JEE Advanced 2013 Paper 1 Question Paper with Solutions

The speed of light in a medium is given by the equation :

$\frac{𝐶}{𝑉}=\frac{{𝜆}_0}{𝜆}=𝜇$

Where :

• V is the velocity of light in the medium
• λ is the wavelength of light in the medium
• λ₀ is the wavelength in free space
• μ is the refractive index of the medium

Given :

$𝜆=\frac{2}{3}{𝜆}_0$

Using the given data :

$\begin{array}{cc}𝑢& =-24\text{ m},\\ \\ {ℎ}_{𝑖}& =\frac{1}{3}{ℎ}_0\end{array}$

To find the magnification (m) :

$𝑚=\frac{{ℎ}_{𝑖}}{{ℎ}_0}=\frac{1}{3}=\frac{𝑣}{𝑢}$

Hence,

$𝑣=\frac{𝑢}{3}$

Given :

$𝑣=8\text{ m},𝑢=-24\text{ m}$

Applying the lens formula :

$\frac{1}{𝑣}-\frac{1}{𝑢}=\frac{1}{𝑓}=(𝜇-1)(\frac{1}{{𝑅}_1}-\frac{1}{{𝑅}_2})$

Since the plane surface has an infinite radius :

$\frac{1}{\infty }=0$

Therefore :

$\frac{1}{8}-\frac{1}{-24}=(1.5-1)(\frac{1}{𝑅}-0)$

$\Rightarrow \frac{1}{6}=\frac{0.5}{𝑅}$

Solving for R :

$\Rightarrow 𝑅=3\text{ m}$

The work done on a particle of mass m by a force

$\overrightarrow{𝐹}=𝐾[\frac{𝑥}{{({𝑥}^2+{𝑦}^2)}^{3/2}}\widehat{𝑖}+\frac{𝑦}{{({𝑥}^2+{𝑦}^2)}^{3/2}}\widehat{𝑗}]$

(with K being a constant with appropriate dimensions), when the particle is taken from the point (a,0) to the point (0,a) along a circular path of radius a about the origin in the x-y plane is:

Let $\overrightarrow{𝑟}=𝑥\widehat{𝑖}+𝑦\widehat{𝑗}$

Therefore, $𝑑\overrightarrow{𝑟}=𝑑𝑥\widehat{𝑖}+𝑑𝑦\widehat{𝑗}$ and $𝑟=\sqrt{\surd {𝑥}^2+{𝑦}^2}$

Work done by the force in moving a particle from A to B is

$\begin{array}{cc}𝑊& ={\int }_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}\overrightarrow{𝐹}\cdot 𝑑\overrightarrow{𝑟}\\ \\ & ={\int }_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}𝐾[\frac{𝑥}{({𝑥}^2+{𝑦}^2{)}^{3/2}}\widehat{𝑖}+\frac{𝑦}{({𝑥}^2+{𝑦}^2{)}^{3/2}}\widehat{𝑗}]\cdot (𝑑𝑥\widehat{𝑖}+𝑑𝑦\widehat{𝑗})\\ \\ & =𝐾{\int }_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}\frac{𝑥𝑑𝑥}{({𝑥}^2+{𝑦}^2{)}^{3/2}}+\frac{𝑦𝑑𝑦}{({𝑥}^2+{𝑦}^2{)}^{3/2}}\\ \\ & =𝐾{\int }_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}\frac{1}{({𝑥}^2+{𝑦}^2{)}^{3/2}}[𝑑\left(\frac{{𝑥}^2}{2}\right)+𝑑\left(\frac{{𝑦}^2}{2}\right)]\\ \\ & =𝐾{\int }_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}\frac{1}{2({𝑥}^2+{𝑦}^2{)}^{3/2}}𝑑({𝑥}^2+{𝑦}^2)\\ \\ & =𝐾{\int }_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}\frac{1}{2{𝑟}^3}𝑑\left({𝑟}^2\right)=𝐾{\int }_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}\frac{2𝑟𝑑𝑟}{2{𝑟}^3}=𝐾{\int }_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}\frac{𝑑𝑟}{{𝑟}^2}(\because 𝑟=\sqrt{\surd {𝑥}^2+{𝑦}^2})\\ \\ & =𝐾{[-\frac{1}{𝑟}]}_{{𝑟}_{𝐵}}^{{𝑟}_{𝐴}}=𝐾[\frac{1}{{𝑟}_{𝐴}}-\frac{1}{{𝑟}_{𝐵}}]\end{array}$

Here, ${𝑟}_{𝐴}=𝑎$ and ${𝑟}_{𝐵}=𝑎$

Therefore, $𝑊=0$

The situation is depicted in the figure below.

