JEE Advanced 2014 Paper 1 Question Paper with Solutions

Work done by the gravitational force is ${W_g} = mgh\cos 180^\circ$

= $-$ mgh = $-$1 $\times$ 10 $\times$ 4 = $-$40 J

Work done by the applied force F

${W_F} = Fd\cos 0^\circ = Fd = 18 \times 5 = 90$ J

According to work-energy theorem

$\Delta$K = W_g + W_F

$\Delta$K = $-$40 J + 90 J = 50 J = (5 $\times$ 10) J

$\therefore$ n = 5

As the film has R₁ = R₂ = R (say), its focal length

${1 \over f} = ({n_1} - 1)\left( {{1 \over R} - {1 \over R}} \right)$


${1 \over f} = 0$ or $f = \infty$. It will not cause refraction.

Refraction will take place at air and glass cylinder. Given u = $\infty$

$- {{{n_1}} \over u} + {{{n_2}} \over v} = {{{n_2} - {n_1}} \over R}$

$- {{{n_1}} \over \infty } + {1.5 \over {{f_1}}} = {{1.5 - 1} \over R}$ or ${f_1} = 3R$

For glass-air refraction, we have

$- {{{n_2}} \over u} + {{{n_1}} \over v} = {{{n_1} - {n_2}} \over R}$

$- {{{n_2}} \over \infty } + {1 \over {{b_2}}} = {{1 - 1.5} \over R}$

or ${f_2} = - 2R$ or $\left| {{f_2}} \right| = 2R$.

When current in wires is in same direction, the magnetic fields due to two wires all in opposite direction.

From $B = {{{\mu _0}} \over {4\pi }}.{{2I} \over r}$, we get

${B_1} = {{{\mu _0}} \over {4\pi }}.2I\left[ {{1 \over {{x_1}}} - {1 \over {({x_0} - {x_1})}}} \right]$

$= {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - {x_1} - {x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$

$= {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - 2{x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$ ..... (1)

When the direction of current in two wires is opposite, field will be in the same direction.

${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{1 \over {{x_1}}} + {1 \over {({x_0} - {x_1})}}} \right]$

${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0} - {x_1} + {x_1}} \over {{x_1}({x_0} - {x_1})}}} \right]$

${B_2} = {{{\mu _0}I} \over {2\pi }}\left[ {{{{x_0}} \over {{x_1}({x_0} - {x_1})}}} \right]$

From ${{m{v^2}} \over r} = qvB$ or $v = {{qBr} \over m}$ or $r = {{mv} \over {qB}}$

$B \propto {1 \over r}$

Therefore, ${{{R_1}} \over {{R_2}}} = {{{B_2}} \over {{B_1}}} = {{{x_0}} \over {({x_0} - 2{x_1})}}$

${{{R_1}} \over {{R_2}}} = {{{x_0}/{x_1}} \over {({x_0}/{x_1}) - (2{x_1}/{x_1})}} = {3 \over {3 - 2}} = 3$

Since A observes B as moving normal to it

${v_B}\cos 30^\circ = {v_A}$

${v_B}{{\sqrt 3 } \over 2} = 100\sqrt 3$

${v_B} = 200$ m/s

Therefore, ${t_0} = {{500} \over {200\sin 30^\circ }} = 5$ s

The current flowing through the capacitor and its charge are given by

$i = {i_0}\cos \omega t$, .... (1)

$Q = \int {idt = \int {{i_0}\cos \omega t\,dt = {{{i_0}} \over \omega }\sin \omega t + k} }$ $= {{{i_0}} \over \omega }\sin \omega t$, ...... (2)

where we have used the initial condition, Q = 0 at t = 0, to get the integration constant k = 0. The charge attains the maximum value Q_max at t = $\pi$ / (2$\omega$),

Q_max = i₀/$\omega$ = 1/500 = 2 $\times$ 10^$-$3 C.

The key is switched from B to D at the time t = 7$\pi$/6$\omega$. Substitute t = 7$\pi$/6$\omega$ in equations (1) and (2) to get current and charge

i = i₀ cos(7$\pi$/6) = cos($\pi$ + $\pi$/6) = $-$ cos($\pi$/6)

$= - \sqrt 3 /2$ A,

Q_i = (i₀/$\omega$) sin(7$\pi$/6) = 2 $\times$ 10^$-$3 sin($\pi$ + $\pi$/6)

= $-$ 1 $\times$ 10^$-$3 C.

