IIT-JEE 2012 Paper 1 Question Paper with Solutions

Let M and l be the mass and length of the rod respectively and m be the mass of the insect. Let the insect be at a distance x from O at any instant of time t.

$\therefore$ x = vt ....... (i)

Angular momentum of the system about O,

$L = \left[ {{{M{l^2}} \over {12}} + m{x^2}} \right]\omega$

$= \left[ {{{M{l^2}} \over {12}} + m{{(vt)}^2}} \right]\omega$ (Using (i))

$= \left[ {{{M{l^2}} \over {12}} + m{v^2}{t^2}} \right]\omega$

As, $\left| {\overrightarrow \tau } \right| = {{dL} \over {dt}} = {d \over {dt}}\left[ {{{M{l^2}} \over {12}} + m{v^2}{t^2}} \right]\omega$

As $\omega$ and v remain constant

$\therefore$ $\left| {\overrightarrow \tau } \right| = 2m\omega {v^2}t$

$\left| {\overrightarrow \tau } \right| \propto t$

Hence, the graph and t is a straight line passing through the (0, 0). Option (b) represents correct plot.

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

${𝑣}_{𝑟𝑚𝑠}=\sqrt{\surd \frac{3𝑘𝑇}{𝑚}}$

where :

• $𝑘$ is the Boltzmann constant
• $𝑇$ is the temperature in Kelvin
• $𝑚$ is the mass of one molecule of the gas

Given that the atomic mass of helium is 4 amu and that of argon is 40 amu, we can find the rms speed ratio by comparing the rms speeds of helium and argon, as temperature $𝑇$ and Boltzmann constant $𝑘$ are the same for both gases :

$\frac{{𝑣}_{𝑟𝑚𝑠}\left(\text{helium}\right)}{{𝑣}_{𝑟𝑚𝑠}\left(\text{argon}\right)}=\sqrt{\surd \frac{{𝑚}_{\text{argon}}}{{𝑚}_{\text{helium}}}}$

Substituting the masses :

${𝑚}_{\text{helium}}=4\text{ amu}$

${𝑚}_{\text{argon}}=40\text{ amu}$

we get :

$\frac{{𝑣}_{𝑟𝑚𝑠}\left(\text{helium}\right)}{{𝑣}_{𝑟𝑚𝑠}\left(\text{argon}\right)}=\sqrt{\surd \frac{40}{4}}=\sqrt{\surd 10}\approx 3.16$

Thus, the ratio of the rms speeds of helium to argon is approximately 3.16.

The correct answer is :

Option D : 3.16

The magnetic flux through a wire loop with $n$ turns, an area vector $\vec{S}$, placed in a uniform magnetic field $\vec{B}$, is given by the equation $\phi = n \vec{B} \cdot \vec{S}$. For a circular loop of radius $r$ that carries a current $i$, the magnetic field at an axial point located a distance $z$ from the center of the loop can be expressed as :

$|\vec{B}|=\frac{\mu_0 i r^2}{2\left(r^2+z^2\right)^{3 / 2}}$

By substituting $r = R$ and $z = \sqrt{3} R$ into the expression for the magnetic field at an axial point of a circular loop, we get the magnetic field $|\vec{B}|$ as follows:

The formula for the magnetic field at an axial point $z$ from a circular loop of radius $r$ carrying a current $i$ is given by:

$B = \frac{\mu_0 i r^2}{2 (r^2 + z^2)^{3/2}}$

Substituting $r = R$ and $z = \sqrt{3} R$ into this formula, we get:

$B = \frac{\mu_0 i R^2}{2 (R^2 + (\sqrt{3} R)^2)^{3/2}}$

Simplify the expression inside the parentheses:

$B = \frac{\mu_0 i R^2}{2 (R^2 + 3R^2)^{3/2}}$

$B = \frac{\mu_0 i R^2}{2 (4R^2)^{3/2}}$

$B = \frac{\mu_0 i R^2}{2 (4^{3/2} R^3)}$

$4^{3/2} = 8$

$B = \frac{\mu_0 i R^2}{2 (8 R^3)}$

$B = \frac{\mu_0 i R^2}{16 R^3}$

$B = \frac{\mu_0 i}{16 R}$

Thus, the magnetic field at the axial point $z = \sqrt{3} R$ is:

