IIT-JEE 2007 Paper 1 Question Paper with Solutions

Given, that x is greater than 2 $\Omega$, when the bridge is balanced then,

$\frac{𝑅}{𝑙}=\frac{𝑥}{100-𝑙}$ ..... (i)

(from metre bridge balanced bridge formula)

$\Rightarrow 100𝑅-𝑅𝑙=𝑙𝑥$

$\Rightarrow 200-2𝑙=𝑙𝑥(\because 𝑅=2\Omega )$

$𝑙=\frac{200}{𝑥+2}$ ..... (ii)

Where, $𝑙$ = length of segment of wire from one end where null point is obtained.

Now, these resistances are interchanged, the Jockey shifts 20 cm. So,

$\frac{𝑥}{𝑙+20}=\frac{2}{80-𝑙}$ ...... (iii)

From equation (iii) we have,

$80𝑥=𝑙(𝑥+2)+40$

$\Rightarrow 80𝑥=\left(\frac{200}{𝑥+2}\right)(𝑥+2)+40$

$\Rightarrow 𝑥=\frac{240}{80}=3\Omega$

$\therefore 𝑥=3\Omega$

About forming an inverted image: A concave mirror will make an upside-down (inverted) image only when the object is placed farther from the mirror than its focal point. This means the distance $𝑥$ from the object to the pole should be bigger than the focal length $𝑓$ (so, $𝑥>𝑓$).

When the object and image overlap: If you place the object at $𝑥=2𝑓$ (twice the focal length), the image will appear at the same spot as the object. At this exact spot, moving your eye side to side does not make the image and object swap positions. This means you cannot see “parallax.”

Understanding parallax with two points: Imagine you have two points in space, $𝐴$ and $𝐵$. $𝐴$ is closer to your eye, $𝐵$ is farther away. If your eye is at point $𝐸$ on the line joining $𝐴$ and $𝐵$, the following image shows this:

How parallax works: If you move your eye in one direction (say, to the left), the point that is closer to you ($𝐴$) will appear to move in the opposite direction (to the right) compared to the background ($𝐵$). The point that is farther away ($𝐵$) will seem to move in the same direction as your eye (to the left).

For example, if you move your eye from $𝐸$ to ${𝐸}_1$ (leftwards), the line from your eye to the far point $𝐵$ changes from $𝐸𝐵$ to ${𝐸}_1𝐵$. The near point $𝐴$ looks like it shifts to the right compared to this line. The line from your eye to the near point $𝐴$ changes from $𝐸𝐴$ to ${𝐸}_1𝐴$, and the far point $𝐵$ now seems to move to the left relative to $𝐴$.

You can try this! Put two fingers in front of your eyes (one closer, one farther), and move your head. You’ll see the closer finger appears to shift more, in the opposite direction you move your eye.

If the two points ($𝐴$ and $𝐵$) are at the same spot (they “coincide”), you won’t see any parallax. There is no shifting effect between them.

What happens in the mirror experiment: In this experiment, when you move your eye to the left, the image of the pin seems to move to the right of the real object pin. This shows the image is between your eye and the object pin. That means the object is closer to the mirror’s pole $𝑃$ than its image is. For this to happen, the object must be between the focal length $𝑓$ and twice the focal length $2𝑓$. In other words, $𝑓<𝑥<2𝑓$.

The acceleration of mass 'm' is due to the force $𝑇\cos 𝜃$, from 2^nd law of motion we have,

$𝑇\cos 𝜃=𝑚𝑎$

$\Rightarrow 𝑎=\frac{𝑇\cos 𝜃}{𝑚}$ ..... (i)

and also, $𝐹=2𝑇\sin 𝜃$

$\Rightarrow 𝑇=\frac{𝐹}{2\sin 𝜃}$ ..... (ii)

From (i) & (ii), we have,

$𝑎=\left(\frac{𝐹}{2\sin 𝜃}\right).\frac{\cos 𝜃}{𝑚}$

$𝑎=\frac{𝐹}{2𝑚\tan 𝜃}$

$\therefore$ $𝑎=\frac{𝐹}{2𝑚\frac{\sqrt{\surd {𝑎}^2-{𝑥}^2}}{𝑥}}$

$\therefore$ $𝑎=\frac{𝐹}{2𝑚}.\frac{𝑥}{\sqrt{\surd {𝑎}^2-{𝑥}^2}}$

Let the radii of the inner and outer cylinders be $r_1$ and $r_2$, respectively. The problem has cylindrical symmetry and Gauss's law can be applied to find the electric field $E$.

