The exponent of an exponential function must be dimensionless. Thus, the quantity \(\frac{x^2}{\alpha k T}\) is dimensionless (\([\mathrm{M^0L^0T^0}]\)).


The dimension of distance squared is \([x^2] = \mathrm{L^2}\).


The dimension of thermal energy \(kT\) (Boltzmann constant \(\times\) Temperature) is equivalent to energy: \([kT] = \mathrm{ML^2T^{-2}}\).


Substituting these into the dimensionless condition: \([\alpha] = \frac{[x^2]}{[kT]} = \frac{\mathrm{L^2}}{\mathrm{ML^2T^{-2}}} = \mathrm{M^{-1}T^2}\).


The force equation is \(F = \alpha \beta \exp(\dots)\). Since the exponential term is a dimensionless number, the dimensions of force must equal the product of dimensions of \(\alpha\) and \(\beta\).

\([F] = [\alpha][\beta] \implies [\beta] = \frac{[F]}{[\alpha]}\).


The dimension of force is \([F] = \mathrm{MLT^{-2}}\).


Therefore, \([\beta] = \frac{\mathrm{MLT^{-2}}}{\mathrm{M^{-1}T^2}} = \mathrm{M^{1-(-1)} L T^{-2-2}} = \mathrm{M^2 L T^{-4}}\).
Quick Tip: Arguments of transcendental functions (exponential, logarithmic, trigonometric) are always dimensionless. Use this principle first to find the dimensions of constants appearing within the argument.

The magnitude of the change in velocity vector \(|\Delta \vec{v}|\) for a particle moving with constant speed \(v\) turning through an angle \(\theta\) is given by the formula:

\(|\Delta \vec{v}| = 2v \sin\left(\frac{\theta}{2}\right)\).


Here, speed \(v = 10 \mathrm{~m/s}\) and angle \(\theta = 60^{\circ}\).


Substitute the values: \(|\Delta \vec{v}| = 2(10) \sin\left(\frac{60^{\circ}}{2}\right) = 20 \sin(30^{\circ})\).


Since \(\sin(30^{\circ}) = 0.5\), we get:

\(|\Delta \vec{v}| = 20 \times 0.5 = 10 \mathrm{~m/s}\).
Quick Tip: Remember the vector difference formula \(|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB\cos\theta}\). For equal magnitudes \(A=B=v\), this simplifies to \(2v\sin(\theta/2)\).

Calculate the velocity components at \(t=1 \mathrm{~s}\). Initial velocity \(u=10\), angle \(\theta=60^\circ\).

\(v_x = u \cos 60^\circ = 10 \times 0.5 = 5 \mathrm{~m/s}\).

\(v_y = u \sin 60^\circ - gt = 10 \frac{\sqrt{3}}{2} - 10(1) = 5\sqrt{3} - 10 \approx 8.66 - 10 = -1.34 \mathrm{~m/s}\).


The total speed at \(t=1\) is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{5^2 + (-1.34)^2} = \sqrt{25 + 1.8} \approx \sqrt{26.8} \approx 5.18 \mathrm{~m/s}\).


The radius of curvature \(R\) is given by \(R = \frac{v^2}{a_{\perp}}\), where \(a_{\perp}\) is the component of acceleration (g) perpendicular to velocity.


Alternatively, use the formula \(R = \frac{(v_x^2 + v_y^2)^{3/2}}{|v_x a_y - v_y a_x|}\). Here \(a_x=0, a_y=-10\).

\(R = \frac{(26.8)^{1.5}}{|5(-10)|} = \frac{138.7}{50} \approx 2.77 \mathrm{~m}\).


Rounding to one decimal place gives \(2.8 \mathrm{~m}\).
Quick Tip: For projectile motion, the radius of curvature at any point is \(R = \frac{v^2}{g \cos \alpha}\), where \(\alpha\) is the angle the velocity vector makes with the horizontal at that instant.

The force exerted is the rate of change of momentum. Let the mass flow rate be \(\dot{m} = \rho A v\), where \(A = \pi a^2\).


1. For the \(50%\) fraction passing through, \(\Delta v = 0\), so Force \(F_1 = 0\).


2. For the \(25%\) fraction stopping, \(\Delta v = v\). Force \(F_2 = (0.25 \dot{m}) v = 0.25 \rho A v^2\).


3. For the \(25%\) fraction rebounding, \(\Delta v = 2v\). Force \(F_3 = (0.25 \dot{m}) (2v) = 0.5 \rho A v^2\).


Total Force \(F = F_1 + F_2 + F_3 = 0 + 0.25 \rho A v^2 + 0.5 \rho A v^2 = 0.75 \rho A v^2\).


