JEE Main 2024 4th April Evening Shift Question Paper with Solutions

Q. value

$$\begin{aligned} & =\{(235.0439)-[39.9054+93.9063+1.0086]\} \times 931 \mathrm{~MeV} \\ & =208 \mathrm{~MeV} \end{aligned}$$

The gravitational force $F$ that an object of mass $m$ placed at a distance $r$ (from the center of Earth) experiences can be calculated using the universal law of gravitation, given by:

$F = \frac{G M m}{r^2}$

Where:

• $G$ is the gravitational constant ($6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2$)
• $M$ is the mass of the Earth (approximately $5.972 \times 10^{24} \text{kg}$)
• $m$ is the mass of the object
• $r$ is the distance from the center of the Earth to the object

However, when the object is at a distance $2R$ from the surface of the Earth, the total distance from the center of the Earth $r'$ becomes $R + 2R = 3R$. This is because the radius of the Earth $R$ is the distance from the Earth's center to its surface, so if the object is $2R$ above the surface, the total distance from the center is $3R$.

At the surface of the Earth, the gravitational force ($F_{earth}$) that acts on an object is given by its weight, which can be calculated using the formula $F_{earth} = m \cdot g$, where $g$ is the acceleration due to gravity on the surface of the Earth. Given that $g = 10 \text{m/s}^2$ and the mass of the body $m = 90 \text{kg}$, we get:

$F_{earth} = 90 \text{kg} \cdot 10 \text{m/s}^2 = 900 \text{N}$

To find the gravitational pull at a distance $2R$ from the Earth's surface, we use the fact that gravitational force varies inversely as the square of the distance from the center of the Earth. Since the distance triples ($3R$ from $R$), the gravitational force becomes $\frac{1}{3^2} = \frac{1}{9}$ of the force at the surface.

Therefore, the gravitational pull $F_{2R}$ on the body when placed at $2R$ from the Earth's surface is:

$F_{2R} = \frac{F_{earth}}{9} = \frac{900 \text{N}}{9} = 100 \text{N}$

Thus, the gravitational pull on the body when it is placed at a distance of $2R$ from the Earth's surface is 100 N. So, the correct option is:

Option C: 100 N

$$\begin{aligned} W & =\int_0^{180} d \tau \cdot d \theta \\ & =m B(\cos 0-\cos 180) \\ & =0.5 \times 8 \times 10^{-2}(2) \\ & =8 \times 10^{-2} \mathrm{~J} \end{aligned}$$

From energy conservation $\rightarrow$

$$\begin{aligned} & m g(R+R \sin 45)=\frac{1}{2} m v^2 \\ & \Rightarrow 10\left(1+\frac{1}{\sqrt{2}}\right) \times 14=\frac{1}{2} v^2 \\ & \Rightarrow 10\left(1+\frac{\sqrt{2}}{2}\right) \times 28=v^2 \\ & \Rightarrow 10(1+0.7) \times 28=v^2 \\ & \Rightarrow v=21.81 \end{aligned}$$

In the given diagram, only the diode given in option (2) is forward biase, so this circuit can be used to measure dynamic resistance of p-n junction diode.

$$\begin{aligned} & u=72 \times \frac{5}{18}=20 \mathrm{~m} / \mathrm{s} \\ & v=0 \\ & t=4 \\ & \Rightarrow \quad 0=20+4 a \\ & \Rightarrow \quad a=-5 \mathrm{~m} / \mathrm{s}^2 \\ & \begin{aligned} \therefore \quad S & =20 \times 4-\frac{1}{2} \times 5 \times 16 \\ & =40 \mathrm{~m} \end{aligned} \end{aligned}$$

In order to determine the most appropriate answer to the question, let's analyze the given Assertion A and Reason R in detail.

Assertion A: Number of photons increases with increase in frequency of light.

This statement is not correct. The number of photons is determined by the intensity (or power) of the light and is given by the formula:

$N = \frac{P}{hf}$

where $N$ is the number of photons per second, $P$ is the power (intensity) of the light, $h$ is Planck's constant, and $f$ is the frequency of the light. As the frequency increases, the energy per photon increases, but it does not necessarily mean that the number of photons increases unless the power of the light also increases proportionally.

Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.

