JEE Main 2023 11th April Morning Shift Question Paper with Solutions

Given the wave equation:

$Y = 10^{-2} \sin 2 \pi(160t - 0.5x + \pi/4)$

Comparing this equation with the general form:

$Y = A \sin(2\pi(ft - kx + \phi))$

We can identify the wave number $k = 0.5\,\mathrm{m}^{-1}$ and the frequency $f = 160\,\mathrm{Hz}$. The wave speed $v$ can be found using the relationship between wave number, wave speed, and frequency:

$v = \frac{\omega}{k} = \frac{2\pi f}{2\pi k}$

Now, we can calculate the wave speed:

$v = \frac{2\pi \times 160}{2\pi \times 0.5} = \frac{160}{0.5}\,\mathrm{m/s}$

$v = 320\,\mathrm{m/s}$

Now, we need to convert the wave speed from meters per second to kilometers per hour:

$v = 320 \frac{\mathrm{m}}{\mathrm{s}} \times \frac{1\,\mathrm{km}}{1000\,\mathrm{m}} \times \frac{3600\,\mathrm{s}}{1\,\mathrm{h}}$

$v = 320 \times \frac{1}{1000} \times 3600\,\mathrm{km/h}$

$v = 1152\,\mathrm{km/h}$

So, the speed of the wave is $1152\,\mathrm{km/h}$.

For a particle executing SHM

$\mathrm{KE}=\frac{1}{2} \mathrm{m} \omega^2\left(\mathrm{A}^2-\mathrm{x}^2\right)$

When $\mathrm{x}=0,$ KE is maximum & when $\mathrm{x}=\mathrm{A},$ KE is zero and KE vs x graph is parabola.

To find the work done by a force during a displacement, we can use the formula:

$W = \int_{x_1}^{x_2} \vec{F} \cdot d\vec{x}$

Here, the force is given by $\vec{F} = (2+3x) \hat{i}$, and we need to find the work done during a displacement from $x = 0$ to $x = 4 \mathrm{~m}$. Since the force is only in the $x$ direction, we can write the integral as:

$W = \int_{0}^{4} (2+3x) dx$

Now we can integrate the function with respect to $x$:

$W = \int_{0}^{4} (2+3x) dx = \int_{0}^{4} 2 dx + \int_{0}^{4} 3x dx$

$W = \left[ 2x \right]_0^4 + \left[ \frac{3}{2}x^2 \right]_0^4$

Now we can plug in the limits of integration:

$W = (2 \cdot 4 - 2 \cdot 0) + \left(\frac{3}{2} \cdot 4^2 - \frac{3}{2} \cdot 0^2 \right)$

$W = 8 + 24$

$W = 32 \mathrm{~J}$

So the work done by the force during the displacement from $x = 0$ to $x = 4 \mathrm{~m}$ is 32 Joules.

To find the number of bullets fired per second, we can use the concept of momentum. When the machine gun fires bullets, it experiences a backward force due to the conservation of momentum. The force applied on the machine gun is used to balance this backward force.

First, let's find the momentum of each bullet:

Momentum = Mass × Velocity

Bullet mass = $10 \mathrm{~g} = 0.01 \mathrm{~kg}$ Bullet velocity = $250 \mathrm{~m/s}$

Momentum per bullet = $0.01 \mathrm{~kg} \cdot 250 \mathrm{~m/s} = 2.5 \mathrm{~kg \cdot m/s}$

Now let's find the momentum per second that needs to be balanced by the applied force:

Force = $125 \mathrm{~N}$

Momentum per second = Force × Time

Since we are considering a time interval of 1 second, the momentum per second is equal to the applied force:

Momentum per second = $125 \mathrm{~N}$

Now, let's find the number of bullets fired per second:

Number of bullets = Momentum per second / Momentum per bullet

Number of bullets = $\frac{125 \mathrm{~N}}{2.5 \mathrm{~kg \cdot m/s}}$

Number of bullets = 50

Therefore, the machine gun fires 50 bullets per second.

To find the speed of light in the denser medium, we can use Snell's Law at the critical angle, where the angle of refraction is $90^{\circ}$. Snell's Law states:

$n_1 \sin{\theta_1} = n_2 \sin{\theta_2}$

where $n_1$ and $n_2$ are the indices of refraction for the denser and rarer media, respectively, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively.

