JEE Main 2023 8th April Evening Shift Question Paper with Solutions

The emf induced in a rod moving through a magnetic field is given by Faraday's law of electromagnetic induction, specifically, in the form of motional emf, which states that:

$\text{emf} = B \cdot L \cdot v$

where:

• (B) is the magnetic field strength,
• (L) is the length of the rod, and
• (v) is the velocity of the rod.

In this case, we are given the emf, (B), and (L), and we need to solve for (v). Rearranging the equation gives:

$v = \frac{\text{emf}}{B \cdot L}$

Substituting the given values:

$v = \frac{0.08 \, \text{V}}{0.4 \, \text{T} \times0.1 \, \text{m}} = 2 \, \text{m/s}$

Potential across

$\mathrm{R}_1=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{C}}$ $=V_A-V_B$    $\left(\because \mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}\right)$

Potential across

$R_2=V_D-V_C$ $=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$    $\left(\because \mathrm{V}_{\mathrm{D}}=\mathrm{V}_{\mathrm{A}}\right)$

Potential across

$R_3=V_D-V_B$ $=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$     $\left(\because \mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}\right)$

It means all the resistance are in parallel

$$\begin{aligned} & \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3} \\ & \Rightarrow \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{20}+\frac{1}{20}+\frac{1}{10} \\ & \Rightarrow \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1+1+2}{20} \\ & \Rightarrow \mathrm{R_{eq}}=\frac{20}{4}=5 \mathrm{k} \Omega \end{aligned}$$

The energy stored in a capacitor can be calculated using the formula:

$U = \frac{1}{2} C V^2$

where:

• (U) is the energy,
• (C) is the capacitance,
• (V) is the voltage.

Initially, the energy stored in the first capacitor is:

$U_{\text{initial}} = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (200 \, \text{V})^2 = 0.012 \, \text{J} = 12 \, \mu\text{J}.$

When the charged capacitor is connected to the uncharged capacitor, the charge will distribute equally between them because they have the same capacitance. Therefore, the final voltage across each capacitor is half of the initial voltage, i.e., 100 V.

The energy in each capacitor after the redistribution is:

$U_{\text{final each}} = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (100 \, \text{V})^2 = 0.003 \, \text{J} = 3 \, \mu\text{J}.$

As there are two capacitors, the total final energy is:

$U_{\text{final total}} = 2 \times U_{\text{final each}} = 2 \times 3 \, \mu\text{J} = 6 \, \mu\text{J}.$

The energy loss is the difference between the initial energy and the final energy:

$\Delta U = U_{\text{initial}} - U_{\text{final total}} = 12 \, \mu\text{J} - 6 \, \mu\text{J} = 6 \, \mu\text{J}.$

When a metal target is bombarded with high-energy electrons, the phenomenon known as X-ray emission occurs. This is due to the excitation of the inner-shell electrons in the metal atoms by the high-energy electrons. When these inner-shell electrons drop back to their original energy levels, they emit energy in the form of X-ray photons.

The efficiency of a Carnot engine is given by:

$\eta = 1 - \frac{T_{c}}{T_{h}}$

where:

• $\eta$ is the efficiency of the Carnot engine,
• $T_{c}$ is the temperature of the cold reservoir,
• $T_{h}$ is the temperature of the hot reservoir.

Note that the temperatures must be in Kelvin for this formula.

Here, the given temperatures are $127^{\circ}C$ and $27^{\circ}C$. Converting these to Kelvin gives:

$T_{h} = 127^{\circ}C + 273.15 = 400.15 K$ $T_{c} = 27^{\circ}C + 273.15 = 300.15 K$

So the efficiency of the Carnot engine is:

$\eta = 1 - \frac{300.15}{400.15} = 0.25$

The efficiency can also be defined as the work done divided by the heat supplied:

$\eta = \frac{W}{Q_{h}}$

where:

• $W$ is the work done by the engine,
• $Q_{h}$ is the heat transferred to the engine by the hot reservoir.