In the case (A), surface charge density $\sigma$ is given to the inner cylinder. Take a cylinder of radius $r\left(r_1<\right. r<r_2$ )="" and="" length="" $l$="" as="" gaussian="" surface.="" apply="" gauss's="" law="" to="" get="" the="" electric="" field<="" p="">

$E(2 \pi r l)=\frac{q_{\mathrm{enc}}}{\epsilon_0}=\frac{\sigma\left(2 \pi r_1 l\right)}{\epsilon_0}, \quad \text { i.e., } \quad E=\frac{\sigma r_1}{\epsilon_0 r} .$

Potential difference between the two cylinders is given by

$\Delta V=-\int_{r_1}^{r_2} E \mathrm{~d} r=-\frac{\sigma r_1}{\epsilon_0} \ln \left(\frac{r_2}{r_1}\right)$

In the case (B), charge density $\sigma$ is given to the outer cylinder. Apply Gauss's law to get $E=0$ (because $q_{\mathrm{enc}}=0$ ) and hence $\Delta V=0$.

In the case (C), uniform linear charge density $\lambda$ is kept along the common axis of cylinders. Apply Gauss's law as in case (A) to get the electric field

$E(2 \pi r l)=\lambda l / \epsilon_0, \quad \text { i.e., } \quad E=\lambda /\left(2 \pi \epsilon_0 r\right) .$

Potential difference between the two cylinders is

$\Delta V=-\int_{r_1}^{r_2} E \mathrm{~d} r=-\frac{\lambda}{2 \pi \epsilon_0} \ln \left(\frac{r_2}{r_1}\right)$

In the case (D) charge density $\sigma$ is given to both the cylinders. Using the argument similar to the case (A), we get, $\Delta V=-\frac{\sigma r_1}{\epsilon_0} \ln \left(\frac{r_2}{r_1}\right)$.

Net charge on sphere = $-20+20$ = 0

When we brought +q charge near neutral conducting sphere induction no produced.

So, net charge on sphere should be zero.

When +q is placed option (d).

Outside the conducting sphere, rearrange of charge taken place on sphere.

Step 1: Initial State (Switch Open)

Let ${𝑞}_1$ and ${𝑉}_1$ be the charge and voltage on capacitor ${𝐶}_1=3𝜇F$. Let ${𝑞}_2$ and ${𝑉}_2$ be the charge and voltage on capacitor ${𝐶}_2=6𝜇F$.

When the switch $𝑆$ is open, the two capacitors are connected in series. This means that the charge stored on each capacitor is the same. So, ${𝑞}_1={𝑞}_2$.

The voltages on the capacitors add up to the battery voltage: ${𝑉}_1+{𝑉}_2=9V$

The charges and voltages on capacitors in series also follow this equation: ${𝐶}_1{𝑉}_1={𝐶}_2{𝑉}_2$ This means the charge on each capacitor is the same since ${𝑞}_1={𝐶}_1{𝑉}_1$ and ${𝑞}_2={𝐶}_2{𝑉}_2$.

Solving these two equations gives:

• ${𝑉}_1=6V$, ${𝑞}_1=3𝜇F\times 6V=18𝜇C$
• ${𝑉}_2=3V$, ${𝑞}_2=6𝜇F\times 3V=18𝜇C$

Total charge on the side connected to $𝑋$ is $-{𝑞}_1+{𝑞}_2=0$. No net charge has moved yet.

Step 2: After Closing the Switch

When the switch is closed, the resistors ${𝑅}_1=3\Omega$ and ${𝑅}_2=6\Omega$ are connected in series with the capacitors and battery.

Total resistance is ${𝑅}_1+{𝑅}_2=9\Omega$.

The current from the battery is: $𝐼=𝑓𝑟𝑎𝑐9V9\Omega =1A$

The voltage drop across each resistor is:

• ${𝑉}_{\prime }^1=𝐼\times {𝑅}_1=1A\times 3\Omega =3V$
• ${𝑉}_{\prime }^2=𝐼\times {𝑅}_2=1A\times 6\Omega =6V$

The new charges on the capacitors are:

• ${𝑞}_{\prime }^1={𝐶}_1\times {𝑉}_{\prime }^1=3𝜇F\times 3V=9𝜇C$
• ${𝑞}_{\prime }^2={𝐶}_2\times {𝑉}_{\prime }^2=6𝜇F\times 6V=36𝜇C$

Final Calculation

The total charge on the plates connected to $𝑋$ is now $-{𝑞}_{\prime }^1+{𝑞}_{\prime }^2=-9+36=27𝜇C$.

This means that $27𝜇C$ of charge flows from $𝑌$ to $𝑋$ when the switch is closed.

a reflected ray and refracted ray and the angle b/w them would be less than (180${}^{\circ }$ - 2$𝜃$).

Iodine and Yttrium are medium sized nuclei and therefore, have more binding energy per nucleon as compared to uranium which has a big nuclei and less binding energy per nucleon. So that Iodine and Yttrium are more stable and therefore possess less energy and less restmass. When uranium nuclei explodes, it will convert into I and Y nuclei having kinetic energies.