Note that negative sign in i indicates the counterclockwise direction of current and negative sign in Q_i indicates the negative charge on the upper plate.

The potential across the capacitor is ${V_C} = {{{Q_i}} \over C} = - {{1 \times {{10}^{ - 3}}} \over {20 \times {{10}^{ - 6}}}}$ = $-$ 50 V (upper plate at lower potential). Immediately after A is connected to D, the potential across the resistor is V_R = 100 V and hence the current through it is i = V_R/R = 100/10 = 10 A (counterclockwise). The potential across the capacitor when it is fully charged is equal to the battery emf. Thus, the charge on the capacitor in fully charged condition is

Q_f = V/C = 50/(20 $\times$ 10^$-$6) = 1 $\times$ 10^$-$3 C.

The sign of Q_f indicates the positive charge on the upper plate. Hence, charge that flows through the battery is

Q = Q_f $-$ Q_i = 1 $\times$ 10^$-$3 $-$ ($-$ 1 $\times$ 10^$-$3) = 2 $\times$ 10^$-$3 C.

Fringe width $\beta = {{\lambda D} \over d}$; $\lambda$₁ = 400 nm; $\lambda$₂ = 400 nm

Since $\lambda$₂ > $\lambda$₁, $\beta$₂ > $\beta$₁

Number of fringes within a distance y is given by

${m_1} = {y \over {{\beta _1}}}$

${m_2} = {y \over {{\beta _2}}}$

Since, $\beta$₂ > $\beta$₁, we get m₂ < m₁.

Position of 3^rd maxima of $\lambda$₂ :

$y' = {D \over d}(3{\lambda _2}) = 1800{D \over d}$

Position of 5^th minima of $\lambda$₁ :

$y'' = {D \over d}\left( {{{9{\lambda _1}} \over 2}} \right) = 1800{D \over d}$

Hence, y' = y''.

Angular fringe width is $\theta = {\beta \over D} = {\lambda \over d}$

Since, $\beta$₂ > $\beta$₁ $\Rightarrow$ $\theta$₂ > $\theta$₁

Given $d \propto {\rho ^a}{S^b}{b^c}$

${M^0}L{T^0} \propto {(M{L^{ - 3}})^a}{(M{T^{ - 3}})^b}{({T^{ - 1}})^c}$

${M^0}L{T^0} \propto {M^{(a + b)}}{L^{ - 3a}}{T^{ - 3b - c}}$

Equating the coefficients, we get

$a + b = 0 - 3a = 1 - 3b - c = 0$

$b = - a$ $a = - {1 \over 3} - c = 3b$

$b = {1 \over 3}c = - 3b \Rightarrow c = 1$

Therefore, $b = {1 \over n} = {1 \over 3} \Rightarrow n = 3$.

Let i₁ and i₂ be the currents as shown in the figure.


Apply Kirchhoff's law in the loop ABCDA and CEFDC to get

${i_1}{R_1} + ({i_1} - {i_2}){R_2} = {V_1}$ ..... (1)

${i_2}{R_3} - ({i_1} - {i_2}){R_2} = {V_2}$ ...... (2)

Multiply equation (1) by R₃ and (2) by R₁ and then subtract to get the current through R₂ as

$({i_1} - {i_2}) = {{{V_1}{R_3} - {V_2}{R_1}} \over {{R_1}{R_3} + {R_2}{R_3} - {R_1}{R_2}}}$.

The current through R₂ becomes zero when V₁R₃ = V₂R₁.

In a thermodynamics process, the heat supplied to the system, the increase in internal energy of the system, and the work done by the system are related by the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$.

The first law of thermodynamics for the process iaf gives

${Q_{iaf}} = {U_{iaf}} + {W_{iaf}} = ({U_f} - {U_i}) + ({W_{ia}} + {W_{af}})$ ..... (1)

Substitute Q_iaf = 500 J, U_i = 100 J, W_ia = 0 (constant volume), and W_af = 200 J in equation (1) to get U_f = 400 J.

In the process ib,

${Q_{ib}} = {U_{ib}} + {W_{ib}} = ({U_b} - {U_i}) + {W_{ib}}$ ..... (2)

Substitute U_b = 200 J, U_i = 100 J, and W_ib = 50 J in equation (2) to get Q_ib = 150 J.

In the process bf,

${Q_{bf}} = {U_{bf}} + {W_{bf}} = ({U_f} - {U_b}) + {W_{bf}}$ ..... (3)

Substitute U_f = 400 J, U_b = 200 J and W_bf = 100 J in equation (3) to get Q_bf = 300 J. Thus, Q_bf/Q_ib = 300/150 = 2.