This statement is correct. According to the photoelectric effect, the maximum kinetic energy of emitted electrons is given by:

$K.E_{\text{max}} = hf - \phi$

where $K.E_{\text{max}}$ is the maximum kinetic energy of the emitted electrons, $h$ is Planck's constant, $f$ is the frequency of the incident radiation, and $\phi$ is the work function of the material. As the frequency of the incident light increases, the kinetic energy of the emitted electrons increases.

Now, let's match the statements with the options given:

Option A: $\mathbf{A}$ is not correct but $\mathbf{R}$ is correct.

This option is correct because Assertion A is incorrect while Reason R is correct.

Option B: $\mathbf{A}$ is correct but $\mathbf{R}$ is not correct.

This option is incorrect because Assertion A is not correct.

Option C: Both $\mathbf{A}$ and $\mathbf{R}$ are correct and $\mathbf{R}$ is the correct explanation of $\mathbf{A}$.

This option is incorrect because Assertion A is not correct.

Option D: Both $\mathbf{A}$ and $\mathbf{R}$ are correct and $\mathbf{R}$ is NOT the correct explanation of $\mathbf{A}$.

This option is incorrect because Assertion A is not correct.

Therefore, the most appropriate answer is:

Option A: $\mathbf{A}$ is not correct but $\mathbf{R}$ is correct.

The displacement of a particle executing Simple Harmonic Motion (SHM) can be expressed as:

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

Where:

$A$ is the amplitude of the SHM,

$\omega$ is the angular frequency,

$t$ is the time,

$\phi$ is the phase constant (phase angle at $t = 0$).

In the given equation, $x = 10 \sin(\omega t + \frac{\pi}{3})$ m, the amplitude $A = 10$ m and the phase constant $\phi = \frac{\pi}{3}$. The time period $T = 3.14$ s is given, from which we can find the angular frequency $\omega$ using the relationship:

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

Substituting the given $T = 3.14$ s:

$\omega = \frac{2\pi}{3.14} \approx 2 \, \text{rad/s}$

To find the velocity of the particle, we differentiate the displacement $x$ with respect to time $t$. The derivative of the displacement gives the velocity:

$v = \frac{dx}{dt}$

So, for $x = 10 \sin(\omega t + \frac{\pi}{3})$:

$v = \frac{d}{dt}[10 \sin(\omega t + \frac{\pi}{3})]$

Applying differentiation, we get:

$v = 10\omega \cos(\omega t + \frac{\pi}{3})$

Plug in the value of $\omega = 2$ rad/s and evaluate it at $t = 0$ to find the initial velocity:

$v = 10 \cdot 2 \cos(2 \cdot 0 + \frac{\pi}{3})$

$v = 20 \cos(\frac{\pi}{3})$

$\cos(\frac{\pi}{3}) = \frac{1}{2}$, therefore:

$v = 20 \cdot \frac{1}{2} = 10 \, \text{m/s}$

Thus, the velocity of the particle at $t = 0$ is $10$ m/s.

$\because \quad E=\frac{1}{2} k A^2$

Here energy depends on $A$ and $k$ and not on mass.

$$\begin{aligned} & m g \sin \theta=\mu \cdot m g \cos \theta \\ & \Rightarrow \tan \theta=\mu=1 \end{aligned}$$

$$\begin{aligned} F & =\left(p_0+\rho g h\right) A \\ & =\left(10^5+1.36 \times 10^4 \times 10 \times \frac{3}{10}\right) \frac{22}{7}\left(\frac{2}{100}\right)^2 \\ & =177 \mathrm{~N} \end{aligned}$$

$$\begin{aligned} & \mu=\sqrt{2} \\ & \sin \theta_C=\frac{1}{\sqrt{2}} \\ & Q_C=45^{\circ} \\ & i=Q_C=45^{\circ} \end{aligned}$$

$$\begin{aligned} & \text { ( } \phi \text { ) lateral displacement }=\frac{t \sin (i-r)}{\cos r} \\ & \sin 45^{\circ}=\sqrt{2} \sin r \\ & \Rightarrow \quad r=30^{\circ} \\ & \therefore \quad d=\frac{4 \sqrt{3} \sin \left(45^{\circ}-30^{\circ}\right)}{\cos 30^{\circ}} \\ & =\frac{4 \sqrt{3} \times \frac{1}{4}}{\frac{\sqrt{3}}{2}}=2 \end{aligned}$$