For largest $𝜆$ in UV $\to$ 122 nm

For $\Delta$E${}_{\min }$,

$\frac{1}{{𝜆}_{𝑙𝑦𝑚𝑎𝑛}}={𝑅}_{𝑐}\left(1-\frac{1}{4}\right)$

[$\because$ $\frac{1}{𝜆}={𝑅}_{𝑐}\left(\frac{1}{{𝑛}_2^2}-\frac{1}{{𝑛}_2^1}\right)$]

$\frac{1}{{𝜆}_{𝑙𝑦𝑚𝑎𝑛}}=\frac{3{𝑅}_{𝑐}}{4}$ ..... (i)

R${}_{𝑐}$ = Rydber constant = 1.097 $\times$ 10${}^7$ m${}^{-1}$

For min $𝜆$ in IR region, n = D $\to$ n = 3

$\frac{1}{{𝜆}_{𝐼𝑅}}=\frac{{𝑅}_{𝑐}}{9}$ ...... (ii)

On dividing eq. (i) by (ii)

$\frac{1}{{𝜆}_{𝑙𝑦𝑚𝑎𝑛}}\times {𝜆}_{𝐼𝑅}=\frac{1}{\frac{3{𝑅}_{𝑐}}{4}}\times \frac{9}{{𝑅}_{𝑐}}=\frac{3{𝑅}_{𝑐}}{4}\times \frac{9}{{𝑅}_{𝑐}}=\frac{27}{4}$

$\therefore$ ${𝜆}_{𝐼𝑅}=\frac{27}{4}\times 122$ nm = 823.5 nm

Case I:

Case II : Tilted at 30${}^{\circ }$.

$𝑁=𝑚𝑔\cos {30}^{\circ }$

$=𝑚𝑔\frac{\sqrt{\surd 3}}{2}$

${𝑓}_2=𝜇𝐵=𝑟𝑚𝑔\frac{\sqrt{\surd 3}}{2}$

${𝑓}_1=\frac{𝑙}{𝑁}=\frac{𝑙}{𝑚𝑔\frac{\sqrt{\surd 3}}{2}}$

Case III : $\Delta$E $=\frac{1}{2}𝑚{𝑣}^2-𝑚𝑔ℎ$ = Decreases

Option (C) is correct because friction is depend on surface.

Case I :

Statement 1 : In the 1^st case the mechanical energy is completely converted into heat energy because of friction.

Case II :

M.E. is converted into heat due to friction but another part of M.E. is retained in the form of P.E. of block. Therefore statement I is correct.

Statement 2 : This is a wrong statement because the coefficient of friction between block and surface does not depend on angle of inclination.

Statement 1 : For an elastic collision, the coefficient of restitution = 1

$𝑒=\mid \frac{{𝑣}_2-{𝑣}_1}{{𝑢}_1-{𝑢}_2}\mid$

$\Rightarrow |{𝑣}_2-{𝑣}_1|=|{𝑢}_1-{𝑢}_2|$

Relative velocity after collision is equal to relative velocity before collision. But in the statement relative speed is given.

Statement 2 : Linear momentum remains conserved in an elastic collision. This statement is true.

Hint :

Relative velocity before collision $=|{𝑢}_1-{𝑢}_2|$

Relative velocity after collision $=|{𝑣}_1-{𝑣}_2|$

Elastic collision $=𝑒=1$

$𝑒=\mid \frac{{𝑣}_1-{𝑣}_2}{{𝑢}_1-{𝑢}_2}\mid =1$

$\Rightarrow |{𝑣}_1-{𝑣}_2|=|{𝑢}_1-{𝑢}_2|$

So statement 1 false

$\sum {𝐹}_{𝑒𝑥𝑡}=0$, So momentum should be conserved.

Statement 2 is true.

For disc A,

$\frac{1}{2}𝑘{𝑥}_2^1=\frac{1}{2}𝐼(2𝜔{)}^2$

$\Rightarrow 𝑘{𝑥}_2^1=2𝐼{𝜔}^2$ ..... (i)

For disc B,

$\frac{1}{2}𝑘{𝑥}_2^2=\frac{1}{2}2𝐼{𝜔}^2$

$\Rightarrow 𝑘{𝑥}_2^2=𝐼{𝜔}^2$ ..... (ii)

On dividing eq. (i) by (ii)

$\frac{𝑘{𝑥}_2^1}{𝑘{𝑥}_2^2}=\frac{2𝐼{𝜔}^2}{𝐼{𝜔}^2}$

$\frac{{𝑥}_1}{{𝑥}_2}=\sqrt{\surd 2}$

When disc B is brought in contact with disc A. Let w' be the final angular velocity of both the disc rotating together.

Applying conservation of angular momentum for the two disc system.