Consider the motion of two balls with respect to rocket.

Distance travelled by ball A from left end of the chamber is

$= {{{u^2}} \over {2a}} = {{{{(0.3)}^2}} \over {2 \times 2}} = {{0.09} \over 4} \approx 0.02$ m

So, collision of two balls will take place very near to left end of the chamber.

For ball B

$S = ut{1 \over 2}a{t^2}$

$- 4 = - 0.2 \times t - {1 \over 2} \times 2 \times {t^2}$

${t^2} + 0.2t - 4 = 0$

$t = {{ - 0.2 \pm \sqrt {{{(0.2)}^2} - 4(1)( - 4)} } \over 2} = {{ - 0.2 \pm \sqrt {0.04 + 16} } \over 2}$

t = 1.9 s, $-$2.1 s

Since t can't be negative

$\therefore$ t = 1.9 s

Nearest integer is 2 s.

Also, from

$S = ut + {1 \over 2}a{t^2}$

${S_A} = 0.3t + {1 \over 2}( - 2){t^2} = 0.3t - {t^2}$

${S_B} = 0.2t + {1 \over 2}(2){t^2} = 0.2t + {t^2}$

$\because$ ${S_A} + {S_B} = 4$

$\Rightarrow$ 0.5 t = 4 or t = 8 s

The displacement of a stationary wave is given by

$y(x,t) = A\sin \left( {{{2\pi x} \over \lambda }} \right)\cos (2\pi vt)$.

The boundary conditions give node at x = 0 and anti-node at x = 3 m i.e.,

y(0, t) = 0, ....... (1)

y(3, t) = $\pm$A. ..... (2)


The fundamental frequency is given by

${v_0} = {v \over \lambda } = {v \over {4l}} = {{100} \over {4(3)}} = {{25} \over 3}$ Hz.

Thus, the waveform will satisfy equations (1) and (2), and the permissible frequencies will be odd multiples of v₀.

Resistance of initially given kettle

$R = \rho {l \over A}$

$= \rho {L \over {\pi {{(d/2)}^2}}} = {{4\rho L} \over {\pi {d^2}}}$

Resistance of two replaced kettles

${R_1} = {{\rho L} \over {\pi {d^2}}}$ and

${R_2} = {{\rho L} \over {\pi {d^2}}}$

So, ${R_1} = {R_2} = {R \over 4}$

If wires are in parallel then equivalent resistance

${R_P} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}} = {R \over 8}$ ....... (1)

If wires are in series then equivalent resistance

${R_S} = {R_1} + {R_2} = {R \over 2}$ ..... (2)

Let V be the applied voltage, m = 0.5 kg be the mass of the water, and S be the specific heat of the water. Initially, the heat produced by a resistance R₁ in time t₁ = 4 min is V²t₁/R₁. This heat is used to raise the temperature of the water by $\Delta$T = 40 K. Thus,

${V^2}{t_1}/{R_1} = mS\Delta T$. ....... (3)

Let t_s and t_p be the time taken to raise the temperature of same amount of water by $\Delta$T when the resistances are connected in series and parallel. Thus,

${V^2}{t_s}/{R_s} = mS\Delta T$ ....... (4)

${V^2}{t_p}/{R_p} = mS\Delta T$ ....... (5)

Divide equation (4) by (3) and use the equation (1) to get ${t_s} = {{{R_s}} \over {{R_1}}}{t_1} = {1 \over 2} \times 4 = 2$ min. Similarly, divide equation (5) by (3) and use the equation (2) to get ${t_p} = {{{R_p}} \over {{R_1}}}{t_1} = {1 \over 8} \times 4 = 0.5$ min.

Minimum length $= {\lambda \over 4} \Rightarrow \lambda = 4l$

Now, v = f $\lambda$ = (244) $\times$ 4 $\times$ l

as l = 0.350 $\pm$ 0.005

$\Rightarrow$ v lies between 336.7 m/s to 346.5 m/s

Now, $v = \sqrt {{{\gamma RT} \over {M \times {{10}^{ - 3}}}}}$, here M is molecular mass in gram $= \sqrt {100\gamma RT} \times \sqrt {{{10} \over m}}$.

