JEE Main 2023 6th April Morning Shift Question Paper with Solutions

The range of transmission (R) of a TV tower (or any radio tower) depends on the height of the tower (h). This relationship is given by the formula:

$R = \sqrt{2hR_E}$

where $R_E$ is the radius of the Earth. This formula comes from the geometry of the situation, considering the curvature of the Earth.

Now, if the height of the tower is increased, this will increase the range of transmission. Specifically, because the height appears under the square root in the formula, the range of transmission is proportional to the square root of the height. So, if the height is multiplied by some factor, the range will be multiplied by the square root of that factor.

In this case, the height of the tower is increased by 21%, which means the new height is $h_2 = 1.21h_1$. If we substitute this into the formula for the range, we get:

$R_2 = \sqrt{2h_2R_E} = \sqrt{2(1.21h_1)R_E} = \sqrt{1.21} \sqrt{2h_1R_E} = 1.1R_1$

This shows that the new range is 1.1 times the original range, or in other words, the range has increased by 10%.

The magnetic field due to a current carrying wire can be calculated using Ampere's law. When the current is uniformly distributed across the cross-section of the wire, the situation will be different inside and outside the wire.

Inside the wire (r < a), the magnetic field is directly proportional to r (the distance from the center of the wire). This is because as you move away from the center of the wire, you enclose more current, so the magnetic field increases linearly with r.

Outside the wire (r > a), all the current in the wire is enclosed, so the magnetic field decreases with increasing r. This is a result of the magnetic field lines spreading out as they move away from the wire.

When the ball is falling with a constant velocity, it means the net force acting on the ball is zero. This is because it's in a state of dynamic equilibrium - the downward force equals the upward force.

The downward force is the gravitational force (weight of the ball), which is given by $F_g = Mg$.

The upward force is the sum of buoyant force and the viscous drag. The buoyant force is the weight of the fluid displaced by the ball, which is given by $F_b = Vg\rho_0 = Mg\rho_0/\rho$ where $V = M/\rho$ is the volume of the ball.

The viscous force, $F_v$, is the force that we need to find.

Since the net force is zero, we have:

$F_g = F_b + F_v$

or

$Mg = Mg\rho_0/\rho + F_v$

which simplifies to

$F_v = Mg - Mg\rho_0/\rho = Mg(1 - \rho_0/\rho)$

The problem requires calculating the moment of inertia of the system consisting of two identical solid spheres fixed at the ends of a light rod. We need to find the moment of inertia about an axis perpendicular to the rod and passing through its midpoint.

First, let’s identify the moment of inertia of each solid sphere about its own center, which is given by the formula:

$I_{\text{sphere}} = \frac{2}{5} m r^2$

Where:

• $m$ is the mass of the sphere = $2 \mathrm{~kg}$
• $r$ is the radius of the sphere = $0.1 \mathrm{~m}$

Substituting the values:

$I_{\text{sphere}} = \frac{2}{5} \times 2 \mathrm{~kg} \times (0.1 \mathrm{~m})^2 = \frac{4}{5} \times 0.01 \mathrm{~kg}~\mathrm{m}^2 = 0.008 \mathrm{~kg}~\mathrm{m}^2$

Now, we need the moment of inertia of the two spheres about the axis passing through the midpoint of the rod. This requires using the parallel axis theorem, which states:

$I_{\text{total}} = I_{\text{sphere}} + m d^2$

Where:

• $d$ is the distance from the center of the sphere to the axis through the rod’s midpoint = $0.2 \mathrm{~m}$

Calculating the additional inertia due to the parallel axis theorem for one sphere:

$I_{\text{parallel}} = m d^2 = 2 \mathrm{~kg} \times (0.2 \mathrm{~m})^2 = 2 \mathrm{~kg} \times 0.04 \mathrm{~m}^2 = 0.08 \mathrm{~kg}~\mathrm{m}^2$

The total moment of inertia for one sphere about the midpoint of the rod is:

$I_{\text{one sphere, total}} = I_{\text{sphere}} + I_{\text{parallel}} = 0.008 \mathrm{~kg}~\mathrm{m}^2 + 0.08 \mathrm{~kg}~\mathrm{m}^2 = 0.088 \mathrm{~kg}~\mathrm{m}^2$

Since there are two identical spheres, the total moment of inertia of the system is:

