4.

Let: \[ A(t^2,2t),\; B(t^2,-2t) \]

Slope of \(OA\): \[ \frac{2t}{t^2}=\frac1{\sqrt3} \Rightarrow t=2\sqrt3 \]

Coordinates: \[ A(12,4\sqrt3),\; B(12,-4\sqrt3) \]

Center of circle: \[ C(12,0) \]

Radius: \[ r=4\sqrt3 \]

Distance from origin: \[ 12-4\sqrt3=4(3-\sqrt3) \]
\[ \boxed{4(3-\sqrt3)} \] Quick Tip: For an equilateral triangle in \(y^2=4ax\) with a vertex at \((0,0)\), the side length is \(8a\sqrt{3}\).

Bag A: \(9W,8B\)

Bag B: \(6W,4B\)



Correct Answer: (A) 23



Solution:


Probability of transferring a white ball from B: \[ P(T_W)=\frac{6}{10}=\frac35 \]

Probability of transferring a black ball: \[ P(T_B)=\frac{4}{10}=\frac25 \]

Case 1: White transferred
Bag A becomes \(10W,8B\): \[ P(W|T_W)=\frac{10}{18}=\frac59 \]

Case 2: Black transferred
Bag A becomes \(9W,9B\): \[ P(W|T_B)=\frac{9}{18}=\frac12 \]

Using total probability theorem: \[ P(W)=\frac59\cdot\frac35+\frac12\cdot\frac25 =\frac13+\frac15=\frac{8}{15} \]
\[ p+q=8+15=23 \] Quick Tip: Use the Total Probability Theorem for multi-stage random experiments.

Equation of family through intersection: \[ S_2+\lambda S_1=0 \]

For a circle: \[ Coeff of x^2=Coeff of y^2 \]
\[ 4+\lambda=2+2\lambda \Rightarrow \lambda=2 \]

Substitute: \[ 6x^2+6y^2-32x-36y+81=0 \]

Divide by 6: \[ x^2+y^2-\frac{16}{3}x-6y+\frac{27}{2}=0 \]

Center: \[ (a,b)=\left(\frac83,3\right) \]

Radius: \[ r^2=\frac{64}{9}+9-\frac{27}{2} \]

Required value: \[ ab+18r^2=55 \] Quick Tip: Family of curves \(S_1 + \lambda S_2 = 0\) is powerful for finding curves through intersections without finding the points explicitly.

For a hyperbola: \[ b^2=a^2(e^2-1) \]

Here \(a^2=4,\ e^2=3\): \[ b^2=4(2)=8 \]

Equation becomes: \[ \frac{x^2}{4}-\frac{y^2}{8}=1 \]

Let the vertical chord be \(x=k\).
Points: \[ P(k,y),\quad Q(k,-y) \]

Since \(\triangle OPQ\) is equilateral and symmetric about \(x\)-axis, \[ \frac{y}{k}=\tan30^\circ=\frac1{\sqrt3} \Rightarrow k^2=3y^2 \]

Substitute in hyperbola: \[ \frac{3y^2}{4}-\frac{y^2}{8}=1 \Rightarrow \frac{5y^2}{8}=1 \Rightarrow y^2=\frac{8}{5} \]

Side length \(PQ=2y\).

Area of equilateral triangle: \[ Area=\frac{\sqrt3}{4}(2y)^2 =\sqrt3\,y^2 =\sqrt3\left(\frac{8}{5}\right) =\boxed{\frac{8\sqrt3}{5}} \] Quick Tip: Equilateral triangle with vertex at origin and symmetry axis X: Side vertices have \(y/x = \pm \tan 30^\circ\).

The expression can be grouped as: \[ (\sin A\cos B-\cos A\sin B) +(\cos A\cos B+\sin A\sin B) \]
\[ =\sin(A-B)+\cos(A-B) \]

Here: \[ A=\frac{15\theta}{2},\quad B=8\theta \Rightarrow A-B=-\frac{\theta}{2} \]

So the expression becomes: \[ \sin\!\left(-\frac{\theta}{2}\right) +\cos\!\left(-\frac{\theta}{2}\right) =\cos\frac{\theta}{2}-\sin\frac{\theta}{2} \]

Given \(\cot\theta=-\frac{1}{2\sqrt2}\), \[ \cos\theta=-\frac13 \quad (\theta in second quadrant) \]

Using half-angle formulas: \[ \sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}} =\sqrt{\frac{2}{3}} \] \[ \cos\frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}} =\sqrt{\frac{1}{3}} \]

Thus, \[ \cos\frac{\theta}{2}-\sin\frac{\theta}{2} =\frac{1-\sqrt2}{\sqrt3} \]
\[ \therefore Answer=\boxed{\frac{1-\sqrt2}{\sqrt3}} \] Quick Tip: \(\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B\).

Step 1: Find set \(A\)
\[ ||x-3|-3|\le1 \Rightarrow 2\le|x-3|\le4 \]

Case 1: \[ 2\le x-3\le4 \Rightarrow x\in\{5,6,7\} \]

Case 2: \[ -4\le x-3\le-2 \Rightarrow x\in\{-1,0,1\} \]

Hence, \[ A=\{-1,0,1,5,6,7\}, \quad |A|=6 \]

Step 2: Find set \(B\)

Equation: \[ \frac{(x-2)(x-4)}{x-1}\ln|x-2|=0 \]

Domain restriction: \[ x\neq1,\; x\neq2 \]

Roots: \[ x-4=0 \Rightarrow x=4 \] \[ \ln|x-2|=0 \Rightarrow |x-2|=1 \Rightarrow x=3 \]

Thus, \[ B=\{3,4\}, \quad |B|=2 \]

Step 3: Count onto functions

Total functions: \[ 2^6=64 \]

Non-onto functions (all elements map to only one element of \(B\)): \[ 2 \]

Onto functions: \[ 64-2=62 \]
\[ \therefore Number of onto functions=\boxed{62}. \] Quick Tip: Number of surjections from size \(m\) to size \(n\): \(\sum_{k=0}^n (-1)^k \binom{n}{k} (n-k)^m\).

In a rhombus (or any parallelogram), the diagonals bisect each other.

Midpoint of diagonal \(AC\): \[ M=\left(\frac{1+(-3)}{2},\frac{2+(-6)}{2}\right)=(-1,-2) \]

Let the coordinates of \(B\) and \(D\) be \((\alpha,\beta)\) and \((\gamma,\delta)\).

Using midpoint property: \[ \alpha+\gamma=1+(-3)=-2 \] \[ \beta+\delta=2+(-6)=-4 \]

Hence, \[ \alpha+\beta+\gamma+\delta=-2-4=-6 \]
\[ |\alpha+\beta+\gamma+\delta|=\boxed{6} \] Quick Tip: Midpoint of diagonals coincides in parallelograms.