At the highest point of the first particle’s trajectory, its speed is given by :

$\text{Speed of the first particle}={𝑢}_0\cos 𝛼$

For the second particle, which was thrown vertically upward, the speed at its highest point is calculated as :

$\text{Speed of the second particle}=\sqrt{\surd {𝑢}_2^0-2𝑔𝐻}$

Where $𝐻$ is the maximum height reached by the second particle :

$𝐻=\frac{{𝑢}_2^0{\sin }^2𝛼}{2𝑔}$

Therefore, the speed of the second particle at the highest point becomes :

$=\sqrt{\surd {𝑢}_2^0-\frac{2𝑔({𝑢}_2^0{\sin }^2𝛼)}{2𝑔}}=\sqrt{\surd {𝑢}_2^0-{𝑢}_2^0{\sin }^2𝛼}={𝑢}_0\cos 𝛼$

Thus, at the highest point :

According to the law of conservation of linear momentum in the horizontal direction :

$𝑚{𝑢}_0\cos 𝛼=2𝑚{𝑣}_1$

$⟹{𝑣}_1=\frac{{𝑢}_0\cos 𝛼}{2}\text{... (i)}$

In the vertical direction :

$𝑚{𝑢}_0\cos 𝛼=2𝑚{𝑣}_2$

$⟹{𝑣}_2=\frac{{𝑢}_0\cos 𝛼}{2}\text{... (ii)}$

Let $𝜃$ be the angle that the composite system makes with the horizontal immediately after the collision. Then :

$\tan 𝜃=\frac{{𝑣}_2}{{𝑣}_1}=1\text{(Using (i) and (ii))}$

$\therefore 𝜃={\tan }^{-1}1={45}^{\circ }=\frac{𝜋}{4}$

To determine the minimum velocity required for the bob of mass $𝑚$ to complete a full circle in the vertical plane, we start with the condition at the bottommost point:

$𝑢=\sqrt{\surd 5𝑔{𝑙}_1}\text{ (at point A)}$

At the highest point B, using the conservation of energy, we get:

$\frac{1}{2}𝑚{𝑢}_2^1=𝑚𝑔\left(2{𝑙}_1\right)+\frac{1}{2}𝑚{𝑣}_2^1$

Simplifying, we have:

${𝑣}_2^1={𝑢}_2^1-4𝑔{𝑙}_1$

Substituting ${𝑢}_1=\sqrt{\surd 5𝑔{𝑙}_1}$:

${𝑣}_2^1=5𝑔{𝑙}_1-4𝑔{𝑙}_1$

${𝑣}_1=\sqrt{\surd 𝑔{𝑙}_1}$

When the bob at point B elastically collides with another identical mass that is initially at rest, the velocities swap due to the nature of elastic collisions. Thus, the second bob attains the velocity:

${𝑢}_2={𝑣}_1=\sqrt{\surd 𝑔{𝑙}_1}$

For the second bob to complete its circular path, the minimum velocity requirement is:

${𝑢}_2=\sqrt{\surd 5𝑔{𝑙}_2}=\sqrt{\surd 𝑔{𝑙}_1}$

Wavelength of light is $\lambda$.

Now, using the relation for intensity, we have

$I = {I_{\max }}{\cos ^2}\left( {{\phi \over 2}} \right)$

${1 \over 2} = {\cos ^2}\left( {{\phi \over 2}} \right) \Rightarrow \cos {\phi \over 2} = {1 \over {\sqrt 2 }}$

$\Rightarrow {\phi \over 2} = {\pi \over 4} \Rightarrow \phi = {\pi \over 2}$

$\Rightarrow \phi = {\pi \over 2},{{3\pi } \over 2},{{5\pi } \over 2},{{7\pi } \over 2}$

$\phi = {{2\pi } \over \lambda }\Delta x \Rightarrow {\pi \over 2} = {{2\pi } \over 2}\Delta x \Rightarrow \Delta x = {\lambda \over 4},{{3\lambda } \over 4},{{5\lambda } \over 4}$

$\Rightarrow \Delta x = (2n + 1){\lambda \over 4}$

Number of nodes = 6

From the given equation, we can see that

$k = {{2\pi } \over \lambda } = 62.8$ m^$-$1

$\therefore$ $\lambda = {{2\pi } \over {62.8}}$ m = 0.1 m

$1 = {{5\lambda } \over 2}$ = 0.25 m

The mid-point of the string is P, an antinode

$\therefore$ maximum displacement = 0.01 m

$\omega$ = 2$\pi$f = 628 s^$-$1

$\therefore$ $f = {{628} \over {2\pi }} = 100$ Hz

But this is fifth harmonic frequency.

$\therefore$ Fundamental frequency f₀

$= {f \over 5} = 20$ Hz.

We have,

$V = {{hf} \over e} - {\phi \over e}$

Slope is h/e.

Slope is the same for both silver and sodium.

Therefore, the ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is 1 : 1.