In this case, we have:

$n_1 \sin{45^{\circ}} = n_2 \sin{90^{\circ}}$

The critical angle is given as $45^{\circ}$, and the speed of light in the rarer medium is given as $3 \times 10^8 \mathrm{~m/s}$. We can find the index of refraction of the rarer medium using the formula:

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

where $c$ is the speed of light in a vacuum ($3 \times 10^8 \mathrm{~m/s}$), and $v$ is the speed of light in the medium.

For the rarer medium, we have:

$n_2 = \frac{3 \times 10^8 \mathrm{~m/s}}{3 \times 10^8 \mathrm{~m/s}} = 1$

Now we can rewrite Snell's Law as:

$n_1 \sin{45^{\circ}} = 1 \cdot 1$

$n_1 \sin{45^{\circ}} = 1$

$n_1 = \frac{1}{\sin{45^{\circ}}}$

Since $\sin{45^{\circ}} = \frac{1}{\sqrt{2}}$, we have:

$n_1 = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$

Now, we can find the speed of light in the denser medium using the formula for the index of refraction:

$v_1 = \frac{c}{n_1}$

$v_1 = \frac{3 \times 10^8 \mathrm{~m/s}}{\sqrt{2}}$

$v_1 = 2.12 \times 10^8 \mathrm{~m/s}$

So, the speed of light in the denser medium is $2.12 \times 10^8 \mathrm{~m/s}$.

The root mean square speed ($v_{rms}$) of a gas is given by the formula:

$v_{rms} = \sqrt{\frac{3kT}{m}}$

where $k$ is the Boltzmann constant, $T$ is the temperature, and $m$ is the molar mass of the gas molecules.

The vessels contain neon (monoatomic), chlorine (diatomic), and uranium hexafluoride (polyatomic) gases. Their molar masses are:

1. Neon: $20.18\,\text{g/mol}$ (monoatomic)
2. Chlorine: $2 \times 35.45 = 70.90\,\text{g/mol}$ (diatomic)
3. Uranium hexafluoride: $238.03 + 6 \times 18.998 = 352.03\,\text{g/mol}$ (polyatomic)

Since the temperature and the Boltzmann constant are the same for all gases, the root mean square speed is inversely proportional to the square root of the molar mass:

$v_{rms} \propto \frac{1}{\sqrt{m}}$

The lighter the gas, the higher its root mean square speed. Comparing the molar masses of the gases, we find that neon is the lightest, followed by chlorine, and uranium hexafluoride is the heaviest. Therefore, the root mean square speeds will be:

$v{rms}(\text{mono}) > v{rms}(\text{dia}) > v_{rms}(\text{poly})$

We are given two temperature scales: the X-scale and the Celsius scale. The relationship between two linear temperature scales can be expressed as follows:

$\frac{X - X_{\text{freeze}}}{X_{\text{boil}} - X_{\text{freeze}}} = \frac{C - C_{\text{freeze}}}{C_{\text{boil}} - C_{\text{freeze}}}$

Here, $X_{\text{freeze}}$ and $X_{\text{boil}}$ are the freezing and boiling points of water on the X-scale, while $C_{\text{freeze}}$ and $C_{\text{boil}}$ are the freezing and boiling points of water on the Celsius scale.

We are given the following values:

X-scale:

$X_{\text{boil}} = 65^{\circ} \mathrm{X}$

$X_{\text{freeze}} = -15^{\circ} \mathrm{X}$

Celsius scale:

$C_{\text{boil}} = 100^{\circ} \mathrm{C}$

$C_{\text{freeze}} = 0^{\circ} \mathrm{C}$

Our goal is to find the equivalent temperature of $-95^{\circ} \mathrm{X}$ on the Fahrenheit scale.

Step 1: Convert $-95^{\circ} \mathrm{X}$ to Celsius:

Use the relationship between X-scale and Celsius scale:

$\frac{-95 - (-15)}{65 - (-15)} = \frac{C - 0}{100 - 0}$

Simplify and solve for C:

$\frac{-80}{80} = \frac{C}{100}$ $C = -100^{\circ} \mathrm{C}$

Step 2: Convert $-100^{\circ} \mathrm{C}$ to Fahrenheit:

Use the conversion formula between Celsius and Fahrenheit:

$T_F = \frac{9}{5}T_C + 32$

Substitute the Celsius temperature:

$T_F = \frac{9}{5} \times (-100) + 32$

$T_F = -180 + 32$

$T_F = -148^{\circ} \mathrm{F}$

So, the equivalent temperature corresponding to $-95^{\circ} \mathrm{X}$ on the Fahrenheit scale is $-148^{\circ} \mathrm{F}$.