According to Bohr's theory, one of the postulates specifies that the angular momentum of an electron in orbit around a nucleus is quantized. This quantization can be expressed by the formula:

$L = n\frac{h}{2\pi}$

Where:

• $L$ is the angular momentum of the electron,
• $n$ is the principal quantum number (or the orbit number in simpler terms), which can take positive integer values (1, 2, 3, ...),
• $h$ is Planck's constant ($6.62607015 \times 10^{-34} m^2 kg / s$), and
• $\frac{h}{2\pi}$ is often denoted as $\hbar$ (h-bar), known as the reduced Planck's constant.

For an electron in the 4th orbit ($n = 4$) of a hydrogen atom, we substitute $n = 4$ into the equation:

$L = 4\frac{h}{2\pi}$

Therefore, the moment of momentum (or angular momentum) of an electron in the $4^{\text{th}}$ orbit of a hydrogen atom is:

$L = 4\frac{h}{2\pi} = 2\frac{2h}{2\pi} = 2\frac{h}{\pi}$

Hence, the correct option is:

Option A: $2 \frac{h}{\pi}$

$$\begin{aligned} E_1 & =\frac{1}{2}\left(\frac{25}{2}\right) \times 10^{-12} \times 144 \\ & =900 \times 10^{-12} \mathrm{~J} \\ E_2 & =\frac{1}{2}\left(6 \times \frac{25}{2} \times 10^{-12}\right)\left(\frac{12}{6}\right)^2=150 \times 10^{-12} \mathrm{~J} \\ \Delta E & =750 \times 10^{-12} \mathrm{~J} \end{aligned}$$

For non-linear polyatomic molecules, both translational and rotational degree of freedom have same value and is equal to 3.

Option D, "Statement I is true but Statement II is false," is the correct choice. Here's an explanation for both statements:

Statement I: True

The contact angle between a solid and a liquid is indeed a measure of the wettability of the solid surface by the liquid. The contact angle is determined by the nature of both the solid and the liquid. It is a function of the interfacial tensions between solid-liquid ($\gamma_{\text{SL}}$), solid-vapor ($\gamma_{\text{SV}}$), and liquid-vapor ($\gamma_{\text{LV}}$). This relationship can be understood through Young's equation:

$\cos \theta = \frac{\gamma_{\text{SV}} - \gamma_{\text{SL}}}{\gamma_{\text{LV}}}$

Where $\theta$ is the contact angle. Therefore, since the interfacial tensions vary with the materials in contact, the contact angle is indeed a property of the materials of both the solid and the liquid.

Statement II: False

The rise of a liquid in a capillary tube is strongly dependent on the inner radius of the tube. This relationship is described by the Jurin's Law, which states the height ($h$) to which a liquid will rise (or fall) in a capillary tube is inversely proportional to the radius ($r$) of the tube, among other factors. The law is given by:

$h = \frac{2\gamma \cos \theta}{\rho g r}$

Where:

• $\gamma$ is the liquid-air surface tension,
• $\theta$ is the contact angle,
• $\rho$ is the density of the liquid,
• $g$ is the acceleration due to gravity,
• $r$ is the radius of the capillary tube.

As seen from the equation, $h$ is inversely proportional to $r$. Therefore, the rise of the liquid indeed depends on the inner radius of the tube, making Statement II false.

$$\begin{aligned} & T=2 \pi \sqrt{\frac{r^3}{G M}} \\ & T^2=\frac{4 \pi^2}{G M}(R+h)^3 \\ & h=\left(\frac{G M T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\ & =\left(\frac{G M \cdot R}{R^2} \cdot \frac{T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\ & =\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{\frac{1}{3}}-R \end{aligned}$$

$$\begin{aligned} & R=\rho \frac{I}{A}=\rho \frac{V}{A^2} \\ & \text { and } \pi r_1^2 I_1=\pi r_2^2 I_2 \\ & A_1 I_1=A_2 I_2 \\ & \text { So } \frac{R_1}{R_2}=\left(\frac{A_2}{A_1}\right)^2 \\ & \Rightarrow \frac{R}{2}=\left(\frac{r_2}{r_1}\right)^4 \\ & \Rightarrow R=32 \end{aligned}$$