$|\vec{B}| = \frac{\mu_0 i}{16 R}$

This magnetic field is directed along the $z$-axis and can be considered uniform at the location of the square loop (since $a \ll R$). Consequently, $\vec{B}$ forms an angle of $45^\circ$ with the area vector $\vec{S}$ (where $|\vec{S}| = a^2$) of the square loop. The magnetic flux through the square loop and the mutual inductance of the loops are given by:

$\begin{aligned} & \phi=n|\vec{B}||\vec{S}| \cos 45^{\circ}=2 \cdot \frac{\mu_0 i}{16 R} \cdot a^2 \cdot \frac{1}{\sqrt{2}}=\frac{\mu_0 i a^2}{2^{7 / 2} R}, \\ & M=\frac{\phi}{i}=\frac{\mu_0 a^2}{2^{7 / 2} R} .\end{aligned}$

B_R = B_T $-$ B_C

R = Remaining portion

T = Total portion and

C = cavity

${B_R} = {{{\mu _0}{I_T}} \over {2a\pi }} - {{{\mu _0}{I_C}} \over {2(3a/2)\pi }}$ ..... (i)

${I_T} = J(\pi {a^2})$

${I_C} = J\left( {{{\pi {a^2}} \over 4}} \right)$

Substituting the values in Eq. (i), we have

${B_R} = {{{\mu _0}} \over {a\pi }}\left[ {{{{I_T}} \over 2} - {{{I_C}} \over 3}} \right]$

$= {{{\mu _0}} \over {a\pi }}\left[ {{{\pi {a^2}J} \over 2} - {{\pi {a^2}J} \over {12}}} \right]$

$= {{5{\mu _0}aJ} \over {12}}$

$\therefore$ $N = 5$

Due to symmetry on upper side and lower side, points P and Q are at same potentials. Similarly, points S and T are at same potentials. Therefore, the simple circuit can be drawn as shown below :

${I_2} = {{12} \over {2 + 2 + 2}} = 2A$

${I_3} = {{12} \over {4 + 4 + 4}} = 1A$

$\therefore$ ${I_1} = {I_2} + {I_3} = 3A$

I_PQ = 0 because V_P = V_Q

Potential drop (from left to right) across each resistance is

${{12} \over 3} = 4V$

$\therefore$ ${V_{MS}} = 2 \times 4 = 8V$

${V_{NQ}} = 1 \times 4 = 4V$

or, ${V_S} < {V_Q}$

Since, the block starts executing simple harmonic motion from extreme position, we have

$x = a\cos \omega t$

At 1 s, we have

$x = 0.2\cos \left( {{\pi \over 3}} \right) = 0.1\,m$

That is, the block is at a distance 5.1 $-$ 0.10 = 5 m from O, which is the range of the bebble as well. Now,

$R = {{{v^2}\sin 2\alpha } \over g}$

$S = {{{v^2}\sin 2(\pi /4)} \over {10}}$

Therefore, $v = \sqrt {50}$ m/s

Alternate Method :

Since, pebble strikes the oscillating mass after 1 sec., its time of flight is $1 \mathrm{~s}$

$$\begin{aligned} & 1=\frac{2 v \sin 45^{\circ}}{\mathrm{g}}=\frac{\sqrt{2} v}{10} \\ & v=\frac{10}{\sqrt{2}}=5 \sqrt{2}=\sqrt{50} \mathrm{~m} / \mathrm{s} \end{aligned}$$

The forces acting on the block are F = 1 N towards the left, weight mg = 0.1 $\times$ 10 = 1 N downwards, normal force N, and the frictional force f. Resolve F and mg along and perpendicular to the inclined plane.

When $\theta$ = 45$^\circ$, the net force that brings the block down is

${F_d} = mg\sin \theta - F\cos \theta = {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = 0$.

Thus, the block is stationary if $\theta = 45^\circ$. When $\theta > 45^\circ$, the force ${F_d} > 0$, and hence the block has a tendency to move down. Thus, the frictional force f acts on the block upwards i.e., towards Q.