$$\begin{aligned} & \mathrm{m}=\tan \theta=\frac{10}{5}=2 \\ & \mathrm{~B}=\mathrm{mt} \\ & \mathrm{B}=2 \mathrm{t} \end{aligned}$$

$$\begin{aligned} & \varepsilon=\left|\frac{\mathrm{d} \phi}{\mathrm{dt}}\right|=\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=\frac{\mathrm{AdB}}{\mathrm{dt}} \\ & \varepsilon=\frac{4 \mathrm{~d}(2 \mathrm{t})}{\mathrm{dt}}=4 \times 2=8 \mathrm{mV} \end{aligned}$$

For equilibrium along the plane

$$\begin{aligned} & \mathrm{mg} \sin \theta=\frac{1}{4 \pi \epsilon_0} \times \frac{\mathrm{q}_0^2}{\left(\mathrm{~h} \operatorname{cosec} 30^{\circ}\right)^2} \\ & \therefore \mathrm{h}^2=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \times \frac{\mathrm{q}_0^2}{\mathrm{mg} \operatorname{cosec} 30^{\circ}} \\ & =9 \times 10^9 \times \frac{\left(2 \times 10^{-6}\right)^2}{0.02 \times 10 \times 2} \\ & \therefore \mathrm{h}=3 \times 10^4 \times \frac{2 \times 10^{-6}}{0.2} \\ & =0.3 \mathrm{~m} \\ & =300 \mathrm{~mm} \end{aligned}$$

To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.

First, let's calculate the heat added ($Q$) to convert 1 kg of water at 100°C into steam at 100°C using the latent heat of vaporization:

$Q = m \times L$

where $m = 1 \mathrm{~kg}$ and $L = 2257 \mathrm{~kJ/kg}$:

$Q = 1 \mathrm{~kg} \times 2257 \mathrm{~kJ/kg} = 2257 \mathrm{~kJ}$

Next, let's calculate the work done ($W$) on the system during the process. The work done is given by:

$W = -P \Delta V$

where $P$ is the atmospheric pressure and $\Delta V$ is the change in volume. We are given that the atmospheric pressure is $P = 1 \times 10^5 \mathrm{~Pa}$, and the change in volume is

$\Delta V = 1.671 \mathrm{~m}^3 - 1.00 \times 10^{-3} \mathrm{~m}^3 = 1.670 \mathrm{~m}^3$.

Now, we can calculate the work done:

$W = -(1 \times 10^5 \mathrm{~Pa})(1.670 \mathrm{~m}^3) = -167000 \mathrm{~J} = -167 \mathrm{~kJ}$

Finally, we can find the change in internal energy ($\Delta U$) using the first law of thermodynamics:

$\Delta U = Q + W$

Substitute the values of $Q$ and $W$:

$\Delta U = 2257 \mathrm{~kJ} - 167 \mathrm{~kJ} = 2090 \mathrm{~kJ}$

The change in internal energy of the system during the process is +2090 kJ.

The current sensitivity ($I_s$) of a moving coil galvanometer is given by:

$I_s = \frac{nBA}{C}$

where $n$ is the number of turns in the coil, $B$ is the magnetic field, $A$ is the area of the coil, and $C$ is the torsional constant of the suspension wire.

We are told that the current sensitivity is increased by 25% by changing only the number of turns and the area of the cross section while keeping the resistance of the galvanometer coil constant. Let the new number of turns be $n'$, and the new area be $A'$.

Since the current sensitivity is increased by 25%, we have:

$I_s' = 1.25I_s = \frac{n' B A'}{C}$

The resistance ($R$) of the galvanometer coil is given by:

$R = \rho \frac{l}{A}$

where $\rho$ is the resistivity of the wire and $l$ is the length of the wire. Since the resistance is kept constant, we can write:

$R' = \rho \frac{l'}{A'} = R$

Now, the voltage sensitivity ($V_s$) of the galvanometer is given by:

$V_s = I_s \times R$

We want to find the percentage change in the voltage sensitivity. Let the new voltage sensitivity be $V_s'$:

$V_s' = I_s' \times R'$

Since $R = R'$, we can write:

$V_s' = 1.25I_s \times R = 1.25V_s$

The percentage change in the voltage sensitivity is:

$\frac{V_s' - V_s}{V_s} \times 100\% = \frac{1.25V_s - V_s}{V_s} \times 100\% = 25\%$