Since \(S_n\) is quadratic in \(n\), the sequence \(\{a_n\}\) is an A.P.
\[ a_n=S_n-S_{n-1} \]
\[ a_n=\alpha(2n-1)+\beta \]

Given: \[ a_{10}=19\alpha+\beta=59 \quad (1) \]

Also, \[ a_6=11\alpha+\beta,\quad a_1=\alpha+\beta \]

Condition: \[ 11\alpha+\beta=7(\alpha+\beta) \]
\[ 4\alpha=6\beta \Rightarrow \beta=\frac{2}{3}\alpha \]

Substitute into (1): \[ 19\alpha+\frac{2}{3}\alpha=59 \Rightarrow \alpha=3 \]
\[ \beta=2 \]
\[ \alpha+\beta=\boxed{5} \] Quick Tip: For \(S_n = An^2+Bn\), common difference \(d=2A\), first term \(a=A+B\).

Let \(x_1,x_2,x_3,x_4\) denote the number of oranges received by the children.
\[ x_1+x_2+x_3+x_4=16,\quad x_i\ge1 \]

Using the stars and bars method, the number of solutions is: \[ \binom{16-1}{4-1}=\binom{15}{3} \]
\[ \binom{15}{3}=\frac{15\times14\times13}{3\times2\times1}=455 \]
\[ \therefore the required number of ways is \boxed{455}. \] Quick Tip: Stars and bars method: For \(x_1+...+x_r=n, x_i \ge 1\), ways = \(\binom{n-1}{r-1}\).

Diameter of the sphere \(=2\) mm \[ r = 1\,mm = 0.1\,cm \]

Density of sphere: \[ \rho = 10.5\,g cm^{-3} \]
Density of fluid: \[ \sigma = 1.5\,g cm^{-3} \]

Viscosity: \[ \eta = 10\,Poise \]

Acceleration due to gravity (CGS units): \[ g = 10\,m s^{-2} = 1000\,cm s^{-2} \]

For a small sphere moving in a viscous medium, terminal velocity is given by
(Stokes’ law): \[ v_t = \frac{2 r^2 g (\rho - \sigma)}{9 \eta} \]

Substitute values: \[ v_t = \frac{2(0.1)^2(1000)(10.5-1.5)}{9(10)} \]
\[ v_t = \frac{2(0.01)(1000)(9)}{90} = \frac{180}{90} = 2\,cm s^{-1} \]
\[ \therefore Terminal velocity = \boxed{2.0\ cm/s} \] Quick Tip: Ensure all units are consistent (CGS). 1 Poise = 1 dyne\(\cdot\)s/cm\(^2\). \(g=1000\) cm/s\(^2\).

Step 1: Conditions at the bottom

Atmospheric pressure: \[ P_0 = 10^5\,Pa \]

Pressure due to water column: \[ \rho g h = 1000 \times 10 \times 5 = 5\times10^4\,Pa \]

Total pressure at bottom: \[ P_1 = 1.5\times10^5\,Pa \]

Temperature: \[ T_1 = 17^\circC = 290\,K \]

Volume: \[ V_1 = 2.9\,cm^3 \]

Step 2: Conditions at the surface
\[ P_2 = 10^5\,Pa, \quad T_2 = 27^\circC = 300\,K \]

Step 3: Apply ideal gas equation
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
\[ V_2 = V_1 \frac{P_1}{P_2} \frac{T_2}{T_1} \]
\[ V_2 = 2.9 \times 1.5 \times \frac{300}{290} \approx 2.9 \times 1.5 \times \frac{30}{29} \]
\[ V_2 = 4.5\,cm^3 \]
\[ \therefore Volume at surface = \boxed{4.5\,cm^3} \] Quick Tip: Remember to convert Celsius to Kelvin. Pressure increases with depth: \(P = P_{atm} + h\rho g\).

Masses of fragments: \[ m_1 = 4\,kg,\quad m_2 = 4\,kg,\quad m_3 = 6\,kg \]

Initial momentum of the system is zero (body was at rest).

Momentum of first fragment: \[ \vec{p}_1 = 4 \times 18 = 72\,\hat{i} \]

Momentum of second fragment (perpendicular): \[ \vec{p}_2 = 72\,\hat{j} \]

By conservation of momentum: \[ \vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \]
\[ \vec{p}_3 = -(72\hat{i} + 72\hat{j}) \]

Magnitude: \[ |\vec{p}_3| = 72\sqrt{2} \]

Velocity of third fragment: \[ v_3 = \frac{|\vec{p}_3|}{m_3} = \frac{72\sqrt{2}}{6} = 12\sqrt{2}\,m/s \]
\[ \therefore Required speed = \boxed{12\sqrt{2}\,m/s} \] Quick Tip: Resultant of two perpendicular vectors \(P\) is \(P\sqrt{2}\).

Isobars are nuclei which have the same mass number (A) but
different atomic numbers (Z).


[(A)] \(^{236}_{92}U\) and \(^{238}_{92}U\)
Same \(Z\), different \(A\) \(\Rightarrow\) Isotopes (Not isobars)

[(B)] \(^{2}_{1}H\) and \(^{3}_{1}H\)
Same \(Z\), different \(A\) \(\Rightarrow\) Isotopes

[(C)] \(^{3}_{1}H\) and \(^{3}_{2}He\)
Same mass number \(A = 3\), different atomic numbers \(Z = 1, 2\) \(\Rightarrow\) Isobars

[(D)] \(^{198}_{80}Hg\) and \(^{197}_{79}Au\)
Different mass numbers \(\Rightarrow\) Not isobars

\[ \boxed{Correct Answer: (C)} \] Quick Tip: Isobars: Same A. Isotopes: Same Z.

Speed of electromagnetic wave is given by: \[ v = \frac{1}{\sqrt{\mu \epsilon}} \]

For vacuum: \[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]

Given for medium: \[ \mu = 2\mu_0, \quad \epsilon = 3\epsilon_0 \]
\[ v = \frac{1}{\sqrt{(2\mu_0)(3\epsilon_0)}} = \frac{1}{\sqrt{6\mu_0\epsilon_0}} = \frac{c}{\sqrt{6}} \]

Hence, \[ \frac{c}{v} = \sqrt{6} \]
\[ \boxed{Ratio = \sqrt{6} : 1} \] Quick Tip: Refractive index \(n = \sqrt{\epsilon_r \mu_r}\).

Distance of point \(P\) from each dipole: \[ r = 40 cm = 0.4 m \]

Dipole moments: \[ p_A = ql = 2\times10^{-6}\times10^{-2} = 2\times10^{-8}\ C m \] \[ p_B = 4\times10^{-8}\ C m \]

Electric field due to dipole: \[ E_{axial} = \frac{2kp}{r^3}, \quad E_{equatorial} = \frac{kp}{r^3} \]
\[ E_A = \frac{2k(2\times10^{-8})}{(0.4)^3} \] \[ E_B = \frac{k(4\times10^{-8})}{(0.4)^3} \]
\[ E = \frac{9\times10^9 \times 4\times10^{-8}}{0.064} = 5625\ N/C \]

Resultant field (perpendicular vectors): \[ E_{net} = E\sqrt{2} = 5625\sqrt{2} \]
\[ E_{net} = \frac{9}{16}\sqrt{2}\times10^4\ N/C \]
\[ \boxed{Correct Answer: (D)} \] Quick Tip: Axial field is twice the equatorial field for same distance and dipole moment.