We can rearrange this equation to solve for $Q_{h}$:

$Q_{h} = \frac{W}{\eta} = \frac{2 \, kJ}{0.25} = 8 \, kJ$

The decay of a radioactive material follows an exponential decay law, which can be expressed as:

$N = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}$

where:

• $N$ is the final amount of the material,
• $N_0$ is the initial amount of the material,
• $t$ is the elapsed time,
• $T$ is the half-life of the material.

From the problem, we know that the material is reduced to $1/8$ of its original amount in 3 days. Therefore, the half-life of the material can be calculated as follows:

$\frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{3}{T}}$

This simplifies to $2^{-3} = 2^{-\frac{3}{T}}$, which gives $T = 1 \, \text{day}$.

Knowing the half-life, we can now find the initial amount of the material. We know that after 5 days, $8 \times 10^{-3} \, \text{kg}$ of the material is left. Therefore, we can write:

$8 \times 10^{-3} \, \text{kg} = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{5}{1}}$

Solving this equation for $N_0$ gives:

$N_0 = 8 \times 10^{-3} \, \text{kg} \cdot 2^{5} = 256 \times 10^{-3} \, \text{kg} = 0.256 \, \text{kg} = 256 \, \text{g}$

The power radiated by a linear antenna is typically determined by its physical length relative to the wavelength of the signal it is transmitting or receiving. The power radiated from a linear antenna of length 'l' is proportional to the square of the ratio of its length to the wavelength, represented as $(\frac{l}{\lambda})^{2}$.

In saturation mode $\mathrm{V}_{\mathrm{CE}}=0$


Now

$$\begin{aligned} & \mathrm{V}_{\mathrm{CC}}-\mathrm{I}_{\mathrm{c}} \mathrm{R}_{\mathrm{c}}=0 ~~~~\text{(from KVL)}\\ & \Rightarrow \mathrm{I}_{\mathrm{b}}=\frac{\mathrm{V}_{\mathrm{CC}}}{\mathrm{R}_{\mathrm{C}}}=\frac{1}{1 \times 10^3}=10^{-3} \mathrm{~A} \end{aligned}$$

Given that

$\beta=100=\frac{I_c}{I_b}$

$\begin{aligned} & \Rightarrow \mathrm{I}_b=\frac{\mathrm{I}_c}{100}=\frac{10^{-3}}{100}=10^{-5} \mathrm{~A} \\ & \Rightarrow \mathrm{I}_b=10 \mu \mathrm{A}\end{aligned}$

The magnetic field at the center of a loop (B₁) is given by

$B_1 = \frac{\mu_0 I}{2r}$

The magnetic field on the axis of the loop at a distance ( r ) from the center (B₂) is given by

$B_2 = \frac{\mu_0 Ir^2}{2(r^2 + d^2)^{3/2}}$

where ( d ) is the distance from the center of the coil along the axis. Since ( d = r ), we get

$B_2 = \frac{\mu_0 I}{4\sqrt{2}r}$

The ratio of $B_1$ to $B_2$ is

$\frac{B_1}{B_2} = \frac{\mu_0 I}{2r} \times \frac{4\sqrt{2}r}{\mu_0 I} = \sqrt{8} : 1$

So, the value of ( x ) is 8.

In the double-slit experiment, the fringe width ($\beta$) is given by the formula :

$ \beta = \frac{\lambda D}{d} $

where:

• $\lambda$ is the wavelength of the light,
• $D$ is the distance between the screen and the double-slit,
• $d$ is the separation between the two slits.

If we are considering a change in wavelength but the fringe width is changing and the setup of the experiment (the values of $D$ and $d$) stays the same, we can see that the fringe width is directly proportional to the wavelength. This is because $D$ and $d$ are constants in this case, so we can write :

$ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} $

Here, the given wavelengths are $\lambda_1 = 400 \, \text{nm}$ and $\lambda_2 = 600 \, \text{nm}$, and the given fringe width for the light of wavelength $400 \, \text{nm}$ is $\beta_1 = 2 \, \text{mm}$. We are asked to find the fringe width $\beta_2$ for the light of wavelength $600 \, \text{nm}$.