$\Rightarrow 𝐼\left(2𝜔\right)+2𝐼\left(𝜔\right)=(𝐼+2𝐼){𝜔}^{\prime }$

$\Rightarrow {𝜔}^{\prime }=\frac{4}{3}𝜔$

Torque on disc A,

${𝑇}_{𝐴}=\frac{\Delta {𝐿}_{𝐴}}{𝑡}=\frac{{𝐿}_{𝑓}-{𝐿}_{𝑖}}{𝑡}=\frac{𝐼\times \frac{4}{3}𝜔-𝐼\times 2𝜔}{𝑡}$

$=\frac{-2𝐼𝜔}{3𝑡}$

Note : The negative sign represents that the torque creates angular retardation.

Loss in kinetic energy

= (K.E.)${}_{𝑖𝑛𝑖𝑡𝑖𝑎𝑙}$ $-$ (K.E.)${}_{𝑓𝑖𝑛𝑎𝑙}$

$=\left[\frac{1}{2}𝐼{\left(2𝜔\right)}^2+\frac{1}{2}\left(2𝐼\right){𝜔}^2\right]-\left[\frac{1}{2}(𝐼+2𝐼){\left(\frac{4}{3}𝜔\right)}^2\right]$

$=[2𝐼{𝜔}^2+𝐼{𝜔}^2]-\left[\frac{3}{2}𝐼\times \frac{4}{3}\times \frac{4}{3}{𝜔}^2\right]$

$=3𝐼{𝜔}^2-\frac{8}{3}𝐼{𝜔}^2=\frac{𝐼{𝜔}^2}{3}$

When the piston is pulled out slowly, the pressure drop produced inside the cylinder is almost instantaneously neutralized by the air entering from outside into the cylinder. So, the pressure inside the cylinder is P${}_0$ throughout the slow pulling process.

Also condition for equilibrium of the piston is

Mg = (P${}_0$ $-$ P)$𝜋$R${}^2$

$\Rightarrow 𝑃=\frac{-𝑀𝑔}{𝜋{𝑅}^2}+{𝑃}_0$

Cylinder is thermally conducting, the temperature remains the same.

$\Rightarrow {𝑃}_0\times (2𝐿\times 𝜋{𝑅}^2)=𝑃𝑦\times 𝜋{𝑅}^2$

$\Rightarrow 𝑦=\frac{{𝑃}_0}{𝑃}\times 2𝐿$

$=\frac{{𝑃}_0}{\left[{𝑃}_0-\frac{𝑀𝑔}{𝜋{𝑅}^2}\right]}\times 2𝐿$

$=\left[\frac{{𝑃}_0\times 𝜋{𝑅}^2}{{𝑃}_0𝜋{𝑅}^2-𝑀𝑔}\right]\times 2𝐿$

At equilibrium, P = P

$\Rightarrow$ P = P${}_0$ + (L${}_0$ $-$ H)$𝜌$g ..... (i)

$\Rightarrow {𝑃}_0𝜋\left(𝜋{𝑅}^2{𝐿}_0\right)=𝑃\left[𝜋{𝑅}^2\right({𝐿}_0-𝐻\left)\right]$

$\Rightarrow 𝑃=\frac{{𝐿}_0{𝑃}_0}{{𝐿}_0-𝐻}$ ..... (ii)

From (i) as (ii)

$\Rightarrow \frac{{𝐿}_0{𝑃}_0}{{𝐿}_0-𝐻}={𝑃}_0+({𝐿}_0-𝐻)𝜌𝑔$

$\Rightarrow {𝐿}_0{𝑃}_0={𝑃}_0({𝐿}_0-𝐻)+({𝐿}_0-𝐻{)}^2𝜌𝑔$

$\Rightarrow 𝜌𝑔({𝐿}_0-𝐻{)}^2+{𝑃}_0({𝐿}_0-𝐻)-{𝐿}_0{𝑃}_0=0$

Reason : Unit of $𝐺{𝑀}_{𝑒}{𝑀}_{𝑠}=𝐹{𝑟}^2=𝑁{𝑚}^2$

$=𝑘𝑔\frac{𝑚}{{𝑠}^2}\times {𝑚}^2$

$=𝐾𝑔{𝑚}^3{𝑠}^{-2}$

Also (volt)(coulomb)(metre) = (Joule)(metre)

= (N $-$ m)(m) = Nm${}^2$ = Kgm${}^3$s${}^{-2}$

A $\to$ P, Q

Reason : ${𝑣}_{𝑟𝑚𝑠}=\sqrt{\surd \frac{3𝑅𝑇}{𝑀}}$

$\Rightarrow {𝑣}_2^{𝑟𝑚𝑠}=\frac{3𝑅𝑇}{𝑀}$

Unit of $\frac{3𝑅𝑇}{𝑀}$ is m${}^2$s${}^{-2}$

Also (farad)(volt)${}^2$(kg)${}^{-1}$ = (joule)(kg${}^{-}1$)