$$\begin{aligned} & w=\frac{-n R}{\gamma-1}(\Delta T) \\ &=\frac{-R}{1 / 2}\left(\frac{T}{\sqrt{2}}-T\right) \\ &=2 R\left(\frac{\sqrt{2} T-T}{\sqrt{2}}\right) \\ &=R T(2-\sqrt{2}) \\ & \therefore \quad T V_\gamma^{-1}=\text { cons. } \\ & T V_\gamma^{-1}=T_f(2 V)^{\gamma-1} \\ & T_f=\frac{T}{\sqrt{2}} \end{aligned}$$

At point $A$

$B_A=\frac{\mu_0 I}{2 \pi r}+\frac{\mu_0(2 I)}{2 \pi(3 r)}$

At point $C$

$$\begin{aligned} & B_C=\frac{\mu_0 I}{2 \pi(3 r)}+\frac{\mu_0(2 I)}{r} \\ & \Rightarrow \frac{B_A}{B_C}=\frac{5}{7} \end{aligned}$$

Both end having same potential, so potential difference between end will be zero.

To solve this problem, we can make use of the concept of center of mass. The center of mass (CM) of a system remains unchanged if the internal forces act within the system without any external force. When one mass moves toward the CM, to keep the CM at the same position, the other mass must move in a way that the product of each mass with its displacement relative to the CM remains constant.

The formula to ensure that the center of mass remains unchanged can be derived from the principle of conservation of momentum or simply by understanding that the weighted average position (considering masses as weights) does not change.

Let's denote: - $x_1$ as the distance moved by $m_1$ towards the CM, - $x_2$ as the distance $m_2$ needs to move towards the CM, - The total mass of the system as $M = m_1 + m_2$.

Since $m_1$ moves towards the CM by 2 cm, we apply the principle that the weighted sum of displacements (taking mass into account) remains 0 to maintain the center of mass at its original position:

$m_1 \cdot x_1 + m_2 \cdot x_2 = 0$

Given that $m_1 = 3 \, \text{kg}$, $m_2 = 2 \, \text{kg}$, and $x_1 = 2 \, \text{cm}$, we substitute these values into the equation:

$3 \cdot 2 + 2 \cdot x_2 = 0$

Solving for $x_2$ gives:

$6 + 2x_2 = 0$

$2x_2 = -6$

$x_2 = -3 \, \text{cm}$

This means the mass $m_2$ should move $3 \, \mathrm{cm}$ towards the center of mass to keep the center of mass of the system at the original position. The negative sign indicates the direction is towards the center of mass, similar to $m_1$'s movement direction in relation to keeping the CM stationary.

When considering the electric flux linked with a cube due to a charge placed at one of its surfaces, it's important to apply Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to $\frac{q_{\text{enc}}}{\epsilon_0}$, where $q_{\text{enc}}$ is the charge enclosed by the surface and $\epsilon_0$ is the permittivity of free space.

In the given scenario, the charge $q$ is placed at the center of one of the surfaces of the cube. Conceptually, we can think of this arrangement as part of a larger situation where if we had a larger, imaginary cube that encompasses the entire setup such that the point charge is at its geometric center, the charge would then be uniformly distributing its flux through all six faces of this larger cube. However, since the actual setup involves only one cube with one face adjacent to the charge, we essentially have only one-half of the total possible geometry through which the flux from this charge can emerge - implying the flux through our actual cube is a fraction of the total flux that would emanate from the charge if it were centrally placed within a larger, encompassing cube.

Therefore, only half of the flux emanating from the charge will pass through the actual cube because the charge is placed directly on one of its surfaces, effectively distributing its influence through 180 degrees (half of the space around it) instead of the full 360 degrees. Hence, the flux linked with the cube will be half of the total flux $\frac{q}{\epsilon_0}$, which is calculated using Gauss's law for a point charge.

Therefore, the correct answer is $\frac{q}{2\epsilon_0}$.