$I_{\text{system}} = 2 \times 0.088 \mathrm{~kg}~\mathrm{m}^2 = 0.176 \mathrm{~kg}~\mathrm{m}^2$

Converting the result to the given form:

$0.176 \mathrm{~kg}~\mathrm{m}^2 = 176 \times 10^{-3} \mathrm{~kg}~\mathrm{m}^2$

So, the moment of inertia of the system about the given axis is:

$176 \times 10^{-3} \mathrm{~kg}~\mathrm{m}^2$

1. The collision frequency (f) is given by the formula :

$f = \sqrt{2} \pi d^2 v n_v$

Where:

- $d$ is the diameter of the molecule,

- $v$ is the average velocity of the molecules, and

- $n_v$ is the number density (number of molecules per unit volume).

2. From this equation, we can see that the collision frequency (f) is directly proportional to the number density ($n_v$), because all the other variables ($d$ and $v$) are constant :

$f \propto n_v$

3. Therefore, the ratio of two different collision frequencies ($f_1$ and $f_2$) is equal to the ratio of their corresponding number densities ($n_{v1}$ and $n_{v2}$):

$\frac{f_1}{f_2} = \frac{n_{v1}}{n_{v2}}$

4. Given that the number density increased from $3 \times 10^{19}$ to $12 \times 10^{19}$, we can substitute these values into the equation to find the ratio of the collision frequencies:

$\frac{f_1}{f_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}}$

5. Simplifying this equation gives:

$\frac{f_1}{f_2} = 0.25$

So, the ratio of the collision frequency of air molecules before and after the increase in number is 0.25.

For $r \leq R, V=\frac{k Q}{R}$, constant everywhere inside the spherical shell.

For $r>R, V=\frac{k Q}{r}$, Decreases with $r$



Therefore for a uniformly thin spherical shell, the electric potential radially away from the centre can be represented as



Hence, option (a) is correct.

In the problem, we are given two resistances, $R_1$ and $R_2$, each with a certain measurement error, $\Delta R_1$ and $\Delta R_2$. These resistances are connected in parallel, and we are asked to find the percentage error in the equivalent resistance of this combination.

The formula for the equivalent resistance $R$ of two resistors $R_1$ and $R_2$ in parallel is:

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$

We want to find the percentage error in $R$, which is given by $(\Delta R / R) \times 100\%$.

In order to find $\Delta R / R$, we differentiate both sides of the above equation with respect to $R$, $R_1$, and $R_2$. This gives us:

$\frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$

We can then solve this equation for $\Delta R / R$:

$\frac{\Delta R}{R} = \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)R$

Substituting the given values, $R_1 = 10 \, \Omega$, $R_2 = 15 \, \Omega$, $\Delta R_1 = \Delta R_2 = 0.5 \, \Omega$, and $R = R_1R_2/(R_1+R_2) = 6 \, \Omega$, we get:

$\frac{\Delta R}{R} = \left(\frac{0.5}{100} + \frac{0.5}{225}\right) \times 6 = \frac{13}{300}$

Finally, to convert this to a percentage, we multiply by 100, giving:

$\frac{\Delta R}{R} \times 100 = \frac{13}{3} = 4.33 \%$

This tells us that the percentage error in the equivalent resistance of the two resistances in parallel is $4.33\%$.

Given, $C_1=2 \mu \mathrm{F} \text { and } K=4$

$\Rightarrow C_1=\frac{\varepsilon_0 A}{\left(\frac{\frac{d}{3}}{K}+\frac{2 d}{3}\right)}=\frac{\varepsilon_0 A}{\left(\frac{d}{3 \times 4}+\frac{2 d}{3}\right)}$

$\Rightarrow 2 \mu F=\frac{\varepsilon_0 A \times 12}{9 d}=\frac{4}{3} \frac{\varepsilon_0 A}{d}$

$\Rightarrow 2 \mu F=\frac{4}{3} \frac{\varepsilon_0 A}{d} \quad\left(\right.$ for $\left.x=\frac{1}{3} d\right)$

$\Rightarrow \frac{\varepsilon_0 A}{d}=\frac{3}{2} \mu \mathrm{F}$

Now,

$$\begin{aligned} & C_2=\frac{\varepsilon_0 A}{\left(\frac{\frac{2 d}{3}}{K}+\frac{d}{3}\right)} \quad\left(\text { for } x=\frac{2 d}{3}\right) \\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\ & \Rightarrow C_2=\frac{\varepsilon_0 A}{\left(\frac{2 d}{12}+\frac{d}{3}\right)} \\ & \Rightarrow C_2=\frac{12 \varepsilon_0 A}{2 d+4 d}=\frac{2 \varepsilon_0 A}{d}=\frac{3}{2} \times 2 \\ & =3 \mu \mathrm{F} \end{aligned}$$

Assertion A: When a body is projected at an angle of $45^{\circ}$, its range is maximum. This is true, and it's a well-established fact in physics. The maximum range of a projectile, assuming no air resistance and flat terrain, is achieved at an angle of $45^{\circ}$.