Electric field at point P due to long uniformly charged solid cylinder is

${E_1} = {{\rho {R^2}} \over {2{\varepsilon _0}(2R)}} = {{\rho R} \over {4{\varepsilon _0}}}$

Electric field at point P due to spherical cavity is

${E_1} = {1 \over {4\pi {\varepsilon _0}}}{{\rho {4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}} \over {{{(2R)}^2}}} = {{\rho R} \over {96{\varepsilon _0}}}$

The electric field at the point P is

$= {E_1} - {E_2}$

$= {{\rho R} \over {4{\varepsilon _0}}} - {{\rho R} \over {96{\varepsilon _0}}} = {{\rho R} \over {4{\varepsilon _0}}}\left[ {1 - {1 \over {24}}} \right] = {{23\rho R} \over {96{\varepsilon _0}}} = {{23\rho R} \over {(16)6{\varepsilon _0}}} = {{23\rho R} \over {16k{\varepsilon _0}}}$

$\therefore$ $k = 6$

Let initial and final kinetic energies of the proton be K_i and K_f and corresponding potential energies be U_i and U_f. When proton is far away from the nucleus (r $\to$ $\infty$), its potential energy is

${U_i} = \mathop {\lim }\limits_{r \to \infty } {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over r} = 0$.

At closest distance, the proton comes to rest momentarily, giving K_f = 0. The potential energy at closest distance is

${U_f} = {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over a}$,

where a is the distance of closest approach. Since electrostatic force is conservative, total energy is conserved i.e., ${K_i} + {U_i} = {K_f} + {U_f}$. Substitute the values to get

${K_i} = {U_f} = {1 \over {4\pi {\varepsilon _0}}}{{120{e^2}} \over a}$

The de-Broglie wavelength of the proton is given by

${\lambda _i} = {h \over {{p_i}}} = {h \over {\sqrt {2{m_p}{K_i}} }} = {h \over e}\sqrt {{{4\pi {\varepsilon _0}a} \over {240{m_p}}}}$

$= 4.2 \times {10^{ - 15}}{\left( {{{10 \times {{10}^{ - 15}}} \over {9 \times {{10}^9} \times 240 \times (5/3) \times {{10}^{ - 27}}}}} \right)^{1/2}}$

= 7 fm.

According to lens maker's formula

${1 \over f} = (n - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

For the first lens

n = 1.5, R₁ = + 14 cm, R₂ = $\infty$

$\therefore$ ${1 \over {{f_1}}} = (1.5 - 1)\left( {{1 \over {14}} - {1 \over \infty }} \right) = {{0.5} \over {14}}$

For the second lens,

n = 1.2, R₁ = $\infty$, R₂ = $-$14 cm

$\therefore$ ${1 \over {{f_2}}} = (1.2 - 1)\left( {{1 \over \infty } - {1 \over { - 14}}} \right) = {{0.2} \over {14}}$

The focal length of the bi-convex lens is

${1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} = {{0.5} \over {14}} + {{0.2} \over {14}} = {{0.7} \over {14}} = {1 \over {20}}$

According to thin lens formula

${1 \over v} - {1 \over u} = {1 \over f}$

Here, u = $-$40 cm

$\therefore$ ${1 \over v} - {1 \over { - 40}} = {1 \over {20}}$ or ${1 \over v} = {1 \over {20}} - {1 \over {40}}$ or v = 40 cm

Let the steady state temperature of the middle plate to T₀. In the steady state, the heat radiated per unit time by the middle plate is equal to the heat received per unit time by it. The middle plate radiates heat from both the surfaces and receives the heat radiated by the first and the third plates.

Stefan-Boltzmann law gives the rate of heat loss and heat gain by the middle plate as

$d{Q_{out}}/dt = \sigma AT_0^4 + \sigma AT_0^4$,

$d{Q_{in}}/dt = \sigma A{(2T)^4} + \sigma A{(3T)^4}$.

The steady state condition, $d{Q_{out}}/dt = d{Q_{in}}/dt$

$\sigma A{(2T)^4} + \sigma A{(3T)^4} = \sigma 2A{({T_0})^4}$

$16{T^4} + 81{T^4} = 2{({T_0})^4}$

$97{T^4} = 2{({T_0})^4}$

${({T_0})^4} = {{97} \over 2}{T^4} = {\left( {{{97} \over 2}} \right)^{1/4}}T$

Angular momentum of a particle about a point is given by:

L = r $\times$ p = m (r $\times$ v)

For L_O

$\left| L \right| = (mvr\sin \theta ) = m(R\omega )(R)\sin 90^\circ$

$= m{R^2}\omega$ = constant

Direction of L_O is always upwards. Therefore, complete L_O is constant, both in magnitude as well as direction.

For L_P

$\left| {{L_P}} \right| = (mvr\sin \theta ) = m(R\omega )(l)\sin 90^\circ$

$= (mRl\omega )$

Magnitude of L_P will remain constant but direction of L_P keeps on changing.