Option A $\frac{q}{2 \epsilon_0}$ is the correct choice.

$$\begin{aligned} \text { Displacement } & =\sqrt{2} R \\ & =\sqrt{8} \mathrm{~km} \end{aligned}$$

To find the power dissipation of the bulb when it's connected to a $100 \mathrm{V}$ supply instead of its rated $200 \mathrm{V}$ supply, we can use the relation between power (P), voltage (V), and resistance (R), which is given by $P = \frac{V^2}{R}$. The resistance of the bulb can be considered constant in this case, allowing us to calculate the change in power dissipation due to the change in voltage.

First, let's find the resistance of the bulb based on its rated conditions:

$P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P}$

Substituting the rated values, we get:

$R = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \, \Omega$

Now, using this resistance, we can find the power dissipation when the bulb is connected to a $100 \mathrm{V}$ supply:

$P = \frac{V^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \, W$

This means the power dissipation of the bulb when connected to a $100 \mathrm{V}$ supply is $12.5 \, W$.

Therefore, the correct answer is Option C: 12.5 W.

Wavelengths are as

Gamma rays $<1 \mathrm{~nm}$

x-ray < $(1-10)$ nm

Infrared $<(700-10^5) \mathrm{~nm}$

Microwave $<(10^5-10^8) \mathrm{~nm}$

$$\begin{aligned} & I_{\max }=\left(\sqrt{4 I_0}+\sqrt{I_0}\right)^2=\left(3 \sqrt{I_0}\right)^2=9 I_0 \\ & I_{\min }=\left(\sqrt{4 I_0}-\sqrt{I_0}\right)^2=I_0 \\ & \frac{I_{\max }}{I_{\min }}=9: 1 \end{aligned}$$

For pure capacitive circuit, $I$ lead by $90^{\circ}$ to $\mathrm{V}$

For pure inductive circuit, $V$ lead by $90^{\circ}$ to $I$

At series LCR resonance, $I$ and $V$ are in same phase.

For LCR series circuit, $V$ and I may suffer some phase difference.

To determine which option is correct based on the principle of homogeneity of dimensions, we need to ensure that both sides of the equation have the same dimensions. The time period ($T$) is measured in units of time ($T$), the gravitational constant ($G$) has units of $\text{m}^3\text{kg}^{-1}\text{s}^{-2}$, mass ($M$) has units of mass ($M$), and the radius of orbit ($r$) has units of length ($L$).

Let's analyze each option:

Option A: $T^2=\frac{4 \pi^2 r}{G M^2}$

The dimensions of the right-hand side of the equation are $\frac{L}{\left(\frac{L^3}{MT^2}\right)M^2}=\frac{L}{L^3M^{-1}T^{-2}M^2}=\frac{L}{L^3T^{-2}}=L^{-2}T^2$, which do not match with $T^2$ (time squared). Thus, Option A is incorrect.

Option B: $T^2=4 \pi^2 r^3$

The dimensions of the right-hand side are $L^3$, which clearly do not match the dimensions $T^2$ of the squared time period. So, Option B is incorrect.

Option C: $T^2=\frac{4 \pi^2 r^3}{G M}$

The dimensions of the right-hand side of the equation are $\frac{L^3}{\left(\frac{L^3}{MT^2}\right)M}=\frac{L^3}{L^3T^{-2}}=T^2$, which match the dimensions of the squared time period $T^2$. Therefore, Option C is correct based on the principle of homogeneity of dimensions.

Option D: $T^2=\frac{4 \pi^2 r^2}{G M}$

The dimensions of the right-hand side are $\frac{L^2}{\left(\frac{L^3}{MT^2}\right)M}=\frac{L^2}{L^3M^{-1}T^{-2}M}=L^{-1}T^2$, which do not match with $T^2$ (time squared). Hence, Option D is incorrect.

Thus, based on the principle of homogeneity of dimensions, Option C is the correct one: $T^2=\frac{4 \pi^2 r^3}{G M}$. This equation also corresponds to Kepler's third law of planetary motion, which relates the orbital period of a planet to its orbital radius, considering the mass of the central body.

So, $Y=A+B$

(d) is aromatic. So it is most stable.

(a) is anti-aromatic. So it is least stable.

$\mathrm{NF}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{S}, \mathrm{HBr}, \mathrm{HCl}$ have non zero dipole moments.