Reason R: For maximum range, the value of $\sin 2\theta$ should be equal to one. This is also true. The range of a projectile, again assuming no air resistance and flat terrain, can be calculated using the formula $R = (v^{2}/g) \cdot \sin(2\theta)$, where $v$ is the initial velocity of the projectile, $g$ is the acceleration due to gravity, and $\theta$ is the launch angle. For the range to be maximized, $\sin(2\theta)$ must be maximized, and the maximum value of $\sin(2\theta)$ is 1. This occurs when $2\theta = 90$ degrees, or $\theta = 45$ degrees, which corresponds to the assertion.

The rate of increase of internal energy of a system can be found using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

In this case, the heat being supplied to the system is 1000 W and the work being done by the system is 200 W.

Therefore, the rate at which the internal energy of the system increases is:

1000 W (heat supplied) - 200 W (work done) = 800 W

So, the correct answer is 800 W.

We know that, $E=\frac{h c}{\lambda}$

$\therefore \lambda \propto \frac{1}{E}$

For minimum wavelength, energy must be maximum which is possible from transition $n=3$ to $n=1$.

The weight of an object on a planet is given by the equation $W = mg$, where $m$ is the mass of the object and $g$ is the acceleration due to gravity.

The acceleration due to gravity on a planet is given by the equation $g = \frac{GM}{R^2}$, where $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.

In this case, the mass of the planet is double that of Earth ($M = 2M_E$), but the density is the same. Density is defined as mass divided by volume, so if the mass is doubled and the density stays the same, the volume must also double.

Since the volume of a sphere (like a planet) is given by the equation $V = \frac{4}{3}\pi R^3$, a doubling of the volume implies that the radius of the planet is $R = 2^{1/3}R_E$.

Substituting these values back into the equation for $g$, we get:

$g_{\text{planet}} = G * \frac{2M_E}{(2^{1/3}R_E)^2} = 2^{1/3} * g_E$

So the weight of the object on the planet is $W_{\text{planet}} = m*g_{\text{planet}} = m2^{1/3} * g_E = 2^{1/3}W_E$

If we consider the given logic circuit, then we get the following output.

$$\begin{array}{lll} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ \hline 1 & 0 & 0 \\ \hline 0 & 1 & 0 \\ \hline 1 & 1 & 0 \\ \hline \end{array}$$

Output $Y$ shows the output of NOR Gate.

Hence, option (a) correct.

According to question,


$\begin{aligned} \mathbf{B}_{\text {net }} & =\frac{\mu_0 i}{2 r} \hat{\mathbf{i}}+\frac{\mu_0 i}{2 r} \hat{\mathbf{j}}=\frac{\mu_0 i}{2 r} \hat{\mathbf{k}} \sqrt{2} \\ & =4 \pi \times 10^{-7} \times \sqrt{2} \times \sqrt{2} \times \frac{1}{2 \times 0.2} \\ & =2 \times 3.14 \times 10^{-6}=628 \times 10^{-8} \mathrm{~T}\end{aligned}$

The de Broglie wavelength of a particle is given by the equation:

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}},$

where:

• $h$ is Planck's constant,
• $p$ is the momentum of the particle,
• $m$ is the mass of the particle, and
• $K$ is the kinetic energy of the particle.

This equation shows that the de Broglie wavelength of a particle is inversely proportional to the square root of its mass and its kinetic energy. This means that the particle with the smallest mass and kinetic energy will have the largest de Broglie wavelength.

Given that the kinetic energies of the electron, alpha particle, and proton are $4K$, $2K$, and $K$, respectively, and knowing that the mass of the electron is less than the mass of the proton and the mass of the proton is less than the mass of the alpha particle, we can infer that the de Broglie wavelength of the electron is greater than the de Broglie wavelength of the proton, which in turn is greater than the de Broglie wavelength of the alpha particle.