For monoatomic gas,

$\gamma$ = 1.67

$\Rightarrow$ $v = 640 \times \sqrt {{{10} \over m}}$

For diatomic gas,

$\gamma = 1.4 \Rightarrow v = 590 \times \sqrt {{{10} \over m}}$

$\therefore$ ${v_{Ne}} = 640 \times {7 \over {10}} = 448$ m/s

${v_{Ar}} = 640 \times {{17} \over {32}} = 340$ m/s

${v_{{O_2}}} = 590 \times {9 \over {16}}=331.8$ m/s

${v_{{N_2}}} = 590 \times {3 \over 5} = 354$ m/s

$\therefore$ Only possible answer is Argon.

The difference between the two measurements by Vernier scale gives elongation of the wire caused by the additional load of 2 kg. In the first measurement, main scale reading is MSR = 3.20 $\times$ 10^$-$2 m and Vernier scale reading is VSR = 20. The least count of Vernier scale is LC = 1 $\times$ 10^$-$5 m. Thus, the first measurement by Vernier scale is

L₁ = MSR + VSR $\times$ LC

= 3.20 $\times$ 10^$-$2 + 20(1 $\times$ 10^$-$5)

= 3.220 $\times$ 10^$-$2 m.

In the second measurement, MSR = 3.20 $\times$ 10^$-$2 m and VSR = 45. Thus, the second measurement by Vernier scale is

L₂ = 3.20 $\times$ 10^$-$2 + 45(1 $\times$ 10^$-$5)

= 3.245 $\times$ 10^$-$2 m.

The elongation of the wire due to force F = 2g is

l = L₂ $-$ L₁ = 0.025 $\times$ 10^$-$2 m.

The maximum error in the measurement of l is $\Delta$l = LC = 1 $\times$ 10^$-$5 m. Young's modulus is given by Y = ${{FL} \over {lA}}$. The maximum percentage error in the measurement of Y is

${{\Delta Y} \over Y} \times 100 = {{\Delta l} \over l} \times 100 = {{1 \times {{10}^{ - 5}}} \over {0.025 \times {{10}^{ - 2}}}} \times 100 = 4\%$.

${i_g}(G + 4990) = V$

$\Rightarrow {6 \over {1000}}(G + 4990) = 30$

$\Rightarrow G + 4990 = {{30,000} \over 6} = 5000$

$\Rightarrow G = 10\,\Omega$

${V_{ab}} = {V_{cd}}$

$\Rightarrow {i_g}G = (1.5 - {i_g})S$


$\Rightarrow {6 \over {1000}} \times 10 = \left( {1.5 - {6 \over {1000}}} \right)S$

$\Rightarrow S = {{60} \over {1494}} = {{2n} \over {249}}$

$\Rightarrow n = {{249 \times 30} \over {1494}} = {{2490} \over {498}} = 5$

In triangle OZP, the distances OZ = R and OP = R sin30$^\circ$ = R/2.


The total torque by the three forces about the centre of mass O is

$\tau$ = 3F| OP | = 3F R/2.

The monent of inertia of the disc about the axis of rotation is $I = {1 \over 2}M{R^2}$. Using $\tau$ = I$\alpha$, we get

$\alpha = {\tau \over I} = {{3F} \over {MR}} = {{3 \times 0.5} \over {1.5 \times 0.5}} = 2$ rad/s².

The angular velocity after t = 1 is given by

$\omega = {\omega _0} + \alpha t = 0 + 2 \times 1 = 2$ rad/s.

Consider the balls and the platform together as a system. There is no external torque on the system about its centre. Hence, angular momentum of the system about its centre is conserved. Initial and final angular momentum of the system are

${L_i} = 0$,

${L_f} = mvr + mvr + I\omega = 2mvr + {1 \over 2}M{R^2}\omega$.

The conservation of angular momentum, L_i = L_f, gives

$\omega = - {{4mvr} \over {M{R^2}}} = - {{4(0.05)(9)(0.25)} \over {0.45{{(0.5)}^2}}} = - 4$ rad/s.

${E_1}(r) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{r^2}}}$

${E_2}(r) = {\lambda \over {2\pi {\varepsilon _0}r}}$

${E_3}(r) = {\sigma \over {2{\varepsilon _0}}}$

At $r = {r_0}$,

${E_1}({r_0}) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}}$

${E_2}({r_0}) = {\lambda \over {2\pi {\varepsilon _0}{r_0}}}$

${E_3}({r_0}) = {\sigma \over {2{\varepsilon _0}}}$

As ${E_1}({r_0}) = {E_2}({r_0}) = {E_3}({r_0})$ (Given)

$\therefore$ ${1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = {\lambda \over {2\pi {\varepsilon _0}{r_0}}} = {\sigma \over {2{\varepsilon _0}}}$

Then

${1 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = {\sigma \over {2{\varepsilon _0}}}$

or $Q = \sigma \pi r_0^2$

Hence, option (a) is incorrect.