Substituting the given values into the proportionality equation, we get :

$ \frac{2 \, \text{mm}}{\beta_2} = \frac{400 \, \text{nm}}{600 \, \text{nm}} $

Solving this equation for $\beta_2$ gives:

$ \beta_2 = 2 \, \text{mm} \times \frac{600 \, \text{nm}}{400 \, \text{nm}} = 3 \, \text{mm} $

The refraction at a spherical surface is governed by the formula:

$\frac{1}{v} - \frac{1}{u} = \frac{n_2 - n_1}{R} n_2$

where:

• (v) is the image distance,
• (u) is the object distance,
• ($n_1$) is the refractive index of the medium where the object is,
• ($n_2$) is the refractive index of the medium where the image is formed,
• (R) is the radius of curvature of the refracting surface.

Here, we have:

• (u = -15) cm (the object is on the same side as the light is coming from, so the distance is taken as negative),
• ($n_1$ = 1.0) (refractive index of the rarer medium),
• ($n_2$ = 1.5) (refractive index of the denser medium),
• (R = -30) cm (the center of curvature is in the denser medium, so the radius is taken as negative).

Substituting these values into the formula, we get:

$\frac{1}{v} - \left(-\frac{1}{15}\right) = \frac{1.5 - 1.0}{-30} \cdot 1.5$

Solving for (v), we get:

$v = \frac{1}{\frac{1}{15} + \frac{1.5}{30}} = -30 \, \text{cm}$

Therefore, the image is formed at a distance of 30 cm from the pole of the surface, on the same side as the object. The negative sign indicates that the image is virtual and is formed on the same side of the surface as the light is coming from.

First, we will use conservation of momentum to find the velocity of the bullet-block system just after the bullet gets embedded into the block.

The initial momentum of the system is given by the momentum of the bullet (as the block is initially at rest), and the final momentum of the system is the combined momentum of the bullet and the block.

Setting initial momentum equal to final momentum:

$m_{\text{bullet}} \cdot v_{\text{bullet}} = (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}$

Solving for ($v_{\text{final}}$):

$v_{\text{final}} = \frac{m_{\text{bullet}} \cdot v_{\text{bullet}}}{m_{\text{bullet}} + m_{\text{block}}}$

Substituting the given values:

$v_{\text{final}} = \frac{0.1 \, \text{kg} \cdot 400 \, \text{m/s}}{0.1 \, \text{kg} + 3.9 \, \text{kg}} = 10 \, \text{m/s}$

Next, we know the block comes to rest after moving 20 m due to friction. The work done by the friction force is equal to the initial kinetic energy of the block (since it comes to rest, the final kinetic energy is 0). The work done by friction is given by the friction force times the distance, and the friction force is equal to the coefficient of friction times the normal force (which is equal to the weight of the block).

So, setting the work done by friction equal to the initial kinetic energy of the block:

$\mu \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot g \cdot d = \frac{1}{2} \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}^2$

Solving for ($\mu$):

$\mu = \frac{\frac{1}{2} \cdot (m_{\text{bullet}} + m_{\text{block}}) \cdot v_{\text{final}}^2}{(m_{\text{bullet}} + m_{\text{block}}) \cdot g \cdot d}$

Substituting the given values:

$\mu = \frac{\frac{1}{2} \cdot (0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot (10 \, \text{m/s})^2}{(0.1 \, \text{kg} + 3.9 \, \text{kg}) \cdot 10 \, \text{m/s}^2 \cdot 20 \, \text{m}} = 0.25$

The Balmer series corresponds to electronic transitions in a hydrogen atom that terminate in the second (n=2) energy level. The spectral lines in the Balmer series are often labeled according to a Greek letter scheme, with H$_\alpha$ corresponding to the n=3 to n=2 transition, H$_\beta$ corresponding to the n=4 to n=2 transition, and so on.

The wavelength of a spectral line in the Balmer series can be calculated using the Rydberg formula:

$ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) $

where $R_H$ is the Rydberg constant for hydrogen, $n$ is the principal quantum number corresponding to the initial energy level, and $\lambda$ is the wavelength of the spectral line.