Therefore, the correct answer is $\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$.

The formula to calculate the radius of an orbit for a hydrogen-like atom/ion is:

$r_n = r_0 \frac{n^2}{Z}$

where:

• $r_n$ is the radius of the nth orbit,
• $n$ is the principal quantum number (the orbit number),
• $r_0$ is the Bohr radius (radius of the first Bohr orbit in the hydrogen atom), and
• $Z$ is the atomic number (the number of protons in the nucleus).

We're dealing with a Li²⁺ ion and we're interested in the fifth orbit ($n = 5$), and given that $r_0$ is 0.51 Å and $Z$ for Li is 3, we can substitute these values into the formula:

$r_5 = 0.51 \times \frac{25}{3} \text{ Å} = 4.25 \text{ Å}$

which is $4.25 \times 10^{-10}$ m, or equivalently $425 \times 10^{-12}$ m when converted to meters.

The pole is vertically submerged in the swimming pool, and the length of the shadow is due to the sunlight that is incident at an angle of $30^{\circ}$ with the surface of water. Hence, the angle of incidence, $i$, is $60^{\circ}$ (since the angle of incidence is measured from the normal to the surface, and the normal is perpendicular to the surface).

Using Snell's law, we find the angle of refraction, $r$, as you correctly did:

$\sin r = \frac{n_{1}}{n_{2}} \sin i = \frac{3}{4} \sin 60^{\circ} = \frac{3 \sqrt{3}}{8}$

This gives $\tan r = \frac{3 \sqrt{3}}{\sqrt{37}}$.

Now, the shadow of the pole in the water forms a right triangle, with the submerged part of the pole as one side, the shadow as the hypotenuse, and the line segment from the water surface to the end of the shadow as the other side. The angle at the water surface is $r$. Thus, we can write the following relationships:

The length of the submerged part of the pole (which I'll call $x$), is given by:

$x = 2.15 \, \text{m} \cdot \cos r$

And the length of the line segment from the water surface to the end of the shadow (which I'll call $y$), is given by:

$y = 2.15 \, \text{m} \cdot \sin r$

The total length of the pole above the water surface is then the sum of $x$ and the depth of the swimming pool (1.5 m), which gives us the equation:

$x \sqrt{3} + 1.5 \, \text{m} \cdot \tan r = 2.15 \, \text{m}$

Solving this for $x$ gives us:

$x = \frac{2.15 \, \text{m}}{\sqrt{3}} - \frac{1.5 \, \text{m} \cdot 3}{\sqrt{37}} = 0.502 \, \text{m} = 50.2 \, \text{cm}$

So, the height of the pole above the water surface is approximately 50 cm.

The resistance ($R$) of a wire can be calculated by the formula:

$R = \rho \frac{L}{A},$

where

• $\rho$ is the resistivity (a property of the material),
• $L$ is the length of the wire, and
• $A$ is the cross-sectional area of the wire.

If the length ($L$) is increased by 20%, $L$ becomes $1.2L$.

If the cross-sectional area ($A$) is reduced by 4%, $A$ becomes $0.96A$.

The new resistance $R'$ is then:

$R' = \rho \frac{1.2L}{0.96A} = 1.25R,$

so the resistance has increased by 25%.

Therefore, the percentage change in the resistance of the metallic wire is 25%.

When a light wave moves from one medium to another, its speed and wavelength may change, but its frequency remains constant because it is determined by the source of the light. This is because frequency depends on the oscillations of the source, which do not change when entering a different medium.

Snell's law, which describes the relationship between the angles of incidence and refraction, can be written as:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2),$

where $n_1$ and $n_2$ are the refractive indices of the two media, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively.

The refractive index of a medium is also related to the speed of light in that medium:

$n = \frac{c}{v},$

where $c$ is the speed of light in a vacuum, and $v$ is the speed of light in the medium.

From the above relations, we see that as light enters a medium with a higher refractive index (and hence a lower speed), its wavelength decreases. The frequency remains the same.

Given that the angle of incidence is $45^\circ$ and the angle of refraction is $30^\circ$, the ratio of the refractive indices is:

$\frac{n_2}{n_1} = \frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \sqrt{2}.$

This indicates that the speed of light (and hence the wavelength) decreases by a factor of $\sqrt{2}$ as it enters the second medium. The frequency remains the same.