Now, ${\lambda \over {2\pi {\varepsilon _0}{r_0}}} = {\sigma \over {2{\varepsilon _0}}}$ or ${r_0} = {\lambda \over {\pi \sigma }}$

Hence, option (b) is incorrect.

At $r = {r_0}/2$,

${E_1}({r_0}/2) = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{{({r_0}/2)}^2}}} = {4 \over {4\pi {\varepsilon _0}}}{Q \over {r_0^2}} = 4{E_1}({r_0})$

or, ${E_1}({r_0}) = {1 \over 4}{E_1}({r_0}/2)$

${E_2}({r_0}/2) = {\lambda \over {2\pi {\varepsilon _0}({r_0}/2)}} = {{2\lambda } \over {2\pi {\varepsilon _0}{r_0}}} = 2{E_2}({r_0})$

or, ${E_2}({r_0}) = {1 \over 2}{E_2}({r_0}/2)$

$\because$${E_1}({r_0}) = {E_2}({r_0})$

$\therefore$ ${1 \over 4}{E_1}({r_0}/2) = {1 \over 2}{E_2}({r_0}/2)$

or, ${E_1}({r_0}/2) = 2{E_2}({r_0}/2)$

Hence, option (c) is correct.

${E_3}({r_0}/2) = {\sigma \over {2{\varepsilon _0}}} = {E_3}({r_0})$

$\because$ ${E_2}({r_0}) = {E_3}({r_0})$

$\therefore$ ${1 \over 2}{E_2}({{{r_0}} \over 2}) = {E_3}({{{r_0}} \over 2})$

or, ${E_2}({r_0}/2) = 2{E_3}({r_0}/2)$

Hence, option (d) is incorrect.

Let A be area of each plate and d is the distance between the plates.

The given capacitor is equivalent to two capacitors in parallel with capacitances

${C_1} = {{K{\varepsilon _0}(A/3)} \over d} = {{K{\varepsilon _0}A} \over {3d}}$

${C_2} = {{{\varepsilon _0}(2A/3)} \over d} = {{2{\varepsilon _0}A} \over {3d}}$

$\therefore$ $C = {C_1} + {C_2}$

$= {{K{\varepsilon _0}A} \over {3d}} + {{2{\varepsilon _0}A} \over {3d}} = {{{\varepsilon _0}A} \over {3d}}(K + 2)$

$\therefore$ ${C \over {{C_1}}} = {{K + 2} \over K}$

Hence, option (d) is correct.

Let V be potential difference between the plates. Then

${E_1} = {V \over d}$ and ${E_2} = {V \over d}$

$\therefore$ ${{{E_1}} \over {{E_2}}} = 1$

Hence, option (a) is correct and option (b) is incorrect.

${Q_1} = {C_1}V = {{K{\varepsilon _0}A} \over {3d}}V$ and ${Q_2} = {C_2}V = {{2{\varepsilon _0}A} \over {3d}}V$

$\therefore$ ${{{Q_1}} \over {{Q_2}}} = {K \over 2}$

Hence, option (c) is incorrect.

As the rod is about to slip, wall and floor exert limiting friction on the ladder.

Case 1 : If $\mu$₁ = 0, $\sum {{{\overrightarrow \tau }_A} = \overrightarrow 0 }$

$mg\left( {{l \over 2}\cos \theta } \right) = {N_1}(l\sin \theta )$

${N_1} = {{mg\cot \theta } \over 2}$

or ${N_1}\tan \theta = {{mg} \over 2}$

and ${N_2} = mg$

Case 2 : If $\mu$₂ = 0, N₁ remain unbalanced and rod can never be in equilibrium.

Case 3 : If $\mu$₁ $\ne$ 0, $\mu$₂ $\ne$ 0

${N_1} = {f_2} = {\mu _2}{N_2}$

${N_2} + {f_1} = mg$

or ${N_2} + {\mu _1}{N_1} = mg$

or ${N_2} + {\mu _1}({\mu _2}{N_2}) = mg$

or ${N_2} = {{mg} \over {1 + {\mu _1}{\mu _2}}}$

The depression of freezing point is an important colligative property, which is influenced by the number of solute particles in a solution. When a solute dissociates or ionizes in a solution, it increases the number of particles in the solution, thereby affecting colligative properties such as the depression of freezing point. The degree of dissociation ($\alpha$) gives us a measure of the extent to which a compound dissociates into its ions. For our case, MX₂ dissociates into one M²⁺ ion and two X^- ions.