To determine the ratio of the densities of the two non-reactive monoatomic ideal gases, we can use the ideal gas law and the given information. The ideal gas law is given by:

$𝑃𝑉=𝑛𝑅𝑇$

Where:

• P is the pressure
• V is the volume
• n is the number of moles
• R is the gas constant
• T is the temperature

Given that the gases are monoatomic and non-reactive, their molecular masses are directly proportional to their atomic masses. Let the atomic masses of the two gases be ${𝑀}_1$ and ${𝑀}_2$, with ${𝑀}_1:{𝑀}_2=2:3$.

The density ($𝜌$) of a gas is given by the formula:

$𝜌=\frac{𝑚}{𝑉}$

Where $𝑚$ is the mass and $𝑉$ is the volume. Using the ideal gas law, we can relate density to pressure and temperature.

From the ideal gas law:

$𝑃𝑉=𝑛𝑅𝑇$

Since $𝑛=\frac{𝑚}{𝑀}$, where $𝑀$ is the molar mass, we have:

$𝑃=\frac{𝑚𝑅𝑇}{𝑀𝑉}$

Rearranging to solve for density ($𝜌=\frac{𝑚}{𝑉}$):

$𝜌=\frac{𝑃𝑀}{𝑅𝑇}$

Thus, the density of a gas is directly proportional to its pressure and molar mass, and inversely proportional to the temperature. Given that the ratio of their partial pressures ${𝑃}_1:{𝑃}_2=4:3$, we can write:

${𝜌}_1=\frac{{𝑃}_1{𝑀}_1}{𝑅𝑇}$

${𝜌}_2=\frac{{𝑃}_2{𝑀}_2}{𝑅𝑇}$

Taking the ratio of the densities:

$\frac{{𝜌}_1}{{𝜌}_2}=\frac{\left(\frac{{𝑃}_1{𝑀}_1}{𝑅𝑇}\right)}{\left(\frac{{𝑃}_2{𝑀}_2}{𝑅𝑇}\right)}=\frac{{𝑃}_1{𝑀}_1}{{𝑃}_2{𝑀}_2}$

Substituting the given ratios, ${𝑃}_1:{𝑃}_2=4:3$ and ${𝑀}_1:{𝑀}_2=2:3$, into the equation, we get:

$\frac{{𝜌}_1}{{𝜌}_2}=\frac{4\times 2}{3\times 3}=\frac{8}{9}$

Therefore, the ratio of their densities is $\frac{8}{9}$, which corresponds to option D. Consequently, the answer is:

Option D: 8 : 9

The electric field at $𝑄$

${\overrightarrow{𝐸}}_{𝑄}=\frac{1}{4𝜋{𝜖}_0}[\frac{\frac{4}{3}𝜋{𝑅}^3{𝜌}_1}{(2𝑅{)}^2}\hat{ı}-\frac{\frac{4}{3}𝜋{𝑅}^3{𝜌}_2}{{𝑅}^2}\hat{ı}]=\overrightarrow{0}$

gives ${𝜌}_1/{𝜌}_2=4$

The electric field at $𝑃$,

${\overrightarrow{𝐸}}_{𝑃}=-\frac{1}{4𝜋{𝜖}_0}[\frac{\frac{4}{3}𝜋{𝑅}^3{𝜌}_1}{(2𝑅{)}^2}\hat{ı}+\frac{\frac{4}{3}𝜋(2𝑅{)}^3{𝜌}_2}{(5𝑅{)}^2}\hat{ı}]=\overrightarrow{0},$

gives ${𝜌}_1/{𝜌}_2=-32/25$.

Duration of pulse, t = 100 ns = 100 $\times$ 10^$-$9 s.

Power of the pulse, P = 30 mW = 30 $\times$ 10^$-$3 W.

Speed of light, c = 3 $\times$ 10⁸ ms^$-$1

The final momentum of the object is

$p = {E \over c} = {{P \times t} \over c} = {{30 \times {{10}^{ - 3}} \times 100 \times {{10}^{ - 9}}} \over {3 \times {{10}^8}}} = {10^{ - 17}}$ kg-ms^$-$1

Considering FBDs of spheres A and B, we get

Here, $𝑥$ is the extension produced in the spring

At equilibrium $\Sigma F_A=0$ and $\Sigma F_B=0$

$\begin{array}{cc}\frac{4}{3}𝜋{𝑅}^3\left(2𝜌\right)𝑔& =𝑘𝑥+\frac{4}{3}𝜋{𝑅}^3𝜌𝑔...........\left(𝑖\right)\\ \\ 𝑘𝑥+\frac{4}{3}𝜋{𝑅}^3\left(2𝜌\right)𝑔& =\frac{4}{3}𝜋{𝑅}^3\left(3𝜌\right)𝑔.........\left(𝑖𝑖\right)\end{array}$

Subtract (ii) from (i), we get

$\begin{array}{cc}-𝑘𝑥& =𝑘𝑥-\frac{4}{3}𝜋R^3\left(2𝜌\right)𝑔\\ \\ 𝑥& =\frac{4𝜋R^3𝜌𝑔}{3𝑘}\end{array}$