Therefore, the correct answer is

$\lambda_2 = \frac{1}{\sqrt{2}} \lambda_1, \quad v_2 = v_1.$

A semiconductor start conduction more as the temperature increases. It means resistance decreases with increase in temperature. So if temperature increases, its resistivity decreases.

Also, $\rho=\frac{m}{n e^2 \tau}$

as temperature increases, $\tau$ decreases but $n$ increases and $n$ is dominant over $\tau$.



$\therefore$ Option (a) is correct.

The work done against the retarding force is indeed equal to the loss in kinetic energy.

The force acting on the particle due to retardation is given by $F = ma = -2mx$.

When we integrate this force over the displacement from $0$ to $x$, we get:

$\Delta KE = W = \int F \cdot dx = \int (-2mx) \, dx = -mx^2$

The negative sign indicates that this is a loss of kinetic energy.

The problem states that the loss in kinetic energy is also given by $\left(\frac{10}{x}\right)^{-n}$ J. Therefore, we have:

$-mx^2 = \left(\frac{10}{x}\right)^{-n}$

Because this is a loss of kinetic energy, we should consider the absolute value. Hence,

$mx^2 = \left(\frac{10}{x}\right)^{-n}$

Substituting the given mass $m = 10 \, \text{g} = 0.01 \, \text{kg}$, we get:

$0.01x^2 = \left(\frac{10}{x}\right)^{-n}$

This simplifies to:

$x^2 = \left(\frac{10}{x}\right)^{-n}$

Comparing the two sides, we can see that $n = 2$.

Since, both spring are in parallel

Combination, therefore, $K_{\text {eq }}=K_1+K_2$

Time period of oscillation, $T=2 \pi \sqrt{\frac{m}{K_{\text {eq }}}}$

$\Rightarrow T=2 \pi \sqrt{\frac{m}{K_1+K_2}}$

Option (a) is correct.

In this problem, the spring is stretched due to the circular motion of the block, so the effective radius of the circular motion becomes the natural length of the spring plus the extension in the spring $(r=0.2+x)$.

The spring force, which is also the centripetal force, is given by $F_c=Kx=m\omega^2 r$, where $K$ is the spring constant, $x$ is the extension of the spring, $m$ is the mass of the block, $\omega$ is the angular velocity, and $r$ is the radius of the circular path.

Plugging in the values and solving for $x$ and $Kx$ gives the extension in the spring as $x = 0.1$ m and the tension in the spring (which is the spring force) as $Kx=7.5 \times 0.1 = 0.75$ N.

Substituting the given values:

$7.5x = 0.1 \cdot (5^2) \cdot (0.2 + x),$

which simplifies to:

$7.5x = 5 \cdot (x + 0.2)$

Solving for $x$ gives $x = 0.1$ m. The tension in the spring is then $kx = 7.5 \cdot 0.1 = 0.75$ N.

So, the correct answer is $0.75$ N.

The average energy density of an electromagnetic wave is equally shared between the electric field and the magnetic field. This means that the average electric energy density is equal to the average magnetic energy density.

The average electric energy density ($u_E$) is given by:

$u_E = \frac{1}{2} \varepsilon_0 E^2$

and the average magnetic energy density ($u_B$) is given by:

$u_B = \frac{1}{2\mu_0} B^2$

where $\varepsilon_0$ is the permittivity of free space, $\mu_0$ is the permeability of free space, $E$ is the electric field strength, and $B$ is the magnetic field strength.

In an electromagnetic wave, $E$ and $B$ are related by the equation:

$E = cB$

where $c$ is the speed of light in vacuum.

Substituting this into the energy density equations gives:

$u_E = \frac{1}{2} \varepsilon_0 (cB)^2$

and

$u_B = \frac{1}{2\mu_0} B^2$

Since $c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$, we can rewrite $u_E$ as:

$u_E = \frac{1}{2} \frac{1}{\mu_0} B^2 = u_B$

Therefore, the ratio of the average electric energy density to the average magnetic energy density is 1.

If the coil is simply moved at uniform or non-uniform speed in a uniform magnetic field without changing the orientation of the coil or the area of the coil enclosed by the magnetic field, the magnetic flux through the coil does not change, and no emf is induced according to Faraday's Law of electromagnetic induction.