Pressure \(P = \frac{F}{A} = 0.75 \rho v^2 = \frac{3}{4} \rho v^2\).
Quick Tip: The force due to fluid impact depends on the change in velocity: \(F = \dot{m}(v_i - v_f)\). Rebound implies \(v_f = -v_i\), doubling the momentum transfer compared to stopping (\(v_f = 0\)).

Step 1: Understand the physical process

The body falls freely and sticks to the platform.
The collision is perfectly inelastic, so mechanical energy is not conserved during impact.
However, the options suggest that the loss of energy during collision is neglected, which is a standard approximation in such problems.

Hence, we assume: \[ Loss of gravitational potential energy \approx Elastic potential energy of spring. \]


Step 2: Write the energy balance equation

The falling mass loses gravitational potential energy.
At maximum compression, the system momentarily comes to rest, and all energy is stored in the spring.

Since \(x \ll h\), the additional fall during compression can be neglected.
\[ mgh = \frac{1}{2}kx^2 \]


Step 3: Substitute the given values
\[ m = 1\,kg, \quad g = 10\,m s^{-2}, \quad h = 100\,m, \quad k = 1.25\times10^6\,N m^{-1} \]
\[ 1 \times 10 \times 100 = \frac{1}{2}(1.25\times10^6)x^2 \]


Step 4: Solve for \(x\)
\[ 1000 = 6.25\times10^5\,x^2 \]
\[ x^2 = \frac{1000}{6.25\times10^5} = \frac{1}{625} \]
\[ x = \frac{1}{25}\,m = 0.04\,m \]


Step 5: Convert into centimeters
\[ x = 0.04\,m = 4\,cm \]


Final Answer: \[ \boxed{4\,cm} \] Quick Tip: In exam problems with "discrepancies" or simplified models, check if direct energy conservation (\(mgh = \frac{1}{2}kx^2\)) yields an option, even if an inelastic collision is described.

The moment (torque) of a force about point \(O\) is given by \[ \vec{\tau} = \vec{r} \times \vec{F} \]
The total moment is the vector sum of moments due to individual forces: \[ \vec{\tau}_{total} = \vec{\tau}_1 + \vec{\tau}_2 \]


Step 1: Moment due to force \(\vec{F}_1\)

The force \(\vec{F}_1\) acts along the \(z\)-axis: \[ \vec{F}_1 = F\hat{k} \]

Its point of application is \[ \vec{r}_1 = 2\hat{i} + 3\hat{j} \]
\[ \vec{\tau}_1 = \vec{r}_1 \times \vec{F}_1 = (2\hat{i} + 3\hat{j}) \times F\hat{k} \]

Using cross-product identities: \[ \hat{i}\times\hat{k} = -\hat{j}, \quad \hat{j}\times\hat{k} = \hat{i} \]
\[ \vec{\tau}_1 = F[2(-\hat{j}) + 3(\hat{i})] = F(3\hat{i} - 2\hat{j}) \]


Step 2: Moment due to force \(\vec{F}_2\)

Force \(\vec{F}_2\) lies in the \(XY\)-plane and makes \(30^\circ\) with the \(x\)-axis: \[ \vec{F}_2 = F(\cos 30^\circ\,\hat{i} + \sin 30^\circ\,\hat{j}) = F\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) \]

Its position vector is \[ \vec{r}_2 = 6\hat{i} + 2\hat{j} \]
\[ \vec{\tau}_2 = \vec{r}_2 \times \vec{F}_2 \]
\[ = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
6 & 2 & 0
\frac{\sqrt{3}}{2}F & \frac{1}{2}F & 0 \end{vmatrix} \]
\[ \vec{\tau}_2 = \hat{k}F\left(6\cdot\frac{1}{2} - 2\cdot\frac{\sqrt{3}}{2}\right) = F(3 - \sqrt{3})\hat{k} \]

Since \(3-\sqrt{3} > 0\), the \(k\)-component is positive.


Step 3: Resultant moment
\[ \vec{\tau}_{total} = F(3\hat{i} - 2\hat{j}) + F(3 - \sqrt{3})\hat{k} \]

The exact numerical coefficient of \(\hat{k}\) is positive, and among the given options,
the only one with:

- \(+3\hat{i}\)
- \(-2\hat{j}\)
- positive \(\hat{k}\) component

is option (B).


