NEET-UG 2026

Time, $t=6\min 40\mathrm{s}$

$$\begin{array}{rl}& =360+40\\ & =400\mathrm{s}\end{array}$$

Distance in new system $d=vt$

$$=1\times 400=400$$

The side of the cube is

$$a=9.0\text{cm}=9.0\times 10^{-2}\text{m}$$

The volume of the cube is

$$V=a^3$$

$$V=(9.0\times 10^{-2}{)}^3{\text{m}}^3$$

$$V=7.29\times 10^{-4}{\text{m}}^3$$

Density is

$$\rho =\frac{\text{mass}}{\text{volume}}$$

$$\rho =\frac{5.580}{7.29\times 10^{-4}}$$

$$\rho =7.654\times 10^3{\text{kg m}}^{-3}$$

Now we apply significant figures.

Mass $5.580\text{kg}$ has $4$ significant figures.

Side $9.0\text{cm}$ has $2$ significant figures.

Since density involves multiplication/division, the final answer should have the least number of significant figures, which is $2$.

So,

$$7.654\times 10^3\approx 7.7\times 10^3{\text{kg m}}^{-3}$$

Hence,

$$X=7.7$$

Correct option: D.

Least count of a vernier calliper

The least count (L.C.) is the smallest measurement a vernier calliper can read. It is given by:

$$\text{L.C.}=1\text{MSD}-1\text{VSD}...(i)$$

It is given that 20 vernier scale divisions (VSD) coincide with 16 main scale divisions (MSD). Each MSD is 1 mm long.

So,

$$20\text{VSD}=16\text{MSD}$$

Dividing both sides by 20, we get

$$1\text{VSD}=\frac{16}{20}\text{MSD}$$

Substitute this value in equation (i):

$$\text{L.C.}=1\text{MSD}-\frac{16}{20}\text{MSD}$$

Simplifying,

$$\text{L.C.}=\frac{4}{20}\text{MSD}$$

Since 1 MSD = 1 mm,

$$\text{L.C.}=\frac{1}{5}\text{mm}=0.2\text{mm}$$

In centimetre, the least count is:

$$\text{L.C.}=0.02\text{cm}$$

During the whole journey, acceleration due to gravity is vertically downward. Therefore, slope of velocity vs time curve should be negative throughout the journey.

∴ Statement (C) is correct

→ There will be large distance for large reaction time

→ Descending order of reaction time

$$\Rightarrow t_B>t_E>t_A>t_D>t_C$$

→ Descending order of distance covered

$$\Rightarrow S_B>S_E>S_A>S_D>S_C$$

$$a=\mu g=0.12\times 10=1.2\mathrm{m}/{\mathrm{s}}^2$$

$$\begin{array}{rl}& F_{\text{net }}=\sqrt{6^2+8^2}\\ & a=\frac{F_{\text{net }}}{m}=\frac{10}{5}=2\mathrm{m}/{\mathrm{s}}^2\\ & \tan \theta =\frac{6}{8}\\ & \theta ={\tan }^{-1}\left(\frac{3}{4}\right)\text{ from }8\mathrm{N}\end{array}$$

$$\begin{array}{rl}& \text{ Power }=\frac{\text{ Work }}{\text{ Time }}\\ & =\frac{mgh}{t}\\ & =\frac{10^3\times 9.8\times 20}{10}=19.6\mathrm{kW}\end{array}$$

$$\begin{array}{rl}& \alpha =\frac{{\omega }_2-{\omega }_1}{\Delta t}=\left(\frac{1200-600}{10}\right)\frac{2\pi }{60}=2\pi \mathrm{rad}/{\mathrm{s}}^2\\ & \begin{array}{ll}\Rightarrow & \text{ Use equation of motion, }{\omega }_2^2={\omega }_1^2+2\alpha \theta \\ \Rightarrow & {\omega }_2=1200\times \frac{2\pi }{60}=40\pi \\ & {\omega }_1=600\times \frac{2\pi }{60}=20\pi \\ & (40\pi {)}^2=(20\pi {)}^2+2\times 2\pi \theta \\ \Rightarrow & 1200{\pi }^2=4\pi \theta \\ \Rightarrow & \theta =300\pi \text{ radian }\\ \Rightarrow & \text{ Number of revolution }=\frac{\theta }{2\pi }=150\end{array}\end{array}$$