(Option A is correct)

If we consider both the spheres as a system, then at equilibrium,

$\begin{array}{cc}\text{ total weight }& =\text{ buoyant force }\\ \text{ Total weight }(A+B)& =\frac{4}{3}𝜋R^3\left(4𝜌\right)𝑔\\ \text{ Total buoyant force }& =\frac{4}{3}𝜋R^3\left(4𝜌\right)𝑔\end{array}$

Hence, sphere A (light sphere) is completely submerged (option D is correct)

Let ${\widehat v_i} = {1 \over 2}\left( {\widehat i + \sqrt 3 \widehat j} \right)$ and ${\widehat v_r} = {1 \over 2}\left( {\widehat i - \sqrt 3 \widehat j} \right)$ be the unit vectors along the incident and the reflected rays. From Snell's law, angle of incidence (say $\theta$) is equal to the angle of reflection. Thus, angle between $- {\widehat v_i}$ and ${\widehat v_r}$ is 2$\theta$.

Using, dot products of vectors, we get

$\cos (2\theta ) = - {\widehat v_i}\,.\,{\widehat v_r} = 1/2$.

Hence, $\theta = 30^\circ$.

To solve this problem, we need to use Hooke's Law, which states that the elongation (or stretch) of a material is directly proportional to the force applied and the length of the material, and inversely proportional to the cross-sectional area and the Young's modulus of the material. Mathematically, this relation can be expressed as:

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

where:

• $\Delta L$ is the elongation
• $F$ is the force applied
• $L$ is the original length of the material
• $A$ is the cross-sectional area of the material
• $Y$ is the Young's modulus of the material

In this problem, the Young's modulus ($Y$) and the force ($F$) are the same for both wires because they are made of the same material (copper) and the forces applied are equal. Therefore, the elongation is dependent on the length and cross-sectional area of each wire.

For the thick wire with length $2L$ and radius $2R$, the cross-sectional area $A_1$ is given by:

$A_1 = \pi (2R)^2 = 4\pi R^2$

For the thin wire with length $L$ and radius $R$, the cross-sectional area $A_2$ is given by:

$A_2 = \pi R^2$

Using the formula for elongation for both wires, we get:

Elongation of the thick wire ($\Delta L_\text{thick}$):

$\Delta L_\text{thick} = \frac{F \cdot 2L}{4\pi R^2 \cdot Y} = \frac{F \cdot 2L}{4\pi R^2 Y} = \frac{FL}{2\pi R^2 Y}$

Elongation of the thin wire ($\Delta L_\text{thin}$):

$\Delta L_\text{thin} = \frac{F \cdot L}{\pi R^2 \cdot Y} = \frac{FL}{\pi R^2 Y}$

To find the ratio of the elongation in the thin wire to that in the thick wire, we divide $\Delta L_\text{thin}$ by $\Delta L_\text{thick}$:

$\text{Ratio} = \frac{\Delta L_\text{thin}}{\Delta L_\text{thick}} = \frac{\frac{FL}{\pi R^2 Y}}{\frac{FL}{2\pi R^2 Y}} = \frac{FL}{\pi R^2 Y} \cdot \frac{2\pi R^2 Y}{FL} = 2$

Therefore, the ratio of the elongation in the thin wire to that in the thick wire is 2. The correct answer is:

Option C: 2.00

Consider the disc and the rings together as a system. Since the external torque ${\vec{\Gamma }}_{\text{ext }}$ is zero, the angular momentum of the system about the vertical axis through $𝑂$ is conserved, which can be expressed as:

${𝐼}_{𝑖}{𝜔}_{𝑖}={𝐼}_{𝑓}{𝜔}_{𝑓}$

Initially, the moment of inertia is due to the disc alone, and its value is:

${𝐼}_{𝑖}=\frac{1}{2}𝑀{𝑅}^2=\frac{1}{2}\times 50\times {0.4}^2=4{\mathrm{kg}\cdot m}^2$

The moment of inertia of a ring with mass $𝑚$ and radius $𝑟$ about an axis perpendicular to its center is $𝑚{𝑟}^2$. Using the theorem of parallel axes, the ring's moment of inertia about the axis of rotation is given by:

${𝐼}_{\text{ring}}=𝑚{𝑟}^2+𝑚{𝑑}^2=2𝑚{𝑟}^2$

where $𝑑=𝑟$ is the distance between the parallel axes. Thus, the moment of inertia of the complete system in the final configuration is:

$\begin{array}{cc}{𝐼}_{𝑓}& =\frac{1}{2}𝑀{𝑅}^2+2𝑚{𝑟}^2+2𝑚{𝑟}^2\\ & =\frac{1}{2}\left(50\right)(0.4{)}^2+2\left(2\right(6.25\left)\right(0.2{)}^2)=5{\mathrm{kg}\cdot m}^2\end{array}$

With no external torque acting on the system, the angular momentum is conserved:

$\begin{array}{cc}{𝐼}_{𝑖}{𝜔}_{𝑖}& ={𝐼}_{𝑓}{𝜔}_{𝑓}\\ 4\times 10& =5{𝜔}_{𝑓}\\ {𝜔}_{𝑓}& =8\mathrm{rad}/s\end{array}$

The situation is as shown in the figure.