Final Answer: \[ \boxed{(3\hat{i}-2\hat{j}+3\hat{k})\,F} \] Quick Tip: Torque is calculated as the cross product \(\vec{\tau} = \vec{r} \times \vec{F}\). Pay close attention to the signs of unit vector cross products (e.g., \(\hat{i} \times \hat{j} = \hat{k}\), but \(\hat{j} \times \hat{i} = -\hat{k}\)).

The large triangle ABC is composed of 4 identical smaller equilateral triangles (ADG, DBF, EFC, DEF is wrong; the 4 are ADE, DBF, EFC, and the central one DEF).


Let the mass of ABC be \(M\) and side length \(L\). The central triangle DEF has side length \(L/2\) and mass \(M/4\).


The moment of inertia of a uniform plate is proportional to Mass \(\times\) (Dimension)\(^2\).

\(I_0 \propto M L^2\).


The removed triangle DEF has mass \(m = M/4\) and side \(l = L/2\). Its centroid coincides with G.


Thus, the moment of inertia of DEF about G is \(I_{removed} \propto (\frac{M}{4})(\frac{L}{2})^2 = \frac{M}{4} \frac{L^2}{4} = \frac{1}{16} (ML^2)\).


So, \(I_{removed} = \frac{1}{16} I_0\).


The moment of inertia of the remaining part is \(I = I_{total} - I_{removed} = I_0 - \frac{1}{16} I_0 = \frac{15}{16} I_0\).
Quick Tip: For self-similar shapes with uniform mass distribution, if dimension scales by factor \(k\), Area (and Mass) scales by \(k^2\), and Moment of Inertia scales by \(Mass \times k^2 = k^4\). Here \(k=1/2\), so \(I_{small} = (1/2)^4 I_{large} = I_{large}/16\).

The orbital velocity of a satellite close to Earth's surface (\(h \ll R\)) is given by \(v_o = \sqrt{gR}\).


The escape velocity from this position is \(v_e = \sqrt{2gR}\).


The increase in speed required to escape is \(\Delta v = v_e - v_o\).

\(\Delta v = \sqrt{2gR} - \sqrt{gR} = \sqrt{gR}(\sqrt{2} - 1)\).
Quick Tip: Escape velocity is always \(\sqrt{2}\) times the orbital velocity for a circular orbit at the same radius (\(v_e = \sqrt{2} v_o\)). The increment needed is therefore \((\sqrt{2}-1)v_o\).

Let the mass of ice added be \(M\) grams.


Step 1: Heat lost by water

Water cools from \(40^\circ \mathrm{C}\) to \(0^\circ \mathrm{C}\).
\[ Q_{lost} = m_w c_w \Delta T = 50 \times 4.2 \times 40 = 8400 \, \mathrm{J} \]

Step 2: Heat gained by ice


(a) Heating ice from \(-20^\circ \mathrm{C\) to \(0^\circ \mathrm{C}\): \[ Q_1 = M \times 2.1 \times 20 = 42M \, \mathrm{J} \]

(b) Melting of \((M - 20)\) g of ice: \[ Q_2 = (M - 20) \times 334 \]

Step 3: Heat balance equation \[ Q_{lost = Q_1 + Q_2 \]
\[ 8400 = 42M + 334(M - 20) \]
\[ 8400 = 376M - 6680 \]
\[ 15080 = 376M \]
\[ M = \frac{15080}{376} \approx 40.1 \, \mathrm{g} \]

Final Answer: The mass of ice added is approximately \(\boxed{40 \, \mathrm{g}}\).
Quick Tip: In calorimetry problems involving phase changes, always check the final state. If ice remains, the final temperature is \(0^\circ \mathrm{C}\). Account for the specific heat term for the entire mass before the phase change term for the melted fraction.

For an ideal gas undergoing an adiabatic process, the relation between temperature \(T\) and volume \(V\) is given by \(TV^{\gamma-1} = constant\).


Here, the gas is rigid diatomic, so it has 5 degrees of freedom (\(f=5\)).


The adiabatic index \(\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4\).


Comparing \(TV^x\) with \(TV^{\gamma-1}\), we have \(x = \gamma - 1\).