Mass of thin wire $=($ Linear mass density $)\times ($ Length $)$

$$\Rightarrow M=mL$$

$L=2\pi r$, where $r=$ radius of circular ring $=\frac{L}{2\pi }$

Using parallel axis theorem,

$$\begin{array}{rl}& I_{y{y}^{\prime }}=I_{CM}+M{r}^2=\frac{M{r}^2}{2}+M{r}^2=\frac{3}{2}M{r}^2\\ & \Rightarrow I_{y{y}^{\prime }}=\frac{3}{2}\times (mL)\times {\left(\frac{L}{2\pi }\right)}^2=\frac{3m{L}^3}{8{\pi }^2}\end{array}$$

W.D. $=U_2-U_1$

$$\begin{array}{rl}& =-\frac{GMm}{R+R}-\left(\frac{-GMm}{R}\right)\\ & =-\frac{GMm}{2R}+\frac{GMm}{R}\\ & =\frac{GMm}{2R}=\frac{mgR}{2}\end{array}$$

A. Young's Modulus $=\frac{\text{ Stress }}{\text{ Strain }}=\frac{FL}{A\Delta L}$

B. Compressibility $=\frac{1}{\text{ Bulk modulus }}=-\frac{1}{\Delta P}\left(\frac{\Delta V}{V}\right)$

C. Bulk Modulus $=-P\left(\frac{V}{\Delta V}\right)$

D. Poisson's Ratio $=\frac{\text{ Lateral strain }}{\text{ Longitudinal strain }}=\frac{\Delta d}{\Delta L}\left(\frac{L}{d}\right)$

$P=P_0+\rho gh$

$$\begin{array}{rl}& \Rightarrow 100\times 10^5=10^5+10^3\times 10\times h\\ & \Rightarrow 10^7=10^5+10^{4}h\\ & \Rightarrow 10^3=10+h\\ & \Rightarrow h=1000-10=990\mathrm{m}\end{array}$$

$v_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}$

For same temperature, $v_{\text{rms }}\propto \frac{1}{\sqrt{M}}$

$$\frac{V_{\mathrm{rms}}^{\mathrm{Ar}}}{V_{\mathrm{rms}}^{\mathrm{Cl}}}=\sqrt{\frac{M_{\mathrm{Cl}}}{M_{\mathrm{Ar}}}}=\sqrt{\frac{70}{40}}=\frac{\sqrt{7}}{2}$$

Using $1^{\text{st }}$ law of thermodynamics for electric heater,

$$\begin{array}{rl}& Q=\Delta U+W\\ & \Rightarrow \frac{dQ}{dt}=\frac{d(\Delta U)}{dt}+\frac{dW}{dt}\\ & \Rightarrow 100\mathrm{W}=\frac{d(\Delta U)}{dt}+75\mathrm{W}\\ & \therefore \frac{d(\Delta U)}{dt}=25\mathrm{W}\end{array}$$

$$\begin{array}{rl}& \text{ Kinetic energy, }K=\frac{1}{2}m{v}^2\\ & \begin{array}{rl}& =\frac{1}{2}m{A}^2{\omega }^2{\cos }^2(\omega t+\varphi )\\ & \therefore K\propto {\cos }^2(\omega t+\varphi )\end{array}\end{array}$$

At equilibrium position

Total energy = K.E

$$\begin{array}{rl}& \frac{1}{2}m{v}^2=0.02\\ & \frac{1}{2}\times 20\times v^2\times 10^{-3}=2\times 10^{-2}\\ & v=\sqrt{2}\\ & v=1.41\mathrm{m}/\mathrm{s}\end{array}$$

Time taken for 30 oscillations $=60\mathrm{s}$

Time period of simple pendulum = Time taken for 1 oscillation

$$\begin{array}{rl}& \Rightarrow T=\frac{60}{30}=2\mathrm{s}\\ & T=2\pi \sqrt{\frac{I}{g}}\Rightarrow I=\frac{g{T}^2}{4{\pi }^2}=\frac{9.8\times 2\times 2}{4\times 9.8}=1\mathrm{m}\end{array}$$

$y(x,t)=2.0\cos 2\pi (10t-0.008x+0.35)$

Total phase

$$\begin{array}{rl}& \varphi =20\pi -2\pi \times 8\times 10^{-3}x+2\pi \times 0.35\\ & \Delta \varphi =k\Delta x\\ & \Delta \varphi =2\pi \times 8\times 10^{-3}\Delta x\\ & =2\pi \times 8\times 10^{-3}\times \left(\frac{100}{2}\right)\\ & =8\pi \times 10^{-1}=0.8\pi \end{array}$$

A. Electrostatic field is zero inside a conductor.

B. Electric field at the surface of a charged conductor $=\frac{\sigma }{{\epsilon }_0}\hat{n}$, depends on surface charge density ( $\sigma$ ).