The positions of all charges are symmetric about the planes $𝑥=+\frac{𝑎}{2}$ and $𝑥=-\frac{𝑎}{2}$. Therefore, the net electric flux crossing the plane $𝑥=+\frac{𝑎}{2}$ is equal to the net electric flux crossing the plane $𝑥=-\frac{𝑎}{2}$.

Similarly, the net electric flux crossing the plane $𝑦=+\frac{𝑎}{2}$ is equal to the net electric flux crossing the plane $𝑦=-\frac{𝑎}{2}$.

According to Gauss's law, the net electric flux crossing the entire region is given by:

$𝜙=\frac{{𝑞}_{\text{inside}}}{{𝜀}_0}=\frac{3𝑞-𝑞-𝑞}{{𝜀}_0}=\frac{𝑞}{{𝜀}_0}$

The charges are symmetrically placed about the planes $𝑧=+\frac{𝑎}{2}$ and $𝑥=+\frac{𝑎}{2}$. Thus, the net electric flux crossing the plane $𝑧=+\frac{𝑎}{2}$ is equal to the net electric flux crossing the plane $𝑥=+\frac{𝑎}{2}$.

Fringe width, $\beta = {{\lambda D} \over d}$

$\therefore$ $\beta \propto \lambda$

As, ${\lambda _R} > {\lambda _G} > {\lambda _B}$ $\therefore$ ${\beta _R} > {\beta _G} > {\beta _B}$

The high pressure pulse is a compression pulse. When this gets reflected from a rigid boundary (closed pipe), the reflected pressure wave has the same phase as the incident wave i.e., a compression pulse is reflected as a compression pulse. In case of open pipe, there is a phase change of 180$^\circ$ when it is reflected by an open end i.e., a compression pulse is reflected as a rarefaction pulse.

To determine Young's modulus using Searle's method, we use the formula:

$𝑌=\frac{4𝑀𝐿𝑔}{𝜋𝑙{𝑑}^2}$

Given:
Length of wire, $𝐿=2\text{ m}$
Diameter of wire, $𝑑=0.5\text{ mm}$
Load, $𝑀=2.5\text{ kg}$
Extension in length, $𝑙=0.25\text{ mm}$

The quantities $𝑑$ and $𝑙$ are measured using a screw gauge and a micrometer, respectively, both having a pitch of 0.5 mm and 100 divisions on their circular scale. The least count, which is the smallest measurement that can be accurately read, is:

$\text{Least count}=\frac{\text{Pitch}}{\text{Total number of divisions on the circular scale}}=\frac{0.5\text{ mm}}{100}=0.005\text{ mm}$

Hence, the least count, $\Delta 𝑑$ and $\Delta 𝑙$, are both 0.005 mm.

Error in measurement of $𝑙$ (extension length) is:

$\frac{\Delta 𝑙}{𝑙}=\frac{0.005}{0.25}=\frac{1}{50}$

Error in measurement of $𝑑$ (diameter) is:

$\frac{\Delta 𝑑}{𝑑}=\frac{0.005}{0.5}=\frac{1}{100}$

Using the formula for Young's modulus, the combined error is:

$𝑌=\frac{4𝑀𝐿𝑔}{𝜋𝑙{𝑑}^2}$

The relative error in $𝑌$ is given by:

$\frac{\Delta 𝑌}{𝑌}=\frac{\Delta 𝑙}{𝑙}+2\frac{\Delta 𝑑}{𝑑}$

As can be seen:

$\frac{\Delta 𝑙}{𝑙}=2\frac{\Delta 𝑑}{𝑑}=\frac{1}{50}$

Therefore, the contributions to the error in the measurement of $𝑌$ from errors in measuring $𝑙$ and $𝑑$ are the same.

Consider a thin spherical shell of radius $𝑅$ with its center at the origin, carrying a uniform positive surface charge density. To understand how the electric field $\mid \overrightarrow{𝐸}\left(𝑟\right)\mid$ and the electric potential $𝑉\left(𝑟\right)$ vary with the distance $𝑟$ from the center, refer to the following details:

Electric Field due to a Uniformly Charged Thin Spherical Shell

Inside the Shell

For $𝑟<𝑅$, the electric field is given by:

${𝐸}_{\text{inside}}=0[𝑟<𝑅]$

On the Surface of the Shell

For $𝑟=𝑅$, the electric field is:

${𝐸}_{\text{surface}}=\frac{1}{4𝜋{𝜀}_0}\frac{𝑄}{{𝑅}^2}$

Outside the Shell

For $𝑟>𝑅$, the electric field is:

${𝐸}_{\text{outside}}=\frac{1}{4𝜋{𝜀}_0}\frac{𝑄}{{𝑟}^2}[𝑟>𝑅]$

Below is a graphical representation of the electric field $𝐸$ variation with the distance $𝑟$ from the center:

Electric Potential due to a Uniformly Charged Thin Spherical Shell

Inside the Shell

For $𝑟<𝑅$, the electric potential is:

${𝑉}_{\text{inside}}=\frac{1}{4𝜋{𝜀}_0}\frac{𝑄}{𝑅}$

On the Surface of the Shell

For $𝑟=𝑅$, the electric potential is:

${𝑉}_{\text{surface}}=\frac{1}{4𝜋{𝜀}_0}\frac{𝑄}{𝑅}$

Outside the Shell

For $𝑟>𝑅$, the electric potential is:

${𝑉}_{\text{outside}}=\frac{1}{4𝜋{𝜀}_0}\frac{𝑄}{𝑟}$

Below is a graphical representation of the electric potential $𝑉$ variation with the distance $𝑟$ from the center:

T = Total portion

R = Remaining portion and

C = Cavity and

For $\theta$ = 90$^\circ$ : $\overrightarrow v \,||\,\overrightarrow E$ and $\overrightarrow v \,||\,\overrightarrow B$

${\overrightarrow F _B} = q(\overrightarrow v \times \overrightarrow B ) = 0$

${\overrightarrow F _E} = q\overrightarrow E$

Therefore, the motion is along y-axis and the particle accelerates.

For $\theta$ = 0$^\circ$ and $\theta$ = 10$^\circ$ : The motion is helical with increasing pitch, along with increasing time (along y-axis), as $p = (v\cos \theta )t$, where v increases due to ${\overrightarrow F _E}$. (For $\theta$ = 0$^\circ$, the change cannot move in a circular path.)

Given that the proton moves at a ${45}^{\circ }$ angle to the vertical, we can derive the electric field (E) as follows :

$𝑞𝐸=𝑚𝑔$ or $𝐸=\frac{𝑚𝑔}{𝑞}$

Since $𝐸=\frac{𝑋}{𝑑}$, we can substitute to find $𝑋$ :

$\frac{𝑋}{𝑑}=\frac{𝑚𝑔}{𝑞}$ or $𝑋=\frac{𝑚𝑔𝑑}{𝑞}$

Where :

• $𝑚=1.67\times {10}^{-27}\mathrm{kg}$


• $𝑔=10m/s^2$


• $𝑑=1\mathrm{cm}=1\times {10}^{-2}m$


• $𝑞=1.6\times {10}^{-19}C$

Substituting these values, we get :

$\begin{array}{cc}𝑋& =\frac{1.67\times {10}^{-27}\times 10\times 1\times {10}^{-2}}{1.6\times {10}^{-19}}V\\ \\ & =\frac{1.67}{1.6}\times {10}^{-9}V\\ \\ & =1\times {10}^{-9}V\end{array}$

For basic amino acids with pH > 7, peptides will exist as cations. For example, when the substituents are basic, that is R₁ = CH₂CONH₂ and R₂ = (CH₂)₄NH₂ or when R₁ = (CH₂)₄NH₂ and R₂ = (CH₂)₄NH₂ or when R₁ = CH₂OH and R₂ = (CH₂)₄NH₂ or when R₁ = (CH₂)₄NH₂ and R₂ = CH₃.

Aqueous solution of CuSO₄ will absorb orange red light and appear blue because blue is the complimentary colour for orange-red.

The structure of aldohexose in D-configuration can be represented as

The van der Waals equation is

$\left( {p + {a \over {{V^2}}}} \right)(V - b) = RT$

but it is given that b = 0. So, the equation reduces to

$\left( {p + {a \over {{V^2}}}} \right)V = RT \Rightarrow pV = - {a \over V} + RT$

Comparing it with a straight line equation we get slope as $-$a. Calculating the slope, we get

${{24.6 - 20.1} \over {3.0 - 0}} = 1.5 \Rightarrow a = 1.5$

There are two ligands in the coordination entity, that is, ammine and aqua, but according to the IUPAC rule, alphabetical order is followed, that is, ammine is named first. Now calculating the oxidation state of Co, the net charge on the coordination entity is +3 as Cl has $-$1 and there are three Cl^$-$ outside the square brackets. So the oxidation state of Co will be III as both ammine and aqua are neutral ligands. So, its correct name will be diamminetetraaquacobalt (III) chloride.