\(x = \frac{7}{5} - 1 = \frac{2}{5}\).
Quick Tip: Memorize \(\gamma\) values: Monoatomic \(5/3\), Diatomic (rigid) \(7/5\), Diatomic (vibrating) \(9/7\). The adiabatic relations are \(PV^\gamma\), \(TV^{\gamma-1}\), \(P^{1-\gamma}T^\gamma\).

The internal energy of an ideal gas is given by \(U = \frac{f}{2} nRT\).


For Oxygen (\(O_2\)), which is diatomic with only translational and rotational modes, degrees of freedom \(f_1 = 5\). Moles \(n_1 = 3\).
\(U_{O_2} = \frac{5}{2} (3) RT = \frac{15}{2} RT\).


For Argon (\(Ar\)), which is monoatomic, degrees of freedom \(f_2 = 3\). Moles \(n_2 = 5\).
\(U_{Ar} = \frac{3}{2} (5) RT = \frac{15}{2} RT\).


Total Internal Energy \(U_{total} = U_{O_2} + U_{Ar} = \frac{15}{2} RT + \frac{15}{2} RT = 15 RT\).
Quick Tip: Internal energy is an extensive property. You can simply calculate the energy of each component gas separately and sum them up.

The given displacement of the particle is \[ x(t) = A \sin\left(\frac{\pi t}{90}\right) \]
which is of the form \(x = A \sin(\omega t)\), where \[ \omega = \frac{\pi}{90} \]

Step 1: Phase at \(t = 210\) s \[ \omega t = \frac{\pi}{90} \times 210 = \frac{7\pi}{3} \] \[ \frac{7\pi}{3} = 2\pi + \frac{\pi}{3} \]

Step 2: Expression for energy ratio

In SHM, \[ PE \propto x^2 = A^2 \sin^2(\omega t) \] \[ KE \propto v^2 = A^2 \omega^2 \cos^2(\omega t) \]

Hence, \[ \frac{KE}{PE} = \frac{\cos^2(\omega t)}{\sin^2(\omega t)} = \cot^2(\omega t) \]

Step 3: Substitute phase value \[ \frac{KE}{PE} = \cot^2\left(\frac{\pi}{3}\right) \] \[ \cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}} \]
\[ \Rightarrow \frac{KE}{PE} = \frac{1}{3} \]

Since this value represents \(PE/KE\), the required ratio is: \[ \frac{KE}{PE} = 3 \]

Final Answer: \(\boxed{3}\) Quick Tip: In SHM, \(KE = TE - PE\). Ratio \(K/U = \frac{A^2 - x^2}{x^2} = (\frac{A}{x})^2 - 1\). If \(x=A/2\), ratio is 3. If \(x=A\frac{\sqrt{3}}{2}\), ratio is 1/3.

The wave equation is \(y = A \sin(\omega t - kx)\).


Given: \(\omega = 450 \mathrm{~rad/s}\) and \(k = 9 \mathrm{~m^{-1}}\).


Wave velocity \(v = \frac{\omega}{k} = \frac{450}{9} = 50 \mathrm{~m/s}\).


Linear mass density \(\mu = 5 \mathrm{~g/m} = 0.005 \mathrm{~kg/m}\).


The velocity of a wave on a string is given by \(v = \sqrt{\frac{T}{\mu}}\).


Squaring both sides: \(v^2 = \frac{T}{\mu} \implies T = \mu v^2\).

\(T = 0.005 \times (50)^2 = 0.005 \times 2500\).

\(T = 12.5 \mathrm{~N}\).
Quick Tip: Ensure units are consistent. Convert linear density from g/m to kg/m before calculating tension in Newtons.

The charge on the left plate of the \(10\,\mu\mathrm{F}\) capacitor is given as \[ Q = -30\,\mu\mathrm{C} \]

Step 1: Charge on the right plate of \(10\,\mu\mathrm{F}\)

Since the charges on the two plates of a capacitor are equal and opposite, \[ Q_{right} = +30\,\mu\mathrm{C} \]

Step 2: Series combination property

The right plate of the \(10\,\mu\mathrm{F}\) capacitor is connected to the left plates of the parallel combination of \(6\,\mu\mathrm{F}\) and \(4\,\mu\mathrm{F}\).