Here, ${\overrightarrow u _1} = 4\widehat i$ ms^$-$1

${\overrightarrow u _2} = 2\left( {\sqrt 3 \widehat i + \widehat j} \right)$ ms^$-$1

When the charged particle enters in an uniform magnetic field, it travels a circular path.

Component of final velocity of particle is in positive y direction.

Centre of circle is present on positive y-axis.

So, magnetic field is present in negative z-direction.

Angle of deviation

$\tan \theta = {{{u_{2y}}} \over {{u_{2x}}}} = {2 \over {2\sqrt 3 }} = {1 \over {\sqrt 3 }}$

$\theta = {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) = 30^\circ = {\pi \over 6}$

$\omega = {\theta \over t}$

$\therefore$ $t = {\theta \over \omega } = {\theta \over {(QB/M)}} = {{M\theta } \over {QB}}$ ($\because$ $\omega = {{QB} \over M}$

$\therefore$ $B = {{M\theta } \over {Qt}} = {{M\pi } \over {6Q \times 10 \times {{10}^{ - 3}}}}$ (.... t = 10 ms = 10^$-$3 s)

$= {{100\pi M} \over {6Q}} = {{50\pi M} \over {3Q}}$

Decay constant, $\lambda = {{\ln 2} \over {{T_{1/2}}}} = {{0.693} \over {1386\,s}} = 5 \times {10^{ - 4}}\,s$

According to radioactive decay, $N = {N_0}{e^{ - \lambda t}}$

${N \over {{N_0}}} = {e^{ - 5 \times {{10}^{ - 4}} \times 80}}$ or ${N \over {{N_0}}} = {e^{ - 0.04}}$

Fraction of nuclei decayed $= {{{N_0} - N} \over {{N_0}}} = 1 - {N \over {{N_0}}}$

$= 1 - {e^{ - 0.04}} = 1 - 0.96 = 0.04 = 4\%$

To determine the speed of a particle after $𝑡=5$ seconds, given a constant power, the following steps should be taken :

Power Equation: Power is the rate at which work is done, and work is the change in kinetic energy :

$𝑃=\frac{𝑑\left(𝑊\right)}{𝑑𝑡}=\frac{𝑑}{𝑑𝑡}\left(\frac{1}{2}𝑚{𝑣}^2\right)$

Given Data :

Mass of the particle ($𝑚$) = 0.2 kg

Constant power ($𝑃$) = 0.5 W

Initial speed (${𝑣}_0$) = 0 m/s

Differentiating Kinetic Energy with Respect to Time :

$𝑃=\frac{𝑑}{𝑑𝑡}\left(\frac{1}{2}𝑚{𝑣}^2\right)=𝑚𝑣\frac{𝑑𝑣}{𝑑𝑡}$

Rearranging to Solve for $𝑣\frac{𝑑𝑣}{𝑑𝑡}$ :

$0.5=0.2\cdot 𝑣\cdot \frac{𝑑𝑣}{𝑑𝑡}$

$\frac{𝑑𝑣}{𝑑𝑡}=\frac{5}{2𝑣}$

Integrating Both Sides :

$𝑣𝑑𝑣=\frac{5}{2}𝑑𝑡$

$\int 𝑣𝑑𝑣=\int \frac{5}{2}𝑑𝑡$

$\frac{{𝑣}^2}{2}=\frac{5𝑡}{2}+𝐶$

Applying Initial Condition (when $𝑡=0$, $𝑣=0$) :

$0=\frac{5\cdot 0}{2}+𝐶$

$𝐶=0$

Solving for $𝑣$ :

$\frac{{𝑣}^2}{2}=\frac{5𝑡}{2}$

${𝑣}^2=5𝑡$

$𝑣=\sqrt{\surd 5𝑡}$

Speed After 5 Seconds :

$𝑣=\sqrt{\surd 5\cdot 5}=\sqrt{\surd 25}=5\text{m/s}$

Thus, the speed of the particle after 5 seconds is 5 m/s.

Let L and A be length and area of cross-section of each block.

