy equivalence principle given by the equation:

$E = mc^2$

where:

- $E$ is the energy

- $m$ is the mass

- $c$ is the speed of light in a vacuum, which is approximately $3 \times 10^8 \mathrm{~m/s}$

Given the mass $m = 1 \mathrm{~g} = 1 \times 10^{-3} \mathrm{~kg}$, we can substitute these values into the equation:

$E = (1 \times 10^{-3} \mathrm{~kg}) \times (3 \times 10^8 \mathrm{~m/s})^2$

Calculating this, we get:

$E = 1 \times 10^{-3} \times 9 \times 10^{16}$

$E = 9 \times 10^{13} \mathrm{~J}$

Next, to convert this energy into electron volts ($\mathrm{eV}$), we use the conversion factor: $1 \mathrm{~J} = 6.242 \times 10^{12} \mathrm{~MeV}$.

Therefore:

$E = 9 \times 10^{13} \mathrm{~J} \times 6.242 \times 10^{12} \mathrm{~MeV/J}$

Calculating this, we get:

$E = 5.6178 \times 10^{26} \mathrm{~MeV}$

Therefore, the energy equivalent of $1 \mathrm{~g}$ of a substance is:

Option A: $5.6 \times 10^{26} \mathrm{~MeV}$

To solve this problem, we consider the buoyancy and weight force acting on the sphere. For the sphere to just float on water, the weight of the water displaced by the sphere must be equal to the weight of the sphere. The volume of water displaced by the sphere is equivalent to the outer volume of the sphere minus the volume of the cavity inside it.

The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$, where $r$ is the radius of the sphere. For the given sphere, its outer radius is $R = \frac{D}{2}$, and the radius of the cavity is $r = \frac{d}{2}$. Therefore, the volume of the sphere excluding the cavity is:

$V_{\text{solid part}} = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3$

$V_{\text{solid part}} = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 - \frac{4}{3}\pi \left(\frac{d}{2}\right)^3$

Relative density ($\sigma$) is defined as the ratio of the density of an object to the density of water. This means the actual density of the sphere is $\sigma \times \text{density of water}$. Since the object just floats, the weight of the displaced water is equal to the weight of the solid part of the sphere (ignoring the cavity), which can be mathematically represented as:

$\text{Weight of solid part} = \text{Weight of displaced water}$

$\sigma \cdot \rho_{\text{water}} \cdot V_{\text{solid part}} \cdot g = \rho_{\text{water}} \cdot V_{\text{displaced water}} \cdot g$

Since the sphere is floating, $V_{\text{displaced water}} = \frac{4}{3}\pi R^3$, the equation simplifies to:

$\sigma \left(\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$

Cancelling out common terms gives:

$\sigma (R^3 - r^3) = R^3$

Given that $\sigma$ is the relative density, we can rearrange the equation to solve for the ratio of $D/d$ or equivalently $R/r$:

$\sigma = \frac{R^3}{R^3 - r^3}$

Solving for $R/r$:

$R^3 (1 - \sigma) + \sigma r^3 = 0$

$\frac{R^3}{r^3} = \frac{\sigma}{\sigma - 1}$

Since $R = \frac{D}{2}$ and $r = \frac{d}{2}$, the ratio of $D/d$ is equal to the ratio of $R/r$, thus:

$\frac{D}{d} = \left(\frac{\sigma}{\sigma - 1}\right)^{1/3}$

This matches Option D:

$\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}$

To understand the impact of placing a dielectric between the plates of the capacitor on the glow of the bulb, we need to consider the properties and behavior of capacitors in an AC circuit.

When a capacitor is connected in series with a bulb in an AC circuit, the impedance of the capacitor plays a significant role in determining the current through the circuit. The impedance $Z_C$ of a capacitor in an AC circuit is given by:

$Z_C = \frac{1}{\omega C}$

where:

• $Z_C$ is the capacitive reactance (impedance of the capacitor).
• $\omega$ is the angular frequency of the AC supply ( $\omega = 2 \pi f$ , where $f$ is the frequency).
• $C$ is the capacitance of the capacitor.