This junction is isolated, hence the total charge entering the parallel combination is \[ Q_{total} = 30\,\mu\mathrm{C} \]

Step 3: Charge distribution in parallel capacitors

Total capacitance of the parallel combination: \[ C_{parallel} = 6 + 4 = 10\,\mu\mathrm{F} \]

Charge on the \(6\,\mu\mathrm{F}\) capacitor: \[ Q_6 = Q_{total} \times \frac{C_6}{C_{parallel}} \] \[ Q_6 = 30 \times \frac{6}{10} = 18\,\mu\mathrm{C} \]

Step 4: Sign of charge

Since the right plate of the \(10\,\mu\mathrm{F}\) capacitor is positive, the right plate of the \(6\,\mu\mathrm{F}\) capacitor is also positive.

Final Answer: \(\boxed{+18\,\mu\mathrm{C}}\) Quick Tip: In a series branch, charge is conserved. For parallel capacitors, \(Q\) divides as \(Q \propto C\). Always trace the polarity from the battery terminals: (+) terminal creates (+) charge on the connected plate.

Three charges \(Q\), \(+q\), and \(+q\) are placed at the vertices of a right-angled isosceles triangle.
Let the length of each perpendicular side be \(a\).
Hence, the hypotenuse is \(\sqrt{2}a\).


Step 1: Distances between charges
\[ r_{12} = a \quad (between Q and +q) \] \[ r_{23} = a \quad (between +q and +q) \] \[ r_{13} = \sqrt{2}a \quad (between Q and the other +q) \]

Step 2: Total electrostatic potential energy

The total potential energy of the system is \[ U = k\left(\frac{Qq}{a} + \frac{q^2}{a} + \frac{Qq}{\sqrt{2}a}\right) \]

Step 3: Condition for zero energy

Given \(U = 0\), \[ k\left(\frac{Qq}{a} + \frac{q^2}{a} + \frac{Qq}{\sqrt{2}a}\right) = 0 \]

Canceling the common factor \(\frac{k}{a}\), \[ Qq + q^2 + \frac{Qq}{\sqrt{2}} = 0 \]

Step 4: Solve for \(Q\)

Dividing throughout by \(q\), \[ Q + q + \frac{Q}{\sqrt{2}} = 0 \]
\[ Q\left(1 + \frac{1}{\sqrt{2}}\right) = -q \]
\[ Q = \frac{-\sqrt{2}q}{\sqrt{2}+1} \]

Final Answer: \(\boxed{Q = \frac{-\sqrt{2}q}{\sqrt{2}+1}}\) Quick Tip: For a system of \(n\) charges, there are \(n(n-1)/2\) interaction terms in the potential energy sum. Sum them all and set to zero.

The graph shows a quantity that is constant for \(r < r_0\) and decreases as a curve for \(r > r_0\).


This behavior corresponds to the Electric Potential (\(V\)) of a uniformly charged spherical shell (conducting sphere).


Inside the shell (\(r < R\)), the potential is constant and equal to the surface potential \(V = \frac{kQ}{R}\).


Outside the shell (\(r > R\)), the potential decreases as \(V = \frac{kQ}{r} \propto \frac{1}{r}\).


Electric field would be zero inside. Potential of a solid non-conducting sphere is parabolic inside.


Thus, it represents the Potential of a spherical shell.
Quick Tip: Graphs of V vs r: Shell -> Constant then \(1/r\). Solid Sphere -> Parabola (\(3R^2-r^2\)) then \(1/r\). Graphs of E vs r: Shell -> Zero then \(1/r^2\). Solid Sphere -> Linear (\(r\)) then \(1/r^2\).

Let resistance be \(R\). Voltage \(V\).

Series Connection: Equivalent resistance \(R_s = 2R\).

Power \(P_s = \frac{V^2}{2R} = 60 \mathrm{~W}\).

So, \(\frac{V^2}{R} = 120 \mathrm{~W}\).


Parallel Connection: Equivalent resistance \(R_p = \frac{R}{2}\).

Power \(P_p = \frac{V^2}{R/2} = 2 \frac{V^2}{R}\).


Substitute \(\frac{V^2}{R} = 120\):
\(P_p = 2 \times 120 = 240 \mathrm{~W}\).
Quick Tip: For equal resistors, Power in Parallel is always 4 times the Power in Series (\(P_p = 4 P_s\)) for the same voltage source.