= N-mkg${}^{-1}$ = kgms${}^{-2}$mkg${}^{-1}$ = m${}^2$s${}^{-2}$

B $\to$ R, S

Reason : $𝐹=𝑞𝑣𝐵\Rightarrow {𝑣}^2=\frac{{𝐹}^2}{{𝑞}^2{𝐵}^2}$

Unit of v${}^2$ is m${}^2$s${}^{-2}$ which to further equal to FV${}^2$kg${}^{-1}$

C $\to$ R, S

Reason : Escape velocity ${𝑣}_{𝑒}=\sqrt{\surd \frac{2𝐺𝑀}{𝑅}}$

${𝑣}_2^{𝑒}=\frac{2𝐺𝑀}{𝑅}$

The unit of $\frac{𝐺𝑀}{𝑅}$ is ${𝑚}^2{𝑠}^{-2}$

D $\to$ R, S

(A) When capacitor is shorted, current will begin to flow and capacitor will start discharging. Because of discharging, magnitude of current in circuit and potential across the two ends will reduce continually until becomes zero. Thermal energy will be generated due to flow of charges.

(B) When wire is moved perpendicularly in magnetic field, due to change in flux associated with wire, a potential difference will be developed across its ends, but not current will flow as circuit is not complete and no heat will be generated.

(C) Since wire is placed in electric field, electrons inside the wire will align in such a way that net electric field inside the metal is zero. Net charges will be developed across the two ends of wire, creating a potential difference. Here, circuit is incomplete, so no current will flow and no heat will be generated.

(D) When a battery is connected, across two ends of wire, there will be constant potential difference. In this case circuit is complete, so current will start flowing and heat will be generated.

Reason : Characteristics x-ray are produced due to transition of electron from one energy level to another. Similarly, the line in the hydrogen spectrum is obtained due to transition of electron from one energy level to another.

A $\to$ P, R

Reason : In photoelectric effect electron from the metal surface are emitted upon the incidence of light of appropriate freaquency.

Note : In $𝛽$-decay, electrons are emitted from the nucleus of atom.

B $\to$ Q, S

Moseley gave a law which related frequency of emitted X-ray with the atomic number of the target material.

$\sqrt{\surd 𝑣}=𝑎(𝑧-𝑏)$

C $\to$ P

In photo electric effect, energy of photons of incident ray gets converted into K.E of emitted electron.

D $\to$ Q

Statement 1

The formula connecting $𝑢,𝑣$ and $𝑓$ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.

✅ True.

This is due to the paraxial approximation—the mirror formula $\frac{1}{𝑓}=\frac{1}{𝑢}+\frac{1}{𝑣}$ holds only when the mirror aperture is small so that rays make small angles with the principal axis. This avoids spherical aberration.

Statement 2

Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces.

❌ False.

The laws of reflection (angle of incidence = angle of reflection) are fundamental and apply to all reflecting surfaces—plane, spherical, or irregular. What changes in large spherical mirrors is that reflected rays do not all converge at a single focus due to geometry, but the law of reflection itself still holds locally at every point on the surface.

Conclusion

• Statement 1: True

• Statement 2: False

✅ Correct Option: C

Option C: Statement 1 is True, Statement 2 is False.

Since, Potential increases

If we give the constant energy release we observe constant wavelength. So the statement 1 is true.

From Einstein equation,

$\frac{ℎ𝑐}{𝜆}=𝑊+𝐾.𝐸$

W = work function (2.4) eV

If we given energy above 2.4 eV it will release photon.

Statement 2 is true.

Hint : (${𝜆}_{𝑟𝑎𝑦}$ $𝜇$ target material) does not depend on V${}_0$ (accelerating potential)

Statement 1 is True

Statement 2 is True, it follows the energy conservatioin.

But statement 2 is not correct explanation of statement 1.

Species having the same number of electrons will have the same bond order.

Number of electrons in CO are 14 (8 from oxygen and 6 from carbon).

NO${}^{-}$ has 16 electrons (7 from nitrogen, 8 from oxygen, add 1 for negative charge).

NO${}^{+}$ has 14 electrons (7 from nitrogen, 8 from oxygen, subtract 1 for positive charge).

CN${}^{-}$ has 14 electrons (7 from nitrogen, 6 from carbon, add 1 for negative charge).

N${}_2$ has 14 electrons (7 from each nitrogen).

$\begin{array}{cc}\text{ (a) }{\mathrm{Na}}_2{O}_2& \to {\mathrm{Na}}^{+}8e^{-}\text{all shells are filled. }\\ O_{2-}^2& \to \begin{array}{ccc}1{𝑠}^2& 2{𝑠}^2& 2{𝑝}^5\\ 1{𝑠}^2& 2{𝑠}^2& 2{𝑝}^5\end{array}]10{𝑒}^{-}\end{array}$

(c) $N_{2}O$

$:N=N\to \ddot{O}:$

all electrons are paired.