C. The interior of a charged conductor cannot have any excess charge in the static situation.

D. At the surface of a charged conductor, the electrostatic field ⟂ surface.

E. The electrostatic potential is constant and can be non-zero everywhere inside a charged conductor.

Sol. Circuit can be draw as,

⇒ Terminal voltage of battery $V=E$ - ir

$$=12-0.6\times 2=10.8\mathrm{V}$$

→ Each side will have resistance of $1\Omega$

→ It is balanced wheat stone bridge

→ No current in the resistance of $2\Omega$

$\to {\left[R_{\text{effective }}\right]}_{AC}=1\Omega$

$$\begin{array}{rl}& \to I=\frac{E}{R_{\text{eff. }}}=\frac{2}{1}\\ & \to I=2\mathrm{A}\end{array}$$

Given:

• Rated power $P_1=400\text{W}$

• Rated voltage $V_1=220\text{V}$

• New voltage $V_2=200\text{V}$

Using the relation

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

First, find the resistance of the heater:

$$R=\frac{V_1^2}{P_1}=\frac{220^2}{400}=\frac{48400}{400}=121\Omega$$

Now find the new power at $200\text{V}$:

$$P_2=\frac{V_2^2}{R}=\frac{200^2}{121}=\frac{40000}{121}\approx 330.6\text{W}$$

So, approximately,

$$P_2\approx 331\text{W}$$

Therefore, the correct answer is:

$$\boxed{\text{Option C: }331\text{W}}$$

$$\begin{array}{rl}{i}_G& =1\mathrm{mA}=0.001\mathrm{A}\\ i_S& =10-i_G\simeq 10\mathrm{A}\end{array}$$

Both shunt resistance and galvanometer are in parallel connection

$$\begin{array}{ll}\therefore & i_S{r}_S=i_G{R}_G\\ \Rightarrow & 10\times r_s=0.001\times 100\\ \Rightarrow & r_s=0.01\Omega \end{array}$$

• Position of null point will not change when galvanometer ( $G$ ) and the cell ( $E$ ) are interchanged.

• There will be no deflection in galvanometer only at balance point.

In an unbalanced meter bridge, if $E$ and $G$ are interchanged mutually, then the deflection in galvanometer may be towards left-side or right-side.

$$\begin{array}{rl}& C_{\mathrm{eq}}=2.5+2.5=5\mu {\mathrm{F}}^{ˋ}\\ & q_1=q_2=q_3=q_4=2.5\times 50=125\mu \mathrm{C}\\ & q_5=2.5\times 50=125\mu \mathrm{C}\end{array}$$

Energy loss $=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}{V}^2$

$$\begin{array}{rl}& =\frac{1}{2}\left(\frac{200\times 200}{400}\right)\times 10^{-12}(100{)}^2\\ & =\frac{1}{2}\times 10^6\times 10^{-12}=0.5\times 10^{-6}\mathrm{J}\end{array}$$

Magnetic field of a circular loop:

$$\begin{array}{rl}& B_0=\frac{{\mu }_{0}Ni}{2R}\\ & i=\frac{2R{B}_0}{{\mu }_{0}N}=\frac{2\times 5\times 10^{-2}\times 3.14\times 10^{-3}}{4\pi \times 10^{-7}\times 100}\\ & i=2.5\mathrm{A}\end{array}$$

Magnetic moment

$$\begin{array}{rl}& M=NiA\\ & =100\times 2.5\times 3.14\times {\left(5\times 10^{-2}\right)}^2\left[\because A=\pi R^2\right]\\ & =2\mathrm{A}{\mathrm{m}}^2\end{array}$$

For a long straight solid wire carrying steady current, which is uniformly distributed across its crosssection, the variation of magnetic field $(B)$ with distance $(r)$ from axis will be

$$\begin{array}{rl}& B=\frac{{\mu }_{0}Ir}{2\pi a^2}\Rightarrow B\propto r,\text{ for }r<a\\ & B=\frac{{\mu }_{0}I}{2\pi r}\Rightarrow B\propto \frac{1}{r},\text{ for }r>a\end{array}$$