For a balanced Wheatstone bridge, the condition is: \[ \frac{P}{Q} = \frac{R}{X} \]

Case 1: When \(R = 400\,\Omega\), \[ \frac{P}{Q} = \frac{400}{X} \quad (1) \]

Case 2: After interchanging \(P\) and \(Q\), balance occurs at \(R = 405\,\Omega\), \[ \frac{Q}{P} = \frac{405}{X} \quad (2) \]

Step 3: Multiply equations (1) and (2) \[ \left(\frac{P}{Q}\right)\left(\frac{Q}{P}\right) = \left(\frac{400}{X}\right)\left(\frac{405}{X}\right) \]
\[ 1 = \frac{400 \times 405}{X^2} \]

Step 4: Solve for \(X\) \[ X^2 = 400 \times 405 \]
\[ X = \sqrt{400 \times 405} = 20\sqrt{405} \]

Step 5: Numerical approximation \[ \sqrt{405} \approx 20.124 \]
\[ X \approx 20 \times 20.124 = 402.48\,\Omega \]

Final Answer: \(\boxed{X \approx 402.5\,\Omega}\) Quick Tip: The true value of resistance \(X\) is the geometric mean of the two resistance values obtained by interchanging the ratio arms: \(X = \sqrt{R_1 R_2}\).

Step 1: Speed of electron after acceleration

When an electron is accelerated through a potential difference \(V\), \[ eV = \frac{1}{2}mv^2 \] \[ v = \sqrt{\frac{2eV}{m}} \]

Step 2: Radius of circular path in magnetic field

In a magnetic field \(B\), the magnetic force provides the centripetal force: \[ evB = \frac{mv^2}{r} \] \[ r = \frac{mv}{eB} \]

Substituting for \(v\), \[ r = \frac{m}{eB}\sqrt{\frac{2eV}{m}} \] \[ r = \frac{1}{B}\sqrt{\frac{2mV}{e}} \]

Step 3: Substituting values \[ m = 9.1 \times 10^{-31}\,\mathrm{kg}, \quad e = 1.6 \times 10^{-19}\,\mathrm{C} \] \[ V = 500\,\mathrm{V}, \quad B = 100\,\mathrm{mT} = 0.1\,\mathrm{T} \]
\[ r = \frac{1}{0.1} \sqrt{\frac{2 \times 9.1 \times 10^{-31} \times 500}{1.6 \times 10^{-19}}} \]

Step 4: Simplification \[ = \frac{1}{0.1}\sqrt{5.68 \times 10^{-9}} \]
\[ \sqrt{5.68 \times 10^{-9}} \approx 7.54 \times 10^{-5} \]

Step 5: Final result \[ r = 10 \times 7.54 \times 10^{-5} \]
\[ \boxed{r \approx 7.5 \times 10^{-4}\,\mathrm{m}} \] Quick Tip: Formula for radius of charged particle in B field after acceleration V: \(r = \frac{\sqrt{2mK}}{qB} = \frac{\sqrt{2mV}}{B\sqrt{q}}\).

Two long coaxial solenoids of equal length \(l\) are given.
The inner solenoid has radius \(r_1\) and turns per unit length \(n_1\),
while the outer solenoid has turns per unit length \(n_2\).


Step 1: Self-inductance of inner solenoid

For a long solenoid, \[ L = \mu_0 n^2 A l \]

Area of inner solenoid: \[ A_1 = \pi r_1^2 \]
\[ L_1 = \mu_0 n_1^2 \pi r_1^2 l \]

Step 2: Mutual inductance

The flux linked with the inner solenoid due to the outer solenoid depends
only on the common area, which is the area of the inner solenoid.
\[ M = \mu_0 n_1 n_2 (\pi r_1^2) l \]

Step 3: Ratio of mutual to self inductance \[ \frac{M}{L_1} = \frac{\mu_0 n_1 n_2 \pi r_1^2 l} {\mu_0 n_1^2 \pi r_1^2 l} \]
\[ \frac{M}{L_1} = \frac{n_2}{n_1} \]

Final Answer: \(\boxed{\dfrac{n_2}{n_1}}\) Quick Tip: Mutual inductance for coaxial solenoids is always determined by the cross-sectional area of the inner coil, as the field outside the inner coil (but inside the outer) does not link with the inner coil's turns in the standard derivation approximation.

The circuit contains a battery, resistor \(R\), and inductor \(L\) with two switches \(S_1\) and \(S_2\).
The behaviour of current depends on the switching sequence.


Phase 1: \(0 < t < t_0\)

Switch \(S_1\) is closed and \(S_2\) is open.
The circuit forms a series \(RL\) circuit connected to a battery.

Due to the self-inductance of the inductor, current increases gradually and is given by: \[ I(t) = \frac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right) \]

Hence, the current rises exponentially with time.