(d) ${\mathrm{KO}}_2\to K^{+}{O}_{-}^2$

$O_{-}^2=\begin{array}{ccc}1{𝑠}^2& 2{𝑠}^2& 2{𝑝}^4\\ 1{𝑠}^2& 2{𝑠}^2& 2{𝑝}^5\end{array}]9{𝑒}^{-}$

Zinc from zinc blende is obtained by first roasting of zinc. After roasting, a reduction process is carried out in the presence of carbon which leads to the formation of free metal that is zinc. The molecular formula of zinc blende is ZnS. Zinc blende is an ore. Any metal can be extracted by following two major steps:

1. Conversion of the ore into metallic oxide

2. Reduction of the metallic oxide into free metal

In order to get zinc from ZnS, first convert the zinc blende (ore) to zinc oxide (metallic oxide). The conversion of ore into metallic oxide can be done by calcination and roasting.

Calcination is generally utilised to convert hydrated oxides or hydroxides and carbonates into respective oxides while roasting is utilised to convert sulphide ores into respective oxides. Zinc blende is sulphide ore. Thus, the roasting process will be carried out to first convert zinc blende to Zinc oxide (ZnO).

2ZnS + 3O${}_2$ $\to$ 2ZnO + 2SO${}_2$

Now, reduction of metallic oxide to free metal is carried out. The reduction of zinc oxide in presence of carbon leads to the formation of zinc metal.

ZnO + C $\to$ Zn + CO

Concentrated HNO${}_3$ and concentrated H${}_2$SO${}_4$ will cause nitration of the ring to introduce $-$NO${}_2$ in one of the rings.

The ring which is adjacent to N atom is more electron rich than the other. This is due to +R effect of NH group.

Further, it will attack on more activated ortho or para positions. Since, ortho attack is not possible due to bulky ring present adjacent to it, there will be an attack on para position.

Alcoholic KOH helps in beta elimination in 1, 2-dibromoethane to form ethene.

The proton of ethene cannot be abstracted by a second molecule of ethene; hence, a much stronger base such as sodium amide is used to carry out another elimination to form ethyne.

Structural Isomers

P = 5 Value electrons

Steric number = 3 + 1 = 4

Geometry - Tetrahedral hybridisation $-$ sp${}^3$

P character $=\frac{3}{4}\times 100=75\%$

Step 1: Formula for Van't Hoff Factor

The Van't Hoff factor ($𝑖$) tells us how particles behave in a solution. We can find it using this formula: $𝑖=\frac{\text{Actual molecular weight}}{\text{Calculated molecular weight}}$.

Step 2: Find Actual Molecular Weight

The real molecular weight of naphthoic acid (C${}_{11}$H${}_8$O${}_2$) is 172 g/mol.

Step 3: Find Calculated Molecular Weight

We use the formula:
$\text{Calculated molecular weight}=\frac{1000\times 𝑤\times {𝐾}_{𝑓}}{𝑊\times \Delta {𝑇}_{𝑓}}$

Here,

$\Delta {𝑇}_{𝑓}$ = Drop in freezing point = 2 K

${𝐾}_{𝑓}$ = Freezing point constant = 1.72 K kg mol${}^{-1}$

$𝑤$ = Mass of naphthoic acid used = 20 g

$𝑊$ = Mass of benzene (solvent) = 50 g

Put the values in:

$=\frac{1000\times 1.72\times 20}{50\times 2}=344$ g/mol

The calculated molecular weight is 344 g/mol.

Step 4: Calculate Van't Hoff Factor

Now, use the formula from Step 1:

$𝑖=\frac{\text{Actual molecular weight}}{\text{Calculated molecular weight}}$

$𝑖=\frac{172}{344}=0.5$

$(\Delta {𝐺}^{\circ }=\Delta {𝐻}^{\circ }-𝑇\Delta {𝑆}^{\circ })$

$=-54.07\times 1000-298\times 10$ J/mol

$=-54070-2980=57050$

$\Delta {𝐺}^{\circ }=-2.303𝑅𝑇{\log }_{10}𝐾$

$-57050=-2.303\times 298\times 8.314{\log }_{10}𝐾$

$\log 𝐾=10$

Statement 1: Boron always forms covalent bonds.

→ Boron has a small size and high ionization energy. It does not easily lose electrons to form ${𝐵}^{3+}$ ions, since removing three electrons requires a lot of energy.

→ Therefore, boron compounds (like BF₃, BCl₃, BH₃, etc.) are covalent in nature.

✅ Statement 1 is True.

Statement 2: The small size of ${𝐵}^{3+}$ favors formation of covalent bond.

→ According to Fajan’s rules, small cations with high charge density have strong polarizing power, which distorts the electron cloud of the anion, leading to covalency.

→ ${𝐵}^{3+}$ is extremely small and highly charged, therefore its compounds are predominantly covalent.

✅ Statement 2 is True.