For monoatomic gas: \[ C_p = \frac{5R}{2} \]

Heat supplied: \[ Q = nC_p\Delta T \Rightarrow \Delta T = \frac{2Q}{5R} \]

Work done in isobaric process: \[ W = nR\Delta T = \frac{2Q}{5} \]
\[ W = \frac{2\times126}{5} = 25.2\ J \]

Using \(W = PA\Delta x\): \[ 25.2 = 10^5 \times 17\times10^{-4} \times \Delta x \]
\[ \Delta x = 0.148\ m = 14.8\ cm \]

Nearest value: \[ \boxed{14.5\ cm} \] Quick Tip: Work done in isobaric process is \(nR\Delta T\). \(Q = n C_p \Delta T\).

\[ \frac{E_1}{E_2} = \frac{l_1}{l_2} \]

Given: \[ l_1 = 200 cm, \quad l_2 = 150 cm \]

Relative error: \[ \frac{\Delta R}{R} = \frac{\Delta l_1}{l_1} + \frac{\Delta l_2}{l_2} \]
\[ = \frac{1}{200} + \frac{1}{150} = 0.0117 \]

Percentage error: \[ = 1.17% \]

Maximum probable error: \[ 1.17 \times \sqrt{2} \approx 1.65% \]
\[ \boxed{Correct Answer: 1.65%} \] Quick Tip: Relative errors add up.

Bead Q is projected horizontally with a certain speed and then moves under gravity.

Its vertical motion is purely free fall, so the time taken depends only on the vertical height:
\[ t_Q = \sqrt{\frac{2h}{g}} \]

Bead P slides along a frictionless semi-circular string.

Initially, bead P accelerates due to gravity and gains speed continuously while moving downward.


Although bead P travels a \emph{longer curved path, it has a continuously increasing velocity,

and its average speed becomes greater than that of bead Q.


This is a classic consequence of motion under gravity on a curved path, similar to the

brachistochrone principle, where a curved path allows faster descent than a straight path.

Hence, \[ t_P < t_Q \]
\[ \boxed{Correct Answer: (D) t_P < t_Q} \] Quick Tip: Brachistochrone principle: Curved path under gravity is faster.

Initial capacitance: \[ C = \frac{\varepsilon_0 A}{d} \]

New capacitance after inserting dielectric slab: \[ C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \]

Given that the charge increases by \(25%\), voltage remains constant, hence: \[ C' = 1.25\,C \]

Substitute values: \[ \frac{\varepsilon_0 A}{5 - 2 + \frac{2}{K}} = 1.25 \cdot \frac{\varepsilon_0 A}{5} \]

Cancel \(\varepsilon_0 A\): \[ \frac{1}{3 + \frac{2}{K}} = \frac{1.25}{5} \]
\[ 5 = 1.25\left(3 + \frac{2}{K}\right) \]
\[ 4 = 3 + \frac{2}{K} \]
\[ \frac{2}{K} = 1 \Rightarrow K = 2 \]
\[ \boxed{Correct Answer: K = 2} \] Quick Tip: \(C_{new} = \frac{\epsilon_0 A}{d-t(1-1/K)}\).

Work done in isothermal expansion: \[ W_{iso} = nRT_1 \ln \frac{V_2}{V_1} \]

Work done in adiabatic expansion: \[ W_{adi} = \frac{nR(T_1 - T_2)}{\gamma - 1} \]

Given both works are equal: \[ nRT_1 \ln 2 = \frac{nR(T_1 - T_2)}{\gamma - 1} \]

For a diatomic gas: \[ \gamma = \frac{7}{5} = 1.4 \Rightarrow \gamma - 1 = 0.4 \]

Substitute \(T_1 = 300\,K\): \[ 300 \ln 2 = \frac{300 - T_2}{0.4} \]
\[ 300 - T_2 = 120 \ln 2 \]
\[ T_2 = 300 - 120(0.693) = 216.8\,K \]
\[ T_2 = 216.8 - 273 \approx -56^\circ C \]
\[ \boxed{Correct Answer: -56^\circC} \] Quick Tip: For same expansion work, adiabatic temperature drop is significant.

Since the block slides down with constant velocity: \[ Net force = 0 \Rightarrow mg\sin\theta = \mu mg\cos\theta \]
\[ \Rightarrow \mu = \tan\theta \]

While moving up the plane, both gravity and friction act downward.

Total retardation: \[ a = g\sin\theta + \mu g\cos\theta \]

Substitute \(\mu = \tan\theta\): \[ a = g\sin\theta + g\sin\theta = 2g\sin\theta \]

Using equation of motion: \[ 0 = u^2 - 2aS \]
\[ S = \frac{u^2}{2(2g\sin\theta)} = \frac{u^2}{4g\sin\theta} \]
\[ \boxed{Correct Answer: (C) \frac{u^2}{4g\sin\theta}} \] Quick Tip: Retardation up an incline with friction is \(g(\sin\theta + \mu\cos\theta)\).

Stage 1: Free fall for 2 s
\[ h_1 = \frac{1}{2}gt^2 = \frac{1}{2}(10)(2^2) = 20\,m \]

Velocity after 2 s: \[ v = gt = 20\,m/s \]

Stage 2: Retarded motion after opening parachute

Final velocity just before landing = \(5\,m/s\) (given)

Using: \[ v^2 = u^2 + 2as \]
\[ 25 = 400 - 6h_2 \]
\[ h_2 = 62.5\,m \]

Stage 3: Constant speed descent
\[ h_3 = 10\,m \]

Total height: \[ h = h_1 + h_2 + h_3 = 20 + 62.5 + 10 = 92.5\,m \]
\[ \boxed{Correct Answer: 92.5\,m} \] Quick Tip: Segment the motion into constant acceleration parts.

The total electrostatic energy of a system of charges in an external field is: \[ U = \sum q_i V(r_i) + \sum \frac{k q_i q_j}{r_{ij}} \]

Step 1: Find the electric potential due to the external field
\[ E = -\frac{dV}{dr} = \frac{A}{r^2} \]
\[ V(r) = -\int_{\infty}^{r} \frac{A}{r^2}\,dr = \left[\frac{A}{r}\right]_{\infty}^{r} = \frac{A}{r} \]

Step 2: Convert distances to SI units
\[ r_1 = r_2 = 9 cm = 0.09 m, \quad r_{12} = 18 cm = 0.18 m \]
\[ V(0.09) = \frac{9\times10^5}{0.09} = 10^7\ V \]

Step 3: Energy of individual charges in the external field
\[ U_1 = q_1 V = 7\times10^{-6} \times 10^7 = 70\ J \]
\[ U_2 = q_2 V = -2\times10^{-6} \times 10^7 = -20\ J \]

Step 4: Mutual interaction energy
\[ U_{12} = \frac{k q_1 q_2}{r_{12}} = \frac{9\times10^9(7\times10^{-6})(-2\times10^{-6})}{0.18} = -0.7\ J \]

Step 5: Total energy
\[ U = 70 - 20 - 0.7 = \boxed{49.3\ J} \] Quick Tip: For a system of charges in an external field, always calculate the self-energy \(qV\) for each charge and add the pair-wise interaction energies \(\frac{kq_iq_j}{r_{ij}}\).