When we place a dielectric between the plates of the capacitor, the capacitance $C$ increases. The capacitance with a dielectric can be described as:

$C' = \kappa C$

where:

• $C'$ is the new capacitance with the dielectric present.
• $\kappa$ is the dielectric constant ( $\kappa > 1$ ).

Since $C' > C$, the new capacitive reactance $Z_C'$ can be given as:

$Z_C' = \frac{1}{\omega C'}$

Because $C' = \kappa C$, we have:

$Z_C' = \frac{1}{\omega \kappa C} = \frac{1}{\kappa} \left( \frac{1}{\omega C} \right) = \frac{Z_C}{\kappa}$

Since $\kappa > 1$, $Z_C' < Z_C$. This means the impedance of the capacitor decreases when a dielectric is placed between its plates. In a series circuit, the overall impedance decreases when the impedance of one component decreases, leading to an increase in the current through the circuit.

Thus, with an increase in current, the bulb will glow brighter. Therefore, the correct option is:

Option C: increases

To find the ratio of the masses $\frac{m_2}{m_1}$ given that the acceleration of the system is $\frac{g}{8}$, where $g$ is the acceleration due to gravity, we need to analyze the forces acting on each mass and apply Newton's second law of motion.

For mass $m_1$, the force acting downward (towards the centre of the Earth) is the gravitational force $m_1g$, and the tension $T$ acts upward. Its equation of motion, considering downward as positive, can be given by:

$m_1g - T = m_1a$

For mass $m_2$, the force acting downward is $m_2g$, but since it is on the opposite side of the pulley, the tension $T$ is upwards (considering upward movement as positive direction), so we have:

$T - m_2g = m_2a$

Since the pulley is light and smooth, the tension $T$ is the same on both sides of the pulley. Also, the system accelerates together, so $a = \frac{g}{8}$.

Adding the two equations to eliminate $T$ gives us:

$m_1g - m_2g = (m_1 - m_2)a$

$g(m_1 - m_2) = (m_1 - m_2)\frac{g}{8}$

Canceling out $g$ and dividing both sides by $m_1 - m_2$ (assuming $m_1 \neq m_2$), we get:

$1 = \frac{1}{8}$

This simplification doesn't align with finding the ratio directly, indicating a mistake in handling the simultaneous equations relation to $a$ and $T$. Let's correct our approach to finding the ratio of the masses based on the acceleration and tension.

Since the acceleration $a = \frac{g}{8}$, and considering the system as a whole, the net force causing the acceleration is the difference in gravitational forces on the two masses. This net force provides the system's entire acceleration. Thus, correctly setting up the equations for the system should give:

$m_2g - m_1g = (m_1 + m_2)\frac{g}{8}$

$g(m_2-m_1) = \frac{g}{8}(m_1 + m_2)$

$8(m_2 - m_1) = m_1 + m_2$

$8m_2 - 8m_1 = m_1 + m_2$

$7m_2 = 9m_1$

$\frac{m_2}{m_1} = \frac{9}{7}$

Therefore, the correct ratio of the masses $\frac{m_2}{m_1}$ is $9:7$, which corresponds to Option C.

To find the total work done by all forces applied on the particle during its displacement, we can use the work-energy theorem which states that the work done by all forces on an object is equal to the change in kinetic energy of the object. So, we first need to find the initial and final kinetic energies of the particle and then calculate the work done.

The velocity of the particle is given by $v = \alpha \sqrt{x}$,

and the kinetic energy $K$ of the particle is given by $K = \frac{1}{2} m v^2$. We can substitute the expression for $v$ into this formula to get the kinetic energy as a function of position $x$:

$K(x) = \frac{1}{2} m (\alpha \sqrt{x})^2 = \frac{1}{2} m \alpha^2 x$

To find the total work done from $x = 0$ to $x = d$, we need to compute the difference in kinetic energy between these two points:

$W = K(d) - K(0)$

At $x = d$,

$K(d) = \frac{1}{2} m \alpha^2 d$

At $x = 0$, since the particle starts from this position,

$K(0) = \frac{1}{2} m \alpha^2 (0) = 0$

So, the work done $W$ is simply the kinetic energy at $x = d$,

$W = \frac{1}{2} m \alpha^2 d - 0 = \frac{1}{2} m \alpha^2 d$

This matches with Option C:

$\frac{m \alpha^2 d}{2}$.