Area under the graph from $t=0$ to $t=20 \mathrm{sec}=200 \mathrm{~m}$

Area under the graph from $\mathrm{t}=20$ to $t=25 \mathrm{sec}=50 \mathrm{~m}$

So distance covered $=(200+50) \mathrm{m}=250 \mathrm{~m}$

Displacement $=(200-50) \mathrm{m}=150 \mathrm{~m}$

$\therefore$ $\frac{\text { distance }}{\text { displacement }}=\frac{250}{150}=\left(\frac{5}{3}\right)$

We are given that the half-life of A is the same as the average life of B. The relationship between half-life ($T_{1/2}$) and the decay constant ($\lambda$) is:

$T_{1/2} = \frac{\ln 2}{\lambda}$

For the average life ($\tau$), the relationship with the decay constant is:

$\tau = \frac{1}{\lambda}$

According to the given information, the half-life of A is equal to the average life of B:

$T_{1/2(A)} = \tau_{B}$

Now, we can substitute the relationships for half-life and average life:

$\frac{\ln 2}{\lambda_{A}} = \frac{1}{\lambda_{B}}$

To find the correct relationship between $\lambda_{A}$ and $\lambda_{B}$, we can rearrange the equation:

$\lambda_{A} = \lambda_{B} \ln 2$

When a monochromatic light is incident on hydrogen atoms in the ground state (n = 1), the hydrogen atoms can absorb energy and transition to higher energy levels. When the atoms return to lower energy levels, they emit radiation of different wavelengths corresponding to the energy differences between the energy levels.

The energy levels of the hydrogen atom are given by the formula:

$E_n = -\frac{13.6 \mathrm{~eV}}{n^2}$

where $E_n$ is the energy of the nth level and $n$ is the principal quantum number.

Since the hydrogen atoms emit radiation of six different wavelengths, there must be six different transitions from the excited states back to lower energy levels.

The six transitions correspond to the following energy level changes:

1. From n = 2 to n = 1
2. From n = 3 to n = 1
3. From n = 3 to n = 2
4. From n = 4 to n = 1
5. From n = 4 to n = 2
6. From n = 4 to n = 3

The highest energy level involved is n = 4. Therefore, the incident light must have a frequency high enough to excite the hydrogen atoms from the ground state (n = 1) to n = 4.

The energy difference between these levels is:

$\Delta E = E_4 - E_1 = \frac{13.6 \mathrm{~eV}}{4^2} - \frac{13.6 \mathrm{~eV}}{1^2} = -0.85 \mathrm{~eV} + 13.6 \mathrm{~eV} = 12.75 \mathrm{~eV}$

The frequency of the incident light is related to the energy difference by the equation:

$\Delta E = h \nu$

where $h$ is the Planck's constant and $\nu$ is the frequency.

Now, we can solve for the frequency:

$\nu = \frac{\Delta E}{h} = \frac{12.75 \mathrm{~eV}}{4.25 \times 10^{-15} \mathrm{~eVs}} = 3 \times 10^{15} \mathrm{~Hz}$

So, the value of $x$ is 3.

The photoelectric effect occurs when light (or more generally, electromagnetic radiation) incident on a metallic surface causes the ejection of electrons from the surface. The energy of the incident photons must be greater than the work function of the metal (denoted by $\phi_0$) for the electrons to be ejected.

The energy of the ejected electrons can be expressed as the difference between the energy of the incident photons and the work function of the metal. The maximum kinetic energy of the ejected electrons is given by:

$K_{max} = h\nu - \phi_0$

where $h$ is the Planck's constant, $\nu$ is the frequency of the incident light, and $\phi_0$ is the work function of the metal.

We can also write the maximum kinetic energy in terms of the stopping potential ($V_0$) and the elementary charge ($e$) of an electron:

$K_{max} = eV_0$

Equating these two expressions for the maximum kinetic energy, we get:

$eV_0 = h\nu - \phi_0$

We can express the frequency $\nu$ in terms of the speed of light $c$ and the wavelength $\lambda$:

$\nu = \frac{c}{\lambda}$

Substituting this expression for frequency in the equation, we get:

$eV_0 = h\frac{c}{\lambda} - \phi_0$

Now, we have two cases:

1. The stopping potential is $V_0$, and the wavelength is $\lambda$.
2. The stopping potential is $\frac{V_0}{4}$, and the wavelength is $2\lambda$.