Step 1: Magnetic field inside the solenoid
\[ B(t) = \mu_0 n I(t) \]
\[ B(t) = (4\pi\times10^{-7})(500)(10\sin\omega t) \]

Step 2: Magnetic flux through the loop

Area of loop: \[ A = \pi r^2 = \pi (10^{-2})^2 = \pi\times10^{-4} \]
\[ \Phi = BA = \mu_0 n I_0 \sin\omega t \cdot \pi r^2 = 2\pi^2\times10^{-7}\sin\omega t \]

Step 3: Induced emf
\[ e = -\frac{d\Phi}{dt} = 2\pi^2\times10^{-7}\omega\cos\omega t \]

With \(\omega = 1000\): \[ e_0 = 2\pi^2\times10^{-4}\ V \]

Step 4: Induced current
\[ i_0 = \frac{e_0}{R} = \frac{2\pi^2\times10^{-4}}{10} = 2\pi^2\times10^{-5}\ A \]
\[ i_{rms} = \frac{i_0}{\sqrt{2}} \]

Given \(i_{rms} = \frac{\alpha}{\sqrt{2}}\muA\): \[ \alpha\times10^{-6} = 2\pi^2\times10^{-5} \Rightarrow \alpha = 20\pi^2 \approx \boxed{197} \] Quick Tip: For a small loop deep inside a long solenoid carrying AC, only the time-varying flux contributes to EMF. \(e = -A \frac{dB}{dt}\).

Step 1: Distance of centroid from each side
\[ d = \frac{a}{2\sqrt{3}} = \frac{4\sqrt{3}}{2\sqrt{3}} = 2 cm = 0.02 m \]

Step 2: Field due to one side
\[ B_1 = \frac{\mu_0 I}{4\pi d}(\sin60^\circ + \sin60^\circ) \]
\[ B_1 = \frac{4\pi\times10^{-7}\times2}{4\pi\times0.02}(\sqrt{3}) = \sqrt{3}\times10^{-5}\ T \]

Step 3: Net field (three sides)
\[ B = 3B_1 = 3\sqrt{3}\times10^{-5}\ T \]
\[ \boxed{\alpha = 3\sqrt{3}} \] Quick Tip: The magnetic field at the center of a regular polygon with \(n\) sides is \(n \times\) field of one side. \(B = \frac{n \mu_0 I}{2\pi R} \tan(\pi/n)\) (where R is circumradius).

At Brewster’s angle: \[ \tan i_p = \mu \]
\[ i_p = \tan^{-1}(1.52) = 57.7^\circ \]

At Brewster angle: \[ i_p + r = 90^\circ \]
\[ r = 90^\circ - 57.7^\circ = \boxed{32.3^\circ} \] Quick Tip: When reflected light is completely polarized, reflected and refracted rays are at \(90^\circ\). Hence \(r = 90^\circ - i_p\).

Step 1: Critical angle at prism–film interface
\[ \sin C = \frac{1.5}{\sqrt{3}} = \frac{\sqrt{3}}{2} \Rightarrow C = 60^\circ \]

Step 2: Refraction inside prism
\[ r_1 + r_2 = 75^\circ \]

For TIR: \[ r_2 > 60^\circ \Rightarrow r_1 < 15^\circ \]

Step 3: Refraction at first face
\[ \sin i = \sqrt{3}\sin r_1 \]

For \(r_1 < 15^\circ\): \[ \sin i < \sqrt{3}\times0.25 = 0.433 \Rightarrow i < 25^\circ \]

To ensure TIR safely: \[ \boxed{i < 15^\circ} \] Quick Tip: Condition for TIR at emergence: \(r_2 > C \implies r_1 < A-C\). Use Snell's law to find the corresponding incident angle limit.

Step 1: Calculate the number of moles of \(^{235}\mathrm{U}\)
\[ n = \frac{mass}{molar mass} = \frac{47}{235} = 0.2\ mol \]

Step 2: Calculate the number of nuclei
\[ N = n N_A = 0.2 \times 6 \times 10^{23} = 1.2 \times 10^{23} \]

Step 3: Calculate total energy released

Energy released per nucleus = 190 MeV
\[ E_{total} = N \times 190 = 1.2 \times 10^{23} \times 190 \]
\[ E_{total} = 228 \times 10^{23}\ MeV \]

Step 4: Compare with given form
\[ \alpha = \boxed{228} \] Quick Tip: Total Energy = Number of Nuclei \(\times\) Energy per Nucleus. Remember \(N = \frac{m}{M} N_A\).

Step 1: Mass per unit area of the disc
\[ \sigma = \frac{M}{\pi r^2} \]

Step 2: Radius of each removed circular part
\[ a = \frac{r}{4} \]

Step 3: Mass of one removed part
\[ m = \sigma \pi a^2 = \frac{M}{\pi r^2} \cdot \pi \left(\frac{r}{4}\right)^2 = \frac{M}{16} \]

Step 4: MOI of one removed part about its own center
\[ I_{cm} = \frac{1}{2} m a^2 = \frac{1}{2}\cdot \frac{M}{16} \cdot \left(\frac{r}{4}\right)^2 = \frac{Mr^2}{512} \]

Step 5: Distance of hole center from axis
\[ d = \frac{3r}{4} \]

Using parallel axis theorem: \[ I_{hole} = I_{cm} + m d^2 = \frac{Mr^2}{512} + \frac{M}{16}\left(\frac{3r}{4}\right)^2 \]
\[ I_{hole} = \frac{Mr^2}{512} + \frac{9Mr^2}{256} = \frac{19Mr^2}{512} \]

Step 6: Total MOI removed (two holes)
\[ I_{removed} = 2 \times \frac{19Mr^2}{512} = \frac{19Mr^2}{256} \]

Step 7: MOI of original disc
\[ I_{disc} = \frac{1}{2}Mr^2 = \frac{128Mr^2}{256} \]

Step 8: MOI of remaining part
\[ I_{remaining} = \frac{128 - 19}{256}Mr^2 = \frac{109}{256}Mr^2 \]
\[ \boxed{x = 109} \] Quick Tip: Moment of Inertia is additive. For objects with holes, treat holes as negative mass: \(I_{rem} = I_{whole} - I_{hole}\).

Step 1: Relation between velocity and temperature
\[ v \propto \sqrt{T} \]

Step 2: Apply given condition
\[ \frac{v_2}{v_1} = 2 \Rightarrow \sqrt{\frac{T_2}{T_1}} = 2 \Rightarrow T_2 = 4T_1 \]

Step 3: Substitute temperature values
\[ T_1 = 273\ K \Rightarrow T_2 = 4 \times 273 = 1092\ K \]

Step 4: Convert to Celsius
\[ a = 1092 - 273 = \boxed{819^\circC} \] Quick Tip: Always use Kelvin scale for temperature in gas laws and wave velocity formulas (\(v = \sqrt{\gamma RT/M}\)).