To determine the atomic number of the metal based on the provided spin-only magnetic moment value, we use the formula for calculating the spin-only magnetic moment:

$\mu = \sqrt{n(n+2)} \mu_{B}$

where $\mu$ is the magnetic moment in Bohr magnetons ($\mu_{B}$), and $n$ is the number of unpaired electrons.

Given that the spin-only magnetic moment value is $3.86 \, \mathrm{BM}$, we can set this value equal to the formula to solve for $n$:

$3.86 = \sqrt{n(n+2)}$

Squaring both sides gives:

$14.8996 = n(n+2)$

Since $n$ must be an integer and this equation doesn't solve neatly for an integer $n$, we look for a value of $n$ that would give a product close to $14.8996$ when plugged into $\sqrt{n(n+2)}$.

Practically, we can test the square root values near $3.86$ for different $n$ to see which one gives a close match. Considering the usual spin-only magnetic moments for common numbers of unpaired electrons:

• $n = 1$, $\mu = \sqrt{3} \approx 1.73 \mu_B$


• $n = 2$, $\mu = \sqrt{8} \approx 2.83 \mu_B$


• $n = 3$, $\mu = \sqrt{15} \approx 3.87 \mu_B$


• $n = 4$, $\mu = \sqrt{24} \approx 4.90 \mu_B$

The value $n = 3$ gives a magnetic moment of approximately $3.87 \mu_B$, which is very close to the given value, $3.86 \mu_B$. Therefore, the metal has 3 unpaired electrons in its d-orbital configuration.

In the +2 oxidation state, metals lose electrons from the s-orbital before the d-orbital. Given that the element is a first row transition metal, starting with scandium (Sc) at atomic number 21 which has an electronic configuration of $[Ar] 3d^1 4s^2$ in its neutral state, we can deduce the configurations:

• $Z = 22$ for Titanium (Ti), which has a neutral configuration of $[Ar] 3d^2 4s^2$. In the +2 oxidation state, it would have an electronic configuration of $[Ar] 3d^2$, resulting in 2 unpaired electrons.


• $Z = 23$ for Vanadium (V), with a neutral configuration of $[Ar] 3d^3 4s^2$. In the +2 state, its configuration would be $[Ar] 3d^3$, presenting 3 unpaired electrons, which matches our calculation.


• $Z = 26$ for Iron (Fe), indicating a neutral configuration of $[Ar] 3d^6 4s^2$. In the +2 state, $[Ar] 3d^6$, which would have 4 unpaired electrons, not matching the calculation.


• $Z = 25$ for Manganese (Mn), with a neutral configuration of $[Ar] 3d^5 4s^2$. In the +2 state, $[Ar] 3d^5$, indicating 5 unpaired electrons, which also does not match our calculation.

So, the atomic number of the metal with a +2 oxidation state and a spin-only magnetic moment value of $3.86 \mathrm{~BM}$ (corresponding to 3 unpaired electrons) is 23, which identifies the metal as Vanadium (V). Therefore,

Option B (23) is the correct answer.

To determine the total sum of oxygen atoms and $\pi$ electrons in vanillin, let's first consider the structure of vanillin. Vanillin is an organic compound with the chemical formula $C_8H_8O_3$. It consists of a benzene ring attached to a methoxy group (-OCH₃), a hydroxyl group (-OH), and an aldehyde group (-CHO).

Firstly, let's count the oxygen atoms:

• One from the methoxy group (-OCH₃)
• One from the hydroxyl group (-OH)
• One from the aldehyde group (-CHO)

Thus, there are a total of 3 oxygen atoms in vanillin.

Next, let's calculate the total number of $\pi$ electrons:

• The benzene ring has 3 double bonds, contributing 6 $\pi$ electrons.
• The aldehyde group has 1 double bond (C=O), contributing 2 $\pi$ electrons.
• The methoxy group and the hydroxyl group do not contribute any $\pi$ electrons since they do not have $\pi$ bonds in their structures.

Therefore, vanillin has a total of 8 $\pi$ electrons from the benzene ring and the aldehyde group.

Adding the number of oxygen atoms (3) and the $\pi$ electrons (8) gives us a total sum of 11.

Thus, the total sum of oxygen atoms and $\pi$ electrons in vanillin is 11.

Commonly used adsorbents in adsorption chromatography are silica gel and alumina.