Let's initially consider 1 mole of MX₂ is present. Since the degree of dissociation ($\alpha$) is 0.5, this means half of the MX₂ dissociates into its ions, and half remains undissociated.

For the dissociation reaction:

$\text{MX}_{2} \rightarrow \text{M}^{2+} + 2\text{X}^{-}$

If the initial amount of MX₂ is 1 mole, then:

• The amount of MX₂ that dissociates = $\alpha = 0.5$ moles
• The amount of M²⁺ formed = $\alpha = 0.5$ moles (since for every mole of MX₂ that dissociates, 1 mole of M²⁺ is formed)
• The amount of X^- formed = $2\alpha = 2 \times 0.5 = 1$ mole (since for every mole of MX₂ that dissociates, 2 moles of X^- are formed)

The Van't Hoff factor (i) quantifies the effect of solute particles on the colligative properties of a solution. It is defined as the ratio of the actual number of particles in solution after dissociation to the number of formula units initially dissolved in the solvent.

$i = \frac{\text{Total number of particles after dissociation}}{\text{Number of moles of solute originally dissolved}}$

Considering the dissociation and the amounts calculated:

• Total number of particles after dissociation = (undissociated MX₂) + (M²⁺) + (X^-) = $(1 - \alpha) + \alpha + 2\alpha$
• Putting the value of $\alpha = 0.5$, we get:

  $i = \frac{(1 - 0.5) + 0.5 + 2(0.5)}{1} = \frac{1 + 1}{1} = 2$

Now, the depression in freezing point ($\Delta T_f$) is directly proportional to the molal concentration of the solute particles and Van't Hoff factor (i):

$\Delta T_f = i \cdot K_f \cdot m$

Where $K_f$ is the cryoscopic constant and $m$ is the molality of the solution.

Without ionic dissociation, the value of $i$ would have been 1 (since the solute would not have dissociated into multiple particles). However, due to the dissociation, the value of $i$ has increased to 2. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation, therefore, would be directly equal to the ratio of the Van't Hoff factors:

$\frac{\text{Observed } \Delta T_f}{\text{In the absence of ionic dissociation}} = \frac{2 \cdot K_f \cdot m}{1 \cdot K_f \cdot m} = \frac{2}{1} = 2$

This ratio shows that due to the ionic dissociation of MX₂ into M²⁺ and X^- ions with a degree of dissociation ($\alpha$) of 0.5, the observed depression in freezing point is twice the value it would have been in the absence of ionic dissociation.

$\mathrm{XeF}_4, \mathrm{BrF}_4^{-},\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right]^{2+},\left[\mathrm{PtCl}_4\right]^{2-}$ are square planar as shown below :



$\mathrm{SF}_4(\mathrm{See}-\mathrm{saw})$ as shown below:


$\mathrm{SiF}_4, \mathrm{BF}_4^{-},\left[\mathrm{FeCl}_4\right]^{2-},\left[\mathrm{CoCl}_4\right]^{2-}$ are tetrahedral as shown below :

(a) It does not undergo self ionisation in water but accepts an electron pair from water, so it behaves as weak monobasic acid

H₃BO₃ + H₂O $\rightleftharpoons$ B(OH)$_4^ -$ + H⁺

Hence, (a) is incorrect.

(b) When treated with 1, 2-dihydroxy or polyhydroxy compounds, they form chelate (ring complex) which effectively remove [B(OH)₄]^$-$ species from solution and thereby produce maximum number of H₃O⁺ or H⁺ ions, i.e., results in increased acidity.

(c) Baric acid crystallises in a layer structure in which planar triangular BO$_3^{3 - }$ ions are bonded together through hydrogen bonds.

(d) In water the pK_a value of H₃BO₃ is 9.25

H₃BO₃ + H₂O $\rightleftharpoons$ B(OH)$_4^ -$ + H⁺ ; pK_a = 9.25

So, it is weak electrolyte in water.