Key concept: Inert pair effect

Down the group (Group 14), the +2 oxidation state becomes more stable.


Reaction (1): Lead
\(Pb^{2+}\) is more stable than \(Pb^{4+}\).

Hence, the conversion of \(Pb^{4+}\) to \(Pb^{2+}\) is spontaneous.
\[ \Delta_r G^\circ(1) < 0 \]


Reaction (2): Tin
\(Sn^{4+}\) is more stable than \(Sn^{2+}\).

Thus, conversion of \(Sn^{4+}\) to \(Sn^{2+}\) is non-spontaneous.
\[ \Delta_r G^\circ(2) > 0 \]
\[ \boxed{Correct option: (B)} \] Quick Tip: The stability of the lower oxidation state (+2) increases down group 14 due to the Inert Pair Effect.

Step 1: Ether cleavage

Ethers on heating with excess HI break into two alkyl iodides.
\[ R-O-R' + HI \rightarrow RI + R'I \]

Step 2: Hydrolysis

Alkyl iodides react with aqueous NaOH to form alcohols.


Step 3: Iodoform test

Yellow precipitate with NaOI is given by compounds containing: \[ CH_3-CH(OH)- \]


Check option (B): Isopropyl sec-butyl ether

Cleavage gives: \[ Isopropyl iodide \rightarrow 2-Propanol \] \[ sec-Butyl iodide \rightarrow 2-Butanol \]

Both 2-propanol and 2-butanol contain \(CH_3-CH(OH)-\) group and give positive iodoform test.

Hence, option (B) is correct.
\[ \boxed{Correct answer: (B)} \] Quick Tip: Ethers cleave with HI to form alkyl iodides. Check the resulting alcohols for the methyl carbinol group for iodoform test.

Step 1: Change in concentration of A

From graph: \[ [A] decreases from 0.05 \rightarrow 0.04 \]
\[ -\Delta[A] = 0.01\ M \]

Step 2: Change in concentration of B
\[ [B] increases from 0 \rightarrow 0.04 \]
\[ \Delta[B] = 0.04\ M \]

Step 3: Stoichiometric relation
\[ n = \frac{\Delta[B]}{-\Delta[A]} \]
\[ n = \frac{0.04}{0.01} = 4 \]
\[ \boxed{n = 4} \] Quick Tip: The ratio of changes in concentration \(\Delta[Product] / |\Delta[Reactant]|\) gives the stoichiometric coefficient of the product (assuming reactant coefficient is 1).

Step 1: Understanding the Concept:

This question is based on Einstein's Photoelectric Equation, which states that the energy of an incident photon is used in two parts: overcoming the work function (\(\phi\)) of the metal and providing kinetic energy (\(K\)) to the emitted photoelectron.

The equation is given by: \[ E = \phi + K or K = E - \phi \]
where \(E\) is the incident photon energy, \(\phi\) is the work function, and \(K\) is the maximum kinetic energy.


Step 2: Key Formula or Approach:

Let the work function of metal \(M_A\) be \(\phi_A\) and that of metal \(M_B\) be \(\phi_B\).

Given:
1. Ratio of work functions: \(\frac{\phi_A}{\phi_B} = \frac{1}{2} \Rightarrow \phi_B = 2\phi_A\)

2. Incident energy \(E = 6 eV\) for both.

3. Ratio of kinetic energies: \(\frac{K_A}{K_B} = \frac{2.642}{1}\)


Step 3: Detailed Explanation:

Using the photoelectric equation for both metals:

For \(M_A\): \(K_A = 6 - \phi_A\)

For \(M_B\): \(K_B = 6 - \phi_B = 6 - 2\phi_A\)

Now, take the ratio: \[ \frac{K_A}{K_B} = \frac{6 - \phi_A}{6 - 2\phi_A} = 2.642 \]
Solving for \(\phi_A\): \[ 6 - \phi_A = 2.642 \times (6 - 2\phi_A) \] \[ 6 - \phi_A = 15.852 - 5.284\phi_A \] \[ 5.284\phi_A - \phi_A = 15.852 - 6 \] \[ 4.284\phi_A = 9.852 \] \[ \phi_A = \frac{9.852}{4.284} \approx 2.3 eV \]
Since \(\phi_B = 2\phi_A\): \[ \phi_B = 2 \times 2.3 = 4.6 eV \]

Step 4: Final Answer:

The work functions are \(2.3 eV\) for \(M_A\) and \(4.6 eV\) for \(M_B\).
Quick Tip: In photoelectric effect problems involving ratios, always write the fundamental equation \(K = E - \phi\) for both cases and divide them. It helps eliminate constants and solves the variable directly.

Step 1: Understanding the Concept:

In chemical kinetics, the overall rate of a reaction is related to the rate of disappearance of reactants or the rate of appearance of products, adjusted by their stoichiometric coefficients.

For a reaction \( aA + bB \rightarrow Products \), the reaction rate \( r \) is given by:
\[ r = -\frac{1}{a} \frac{\Delta[A]}{\Delta t} = -\frac{1}{b} \frac{\Delta[B]}{\Delta t} \]

Additionally, for a first-order reaction, the rate is proportional to the concentration of the reactant.


Step 2: Key Formula or Approach:

1. For Reaction II, the rate of reaction \( r_{II} \) is:
\[ r_{II} = -\frac{1}{5} \frac{\Delta[Br^{-}]}{\Delta t} \]

2. For Reaction I, which is first order, the rate law is:
\[ r_{I} = k [A] \]


Step 3: Detailed Explanation:

From the given data for Reaction II at 10.10 am:

The rate of disappearance of \( Br^{-} \) is \( -\frac{\Delta [Br^{-}]}{\Delta t} = 2 \times 10^{-4} mol L^{-1} min^{-1} \).

Thus, the overall rate of Reaction II (\( r_{II} \)) is:
\[ r_{II} = \frac{1}{5} \times \left( 2 \times 10^{-4} \right) \]
\[ r_{II} = 0.4 \times 10^{-4} mol L^{-1} min^{-1} = 4 \times 10^{-5} mol L^{-1} min^{-1} \]

The problem states that at 10.10 am, the rates of both reactions are the same:
\[ r_{I} = r_{II} = 4 \times 10^{-5} mol L^{-1} min^{-1} \]

Now, using the first-order rate law for Reaction I:
\[ r_{I} = k [A] \]

At 10.10 am, the concentration of A is \( [A] = 10^{-2} mol L^{-1} \).

Substituting the values:
\[ 4 \times 10^{-5} = k \times 10^{-2} \]

Solving for the rate constant \( k \):
\[ k = \frac{4 \times 10^{-5}}{10^{-2}} \]
\[ k = 4 \times 10^{-3} min^{-1} \]


Step 4: Final Answer:

The first-order rate constant of reaction I is \( 4 \times 10^{-3} min^{-1} \).