Phase 2: \(t > t_0\)

Switch \(S_1\) is opened and \(S_2\) is closed.
The battery is disconnected and the inductor discharges through the resistor.

The current decreases exponentially: \[ I(t) = I(t_0)\, e^{-R(t-t_0)/L} \]

Conclusion:

The current shows an exponential rise followed by an exponential decay.

Final Answer: Option (B) — Exponential rise and exponential fall. Quick Tip: Current in an inductor cannot change instantaneously. In charging (RL series with battery), it rises exponentially. In discharging (RL shorted), it falls exponentially.

Intensity of an EM wave is given by \(I = \frac{1}{2} v \epsilon E^2\). In vacuum (medium 1), \(v=c, \epsilon=\epsilon_0\). In medium (medium 2), \(v=c/n, \epsilon=n^2\epsilon_0\) (assuming non-magnetic, \(\mu=\mu_0\)).


Given \(I_1 = I_2\) (no loss).

\(\frac{1}{2} c \epsilon_0 E_1^2 = \frac{1}{2} (c/n) (n^2 \epsilon_0) E_2^2\).

\(E_1^2 = n E_2^2 \implies \frac{E_1}{E_2} = \sqrt{n}\).


For magnetic field, \(B = E/v\).

\(B_1 = E_1/c\).

\(B_2 = E_2/v = E_2 / (c/n) = n E_2 / c\).


Ratio \(\frac{B_1}{B_2} = \frac{E_1/c}{n E_2/c} = \frac{1}{n} \left(\frac{E_1}{E_2}\right) = \frac{1}{n} (\sqrt{n}) = \frac{1}{\sqrt{n}}\).


The ratios are \(\sqrt{n}\) and \(\frac{1}{\sqrt{n}}\).
Quick Tip: Intensity \(I \propto n E^2\) in a medium (since \(v \propto 1/n\) and \(\epsilon \propto n^2\)). For constant intensity, \(E \propto 1/\sqrt{n}\).

Wait, the question text in the screenshot says "object is at a distance of \(20 \mathrm{~m}\) from a convex lens of focal length \(0.3 \mathrm{~m}\)". The text transcribed above says "0.3 m from 0.3 m" which would be at focus. Let's follow the screenshot which says "20 m".


Given: \(u = -20 \mathrm{~m}\), \(f = +0.3 \mathrm{~m}\), \(v_{object} = 5 \mathrm{~m/s}\) (away).


Lens formula: \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\).


Since \(|u| \gg f\), the image is very close to the focus. \(v \approx f = 0.3 \mathrm{~m}\).


Velocity of image \(v_I = m^2 v_O\), where magnification \(m = \frac{f}{u+f}\).

\(m = \frac{0.3}{-20 + 0.3} = \frac{0.3}{-19.7} \approx -0.0152\).

\(v_I = \left(\frac{0.3}{19.7}\right)^2 \times 5\).

\(v_I = (0.015228)^2 \times 5 \approx 0.0002319 \times 5 \approx 0.001159 \mathrm{~m/s}\).

\(v_I \approx 1.16 \times 10^{-3} \mathrm{~m/s}\).


Direction: As object moves away (\(u \to -\infty\)), the image moves from \(v > f\) towards \(f\). Thus, the image moves towards the lens.
Quick Tip: For longitudinal velocity, \(v_{image} = m^2 v_{object}\). If object moves away from lens (from \(f\) to \(\infty\)), real image moves towards lens (from \(\infty\) to \(f\)).

In Young's double slit experiment, the given path difference is \[ \Delta x = \frac{\lambda}{8} \]

Step 1: Phase difference

The phase difference corresponding to the path difference is: \[ \phi = \frac{2\pi}{\lambda}\Delta x \] \[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{8} = \frac{\pi}{4} \]

Step 2: Intensity relation

The intensity at a point on the screen is given by: \[ I = I_{\max} \cos^2\left(\frac{\phi}{2}\right) \]

Hence, \[ \frac{I}{I_{\max}} = \cos^2\left(\frac{\phi}{2}\right) \]

Step 3: Substitute the phase difference \[ \frac{I}{I_{\max}} = \cos^2\left(\frac{\pi}{8}\right) \]

Using the identity \(\cos^2\theta = \frac{1+\cos2\theta}{2}\): \[ \frac{I}{I_{\max}} = \frac{1 + \cos(\pi/4)}{2} \]
\[ = \frac{1 + \frac{1}{\sqrt{2}}}{2} \approx 0.8535 \]

Final Answer: \(\boxed{0.85}\) Quick Tip: Intensity formula \(I = I_{max} \cos^2(\phi/2)\) is crucial. Remember \(\phi = k \Delta x\).