$\therefore$ Thermal resistance of block 1 is

$\therefore$ ${R_1} = {L \over {\kappa A}}$

Thermal resistance of block 2 is

${R_2} = {L \over {2\kappa A}}$

In configuration I, two blocks are connected in series. So, their equivalent thermal resistance is

${R_S} = {R_1} + {R_2} = {L \over {\kappa A}} + {L \over {2\kappa A}} = {3 \over 2}{L \over {\kappa A}}$ ...... (i)

$\therefore$ Rate of heat flow in configuration I is

$\therefore$ ${Q \over t} = {{{T_1} - {T_2}} \over {{R_s}}}$ ..... (ii)

In configuration II, two blocks are connected in parallel. So, their equivalent thermal resistance is

${1 \over {{R_P}}} = {1 \over {{R_1}}} + {1 \over {{R_2}}} = {1 \over {{L \over {\kappa A}}}} + {1 \over {{L \over {2\kappa A}}}} = {{3\kappa A} \over L}$ ...... (iii)

${R_P} = {1 \over 3}{L \over {\kappa A}}$

$\therefore$ Rate of same amount of heat flow in configuration II is

$\therefore$ ${Q \over {t'}} = {{{T_1} - {T_2}} \over {{R_P}}}$ ...... (iv)

Divide (ii) by (iv), we get

${{t'} \over t} = {{{R_P}} \over {{R_S}}} = {1 \over 3}{L \over {\kappa A}} \times {{2\kappa A} \over {3L}} = {2 \over 9}$ (Using (i) and (iii))

$t' = {2 \over 9}t = {2 \over 9} \times 9s = 2s$

From the given data :

• One main scale division (MSD) is $5.15\text{ cm}-5.10\text{ cm}=0.05\text{ cm}$.


• One Vernier scale division (VSD) is $\frac{2.45\text{ cm}}{50}=0.049\text{ cm}$.

The least count (LC) of the Vernier calipers is:

$\text{LC}=1\text{ MSD}-1\text{ VSD}=0.05\text{ cm}-0.049\text{ cm}=0.001\text{ cm}$

Given the main scale reading (MSR) is 5.10 cm and the Vernier scale reading (VSR) is 24. The diameter $𝐷$ of the cylinder is calculated as:

$\begin{array}{cc}𝐷& =\text{MSR}+\text{VSR}\times \text{LC}\\ \\ & =5.10\text{ cm}+24\times 0.001\text{ cm}\\ \\ & =5.124\text{ cm}\end{array}$

The electronic configurations of the central metal ions are as follows:

The number of unpaired electrons is P = 5, Q = 3 and R = 4. From the relation $\mu = \sqrt {n(n + 2)}$, where n is the number of unpaired electrons, we have the order of spin only magnetic moment as Q < R < P.

The reaction involved is

The number of carboxylic acid groups in the product is 2.

This is a first-order reaction because t_75% = 2 $\times$ t_50%. The graph shows that the order with respect to Q is 0, so we can write the rate expression as

Rate = k[P]¹[Q]⁰

The reaction involved is 4HNO₃ $\to$ 4NO₂ + 2H₂O + O₂. The yellow-brown color appears due to the formation of NO₂.

The correct answer is Option A: Ag, Cu and Pb. Here's why :

Sulphide ores are minerals where the metal is combined with sulfur. Many metals are found in this form due to the chemical reactivity of sulfur.

• Ag (Silver): Silver is often found as the sulfide mineral argentite (Ag₂S).
• Cu (Copper): Copper is commonly found as chalcopyrite (CuFeS₂) and chalcocite (Cu₂S).
• Pb (Lead): Lead is extracted from the sulfide mineral galena (PbS).

Let's look at why the other options are incorrect :

• Option B: Ag, Mg and Pb - Magnesium (Mg) is typically found in oxide ores, not sulfide ores.
• Option C: Ag, Cu and Sn - Tin (Sn) is primarily found in oxide ores, not sulfide ores.
• Option D: Al, Cu and Pb - Aluminum (Al) is mainly extracted from bauxite ore, an oxide mineral.

The bulky groups cause steric hindrance in the formation of transition state. Therefore, higher homologues (Q) are less reactive than lower homologues (P). In compound (S), the transition state is highly stabilized by Ph-(C=O)- group, so it has highest rate of reaction towards S_N2.

The relative reactivities towards S_N2 reaction S > P > R > Q.

Comparatively phenol is a weaker acidic than carbonic acid (H₂CO₃), and hence does not liberate CO₂ on treatment with aqueous NaHCO₃ solution. The rest are more acidic than H₂CO₃, and so liberate CO₂.

When dealing with solutions and thermodynamics, particularly with ideal solutions, several key aspects must be considered to understand the behavior of the system and its interactions with the surroundings. Let's analyze each option in the context of forming an ideal solution from benzene and naphthalene at room temperature:

Option A: $\Delta G$ is positive

For an ideal solution, the Gibbs free energy change ($\Delta G$) upon mixing can be described as follows:

$\Delta G = \Delta H - T\Delta S$

An ideal solution has the characteristic that there is no enthalpy change upon mixing ($\Delta H = 0$) because the intermolecular forces between unlike molecules are assumed to be equal to those between like molecules. Additionally, when two or more components form an ideal solution, the entropy of the system ($\Delta S$) increases due to the mixing of different molecules, which is a manifestation of increased randomness or disorder.