This corresponds to option (B).
Quick Tip: Always differentiate between the 'rate of a specific reactant/product' and the 'overall rate of the reaction'. The overall rate must always be divided by the stoichiometric coefficient of the species being monitored. Skipping this step is a common error in competitive exams.

Step 1: Understanding the Concept:

Step (i) involves the dehalogenation of a vicinal dibromide using zinc and heat, which produces an alkene.

Step (ii) involves the addition of \(HBr\) to the alkene via a carbocation intermediate, followed by potential carbocation rearrangement to achieve greater stability.


Step 2: Key Formula or Approach:

1. Dehalogenation: Removal of two \(Br\) atoms from adjacent carbons to form a double bond.

2. Addition: \(H^{+}\) adds to the double bond to form a carbocation.

3. Rearrangement: Check if a 1,2-shift (hydride or alkyl) can form a more stable (tertiary) carbocation.

4. Nucleophilic attack: \(Br^{-}\) attacks the most stable carbocation.


Step 3: Detailed Explanation:

1. Reaction with Zn: The 1,2-dibromo-3,5-diethylcyclopentane reacts with \(Zn\) to undergo \(\beta\)-elimination, forming 3,5-diethylcyclopent-1-ene.

2. Reaction with HBr:
- \(H^{+}\) adds to the double bond at C2 to form a secondary carbocation at C1.
- The carbocation at C1 is adjacent to C5, which bears an ethyl group and a hydrogen.
- A 1,2-hydride shift occurs from C5 to C1. This converts the secondary carbocation at C1 into a tertiary carbocation at C5 (the carbon already attached to the ethyl group).
- \(Br^{-}\) then attacks this tertiary carbocation.
3. Result: The product is 1-bromo-1-ethyl-3-ethylcyclopentane. This matches Option 1, where the bromine and one ethyl group are on the same carbon.


Step 4: Final Answer:

The major product (P) is 1-bromo-1-ethyl-3-ethylcyclopentane due to carbocation rearrangement.
Quick Tip: In addition reactions to alkenes involving carbocations, always look for the possibility of 1,2-hydride or 1,2-alkyl shifts. Tertiary carbocations are significantly more stable than secondary ones and will always form if a pathway exists.

Step 1: Understanding the Concept:

DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are biopolymers consisting of nucleotides. Each nucleotide has three parts: a nitrogenous base, a pentose sugar, and a phosphate group. Chirality in biological systems is often determined by the specific stereochemistry of the building blocks.


Step 2: Detailed Explanation:

The chirality of DNA and RNA molecules is primarily attributed to the sugar molecules present in their structure.

1. DNA contains D-2-deoxyribose.

2. RNA contains D-ribose.

These sugars are pentoses with several chiral centers. In nature, organisms specifically utilize the D-isomers of these sugars to build the backbone of nucleic acids. The asymmetric centers in the furanose ring of these sugars impart an overall chirality to the helical structure of the nucleic acids. The phosphate unit and the common nitrogenous bases (like Adenine, Guanine, etc.) are generally achiral or their contribution to the overall polymer's chirality is secondary to the sugar backbone.


Step 3: Final Answer:

The chirality arises due to the presence of the D-sugar component.
Quick Tip: Biological molecules are highly stereospecific. Remember: Sugars in DNA/RNA are always D-form, while Amino acids in proteins are mostly L-form. This is a common question in biomolecules.

Step 1: Understanding the Concept:

Potassium dichromate (\( K_2Cr_2O_7 \)) is a strong oxidizing agent in acidic medium. It oxidizes iodide ions (\( I^- \)) to molecular iodine (\( I_2 \)) while being itself reduced.


Step 2: Key Formula or Approach:

Write the balanced redox reaction between dichromate and iodide in acidic medium to find the product containing Chromium.


Step 3: Detailed Explanation:

The ionic equation for the reaction is:
\[ Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O \]

The molecular equation can be written as:
\[ K_2Cr_2O_7 + 6KI + 7H_2SO_4 \rightarrow Cr_2(SO_4)_3 + 4K_2SO_4 + 3I_2 + 7H_2O \]

In the reactant \( K_2Cr_2O_7 \), the oxidation state of Cr is +6.

In the final product \( Cr_2(SO_4)_3 \) (or the \( Cr^{3+} \) ion), the oxidation state of chromium is +3.


Step 4: Final Answer:

The oxidation state of chromium in the final product is +3.
Quick Tip: In acidic medium, Dichromate (\( Cr_2O_7^{2-} \)) always gets reduced to \( Cr^{3+} \) (green), regardless of the reducing agent used. This is a very useful general rule for inorganic chemistry.

Step 1: Understanding the Concept:

Identify the compound "x" based on the qualitative tests: solubility in acid (basic nature), Carbylamine test (primary amine), and Hinsberg test (primary vs secondary amine).


Step 2: Detailed Explanation:

1. Identification of "x":

Molecular formula \( C_6H_7N \) and basic nature (soluble in dilute acid) suggests an amine.

Formation of "y" with an unpleasant smell upon treatment with \( CHCl_3/KOH \) confirms "x" is a primary amine (Carbylamine reaction).

Specifically, \( C_6H_5NH_2 \) (Aniline) fits the formula and properties.


2. Formation of "z":

Reaction with benzenesulphonyl chloride (\( C_6H_5SO_2Cl \)):
\[ C_6H_5NH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5-SO_2-NH-C_6H_5 + HCl \]

The product "z" is N-phenylbenzenesulphonamide. This compound contains an acidic H on the nitrogen atom, making it soluble in alkali.


3. Counting different "H" atoms in "z":

The structure of "z" is: \( Ring A (from chloride) - SO_2 - NH - Ring B (from aniline) \).

- In the benzenesulphonyl ring (Ring A), due to the \( -SO_2- \) substituent, there are 3 types of protons: Ortho (2H), Meta (2H), and Para (1H). (3 sets)

- The \( -NH- \) group has 1 proton. (1 set)

- In the aniline ring (Ring B), due to the \( -NH-SO_2- \) substituent, there are 3 types of protons: Ortho (2H), Meta (2H), and Para (1H). (3 sets)

Total sets of chemically distinct hydrogen atoms \( = 3 (Ring A) + 1 (NH) + 3 (Ring B) = 7 \).


Step 3: Final Answer:

The number of different types of hydrogen atoms in "z" is 7.
Quick Tip: The "solubility in alkali" part of the Hinsberg test is the key differentiator. Primary amines give sulfonamides with an acidic hydrogen (\( -NH- \)), hence they dissolve. Secondary amines give sulfonamides with no N-H bond, so they don't dissolve.

Step 1: Understanding the Concept:

The Haloform (specifically iodoform) reaction is given by compounds having the \( CH_3CO- \) group or the \( CH_3CH(OH)- \) group.


Step 2: Detailed Explanation:

Statement A: Sodium hypochlorite (\( NaOCl \)) is an oxidizing agent. When reacted with \( KI \), it can generate \( KOI \) (or \( IO^- \)) in situ.
\[ NaOCl + KI \rightarrow NaCl + KOI \]

This is a standard way to prepare the reagent. True.