For the first case, we use the photoelectric effect equation as is:

$eV_0 = \frac{hc}{\lambda} - \phi_0$

For the second case, we replace $V_0$ with $\frac{V_0}{4}$ and $\lambda$ with $2\lambda$:

$\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi_0$

Now we have two equations:

(1) $eV_0 = \frac{hc}{\lambda} - \phi_0$

(2) $\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi_0$

We can rewrite equation (1) as:

$\frac{eV_0}{4} = \frac{hc}{4\lambda} - \frac{\phi_0}{4}$

Now we can equate the two expressions for $\frac{eV_0}{4}$:

$\frac{hc}{4\lambda} - \frac{\phi_0}{4} = \frac{hc}{2\lambda} - \phi_0$

Now we isolate the terms containing $\phi_0$:

$\frac{3\phi_0}{4} = \frac{hc}{4\lambda} - \frac{hc}{2\lambda}$

$\frac{3\phi_0}{4} = \frac{hc}{4\lambda}$

Now we can write the work function $\phi_0$ in terms of the threshold wavelength $\lambda_0$:

$\phi_0 = \frac{hc}{\lambda_0}$

Substituting the expression for the work function $\phi_0$ in terms of the threshold wavelength $\lambda_0$ into the previous equation, we get:

$\frac{3}{4} \frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$

Now we can cancel out the common terms $hc$ from both sides:

$\frac{3}{4\lambda_0} = \frac{1}{4\lambda}$

Next, we can cross-multiply to solve for $\lambda_0$:

$3\lambda = \lambda_0$

Thus, the threshold wavelength for this metallic surface is $3\lambda$.

$\mathrm{E}_2-\mathrm{E}_1=8-4=4 \mathrm{~V}$

$$\begin{aligned} & \frac{1}{3}+\frac{1}{6}=\frac{1}{2}=\frac{1}{R} \\ & R=2 \Omega \end{aligned}$$




$I=\frac{4}{8}=0.5 \mathrm{~A}$



$$\begin{aligned} & \mathrm{I}_1=\left(\frac{6}{3+6}\right) \times 0.5 \\ & \mathrm{I}_1=\frac{2}{3} \times 0.5=\frac{1}{3} \mathrm{~A} \end{aligned}$$

$I_1=\frac{x}{3}=\frac{1}{3} \therefore x=1$

Statement I is correct. Astronomical unit (AU), parsec (pc), and light year (ly) are indeed units for measuring astronomical distances.

However, Statement II is incorrect. The correct order of these units is:

$\mathrm{AU} < \mathrm{ly} < \mathrm{Parsec} (\mathrm{Pc})$

1 AU is the average distance from the Earth to the Sun, which is about 93 million miles or 150 million kilometers.

1 light year is the distance that light travels in one year in a vacuum, which is approximately 9.461 trillion kilometers (5.878 trillion miles).

1 parsec is equal to about 3.26 light years, or approximately 30.9 trillion kilometers (19.2 trillion miles). Parsec (parallax of one arcsecond) is the distance at which an arc of length $1 \mathrm{Au}$ subtends an angle of one arcsecond $(1 \mathrm{Pc}=3.26 \mathrm{ly})$.



So, the correct answer is Statement I is correct but Statement II is incorrect.

Let's find the focal length of the lens in air and in the liquid. We will use the lens maker's formula:

$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

where $f$ is the focal length, $\mu$ is the refractive index of the lens material, and $R_1$ and $R_2$ are the radii of curvature of the lens surfaces.

Since the lens is convex, both surfaces have the same radius of curvature (positive), so $R_1 = R_2 = 20 \mathrm{~cm}$.

First, let's find the focal length of the lens in air:

$\frac{1}{f_\text{air}} = (1.8 - 1)\left(\frac{1}{20} - \frac{1}{20}\right) = 0.8\left(\frac{1}{20}\right)$

$f_\text{air} = \frac{1}{0.8\left(\frac{1}{20}\right)} = 25 \mathrm{~cm}$

Now, let's find the focal length of the lens in the liquid. The relative refractive index of the lens with respect to the liquid is:

$\mu_\text{rel} = \frac{1.8}{1.5} = 1.2$

$\frac{1}{f_\text{liquid}} = (1.2 - 1)\left(\frac{1}{20}\right)$

$f_\text{liquid} = \frac{1}{0.2\left(\frac{1}{20}\right)} = 100 \mathrm{~cm}$

The power of a lens is given by:

$P = \frac{1}{f}$

Now, we can find the ratio of the power of the lens in air to its power in the liquid:

$\frac{P\text{air}}{P\text{liquid}} = \frac{f\text{liquid}}{f\text{air}} = \frac{100}{25} = 4$

So, the ratio of the power of the lens in air to its power in the liquid is $x:1$, where $x = 4$.