Since $\Delta H = 0$ and $\Delta S > 0$ for an ideal solution, applying these to the Gibbs free energy equation yields:

$\Delta G = 0 - T\Delta S = -T\Delta S$

Given that $T > 0$ and $\Delta S > 0$, $\Delta G$ will ultimately be negative, indicating a spontaneous process. Therefore, Option A is false.

Option B: $\Delta S_{system}$ is positive

As mentioned earlier, the entropy of the system increases in the formation of an ideal solution. This is because the different molecules of benzene and naphthalene mix, increasing the randomness or disorder of the system. Therefore, Option B is true.

Option C: $\Delta S_{surroundings}$ = 0

In the formation of an ideal solution, there is no heat exchange with the surroundings since $\Delta H = 0$. Without a heat exchange, there is no change in the entropy of the surroundings, as entropy change in the surroundings is generally associated with heat exchange according to the relation $\Delta S = \frac{q}{T}$ for reversible processes. Therefore, Option C is true.

Option D: $\Delta H$ = 0

In the context of an ideal solution, it is assumed that the enthalpy change ($\Delta H$) is zero because the energy required to break interactions among similar molecules is exactly balanced by the energy released from the formation of new interactions between different types of molecules. This implies no net heat effect. Therefore, Option D is true.

In summary, Options B, C, and D are true, while Option A is false for the ideal solution of benzene and naphthalene at room temperature.

Methylene blue adsorbs onto activated charcoal from its aqueous solution at 25°C in a process known as physical adsorption (physisorption).

(i) Physical adsorption of methylene blue on activated charcoal can occur at all temperatures, but it is more pronounced at lower temperatures. This means Option (A) is incorrect.

(ii) In the bulk, forces between molecules are balanced. However, molecules on the surface experience unbalanced or residual attractive forces, leading to higher surface energy. These unbalanced forces are responsible for van der Waals attractions between the adsorbent (activated charcoal) and the adsorbate (methylene blue). The formation of these weak attractions reduces the energy of surface molecules, resulting in a decrease in the system's enthalpy. Therefore, Option (B) is correct.

(iii) The extent of physisorption decreases as temperature increases because the weak van der Waals forces weaken at higher temperatures, reducing the adsorbate's ability to adhere to the adsorbent surface. Consequently, Option (C) is incorrect.

(iv) Physical adsorption is a reversible process. As the temperature decreases, the attractive forces between the adsorbent and adsorbate strengthen, increasing the extent of physisorption. Thus, Option (D) is incorrect.

To solve this problem, we will use the de Broglie wavelength formula, which is given by

$\lambda = \frac{h}{mv}$

where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, $m$ is the mass of the particle, and $v$ is the velocity of the particle.

However, it's more relevant to use the form of the de Broglie equation that involves temperature, given that kinetic energy ($KE$) at a particular temperature is linked to the velocity of the gas particles. The kinetic energy for a gas particle can be calculated using the equation:

$KE = \frac{3}{2}k_{B}T$

where $k_B$ is the Boltzmann constant and $T$ is the temperature in Kelvin. The velocity $v$ of a gas particle can be related to its kinetic energy by the equation:

$KE = \frac{1}{2}mv^2$

Solving for $v$ gives:

$v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{3k_BT}{m}}$

Substituting $v$ in the de Broglie wavelength formula gives:

$\lambda = \frac{h}{m\sqrt{\frac{3k_BT}{m}}} = \frac{h}{\sqrt{3mk_BT}}$

Given that we are comparing helium (He) and neon (Ne) gases, the relative de Broglie wavelengths can be related by the mass of the particles and their temperatures. For helium and neon, respectively, this becomes:

$\lambda_{He} = \frac{h}{\sqrt{3m_{He}k_BT_{He}}}$

$\lambda_{Ne} = \frac{h}{\sqrt{3m_{Ne}k_BT_{Ne}}}$

Given the temperatures are –73°C for He and 727°C for Ne, we first convert these temperatures to Kelvin:

$T_{He} = 200K \quad (\text{-73°C + 273})$

$T_{Ne} = 1000K \quad (\text{727°C + 273})$

The ratio of the de Broglie wavelength of He to Ne, $\frac{\lambda_{He}}{\lambda_{Ne}}$, would thus be:

$\frac{\lambda_{He}}{\lambda_{Ne}} = \frac{\frac{h}{\sqrt{3m_{He}k_BT_{He}}}}{\frac{h}{\sqrt{3m_{Ne}k_BT_{Ne}}}} = \sqrt{\frac{m_{Ne}T_{Ne}}{m_{He}T_{He}}}$

Substituting the masses ($m_{He}=4$ a.m.u. and $m_{Ne}=20$ a.m.u., with $1$ a.m.u. approximately equal to $1.660539 \times 10^{-27}$ kg, though this conversion is not strictly necessary for a ratio where units cancel out) and the temperatures:

$\frac{\lambda_{He}}{\lambda_{Ne}} = \sqrt{\frac{20 \times 1000}{4 \times 200}} = \sqrt{\frac{20000}{800}} = \sqrt{25} = 5$

Therefore, the value of $M$, which is the multiplier for the de Broglie wavelength of He gas at –73°C compared to that of Ne at 727°C, is 5.