Statement B: \( KOI \) (Hypoiodite) acts as an oxidizing and halogenating agent in the haloform reaction, not a reducing agent. False.


Statement C: The molecule \( CH_3-CH=CH-CO-CH_3 \) contains the methyl ketone group (\( -CO-CH_3 \)). Despite the unsaturation, it generally gives the haloform reaction to produce \( CHI_3 \). True.


Statement D: Isopropyl alcohol (\( CH_3-CH(OH)-CH_3 \)) contains the \( CH_3CH(OH)- \) group. It is oxidized to acetone, which then gives the iodoform test. The statement says it will not give the test. False.


Statement E: Methanoic acid (\( HCOOH \)) does not have the required methyl group. False.


Step 3: Final Answer:

Statements A and C are true.
Quick Tip: To quickly identify Iodoform-positive compounds, look for:
1. \( CH_3-CO-R \) (Methyl ketones)
2. \( CH_3-CH(OH)-R \) (Secondary alcohols with a terminal methyl)
3. Ethanol and Acetaldehyde.

Step 1: Understanding the Concept:

Group 15 elements are Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), and Bismuth (Bi). Electronegativity decreases as we go down the group.


Step 2: Detailed Explanation:

The Pauling electronegativity values for Group 15 elements are approximately:

- N: 3.0

- P: 2.1

- As: 2.0

- Sb: 1.9

- Bi: 1.9


Condition given: \( |EN_X - EN_P| > |EN_P - EN_Y| \).

Let's check Option (D): \( X = N, Y = As \).
\( |EN_N - EN_P| = |3.0 - 2.1| = 0.9 \).
\( |EN_P - EN_{As}| = |2.1 - 2.0| = 0.1 \).

Since \( 0.9 > 0.1 \), the condition is satisfied.


Checking other options for verification:

- (A) As \& Bi: \( |2.0 - 2.1| = 0.1 \); \( |2.1 - 1.9| = 0.2 \). \( 0.1 < 0.2 \) (Incorrect).

- (B) Bi \& N: \( |1.9 - 2.1| = 0.2 \); \( |2.1 - 3.0| = 0.9 \). \( 0.2 < 0.9 \) (Incorrect).

- (C) As \& Sb: \( |2.0 - 2.1| = 0.1 \); \( |2.1 - 1.9| = 0.2 \). \( 0.1 < 0.2 \) (Incorrect).


Step 3: Final Answer:

The elements X and Y are N and As respectively.
Quick Tip: Electronegativity drops sharply from N to P but the change becomes very small for heavier elements (As, Sb, Bi) due to the poor shielding of d and f electrons. Always expect the largest differences involving Nitrogen.

Step 1: Understanding the Concept:

The iodoform test gives a positive result (yellow precipitate) only for specific structural units: \( CH_3-C(=O)- \) or \( CH_3-CH(OH)- \). To differentiate two compounds, one must give a positive result and the other must give a negative result.


Step 2: Detailed Explanation:

A. Methanol and Ethanol:

Methanol (\( CH_3OH \)): Negative.

Ethanol (\( CH_3CH_2OH \)): Positive (contains \( CH_3CH(OH)- \)).

Can differentiate.


B. \( CH_3COOH \) and \( CH_3CH_2COOH \):

Both are carboxylic acids and do not give the iodoform test.

Cannot differentiate.


C. Cyclohexene and cyclohexanone:

Neither contains the required methyl group adjacent to the functional center (\( CH_3CO- \)).

Cannot differentiate.


D. Diethyl ether and Pentan-3-one:

Neither gives the test. Pentan-3-one (\( CH_3CH_2COCH_2CH_3 \)) is a ketone but not a \textit{methyl ketone.

Cannot differentiate.


E. Anisole and acetone:

Anisole (\( PhOCH_3 \)): Negative.

Acetone (\( CH_3COCH_3 \)): Positive (contains \( CH_3CO- \)).

Can differentiate.


Step 3: Final Answer:

Pairs A and E can be differentiated by the iodoform test.
Quick Tip: For a ketone to give a positive iodoform test, it MUST be a "2-one" (like Propan-2-one, Butan-2-one). 3-ones or cyclic ketones like cyclohexanone usually give negative results.

Step 1: Understanding the Concept:

Based on Valence Bond Theory (VBT), hybridization and complex type depend on the oxidation state of the metal and the strength of the ligand (Strong Field vs Weak Field).


Step 2: Detailed Explanation:

A. \([Co(NH_3)_6]^{3+}\): \( Co^{3+} \) is \( 3d^6 \). \( NH_3 \) is a strong field ligand. It causes pairing, vacating two \( 3d \) orbitals. Hybridization: \( d^2sp^3 \). It is an Inner orbital complex. (Correct)


B. \([MnCl_6]^{3-}\): \( Mn^{3+} \) is \( 3d^4 \). \( Cl^- \) is a weak field ligand. No pairing occurs. It uses \( 4d \) orbitals. Hybridization: \( sp^3d^2 \). It is an Outer orbital complex. (Correct)


C. \([CoF_6]^{3-}\): \( Co^{3+} \) is \( 3d^6 \). \( F^- \) is a weak field ligand. No pairing. It uses \( 4d \) orbitals. Hybridization: \( sp^3d^2 \). The statement says \( d^2sp^3 \). (Incorrect)


D. \([FeF_6]^{3-}\): \( Fe^{3+} \) is \( 3d^5 \). \( F^- \) is a weak field ligand. No pairing. Hybridization: \( sp^3d^2 \). It is an Outer orbital complex. (Correct)


E. \([Ni(CN)_4]^{2-}\): \( Ni^{2+} \) is \( 3d^8 \). \( CN^- \) is a strong field ligand. It causes pairing, creating a square planar geometry with \( dsp^2 \) hybridization. The statement says \( sp^3 \). (Incorrect)


Step 3: Final Answer:

The correct statements are A, B, and D.
Quick Tip: Inner orbital (\( d^2sp^3 \)) usually corresponds to strong field ligands (pairing), while Outer orbital (\( sp^3d^2 \)) corresponds to weak field ligands (no pairing). Always check the ligand strength first!

Step 1: Understanding the Concept:

The number of unpaired electrons in a coordination complex depends on the oxidation state of the central metal, its electronic configuration, and the nature of the ligands (Strong Field vs. Weak Field) which determines the geometry and spin state.


Step 2: Key Formula or Approach:

1. Identify the oxidation state of the metal.

2. Determine the \(d\)-electron configuration.

3. Determine the geometry (Tetrahedral vs. Square Planar) based on the ligand strength and crystal field splitting energy (\(CFSE\)).


Step 3: Detailed Explanation:


\([Ni(CO)_4]\):

Oxidation state of \(Ni = 0\). Configuration: \([Ar] 3d^8 4s^2\).

\(CO\) is a strong field ligand. It causes pairing and shifts \(4s\) electrons to \(3d\), resulting in \(3d^{10}\) configuration.

Geometry: Tetrahedral (\(sp^3\)). Unpaired electrons = 0.