Furthermore, Statement 2 gives the correct reason why boron compounds are covalent — due to the small size and high charge of ${𝐵}^{3+}$.

✅ Correct Option: A

Answer: Option A — Both statements are true, and Statement 2 is the correct explanation for Statement 1.

H${}_3$BO${}_3$ (orthoboric acid) is a weak Lewis acid.

H${}_3$BO${}_3$ + H${}_2$O $\rightleftharpoons$ B(OH)${}_{-}^4$ + H${}^{\oplus }$

It does not donate proton rather it accepts OH${}^{-}$ form water.

So, Statement 1 is True, Statement 2 is False.

We know that hydrogen are of two types :

1. Intermolecular hydrogen bonding

2. Intramolecular hydrogen bonding

In any organic compound, if there is ortho substituent and hydrogen bonding is possible in that compound then always the bonding will be intramolecular hydrogen bonding. If the substituent is at para position, then intermolecular hydrogen bonding exists. Therefore, we can say that o-Hydroxybenzoic acid will show intramolecular benzoic acid and p-hydroxybenzoic acid will show intermolecular hydrogen bonding.

The strength of intermolecular hydrogen bonding is more than intramolecular hydrogen bonding. Therefore we can say that p-hydroxybenzoic acid because of intermolecular hydrogen bonding has a higher boiling point than o-hydroxybenzoic acid. Therefore, statement 1 is false but statement 2 is correct.

Now, if we look at the reason it states that o-hydroxybenzoic acid has intramolecular hydrogen bonding. From the above explanation we can say that this statement is true.

In colloidal surfactants, critical micelle concentration (CMC) is the concentration of surfactants above which micelles. Once micelles are formed, the polar surfactants are not free to conduct electricity. Also, a micelle is much larger than a monomer it diffuses more slowly through solution and so is a less efficient charge carrier.

Argon is an inert gas; hence, less reactive. It shield metal during welding. Argon is frequently blended with carbon dioxide (CO${}_2$), hydrogen (H${}_2$), helium (He) or oxygen (O${}_2$) to enhance the arc characteristics or facilitate metal transfer in gas metal arc welding.

Hybridisation, H = $\frac{1}{2}$(V + X $-$ C + A) = $\frac{1}{2}$ (8 + 0 $-$ 0 + 0 ) = 4. So, sp${}^3$ hybridised and has tetrahedral structure.

No. of lone pair = H $-$ X $-$ D = 4 $-$ 0 $-$ 3 = 1. As lone pair is present the electron pair geometry will be distorted tetrahedral and structure is pyramidal.

H = hybridisation no. (if 2 then sp, if 3 then sp${}^2$ etc.)

V = valence electrons of central atom.

X = no. of monovalent atom.

C = cationic charge, A = anionic charge.

D = no. of divalent atom

6XeF${}_4$ + 12H${}_2$O $\to$ 4Xe + 2XeO${}_3$ + 24HF + 3O${}_2$

XeF${}_6$ + 3H${}_2$O $\to$ XeO${}_3$ + 6HF

XeF${}_4$ and XeF${}_6$ are expected to be oxidizing because in both the reactions XeF${}_4$ and XeF${}_6$ are themselves reduced to xenon (Xe) and oxidises water to oxygen.

Moles of NaCl = 4 $\times$ 0.5 = 2 moles (Since 500 ml = 0.5 L)

2 moles of NaCl = 2 moles of Cl${}^{-}$

2Cl${}^{-}$ $\to$ Cl${}_2$ + 2e${}^{-}$

This equation clears that 2 moles of Cl${}^{-}$ yields 1 mole of Cl${}_2$ gas

Therefore, 1 mole of Cl${}_2$ gas is evolved.

Number of moles of Na${}^{+}$ = 2

2NaCl = 2Na

2Na + 2Hg $\to$ 2Na(Hg)

By electrolysis we get 2 moles of sodium which can combine with exactly 2 moles of mercury to give 2 moles of amalgam.

$\therefore$ Maximum weight of Na amalgam (assuming equimolar Na and Hg) = 46 + 400 = 446 g

Na${}^{+}$ + e${}^{-}$ $\to$ Na

Moles of Na discharged at cathode = 2

$\therefore$ The number of electron required for this purpose = 2 moles

$\therefore$ Total charge required = 2 faraday = 2 $\times$ 96500 = 193000 coulombs.

(A) The complex [Co(NH${}_3$)${}_4$(H${}_2$O)${}_2$]Cl${}_2$ shows geometrical isomers (cis-trans isomers). It is paramagnetic due to presence of unpaired electrons. The central metal (Co) ion has +2 oxidation state.

(B) The complex [Pt(NH${}_3$)${}_2$Cl${}_2$] shows geometrical isomers (cis-trans isomers). The complex is diamagnetic as all the electrons are paired. The Pt metal ion has +2 oxidation state.