The structure of melamine is :

Each nitrogen has one lone pair of electrons. Number (no.) of nitrogen in a molecule $=6$

Total no. of lone pairs in melamine

$$\begin{aligned} & =\text { No. of nitrogen } \times \text { lone pair } \\ & =6 \times 1=6 \end{aligned}$$

Hence, total number of lone pair on nitrogen is 6.

To find the standard enthalpy of combustion of glucose, we will first write the balanced chemical equation for the combustion of glucose. The chemical formula for glucose is $C_6H_{12}O_6$. Combustion involves the reaction of a substance with oxygen to produce carbon dioxide (CO₂) and water (H₂O) typically. The reaction for the combustion of glucose can be represented as:

Each molecule of glucose (C6H12O6) produces six molecules of CO₂ and six molecules of H₂O. Now, we use the given standard enthalpies of formation (ΔH_f°) for these substances to find the enthalpy of the reaction. The enthalpy of a reaction can be calculated using the formula:

For the given reaction:

• ΔH_f° of CO₂(g) = –400 kJ/mol
• ΔH_f° of H₂O(l) = –300 kJ/mol
• ΔH_f° of glucose(s) = –1300 kJ/mol

The sum of the standard enthalpies of formation of the products:

The sum of the standard enthalpies of formation of the reactants (glucose as only part contributing with enthalpy value other than zero):

Therefore, the standard enthalpy of combustion of glucose is:

So, -2900 kJ of energy is released per mole of glucose combusted at standard conditions. The standard enthalpy of combustion per gram of glucose can be calculated using the molar mass of glucose. Glucose has a molar mass of approximately 180 g/mol.

Thus, the correct answer is -16.11 kJ per gram of glucose, which corresponds to option C –16.11 kJ.

${[Co{(N{H_3})_4}C{l_2}]^ + }$ and ${[Pt{(N{H_3})_2}({H_2}O)Cl]^ + }$ are octahedral and square planar complexes, which show geometrical isomerism. $[Pt{(N{H_3})_3}(N{O_3})]Cl$ and $[Pt{(N{H_3})_3}Cl]Br$ show ionisation isomerism.

The reactions involved are

(A)

(B)

(C) ${(N{H_4})_2}C{O_3}\buildrel \Delta \over \longrightarrow 2N{H_3} + C{O_2} + {H_2}O$

(D)

Let's break down this problem step-by-step:

Understanding the Reaction:

The hydrolysis of methyl acetate (MeCOOCH₃) is catalyzed by both strong acids (like HX) and weak acids (like HA). The reaction proceeds as follows:

$𝑀𝑒𝐶𝑂𝑂𝐶{𝐻}_3+{𝐻}_2𝑂\underset{𝑜𝑟𝐻𝐴}{\overset{{𝐻}^{+}𝑜𝑟𝐻𝑋}{\to \leftarrow }}𝑀𝑒𝐶𝑂𝑂𝐻+𝐶{𝐻}_3𝑂𝐻$

Rate of Reaction and Acid Strength:

The rate of acid-catalyzed hydrolysis depends on the concentration of H⁺ ions.

• Strong acids ionize completely. So, a 1M solution of HX provides 1M H⁺ ions.


• Weak acids ionize partially. So, a 1M solution of HA will provide significantly less than 1M H⁺ ions.

Information from the Problem:

We are told that the initial rate of hydrolysis with HA is 1/100^th of that with HX. Since the rate is directly proportional to [H⁺], this means the [H⁺] from HA is 1/100^th of that from HX.

Calculating K_a:

1. [H⁺] from HX: Since HX is a strong acid and fully ionizes, [H⁺] = 1M.


2. [H⁺] from HA: This is 1/100^th the [H⁺] from HX, so [H⁺] = (1/100) \* 1M = 10^-2 M.


3. Setting up the K_a expression:

${𝐾}_{𝑎}=\frac{\left[{𝐻}^{+}\right]\left[{𝐴}^{-}\right]}{\left[𝐻𝐴\right]}$

Since the initial concentration of HA is 1M and it ionizes to a small extent, we can approximate:

• [HA] ≈ 1 M


• [H⁺] = [A^-] = 10^-2 M

1. Calculating K_a:

${𝐾}_{𝑎}=\frac{\left({10}^{-2}\right)\left({10}^{-2}\right)}{1}=1\times {10}^{-4}$