(a) The compound is monocyclic but non-planar due to presence of two $s p^3$ hybridised carbon atoms. As a result, delocalisation of pi $(\pi)$ electrons (or a conjugate system) is not possible. The compound is non-aromatic.

(b) The compound is monocyclic and planar. Carbon atoms are $s p^2$ hybridised and the pi $(\pi)$ electrons are conjugated, i.e., there is delocalisation of pi $(\pi)$ electrons. It follows $4 n$$\pi$ electron system with $4 \pi$ electrons. This makes the compound anti-aromatic and least stable.

(c) The compound is monocyclic and planar. The carbons including carbonyl carbon are $s p^2$ hybridised. There is no extended delocalisation of pi ( $\pi$ ) electrons (due to exocyclic carbonyl double bond). It also follows $4 n \pi$ electron system with $4 \pi$ electrons inside the ring. This makes compound anti-aromatic and least stable.

(d) The compound is monocyclic and planar. The all carbons including carbonyl carbons are $s p^2$ hybridised. Though, there is no extended delocalisation of electron inside the ring, but ring has $(4 n+2) \pi$ electrons, i.e., $6 \pi$ electrons. This makes compound aromatic and most stable.

Both phenol and benzoic acid are soluble in NaOH, so they cannot be separated.

C₆H₅OH + NaOH $\to$ C₆H₅ONa + H₂O

C₆H₅COOH + NaOH $\to$ C₆H₅COONa + H₂O

Benzel alcohol C₆H₅CH₂OH is not soluble in NaOH and NaHCO₃, while benzoic acid is soluble in both, so these can be separated. Both C₆H₅OH and C₆H₅CH₂OH are insoluble in NaHCO₃, so they cannot be separated. C₆H₅CH₂COOH is soluble in NaOH and NaHCO₃, whereas benzyl alcohol is not soluble in NaOH and NaHCO₃. So, they can be separated.

Only HF does not react with AgNO₃. All others react with AgNO₃ to give precipitate of AgX which is soluble in Na₂S₂O₃

$HX + AgN{O_3} \to AgX \downarrow + HN{O_3}(X = Cl,Br,I)$

$AgX + N{a_2}{S_2}{O_3} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{Soluble} \, + NaX$

The structures of the given compounds are as follows. Only aspirin has $-$COOH group.

As entropy is a state function and is additive

$\Delta {S_{X \to Z}} = \Delta {S_{X \to Y}} + \Delta {S_{Y \to Z}}$

On moving from Y to Z, the work done is zero as the volume is kept constant (isochoric process), so

${W_{X \to Y \to Z}} = {W_{X \to Y}}$

The reactions involved in the transformation are:

In HNO₃, the oxidation state of N is +5 while in NO, it is +2. In N₂, it is zero and in NH₄Cl, the oxidation state of nitrogen is $-$3. So, the decreasing order of oxidation state of nitrogen is

HNO₃ > NO > N₂ > NH₄Cl

8 X atoms present at the corners.

Atoms contribute to 1 unit cell $= {1 \over 8} \times 8 = 1$

6 X atoms present at the face centres.

Atoms contribute to 1 unit cell $= 6 \times {1 \over 2} = 3$

Total X atoms $= 3 + 1 = 4$

4 M atoms present at edge centres.

Atoms present in 1 unit cell $= 4 \times {1 \over 4} = 1$

1 M atom present at body centre and it contribute completely to 1 unit cell.

Thus, total M atoms in one unit cell $= 1 + 1 = 2$

Ratio is $M:X::2:4::1:2$

Thus, empirical formula is $M{X_2}$.

To calculate the volume of the 29.2% (w/w) HCl stock solution needed to prepare a 200 mL solution of 0.4 M HCl, we need to use several steps involving concentration and density conversions.

First, we calculate the mass of HCl that is contained in the 200 mL of a 0.4 M solution:

$Mass = Molarity \times Volume \times Molecular\ Weight$

$Mass = 0.4 \ mol/L \times 0.200 \ L \times 36.5 \ g/mol$

Note that we convert the volume from mL to L to match the units of molarity (mol/L).

Now, we calculate it:

$Mass = (0.4 \times 0.200 \times 36.5) \ g$

$Mass = 0.08 \times 36.5 \ g$