Using this formula, the wavelength of the H$_\alpha$ line is:

$ \frac{1}{\lambda_{\alpha}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) $

And the wavelength of the H$_\beta$ line is:

$ \frac{1}{\lambda_{\beta}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) $

Therefore, the ratio of the wavelengths of the H$_\alpha$ and H$_\beta$ lines is:

$ \frac{\lambda_{\alpha}}{\lambda_{\beta}} = \frac{\left( \frac{1}{2^2} - \frac{1}{4^2} \right)}{\left( \frac{1}{2^2} - \frac{1}{3^2} \right)} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{27}{20} $

So, comparing with the given ratio $\frac{x}{20}$, we find that $x = 27$.

The Q factor (quality factor) in a series RLC circuit can be given by the formula:

$Q = \frac{X_L}{R} = \frac{\omega L}{R}$

where ($X_L$) is the inductive reactance, (R) is the resistance, (L) is the inductance, and ($\omega$) is the angular frequency.

In a series RLC circuit at resonance, the resonant frequency (f) is given by

$f = \frac{1}{2\pi\sqrt{LC}}$

or equivalently, the angular frequency ($\omega$) at resonance is

$\omega = \frac{1}{\sqrt{LC}}$

Substituting (L = 1H) and ($C = 6.25 \mu F = 6.25 \times 10^{-6} F$) into the equation for (\omega) gives

$\omega = \frac{1}{\sqrt{1H \times 6.25 \times 10^{-6}F}}$ = 400 rad/s

Substituting ($\omega = 400 rad/s$), (L = 1H), and ($R = 100\Omega$) into the equation for the Q factor gives

$Q = \frac{\omega L}{R} = \frac{400 rad/s \times 1H}{100\Omega} = 4$

So, the Q factor of the circuit is 4.

Using the formula:

$I = n e A v$

where $I$ is the current, $n$ is the number density of free electrons, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $v$ is the drift speed of the electrons.

We can isolate $v$ to find:

$v = \frac{I}{n e A}$

Substituting the given values:

$v = \frac{3.2 \, \text{A}}{8 \times 10^{28} \, \text{m}^{-3} \times 1.6 \times 10^{-19} \, \text{C} \times 2 \times 10^{-6} \, \text{m}^2}$

This simplifies to:

$v = \frac{3.2}{16 \times 1.6 \times 10^3} \, \text{ms}^{-1} = 125 \times 10^{-6} \, \text{ms}^{-1}$

So, the drift speed of the electrons is indeed $125 \times 10^{-6} \, \text{ms}^{-1}$.

The fundamental frequency (also known as the first harmonic) of a vibrating string is given by the formula:

$f = \frac{v}{2L}$

where:

• (f) is the frequency,
• (v) is the speed of the wave in the string, and
• (L) is the length of the string.

In this case, the speed of the wave in the string stays the same because it depends on the properties of the string and the tension in it, which we can assume to be constant.

We can write the equation for the fundamental frequency of the original string and the shorter string:

$f_1 = \frac{v}{2L_1}$

$f_2 = \frac{v}{2L_2}$

where:

• $(f_1 = 120 \, \text{Hz})$ and $(L_1 = 90 \, \text{cm})$ for the original string, and
• $(f_2 = 180 \, \text{Hz})$ and $(L_2)$ is what we're trying to find for the shorter string.

We can set up a ratio of these two equations:

$\frac{f_1}{f_2} = \frac{L_2}{L_1}$

Substituting in the given values, we get:

$\frac{120 \, \text{Hz}}{180 \, \text{Hz}} = \frac{L_2}{90 \, \text{cm}}$

Solving for ($L_2$) gives:

$L_2 = 90 \, \text{cm} \times \frac{120 \, \text{Hz}}{180 \, \text{Hz}} = 60 \, \text{cm}$

So, the length of the string producing a fundamental frequency of 180 Hz will be 60 cm.

Area under velocity time graph gives displacement of body in given time.

Area under acceleration time graph gives change in velocity in the given time.

So Statement I false but Statement II True

After time $t$, the angular displacement will be

$\theta=\omega t$

Total angular displacement from $x$-axis.