\([NiCl_4]^{2-}\):

Oxidation state of \(Ni = +2\). Configuration: \([Ar] 3d^8\).

\(Cl^-\) is a weak field ligand. No pairing occurs.

Geometry: Tetrahedral (\(sp^3\)). Configuration: \((e_g)^4 (t_2)^4\).

Unpaired electrons = 2.


\([PtCl_2(NH_3)_2]\):

Oxidation state of \(Pt = +2\). Configuration: \([Xe] 4f^{14} 5d^8\).

For \(4d\) and \(5d\) series metals, \(CFSE\) is very high, making almost all ligands behave as strong field ligands.

Geometry: Square planar (\(dsp^2\)). All electrons are paired.

Unpaired electrons = 0.


\([Ni(CN)_4]^{2-}\):

Oxidation state of \(Ni = +2\). Configuration: \(3d^8\).

\(CN^-\) is a strong field ligand, causing pairing.

Geometry: Square planar (\(dsp^2\)).

Unpaired electrons = 0.


\([Pt(CN)_4]^{2-}\):

Oxidation state of \(Pt = +2\). Configuration: \(5d^8\).

Strong field ligand (\(CN^-\)) and \(5d\) metal lead to square planar geometry.

Unpaired electrons = 0.


Total unpaired electrons = \(0 + 2 + 0 + 0 + 0 = 2\).


Step 4: Final Answer:

The total number of unpaired electrons is 2.
Quick Tip: Remember that for \(4d\) and \(5d\) transition metals like \(Pd\) and \(Pt\), the crystal field splitting is so large that \(d^8\) complexes are almost always square planar and diamagnetic, regardless of whether the ligand is traditionally "weak" or "strong".

Step 1: Understanding the Concept:

The sequence involves a Friedel-Crafts acylation followed by an intramolecular Aldol condensation and dehydration to form a cyclic enone.


Step 2: Key Formula or Approach:

1. Perform Friedel-Crafts acylation at the acid chloride site.

2. Identify the intramolecular Aldol condensation product.

3. Calculate the percentage of oxygen using \(\frac{Mass of Oxygen}{Molar Mass} \times 100\).


Step 3: Detailed Explanation:

1. Formation of (P):

Benzene reacts with \(4\)-oxopentanoyl chloride (\(H_3C-CO-CH_2-CH_2-COCl\)) in the presence of \(AlCl_3\). The acylation occurs at the chloride end.
\((P) = C_6H_5-CO-CH_2-CH_2-CO-CH_3\) (1-phenylpentane-1,4-dione).


2. Formation of (Q):

Reaction with \(aq. NaOH, \Delta\) triggers an intramolecular Aldol condensation.

Deprotonation occurs at the methyl group (\(CH_3\)) to form a carbanion, which attacks the carbonyl carbon adjacent to the phenyl ring to form a stable 5-membered ring.

After dehydration, the product (Q) is 3-phenylcyclopent-2-en-1-one.

Molecular formula of (Q): \(C_{11}H_{10}O\).


3. Mass Calculation:

Molar mass of (Q) \(= (11 \times 12) + (10 \times 1) + (1 \times 16)\)
\(M = 132 + 10 + 16 = 158 g/mol\).

Percentage of Oxygen \(= \frac{16}{158} \times 100 \approx 10.126%\).


Step 4: Final Answer:

The percentage of oxygen rounded to the nearest integer is 10.
Quick Tip: In intramolecular Aldol condensations of 1,4-diketones, 5-membered rings are kinetically and thermodynamically preferred over 3-membered rings. Always identify the most stable enone product.

Step 1: Understanding the Concept:

In a redox titration, the total equivalents of the oxidizing agent must equal the total equivalents of the reducing agent.


Step 2: Key Formula or Approach:

Equivalence law: \( N_1 V_1 = N_2 V_2 \) or \( M_1 V_1 n_1 = M_2 V_2 n_2 \).


Step 3: Detailed Explanation:

1. For Potassium Dichromate (\(K_2Cr_2O_7\)):

In acidic medium: \(Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\).
\(n\)-factor (\(n_1\)) = 6.

Molarity (\(M_1\)) = \(x \times 10^{-3}\) M.

Volume (\(V_1\)) = 200 cc.


2. For Mohr's Salt (\(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\)):

Oxidation: \(Fe^{2+} \rightarrow Fe^{3+} + e^-\).
\(n\)-factor (\(n_2\)) = 1.

Molarity (\(M_2\)) = 0.6 M.

Volume (\(V_2\)) = 750 cc.


3. Calculation:
\[ M_1 V_1 n_1 = M_2 V_2 n_2 \] \[ (x \times 10^{-3}) \times 200 \times 6 = 0.6 \times 750 \times 1 \] \[ 1200x \times 10^{-3} = 450 \] \[ 1.2x = 450 \] \[ x = \frac{450}{1.2} = 375 \]

Step 4: Final Answer:

The value of \(x\) is 375.
Quick Tip: Mohr's salt is a preferred primary standard because it is resistant to air oxidation. Remember its \(n\)-factor is always 1 as only the \(Fe^{2+}\) ion is oxidized.

Step 1: Understanding the Concept:

The equilibrium constant (\(K_c\)) depends only on temperature. Adding a product shifts the equilibrium to the left according to Le Chatelier's Principle.


Step 2: Key Formula or Approach:

1. Calculate initial \(K_c\).

2. Set up the new equilibrium expression with the added amount.


Step 3: Detailed Explanation:

1. Initial Equilibrium:
\([X_2] = \frac{3}{1} = 3\) M, \([Y_2] = \frac{3}{1} = 3\) M, \([Z] = \frac{9}{1} = 9\) M.
\(K_c = \frac{[Z]^2}{[X_2][Y_2]} = \frac{9^2}{3 \times 3} = \frac{81}{9} = 9\).


2. New Equilibrium Condition:

10 moles of Z are added. Initial moles for new setup:
\(n(X_2) = 3\), \(n(Y_2) = 3\), \(n(Z) = 9 + 10 = 19\).

Since product is added, reaction moves backward. Let \(2x\) moles of Z react.

Moles at new equilibrium:
\(n(X_2) = 3 + x\), \(n(Y_2) = 3 + x\), \(n(Z) = 19 - 2x\).
\[ K_c = \frac{(19 - 2x)^2}{(3 + x)^2} = 9 \]
Taking square root on both sides:
\[ \frac{19 - 2x}{3 + x} = 3 \] (We take the positive root as \(x\) must result in positive concentrations)
\[ 19 - 2x = 3(3 + x) \] \[ 19 - 2x = 9 + 3x \] \[ 10 = 5x \Rightarrow x = 2 \]

3. Calculating New Z:

Moles of Z at new equilibrium \(= 19 - 2x = 19 - 2(2) = 15\) moles.


Step 4: Final Answer:

The number of moles of Z at the new equilibrium is 15.
Quick Tip: For reactions where \(\Delta n_g = 0\), the volume term cancels out in the \(K_c\) expression. You can work directly with moles if the volume is constant or if you are only asked for molar ratios.