Induced emf across the shorter side

$$\begin{array}{rl}& E_{\text{induced }}=BVl[\because \vec{B}\perp \vec{V}\perp \vec{I}]\\ & =0.3\times 2\times 10^{-2}\times 3\times 10^{-2}\\ & =1.8\times 10^{-4}\mathrm{V}\\ & =1.8\times 10^{-4}\text{ volt }\end{array}$$

Resonance frequency

$$\begin{array}{rl}& f_0=\frac{1}{2\pi \sqrt{LC}}\\ & =\frac{1}{2\pi \sqrt{1\times 10^{-7}\times 10^{-3}}}\\ & =15.9\mathrm{kHz}\end{array}$$

Alternating current,

$$i=i_{\text{peak }}\sin (\omega t)$$

where $\omega =2\pi f=2\pi \times 60=120\pi \mathrm{rad}/\mathrm{s}$

$$\begin{array}{rl}& \Rightarrow i=5\sin (120\pi t)\\ & \Rightarrow 5=5\sin (120\pi t)\\ & \Rightarrow \sin (120\pi t)=1=\sin \left(\frac{\pi }{2}\right)\\ & \therefore 120\pi t=\frac{\pi }{2}\\ & \Rightarrow t=\frac{1}{240}\mathrm{s}\end{array}$$

$$\begin{array}{rl}& i=e\\ & \text{ Equation of prism }\\ & \begin{array}{rl}& i+e=A+\delta \\ & 2i-A=\delta \\ & 2(50)-60=\delta \\ & \delta =40^{\circ }\end{array}\end{array}$$

$$F_2\text{ is the second principal focus ⇒ It is the virtual image position for object at infinity. }$$

$F_1$ is the first principal focus ⇒ It is the virtual object position for which image is formed at infinity. The best appropriate answer is option (2), although it should be second principal focus.

$I=I_0{\cos }^2\frac{\Delta \varphi }{2}[I_0\to$ maximum intensity $]$

$$\begin{array}{rl}& I=I_0{\cos }^2\frac{K\Delta x}{2}\\ & K=I_0{\cos }^2\left(\frac{2\pi }{\lambda }\times \frac{\lambda }{2}\right)\\ & K=I_0\\ & K_1=I_0{\cos }^2\left(\frac{2\pi }{\lambda }\times \frac{\lambda }{3\times 2}\right)=I_0{\cos }^2\left(\frac{\pi }{3}\right)\\ & K_1=\frac{I_0}{4}\\ & \frac{K}{K_1}=4\Rightarrow K_1=\frac{K}{4}\end{array}$$

In interference and diffraction there is no loss of energy, the energy gets redistributed. Interference and diffraction both are exhibited in light as well as sound waves.

$$\begin{array}{rl}& \frac{K{Q}^2}{2r}=3.4\mathrm{eV}\\ & \frac{9\times 10^9\times {\left(1.6\times 10^{-19}\right)}^2}{2\times 3.4\times 1.6\times 10^{-19}}=r\\ & 2.1176\times 10^{-10}=r\\ & r=2.1\times 10^{-10}\mathrm{m}\end{array}$$

As we know,

Size of nucleus, $r=r_0(A{)}^{\frac{1}{3}}$

$$\begin{array}{rl}& \Rightarrow r^3=r_0^3\cdot A\\ & \Rightarrow V\propto A\left(\because V=\frac{4}{3}\pi r^3\right)\end{array}$$

So, option $\mathrm{A}\to$ wrong and $\mathrm{B}\to$ correct while the difference between the actual mass of nucleus and its constituents is called the mass defect.

∴ option $\mathrm{C}\to$ wrong and $\mathrm{D}\to$ correct

Given

$\rho =2.29\times 10^{17}\mathrm{kg}/{\mathrm{m}}^3$, mass $m=19.926\times 10^{-27}\mathrm{kg}$

$R_0=1.2\times 10^{-15}\mathrm{m}$ and $R=R_0{A}^{\frac{1}{3}}$

Now use volume $=\frac{\text{ Mass }}{\text{ Density }}$

$\Rightarrow \frac{4}{3}\pi R^3=\frac{M}{\rho }$

$\Rightarrow \frac{4}{3}\pi {\left[R_0(A{)}^{1/3}\right]}^3=\frac{M}{\rho }$

$\Rightarrow \frac{4}{3}\pi R_0^{3}A=\frac{M}{\rho }$

$\Rightarrow A=\frac{M}{\rho }\times \frac{3}{4\pi \times R_0^3}=\frac{19.926\times 10^{-27}\times 3}{2.29\times 10^{17}\times 12.56\times {\left(1.2\times 10^{-15}\right)}^3}\approx 12$