Let's begin by understanding the equations related to simple harmonic motion (SHM). For a particle executing SHM, the position $x$, velocity $v$, and acceleration $a$ are given by the following equations:

1. Position: $x = A \cos(\omega t + \phi)$

2. Velocity: $v = -A \omega \sin(\omega t + \phi)$

3. Acceleration: $a = -A \omega^2 \cos(\omega t + \phi)$

Here, $A$ is the amplitude of the motion, $\omega$ is the angular frequency, and $\phi$ is the phase constant.

Given the magnitudes at a certain instant:

$x = 4 \, \mathrm{m}$

$v = 2 \, \mathrm{ms}^{-1}$

$a = 16 \, \mathrm{ms}^{-2}$

Using the acceleration equation:

$a = -A \omega^2 \cos(\omega t + \phi)$

Since we’re given the magnitude of the acceleration, we remove the negative sign:

$16 = A \omega^2 \cos(\omega t + \phi)$

Using the position equation:

$x = A \cos(\omega t + \phi)$

We already know $x = 4 \, \mathrm{m}$, so:

$4 = A \cos(\omega t + \phi)$

From these two equations, we know:

$A \omega^2 \cos(\omega t + \phi) = 16$

$A \cos(\omega t + \phi) = 4$

Therefore:

$A \omega^2 \cdot 4/A = 16$

$4 \omega^2 = 16$

$\omega^2 = 4$

$\omega = 2 \, \mathrm{rad/s}$

Next, using the velocity equation:

$v = -A \omega \sin(\omega t + \phi)$

Again, we consider the magnitude:

$2 = A \cdot 2 \sin(\omega t + \phi)$

$2 = 2A \sin(\omega t + \phi)$

$\sin(\omega t + \phi) = \dfrac{1}{A}$

We know from the position equation that:

$\cos(\omega t + \phi) = \dfrac{4}{A}$

Using the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we get:

$\left(\dfrac{1}{A}\right)^2 + \left(\dfrac{4}{A}\right)^2 = 1$

$\dfrac{1}{A^2} + \dfrac{16}{A^2} = 1$

$\dfrac{17}{A^2} = 1$

$A^2 = 17$

$A = \sqrt{17} \, \mathrm{m}$

Therefore, the amplitude of the motion is $\sqrt{17} \, \mathrm{m}$, meaning $x$ is 17.

$$\begin{aligned} & m g \times \frac{L}{2} \cos \theta=N \times L \cos \theta \\ & \Rightarrow \quad N=\frac{m g}{2}=\frac{W}{2} \end{aligned}$$

The wavelength of light emitted by a Light Emitting Diode (LED) fabricated using a semiconducting material can be determined by the energy band gap of the material. The energy of the photon emitted, which corresponds to the band gap energy, is given by the equation:

$E = \frac{hc}{\lambda}$

Where:

• $E$ is the energy of the emitted photon (in joules when $h$ and $c$ are in SI units), corresponding to the band gap energy of the material.
• $h$ is Planck's constant ($6.626 \times 10^{-34} \, \mathrm{m^2\,kg/s}$).
• $c$ is the speed of light in vacuum ($3.0 \times 10^{8} \,\mathrm{m/s}$).
• $\lambda$ is the wavelength of the emitted light (in meters).

However, since the energy band gap given is in electronvolts (eV), and we are looking for the wavelength in nanometers (nm), we can use the energy formula directly in terms of eV and then do the unit conversion conveniently. The conversion between energy (in eV) and wavelength (in nm) without needing to convert eV to Joules is facilitated by the equation:

$\lambda(\mathrm{nm}) = \frac{1240}{E(\mathrm{eV})}$

Here, 1240 nm·eV is a conversion factor used for directly converting energy in eV to wavelength in nm.

Given the band gap energy of GaAs is $1.42 \mathrm{~eV}$, the wavelength ($\lambda$) of light emitted can be found as:

$\lambda = \frac{1240}{1.42} = 873.24 \mathrm{~nm}$

Thus, the wavelength of light emitted from the LED is approximately 873.24 nm, which is closest to:

Option B: 875 nm.