$\theta_{\text {total }}=\omega t+\frac{\pi}{6} \quad\quad\left(\because 30^{\circ}=\frac{\pi}{6}\right)$

Now, OP has two component

The horizontal component will be the projection along $x$-axis

$=r \cos \left(\theta_{\text {Total }}\right)=r \cos \left(\omega t+\frac{\pi}{6}\right)$

The increase in kinetic energy can be found using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

The work done by a force is given by the equation:

$W = F \cdot d$

where ( F ) is the force and ( d ) is the distance over which the force is applied.

However, we don't have the distance in this problem. But we do know that the force is applied for a time of 5 seconds, and that the initial momentum of the body is 10 kg m/s. We can use these facts to find the work done.

First, we can use the equation for force, ( F = ma ), to find the acceleration of the body:

$a = \frac{F}{m} = \frac{2 \, \text{N}}{5 \, \text{kg}} = 0.4 \, \text{m/s}^2$

Then, we can use the equation for distance in uniformly accelerated motion, ( $d = v_i t + \frac{1}{2} a t^2$), where ( $v_i$ ) is the initial velocity of the body. We can find ( $v_i$) from the initial momentum and the mass of the body:

$v_i = \frac{p}{m} = \frac{10 \, \text{kg m/s}}{5 \, \text{kg}} = 2 \, \text{m/s}$

Substituting ( $v_i$ ), ( a ), and ( t ) into the equation for ( d ) gives:

$d = 2 \, \text{m/s} \cdot 5 \, \text{s} + \frac{1}{2} \cdot 0.4 \, \text{m/s}^2 \cdot (5 \, \text{s})^2 = 10 \, \text{m} + 5 \, \text{m} = 15 \, \text{m}$

Finally, we can substitute ( F ) and ( d ) into the equation for work to find the increase in kinetic energy:

$\Delta KE = W = F \cdot d = 2 \, \text{N} \cdot 15 \, \text{m} = 30 \, \text{J}$

So, the increase in the kinetic energy of the body is 30 J.

The acceleration due to gravity ($g$) at the surface of Earth is approximately $9.81 \, \text{m/s}^2$. This acceleration decreases as we move away from the Earth's surface because gravity is a force that attracts objects towards the center of the Earth. This force decreases with the square of the distance from the center of the Earth, due to the inverse square law.

However, if we are at a height ($h$) much less than the radius of the Earth ($R$), we can use a linear approximation to calculate the new acceleration due to gravity ($g'$) at this height. This is because for small heights compared to the radius of Earth, the decrease in $g$ can be approximated to be linear. This is represented by the following formula:

$g' = g\left(1 - \frac{2h}{R}\right)$

Here's how this formula is derived:

The force of gravity ($F$) is given by the universal law of gravitation:

$F = G \frac{m_1 m_2}{d^2}$

where:

• $G$ is the gravitational constant,
• $m_1$ and $m_2$ are the masses of the two objects (in this case, the Earth and the object we're considering),
• $d$ is the distance between the centers of the two objects.

If we consider an object of mass $m$ at the surface of Earth, the force it experiences due to gravity is:

$F = G \frac{mM}{R^2} = mg$

where:

• $M$ is the mass of Earth,
• $R$ is the radius of Earth,
• $g = \frac{GM}{R^2}$ is the acceleration due to gravity at the Earth's surface.

Now, if the object is at a height $h$ above the Earth's surface, the force it experiences is:

$F' = G \frac{mM}{(R + h)^2}$

We can write $(R + h)^2$ as $R^2 (1 + \frac{h}{R})^2$, and because $h << R$, we can use the binomial approximation $(1 + x)^2 \approx 1 + 2x$ for small $x$ to get:

$F' \approx G \frac{mM}{R^2} \left(1 - \frac{2h}{R}\right) = mg \left(1 - \frac{2h}{R}\right)$

Equating the forces $F = mg$ and $F' = mg'$ gives us:

$g' = g \left(1 - \frac{2h}{R}\right)$

The photoelectric effect is the phenomenon of emission of electrons (or photoelectrons) from the surface of a metal when it is illuminated by light of sufficient energy. The observations from the photoelectric effect led to the development of quantum theory.