A. $E=hv$ is energy of photon

B. Diffraction and interference confirm wave nature of light

C. $\lambda =\frac{h}{p}$ is de Broglie wavelength of particle.

D. Compton effect confirms particle nature of light.

For incident radiation having wavelength ( $\lambda$ ), photoelectric effect doesn't occur when $\frac{hc}{\lambda }<$ work-function

$$\begin{array}{rl}& \Rightarrow \lambda >\frac{hc}{W_0}\\ & \Rightarrow \lambda >\frac{6.6\times 10^{-34}\times 3\times 10^8}{6.6\times 1.6\times 10^{-19}}\\ & \Rightarrow \lambda >\frac{3\times 10^{-7}}{1.6}\\ & \Rightarrow \lambda >\frac{300}{1.6}\mathrm{nm}\\ & \Rightarrow \lambda =187.5\mathrm{nm}\end{array}$$

∴ Option (3) 200 nm is correct.

For ideal diode, forward resistance $=0$ and reverse biased resistance $=\infty$

$\therefore$ Circuit can be redrawn as,

$$\begin{array}{rl}I& =\frac{10}{2}+\frac{10}{4}\\ & =\frac{30}{4}=\frac{15}{2}\mathrm{A}\end{array}$$

Voltage drop will be across the diode, when it will be in reverse bias 1

In positive half cycle it will be in reverse biased

$V-I$ characteristics of a forward-biased junction diode

When forward bias voltage increases beyond threshold voltage, diode current increases significantly. This is not reverse saturation current.

$$\begin{array}{rl}& \text{ Mole of urea }=\frac{5.4}{60}=0.09\\ & \begin{array}{rl}\text{ Number of hydrogen atoms }& =0.09\times 4\times 6.022\times 10^{23}\\ & =2.1679\times 10^{23}\\ & \simeq 2.168\times 10^{23}\end{array}\end{array}$$

The reaction of carbon dioxide with hot coke is:

$$\mathrm{C}{\mathrm{O}}_2+\mathrm{C}\to 2CO$$

At the same temperature and pressure, volumes of gases are in the same ratio as the number of moles.

Let $x\mathrm{d}{\mathrm{m}}^3$ of $\mathrm{C}{\mathrm{O}}_2$ react.

Then according to the equation:

• $x\mathrm{d}{\mathrm{m}}^3$ of $\mathrm{C}{\mathrm{O}}_2$ is consumed

• $2x\mathrm{d}{\mathrm{m}}^3$ of $\mathrm{CO}$ is formed

Initially, volume of $\mathrm{C}{\mathrm{O}}_2=1\mathrm{d}{\mathrm{m}}^3$

So after reaction:

• Unreacted $\mathrm{C}{\mathrm{O}}_2=(1-x)\mathrm{d}{\mathrm{m}}^3$

• Formed $\mathrm{CO}=2x\mathrm{d}{\mathrm{m}}^3$

Total final volume is given as $1.4\mathrm{d}{\mathrm{m}}^3$

So,

$$(1-x)+2x=1.4$$

$$1+x=1.4$$

$$x=0.4$$

Now,

$$\text{Volume of CO}=2x=0.8\mathrm{d}{\mathrm{m}}^3$$

\text{Volume of CO_2} = 1 - x = 0.6 \,\mathrm{dm^3}

Hence, the gaseous mixture contains:

$$0.8\mathrm{d}{\mathrm{m}}^3\mathrm{of}\mathrm{CO}\text{and}0.6\mathrm{d}{\mathrm{m}}^3\mathrm{of}\mathrm{C}{\mathrm{O}}_2$$

So the correct option is Option B.

1. Principal Quantum Number ($n$): This number tells us the principal energy shell. In the notation of an orbital (like 3d or 4s), the number at the beginning is the value of $n$.

2. Azimuthal Quantum Number ($l$): This number defines the subshell and the shape of the orbital. We use specific letters to denote the value of $l$:

   • If $l=0$, the subshell is 's'.

   • If $l=1$, the subshell is 'p'.

   • If $l=2$, the subshell is 'd'.

   • If $l=3$, the subshell is 'f'.