The power delivered to the houses is given as 60 kW. This power is supplied at a voltage of 120 V. The power consumed in a resistive load can be found using the formula $P = V^2/R$, where P is the power, V is the voltage, and R is the resistance.

We can rearrange this formula to solve for the resistance:

$R = V^2/P$

Substituting the given values gives:

$R = (120 \, \text{V})^2 / 60,000 \, \text{W} = 0.24 \, \Omega$

Since we want the resistance in milliohms (mΩ), we can convert this to milliohms by multiplying by 1000:

$R = 0.24 \, \Omega \times 1000 = 240 \, m\Omega$

So, the value of resistive load required in the secondary circuit is 240 mΩ.

The assertion (A) is true: Earth does have an atmosphere, while the Moon does not have a significant atmosphere.

The reason (R) is also true: The escape velocity on the Moon is indeed smaller than that on Earth. The escape velocity is the minimum velocity an object must have to escape the gravitational pull of a planet or moon. A smaller escape velocity means it's easier for particles (such as the particles that make up an atmosphere) to escape into space.

Additionally, the reason (R) is a correct explanation for the assertion (A). The fact that the Moon's escape velocity is smaller than the Earth's is a major reason why the Moon doesn't have a significant atmosphere. Over time, particles that could have made up an atmosphere have escaped the Moon's gravitational pull and dispersed into space.

$2\left[\stackrel{+1}{\mathrm{Au}}(\mathrm{CN})_2\right]^{-}+\stackrel{0}{\mathrm{Z}} \mathrm{n}(\mathrm{s}) \longrightarrow 2 \stackrel{0}{\mathrm{Au}}+\left[\stackrel{+2}{\mathrm{Zn}}(\mathrm{CN})_4\right]^{-2}$

$\mathrm{Zn}$ displaced $\mathrm{Au}^{+}$

Reduction and Oxidation both are taking place.

Given the balanced chemical equation:

$3 \mathrm{BaCl}_2 + 2 \mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ba}_3\mathrm{(PO}_4)_2 + 6 \mathrm{NaCl}$

We can see that 3 moles of $\mathrm{BaCl}_2$ react with 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$ to produce 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$.

Given that you have 5 moles of $\mathrm{BaCl}_2$ and 2 moles of $\mathrm{Na}_3\mathrm{PO}_4$, let's calculate the maximum number of moles of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$ that can be formed:

From $\mathrm{BaCl}_2$, using the molar ratio from the balanced equation:

$5 \text{ moles } \mathrm{BaCl}_2 \times \frac{1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2}{3 \text{ moles } \mathrm{BaCl}_2} = \frac{5}{3} \text{ moles } \mathrm{Ba}_3\mathrm{(PO}_4)_2$

From $\mathrm{Na}_3\mathrm{PO}_4$, using the molar ratio from the balanced equation:

$2 \text{ moles } \mathrm{Na}_3\mathrm{PO}_4 \times \frac{1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2}{2 \text{ moles } \mathrm{Na}_3\mathrm{PO}_4} = 1 \text{ mole } \mathrm{Ba}_3\mathrm{(PO}_4)_2$

The limiting reactant is $\mathrm{Na}_3\mathrm{PO}_4$ because it produces 1 mole of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$, which is less than the $\frac{5}{3}$ moles produced by $\mathrm{BaCl}_2$.

Therefore, the maximum number of moles of $\mathrm{Ba}_3\mathrm{(PO}_4)_2$ formed is 1 mole. Rounding to the nearest integer, the answer is 1.

A compound is formed by two elements $\mathrm{X}$ and $\mathrm{Y}$. The element $\mathrm{Y}$ forms a cubic close-packed (CCP) arrangement, and element $\mathrm{X}$ occupies one third of the tetrahedral voids.

In a CCP structure, there are 4 atoms of $\mathrm{Y}$ in a unit cell. This means there are 8 tetrahedral voids in the CCP structure.

Since $\mathrm{X}$ occupies one third of the tetrahedral voids, there are $\frac{1}{3} \times 8 = \frac{8}{3}$ atoms of $\mathrm{X}$ in the formula unit.

The ratio of $\mathrm{X}$ to $\mathrm{Y}$ in the formula unit is $\frac{8}{3} : 4 = \frac{2}{3} : 1$. Multiplying both parts of the ratio by 3, we get $2 : 3$.

So, the formula of the compound is $\mathrm{X_2Y_{3}}$