According to the principles of the photoelectric effect:

A. The photocurrent (number of photoelectrons ejected per unit time) is indeed proportional to the intensity of the incident radiation. More intense light means more photons hitting the surface and thus more electrons being ejected.

B. The maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light but rather on its frequency. Increasing the intensity of light increases the number of photoelectrons (current) but does not increase their maximum kinetic energy.

C. The maximum kinetic energy of the photoelectrons does indeed depend on the frequency of the incident light. If the frequency of the incident light is below a certain threshold frequency specific to the metal, no photoelectrons are emitted regardless of the intensity of the light. Above this threshold, the maximum kinetic energy of the photoelectrons increases linearly with the frequency of the light.

D. The emission of photoelectrons does not require a minimum threshold intensity of incident radiation, but rather a minimum threshold frequency.

E. The maximum kinetic energy of the photoelectrons is not independent of the frequency of the incident light, but rather depends on it.

Therefore, only statements A and C are correct.

The electric potential (V) at a distance (r) from a point charge (Q) is given by the formula:

$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$

In this case, we know (V) and (Q), and we're asked to solve for (r). We can rearrange the formula to solve for (r):

$r = \frac{1}{4\pi\epsilon_0} \frac{Q}{V}$

Substituting the given values into this equation gives:

$r = \frac{9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2}}{1} \frac{5 \times 10^{-9} \, \text{C}}{50 \, \text{V}} = 0.9 \, \text{m}$

So, the distance of the point P from the point charge is 0.9 meters. This corresponds to 90 cm.

Let's analyze the SI units of each quantity from List I:

1. Torque: Torque (τ) is given by the cross product of the radius (r) and the force (F). Therefore, its SI units are Newton meter (Nm), which translates to $\mathrm{ML^2T^{-2}}$ in fundamental units.




2. Stress: Stress is force per unit area. The SI unit for force is the Newton (N) and for area is meter squared (m²). Therefore, the SI unit for stress is Pascal (Pa), which translates to $\mathrm{ML^{-1}T^{-2}}$ in fundamental units.




3. Pressure gradient: The pressure gradient is the rate of increase or decrease in pressure. It has units of pressure per distance. In SI units, that's Pascal per meter (Pa/m), which translates to $\mathrm{ML^{-2}T^{-2}}$ in fundamental units.




4. Coefficient of viscosity: This is a measure of a fluid's resistance to shear or flow, and its SI units are the Pascal second (Pa.s), which translates to $\mathrm{ML^{-1}T^{-1}}$ in fundamental units.

Therefore, the correct matches are:

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

If we take the velocity (v) of the satellite to be $v = \sqrt{GM/r}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the satellite, then the orbital angular momentum (L) is:

$L = mvr = m \sqrt{GMr}$

If the distance (r) from the Earth's center is increased by a factor of 8, the new angular momentum $L'$ is:

$L' = m \sqrt{GM(8r)} = m \sqrt{8GMr} = 2 \sqrt{2} L$

However, the distance from the earth's surface is given, so we have to take into account the radius of the Earth ($R$) in our calculations. When the height from the earth's surface is increased eight times, the distance from the earth's center is $r = R + 8h$, which is approximately $9R$ (because the height of the satellite above the Earth is generally much less than the radius of the Earth).

So, the new angular momentum (L'') is:

$L'' = m \sqrt{GM(9R)} = 3 \sqrt{GM(R)} = 3L$

A hydraulic lift works based on Pascal's principle, which states that the pressure applied at one point in an incompressible fluid is transmitted equally in all directions.

The force exerted by the car on the hydraulic fluid is equal to the weight of the car, which is $F = mg$, where $m$ is the mass of the car and $g$ is the acceleration due to gravity.

Substituting the given values, we get:

$F = 5000 \, \text{kg} \times 10 \, \text{m/s}^2 = 50000 \, \text{N}$

The pressure exerted by the car on the hydraulic fluid is equal to the force divided by the area over which the force is distributed, which is $P = \frac{F}{A}$, where $A$ is the cross-sectional area of the cylinder carrying the load.