(C) The complex [Co(H${}_2$O)${}_5$Cl]Cl cannot show geometrical isomerism. It is paramagnetic due to presence of unpaired electrons. The cobalt ion has +2 oxidation state.

(D) The complex [Ni(H${}_2$O)${}_6$]Cl${}_2$ is paramagnetic in nature due to presence of unpaired electrons. In [Ni(H${}_2$O)${}_6$]Cl${}_2$, oxidation state of nickel is +2

Ni${}^{2+}$ : [Ar] 3d${}^8$ 4s${}^2$;

2 unpaired electrons.

(A) Cellulose - a natural polymer of $𝛼$-D-glucose, linked by glycoside linkage.

(B) Nylon-6, 6 is a synthetic polymer of adipic acid and 1, 6-diaminohexane. The diacid is linked with diamine through amide linkage.

(C) Protein- a natural polymer of $𝛼$-amino acids where individual amino acid units are linked by amide linkage.

(D) Sucrose- has glycoside linkage, a disaccharide

(A) $𝑍=𝑃{𝑉}_{𝑚}/𝑅𝑇$ at high pressure and low temperature. The value of Z tends towards infinity at very high pressure such as 200 bar.

Equation $(𝑃+𝑎{𝑛}^2/{𝑉}^2)(𝑉-𝑛𝑏)=𝑛𝑅𝑇$ reduces to $𝑃(𝑉-𝑛𝑏)=𝑛𝑅𝑇$.

(B) For hydrogen gas value of Z = 1 at P = 0 and it increase continuously on increasing pressure. At such low pressure the gas behaves ideally, the van der Waals equation is reduced to :

$𝑃𝑉=𝑛𝑅𝑇$

(C) As can be seen from the plot of compressibility factor (Z) versus pressure (in bar), the compressibility factor for carbon dioxide gas is less than 0 at 1 atm (or 1.013 bar). This is because CO${}_2$ molecules have larger attractive forces, under normal conditions.

(D) $𝑍=𝑃{𝑉}_{𝑚}/𝑅𝑇$, at very large molar volume Z $\ne$ 1.

The van Der Waals equation for real gas:

$\left(𝑃+\frac{𝑎{𝑛}^2}{{𝑉}^2}\right)(𝑣-𝑛𝑏)=𝑛𝑅𝑇$

All very large molar volume, $\frac{𝑎{𝑛}^2}{{𝑉}^2}$ almost neglibible, hence, equation is reduced to

$𝑃(𝑣-𝑛𝑏)=𝑛𝑅𝑇$.

Since $𝛼,𝛽$ are the roots of

${𝑥}^2-𝑝𝑥+𝑟=0$

$\therefore$ $𝛼+𝛽=𝑝$ and $𝛼𝛽=𝑟$

We know that

$\frac{𝛼}{2}$ and 2$𝛽$ are the roots of the equation ${𝑥}^2-𝑞𝑥+𝑟=0$

$\therefore$ $\frac{𝛼}{2}+2𝛽=𝑞$ and $\frac{𝛼}{2}\times 2𝛽=𝑟$

Solving $𝛼+𝛽=𝑝$ and $\frac{𝛼}{2}+2𝛽=𝑞$

We get $𝛼=\frac{2}{3}(2𝑝-𝑞)$ and $𝛽=\frac{1}{3}(2𝑞-𝑝)$

$𝛼𝛽=𝑟$

$𝑟=\frac{2}{9}(2𝑝-𝑞)(2𝑞-𝑝)$

We have,

$\underset{𝑡\to 𝑥}{lim}\frac{{𝑡}^2𝑓\left(𝑥\right)-{𝑥}^2𝑓\left(𝑡\right)}{𝑡-𝑥}=1$

$\Rightarrow \underset{𝑡\to 𝑥}{lim}\frac{2𝑡𝑓\left(𝑥\right)-{𝑥}^2{𝑓}^{\prime }\left(𝑡\right)}{1-0}=1$

$\Rightarrow 2𝑥𝑓\left(𝑥\right)-{𝑥}^2{𝑓}^{\prime }\left(𝑥\right)=1$

$\Rightarrow \frac{{𝑥}^2{𝑓}^{\prime }\left(𝑥\right)-2𝑥𝑓\left(𝑥\right)}{{𝑥}^4}=\frac{-1}{{𝑥}^4}$

$\Rightarrow \frac{𝑑}{𝑑𝑥}\left(\frac{𝑓\left(𝑥\right)}{{𝑥}^2}\right)=\frac{-1}{{𝑥}^4}$ ...... (i)

On integrating both sides of this equation, we get

$\frac{𝑓\left(𝑥\right)}{{𝑥}^2}=\frac{1}{3{𝑥}^3}+𝑐$

Now, since

$𝑓\left(1\right)=1\Rightarrow 𝑐=\frac{2}{3}$