The designation for an orbital is given by writing the value of $n$ followed by the letter for the subshell (determined by $l$).

Now, let's look at each item from List I.

A. For $n=2$ and $l=1$:

• The principal quantum number $n$ is 2.

• The azimuthal quantum number $l$ is 1, which corresponds to the 'p' subshell.

• Therefore, the orbital is designated as 2p.

• This matches with II in List II.

B. For $n=4$ and $l=0$:

• The principal quantum number $n$ is 4.

• The azimuthal quantum number $l$ is 0, which corresponds to the 's' subshell.

• Therefore, the orbital is designated as 4s.

• This matches with III in List II.

C. For $n=5$ and $l=3$:

• The principal quantum number $n$ is 5.

• The azimuthal quantum number $l$ is 3, which corresponds to the 'f' subshell.

• Therefore, the orbital is designated as 5f.

• This matches with IV in List II.

D. For $n=3$ and $l=2$:

• The principal quantum number $n$ is 3.

• The azimuthal quantum number $l$ is 2, which corresponds to the 'd' subshell.

• Therefore, the orbital is designated as 3d.

• This matches with I in List II.

So, the correct matches are:

• A → II

• B → III

• C → IV

• D → I

Looking at the options, we can see that this combination corresponds to Option C.

The correct answer is Option C: A-II, B-III, C-IV, D-I.

Energy = Power × Time

$$=150\mathrm{watt}\times 1\mathrm{s}$$

$$=150\mathrm{J}$$

Only $8\%$ of this energy is converted into light energy.

$$\text{Energy as light}=\frac{150\times 8}{100}=12\mathrm{J}$$

We know that total light energy is equal to the number of photons multiplied by the energy of one photon.

$$E=n\times \text{Energy of one photon}$$

Energy of one photon = $4.42\times 10^{-19}\mathrm{J}$

$$n=\frac{E}{\text{Energy of one photon}}=\frac{12}{4.42\times 10^{-19}}$$

$$n=2.715\times 10^{19}$$

Hence, the bulb emits approximately $2.7\times 10^{19}$ photons per second.

$$\begin{array}{rl}& {\mathrm{K}}_{\mathrm{P}}={\mathrm{K}}_{\mathrm{C}}(\mathrm{RT}{)}^{\Delta {\mathrm{n}}_{\mathrm{g}}}\\ & \text{ For }{\mathrm{K}}_{\mathrm{P}}={\mathrm{K}}_{\mathrm{C}}\left(\Delta {\mathrm{n}}_{\mathrm{g}}=0\right)\\ & {\mathrm{K}}_{\mathrm{P}}\ne {\mathrm{K}}_{\mathrm{C}}\left(\Delta {\mathrm{n}}_{\mathrm{g}}\ne 0\right)\end{array}$$

$$\begin{array}{rl}(1)& {\mathrm{H}}_2\mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\rightleftharpoons {\mathrm{H}}_2(\mathrm{g})+{\mathrm{CO}}_2(\mathrm{g})\\ & \Delta {\mathrm{n}}_{\mathrm{g}}=0\left({\mathrm{K}}_{\mathrm{P}}={\mathrm{K}}_{\mathrm{C}}\right)\end{array}$$

$$\begin{array}{rl}(2)& {\mathrm{N}}_2(\mathrm{g})+3{\mathrm{H}}_2(\mathrm{g})\rightleftharpoons 2{\mathrm{NH}}_3(\mathrm{g})\\ & \Delta {\mathrm{n}}_{\mathrm{g}}=-2\left({\mathrm{K}}_{\mathrm{P}}\ne {\mathrm{K}}_{\mathrm{C}}\right)\end{array}$$

$$\begin{array}{rl}(3)& {\mathrm{H}}_2(\mathrm{g})+{\mathrm{I}}_2(\mathrm{g})\rightleftharpoons 2\mathrm{HI}(\mathrm{g})\\ & \Delta {\mathrm{n}}_{\mathrm{g}}=0\left({\mathrm{K}}_{\mathrm{P}}={\mathrm{K}}_{\mathrm{C}}\right)\end{array}$$

$$\begin{array}{rl}(4)& {\mathrm{N}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})\rightleftharpoons 2\mathrm{NO}(\mathrm{g})\\ & \Delta {\mathrm{n}}_{\mathrm{g}}=0\left({\mathrm{K}}_{\mathrm{P}}=\mathrm{Kc}\right)\end{array}$$