However, the given area is in cm², so we need to convert it to m². We know that 1 m² = 10,000 cm², so:

$A = 250 \, \text{cm}^2 \times \frac{1 \, \text{m}^2}{10000 \, \text{cm}^2} = 0.025 \, \text{m}^2$

Substituting the values of force and area into the formula for pressure, we get:

$P = \frac{50000 \, \text{N}}{0.025 \, \text{m}^2} = 2 \times 10^6 \, \text{Pa}$

The kinetic energy of an ideal gas is given by the equation:

$KE = \frac{3}{2} kT$

where (k) is Boltzmann's constant and (T) is the absolute temperature in kelvins. Therefore, the kinetic energy of a gas is directly proportional to its temperature.

If the kinetic energy doubles, the temperature must also double. The original temperature is given as ($27^\circ C$), which is equal to (300 K) in absolute terms. Therefore, the final temperature ($T_f$) in kelvins is:

$T_f = 2 \cdot 300 K = 600 K$

Converting this back to degrees Celsius gives:

$T_f = 600K - 273 = 327 ^\circ C$

Both Assertion A and Reason R are correct, and R is indeed the correct explanation of A.

Electromagnets are commonly made of soft iron because of its high permeability, which allows it to easily magnetize in response to an external magnetic field. Its low retentivity is also desirable because it allows the magnetization to be easily reversed or removed once the external field is removed. This combination of properties makes soft iron ideal for electromagnets, which require rapid and efficient changes in magnetization.

Total initial kinetic energy

$=\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \mathrm{I} \omega^2$

$\mathrm{v}=\mathrm{R} \omega$ (for pure rolling)

$\mathrm{K} . E C=\frac{1}{2} m \mathrm{v}^2+\frac{1}{2} \times \frac{2}{3} m \mathrm{R}^2 \times \frac{\mathrm{v}^2}{\mathrm{R}^2}=\frac{5}{6} m \mathrm{v}^2$

Energy remains conserve during whole journey.

$\mathrm{K}_{\cdot} \mathrm{E}_i+\mathrm{P.E}_{\cdot i}=\mathrm{K}_{\cdot} \mathrm{E}_f+\mathrm{P.E}_{\cdot f}$

$$\begin{aligned} & \Rightarrow \frac{5}{2} m \mathrm{v}^2=m g \mathrm{H} ~~~~~~~(\because {K.E.}_f=0)\\ & \Rightarrow \mathrm{H}=\frac{5}{6} \times \frac{\mathrm{v}^2}{g} \\ & =\frac{5 \times(3)^2}{6 \times 10} \\ & =\frac{15}{20} \mathrm{~m}=0.75 \mathrm{~m}=75 \mathrm{~cm} \end{aligned}$$

The change in length of the rod when it is heated is given by the equation:

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

where

• $\Delta L$ is the change in length,
• $L_0$ is the original length,
• $\alpha$ is the coefficient of linear expansion, and
• $\Delta T$ is the change in temperature.

Substituting the given values:

$\Delta L = 1 \, \text{m} \cdot 10^{-5} \, \text{K}^{-1} \cdot 200 \, \text{K} = 0.002 \, \text{m}$

The rod is not allowed to extend or bend, so a stress is created in the rod. This stress can be calculated using Young's modulus (Y), which is the ratio of the stress (force per unit area, F/A) to the strain (change in length per unit length, $\Delta L / L_0$):

$Y = \frac{F/A}{\Delta L / L_0}$

Rearranging for F gives:

$F = Y \cdot A \cdot \frac{\Delta L}{L_0}$

Substituting the given values:

$F = 2 \times 10^{11} \, \text{N/m}^2 \cdot 10^{-4} \, \text{m}^2 \cdot \frac{0.002 \, \text{m}}{1 \, \text{m}} = 4 \times 10^{4} \, \text{N}$

So the compressive tension produced in the rod is $4 \times 10^{4} \, \text{N}$.