To understand which of the given reagents will give copper metal upon heating with $Cu_2S$, let's discuss each one based on their chemical reactions during the process:

Option A: CuFeS₂

$CuFeS_2$ (Chalcopyrite) when heated with $Cu_2S$, it can react to form copper metal as part of the copper extraction process. However, the direct reaction between $CuFeS_2$ and $Cu_2S$ is more involved and typically requires a series of steps, including smelting and conversion to get to the copper metal. Therefore, while $CuFeS_2$ is a source of copper, it's not directly because of its reaction with $Cu_2S$, but through a complex metallurgical process.

Option B: CuO

$$\begin{aligned} & 4 \mathrm{CuO} \xrightarrow{\Delta} 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{O}_2 \\ & \mathrm{Cu}_2 \mathrm{~S}+2 \mathrm{Cu}_2 \mathrm{O} \xrightarrow{\Delta} 6 \mathrm{Cu}+\mathrm{SO}_2 \end{aligned}$$

Option C: Cu₂O

Copper(I) oxide ($Cu_2O$) can also react with $Cu_2S$ to produce copper metal directly. The equation for this reaction is:

$2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2$

Here, $Cu_2O$ serves as the oxidizing agent, reducing $Cu_2S$ to copper metal, while itself being reduced to copper, and sulfur is oxidized to sulfur dioxide. This reaction is similar in principle to the reaction with $CuO$, but involves copper(I) oxide instead.

Option D: CuSO₄

$\begin{array}{r}\mathrm{CuSO}_4 \xrightarrow{720^{\circ} \mathrm{C}} \mathrm{CuO}+\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \\ 4 \mathrm{CuO} \xrightarrow{1100^{\circ} \mathrm{C}} 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{O}_2 \\ 2 \mathrm{Cu}_2 \mathrm{O}+\mathrm{Cu}_2 \mathrm{~S} \xrightarrow{\Delta} 6 \mathrm{Cu}+\mathrm{SO}_2\end{array}$

Steric effect of halogens are as follows Cl₂ < Br₂ < I₂

Electronic effect of phenolic group directs the approaching electrophile towards ortho and para positions. Tertiary butyl group has large size so it causes steric effect around aromatic nucleus. On the basis of above factors the products of the given reactions are as follows


Hence, orientation in electrophilic substitution reaction is decided by

(a) The steric effect of the halogen

(b) The steric effect of the tert-butyl group

(c) The electronic effect of the phenolic group

So, (a), (b) and (c) are correct choices.

In a galvanic cell, also known as a voltaic cell, a salt bridge plays a crucial role in maintaining electrical neutrality within the internal circuit, which is critical for the ongoing electrochemical reaction and thus the production of electrical energy. Here, we will analyze each of the given options in the context of the function of a salt bridge in a galvanic cell:

Option A: does not participate chemically in the cell reaction.

This statement is correct. The salt bridge in a galvanic cell does not directly participate in the chemical reactions occurring at the electrodes. Its primary function is to complete the electrical circuit between the cathode and anode by allowing the transfer of ions. The salt bridge contains a salt solution (usually KNO₃, KCl, or NH₄NO₃), where the ions migrate to oppose and balance the charge buildup due to the migration of electrons through the external circuit.

Option B: stops the diffusion of ions from one electrode to another.

This option can be misleading. The salt bridge does not stop the diffusion of ions from diffusing from one electrode to another but rather provides a pathway for ions to flow back and forth, which is essential to maintain the charge balance. Hence, this statement is not the best descriptor of the salt bridge's function.

Option C: is necessary for the occurrence of the cell reaction.

This option is correct. A salt bridge is necessary for the occurrence of the cell reaction because it maintains electrical neutrality in the electrochemical cell. Without the salt bridge, the flow of electrons through the external circuit would soon cease as the solutions in the anode and cathode compartments become respectively positively and negatively charged, which would stop the electrochemical reaction.

Option D: ensures mixing of the two electrolytic solutions.

This statement is incorrect. The purpose of the salt bridge is not to mix the two electrolytic solutions; in fact, it prevents them from mixing, which could otherwise result in a direct neutralization reaction that could interfere with the proper functioning of the cell. The ions in the salt bridge only migrate enough to balance the charges in the separate solutions.

In summary, the correct answers are Option A ("does not participate chemically in the cell reaction") and Option C ("is necessary for the occurrence of the cell reaction").

To determine the total number of black-colored sulphides among the given compounds, we will need to identify the color properties of the sulphides of lead (PbS), copper (CuS), mercury (HgS), manganese (MnS), silver (Ag₂S), nickel (NiS), cobalt (CoS), bismuth (Bi₂S₃), and tin (SnS₂). The colors of these sulphides are characteristic of their chemical composition.