To find the energy contained in a volume of an electromagnetic wave, we need to calculate the energy density and then multiply it by the volume.

For an electromagnetic wave, the energy density $u$ is given by:

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

where $\varepsilon_0$ is the vacuum permittivity and $E$ is the electric field amplitude.

First, let's find the amplitude of the electric field. In the given equation:

$\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}\right) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}$

The amplitude $E$ is $20 \mathrm{~N/C}$.

Now, we can find the energy density:

$u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2} \cdot (20 \mathrm{~N/C})^2$

$u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \cdot 400$

$u = 1.77 \times 10^{-9} \mathrm{J/m^3}$

The energy contained in a volume of $5 \times 10^{-4} \mathrm{m^3}$ will be:

$E_\text{total} = u \cdot V$

$E_\text{total} = 1.77 \times 10^{-9} \mathrm{J/m^3} \cdot 5 \times 10^{-4} \mathrm{m^3}$

$E_\text{total} = 8.85 \times 10^{-13} \mathrm{J}$

Given:

1. When tension $T_1 = 100 \mathrm{~N}$, extension $= l_1 - l_0$.
2. When tension $T_2 = 120 \mathrm{~N}$, extension $= l_2 - l_0$.

Now, let's write the equations using Hooke's Law:

$100 = k(l_1 - l_0)$

$120 = k(l_2 - l_0)$

Divide the first equation by the second equation:

$\frac{5}{6} = \frac{l_1 - l_0}{l_2 - l_0}$

Given the relationship between $l_1$ and $l_2$:

$10l_2 = 11l_1$

Now, let's solve for $l_0$:

$5l_2 - 5l_0 = 6l_1 - 6l_0$

$l_0 = 6l_1 - 5l_2$

Substitute the relationship between $l_1$ and $l_2$:

$l_0 = 6l_1 - 5\left(\frac{11l_1}{10}\right)$

$l_0 = 6l_1 - \frac{11l_1}{2}$

$l_0 = \frac{l_1}{2}$

Therefore, the natural length of the wire is $\frac{1}{x}l_1 = \frac{2}{1}l_1 = 2l_1$.

The value of $x$ is $2$.

When a coin is placed on a rotating table and is just about to slip, the centrifugal force acting on the coin equals the maximum static friction force. Let's denote the mass of the coin as $m$, the initial angular velocity as $\omega_1$, and the final angular velocity as $\omega_2$.

Initially, when the coin is placed at a distance of 1 cm from the center, the centrifugal force acting on the coin is:

$F_1 = m r_1 \omega_1^2$

where $r_1 = 1 \mathrm{~cm}$.

When the angular velocity is halved ($\omega_2 = \frac{1}{2}\omega_1$), the centrifugal force acting on the coin when it just slips is:

$F_2 = m r_2 \omega_2^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2$

Since the coin is just about to slip in both cases, the maximum static friction force remains the same. Therefore, we can equate the centrifugal forces:

$m r_1 \omega_1^2 = m r_2 \left(\frac{1}{2}\omega_1\right)^2$

Canceling the mass $m$ and the initial angular velocity $\omega_1^2$ from both sides, we get:

$r_1 = r_2 \left(\frac{1}{2}\right)^2$

Now, we can solve for $r_2$:

$r_2 = \frac{r_1}{\left(\frac{1}{2}\right)^2} = \frac{1 \mathrm{~cm}}{\frac{1}{4}} = 4 \mathrm{~cm}$

So, the coin will just slip when placed at a distance of 4 cm from the center when the angular velocity is halved.

The acceleration due to gravity at the surface of a planet can be expressed as:

$g \propto \rho R$

Now let's find the ratio of acceleration due to gravity at the surfaces of planets A and B:

$\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B}$

Given that the densities are $\rho$ and $\frac{\rho}{3}$ and the radii are $R$ and $4R$ for planets A and B, respectively, we have:

$\frac{g_A}{g_B} = \frac{\rho \cdot R}{\left(\frac{\rho}{3}\right) \cdot (4R)} = \frac{3}{4}$

So, the ratio of acceleration due to gravity at their surfaces is 3 : 4