The equation of the trajectory given is $y = x - \frac{x^2}{20}$.

This is a parabola, and it represents the path of the projectile.

The maximum height of the projectile corresponds to the vertex of the parabola.

The x-coordinate of the vertex for a parabola given by $y = ax^2 + bx + c$ is $-b/2a$.

In this case, $a = -1/20$ and $b = 1$, so the x-coordinate of the vertex is:

$x_{\text{vertex}} = -\frac{b}{2a} = -\frac{1}{2 \times (-1/20)} = 10$

Substituting this into the equation of the trajectory gives the y-coordinate of the vertex, which is the maximum height:

$y_{\text{max}} = 10 - \frac{10^2}{20} = 10 - 5 = 5$

So, the maximum height attained by the projectile is 5 m.

Phenolphthalein is a commonly used acid/base indicator that can be synthesized from phthalic anhydride and phenol (carbolic acid) in the presence of concentrated sulfuric acid (H₂SO₄). Therefore, compound 'X' is phenol and compound 'Y' is phenolphthalein.

For good quality cement, the ratio of lime (CaO) to the sum of the oxides of silicon (SiO₂), aluminum (Al₂O₃), and iron (Fe₂O₃) should be approximately 2:1.

This proportion ensures that the cement has good strength and setting properties, as well as resistance to chemical attack.

IUPAC name of the following compound is


2-Methyl-5-oxohexanoic acid

Urea is a negative catalyst, which means that it slows down the rate of a chemical reaction. In the case of hydrogen peroxide, urea absorbs the light that would otherwise cause it to decompose. This prevents the reaction from happening, or at least slows it down significantly.

In the Hall-Héroult process, graphite anodes are used to reduce alumina (Al₂O₃). During the electrolytic reduction, Al³⁺ ions are reduced at the cathode to form aluminum, while at the anode, oxide ions are oxidized to form oxygen gas. The oxygen gas then reacts with the carbon in the graphite anodes to produce carbon dioxide. This means that graphite is consumed during the process, and the anodes have to be replaced periodically.

So, in the context of the question, graphite (Option C) is the correct answer, as it is the material of the anode used for the electrolytic reduction of Al₂O₃ in the Hall-Héroult process.

The van der Waals constant 'a' is a measure of the strength of the intermolecular forces in a gas. Larger molecules and molecules with stronger intermolecular forces will have larger 'a' constants.

Here, we are comparing argon (Ar), methane (CH₄), water (H₂O), and benzene (C₆H₆).

Argon (Ar) is a noble gas and is monoatomic. Methane (CH₄) is a small molecule with weak intermolecular forces. Water (H₂O) is a small molecule but has strong hydrogen bonding, making its intermolecular forces stronger than those in methane. Benzene (C₆H₆) is a larger molecule than the others and has relatively strong intermolecular forces due to its larger size and polarizability.

So, in the increasing order of van der Waals constant 'a', it should be:

Argon (Ar) < Methane (CH₄) < Water (H₂O) < Benzene (C₆H₆)

Most of the chemical reactions have energy barrier.

A positive catalyst in a chemical reaction increases the rate of reaction by lowering the energy barrier. It does not alter the energy of reactant as well as product.

The coagulating power of an electrolyte is inversely proportional to its coagulating value. That is,

$\text{Coagulating power} = \frac{1}{\text{Coagulating value}}$

Therefore, the coagulating power of AlCl₃ is $\frac{1}{0.09}$

and the coagulating power of NaCl is $\frac{1}{50.04}$

We are asked to find the factor by which the coagulating power of AlCl₃ is greater than the coagulating power of NaCl. This factor is given by the ratio

$x = \frac{\text{Coagulating power of AlCl₃}}{\text{Coagulating power of NaCl}} = \frac{\frac{1}{0.09}}{\frac{1}{50.04}} = \frac{50.04}{0.09}$

$x = \frac{50.04}{0.09} \approx 556.22$

So, the coagulating power of AlCl₃ is approximately 556 times the coagulating power of NaCl.

However, since the problem asks for the nearest integer, the answer would be 556.