• PbS (Lead Sulphide): Lead Sulphide appears as a dark, metallic gray to black color. It is commonly known as galena in its mineral form.
• CuS (Copper Sulphide): Copper Sulphide can be found in various shades, including a dark, nearly black color, known as covellite.
• HgS (Mercury Sulphide): Mercury Sulphide exists in two forms - cinnabar (red) and metacinnabar (black). So, HgS in its metacinnabar form is black.
• MnS (Manganese Sulphide): Manganese Sulphide often appears in a green to pale pink color, but it is not typically black.
• Ag₂S (Silver Sulphide): Silver Sulphide is black. It is commonly known for tarnishing silver into a black color.
• NiS (Nickel Sulphide): Nickel Sulphide can vary in color, but the most common form is a dark bronze to black color.
• CoS (Cobalt Sulphide): Cobalt Sulphide can exhibit a variety of colors, from pink to black depending on the specific conditions and forms. However, the most commonly encountered form, linnaeite (Co₃S₄), can appear gray to black.
• Bi₂S₃ (Bismuth Sulphide): Bismuth Sulphide is characterized by a dark brown color, which can appear nearly black.
• SnS₂ (Tin(IV) Sulphide): Tin(IV) Sulphide, or cassiterite, is usually yellow to brown, not black.

Based on the descriptions, the sulphides that are black are PbS, CuS, HgS (metacinnabar form), Ag₂S, NiS, and potentially CoS (though CoS can vary and might not always be classified as black, but we will count it here given its common dark forms). Thus, the total number of black-colored sulphides is 6 (including CoS). If we are strictly defining black without including the darkest variations of other colors, we would primarily count PbS, CuS, HgS in its metacinnabar form, Ag₂S, and NiS, with CoS being a conditional addition based on form.

As the vessel is thermally insulated (so, q = 0). Therefore,

$\Delta$U = 0, $\Delta$T = 0 $\Rightarrow$ T₁ = T₂

According to combined gas law equation,

${{{P_2}{V_2}} \over {{T_2}}} = {{{P_1}{V_1}} \over {{T_1}}} \Rightarrow {P_2}{V_2} = {P_1}{V_1}$

As n = 4, so l = 0, 1, 2, 3 which implies the orbitals are 4s, 4p, 4d and 4f.

Now, |m_l | = 1 implies m_l = +1 and −1. Therefore, l can be 3, 2, 1, as for l = 0, m_l = 0.

For l = 1, m_l = −1, 0, +1

For l = 2, m_l = −2, −1, 0, +1, +2

For l = 3, m_l = −3, −2, −1, 0, +1, +2, +3

Therefore, there are six possible orbitals with |m_l | = 1 . Also, given that m_s = −1/2 so six electrons are possible with m_s = −1/2.

(a) Ice floats in water due to the low density of ice as compare to water which is due to open cage like structure (formed by intermolecular H-bonding).

(b) Basic strength of RNH₂ > R₃N. It is also explained by hydrogen bonding.


(c)
(d) Dimerisation of acetic acid in benzene is due to intermolecular hydrogen bonding.

Chemical reaction and product formed after hydrolysis of given peptide can be represented as


(A) is glycine which is only naturally occurring amino acid. While (B), (C) and (D) are not the naturally occurring amino acids. Hence, correct integer is (1).

Understanding the Concepts

Molarity (M): Moles of solute per liter of solution.

Molality (m): Moles of solute per kilogram of solvent.

Density: Mass per unit volume.

Steps to Calculate Molality

1. Find the Mass of Solute (H₂X):

• Molarity = moles of solute / volume of solution (in liters)


• 3.2 M solution means 3.2 moles of H₂X are present in 1 liter of solution.


• Mass of H₂X = moles $\times$ molar mass = 3.2 moles $\times$ 80 g/mole = 256 g

1. Find the Mass of Solvent:

• Assuming no change in volume, the volume of the solution remains 1 liter.


• Density = mass / volume


• Mass of solvent = density $\times$ volume = 0.4 g/mL $\times$ 1000 mL = 400 g

1. Convert Mass of Solvent to Kilograms:

• 1 kg = 1000 g


• Mass of solvent = 400 g $\times$ (1 kg / 1000 g) = 0.4 kg

1. Calculate Molality:

• Molality = moles of solute / mass of solvent (in kg)


• Molality = 3.2 moles / 0.4 kg = 8 mol/kg

Answer:

The molality of the 3.2 molar solution is 8 mol/kg.