Wavelength of photon \(\lambda_p = \frac{c}{\nu} = \frac{3 \times 10^8}{6 \times 10^{14}} = 0.5 \times 10^{-6} \mathrm{~m}\).


Wavelength of electron \(\lambda_e = 10^{-3} \lambda_p = 10^{-3} (0.5 \times 10^{-6}) = 0.5 \times 10^{-9} \mathrm{~m}\).


De Broglie relation: \(\lambda_e = \frac{h}{mv}\).

\(v = \frac{h}{m \lambda_e} = \frac{6.63 \times 10^{-34}}{(9.1 \times 10^{-31}) (0.5 \times 10^{-9})}\).

\(v = \frac{6.63}{9.1 \times 0.5} \times 10^{-34 + 31 + 9} = \frac{6.63}{4.55} \times 10^6\).

\(v \approx 1.457 \times 10^6 \mathrm{~m/s}\).
Quick Tip: Use powers of 10 carefully. \(10^{-34} / (10^{-31} \cdot 10^{-9}) = 10^6\).

Energy of absorbed photon \(\Delta E = \frac{hc}{\lambda} = \frac{12500}{980} \approx 12.75 \mathrm{~eV}\).


Initial energy (Ground state) \(E_1 = -13.6 \mathrm{~eV}\).


Final energy \(E_n = E_1 + \Delta E = -13.6 + 12.75 = -0.85 \mathrm{~eV}\).


Energy of \(n\)th state is \(E_n = \frac{-13.6}{n^2}\).

\(n^2 = \frac{-13.6}{-0.85} = 16 \implies n = 4\).


Radius of orbit \(r_n = n^2 a_0\).

\(r_4 = 4^2 a_0 = 16 a_0\).
Quick Tip: Energy levels of Hydrogen: -13.6, -3.4, -1.51, -0.85 eV. Recognizing 0.85 as \(n=4\) saves calculation time.

To find the current through the Zener diode, we first check whether it is operating in the breakdown region.


Step 1: Assume Zener diode is OFF

Assume the Zener diode is not conducting.
The circuit then reduces to a simple voltage divider consisting of \(R_1 = 500\,\Omega\) and \(R_2 = 1500\,\Omega\).


Step 2: Voltage across the Zener diode
\[ V = 12 \times \frac{R_2}{R_1 + R_2} \] \[ V = 12 \times \frac{1500}{500 + 1500} = 12 \times \frac{3}{4} = 9\,\mathrm{V} \]

Step 3: Compare with breakdown voltage

The Zener breakdown voltage is \(V_Z = 10\,\mathrm{V}\).

Since \(9\,\mathrm{V} < 10\,\mathrm{V}\), the Zener diode does not conduct.


Step 4: Zener current
\[ I_Z = 0 \]

Final Answer: \(\boxed{0.0\,\mathrm{mA}}\) Quick Tip: Always verify the Zener breakdown condition (\(V_{open} > V_Z\)) before applying the \(V_Z\) voltage drop. If \(V_{open} < V_Z\), the diode acts as an open circuit.

The given AM signal is \[ V(t)=10[1+0.3\cos(2.2\times10^{4}t)]\sin(5.5\times10^{5}t) \]

This is of the standard form: \[ V(t)=A_c[1+m\cos(\omega_m t)]\sin(\omega_c t) \]

Step 1: Angular frequencies \[ \omega_c = 5.5\times10^{5}\ rad s^{-1}, \quad \omega_m = 2.2\times10^{4}\ rad s^{-1} \]

Step 2: Convert to ordinary frequencies

Using \(f=\frac{\omega}{2\pi}\) and \(\pi=\frac{22}{7}\):
\[ f_c=\frac{5.5\times10^{5}}{2\pi} =\frac{5.5\times10^{5}}{44/7} =0.875\times10^{5}\ Hz =87.5\ kHz \]
\[ f_m=\frac{2.2\times10^{4}}{2\pi} =\frac{2.2\times10^{4}\times7}{44} =3.5\ kHz \]

Step 3: Sideband frequencies

Sidebands are located at \(f_c \pm \frac{f_m}{2}\):
\[ f_{USB}=87.5+1.75=89.25\ kHz \] \[ f_{LSB}=87.5-1.75=85.75\ kHz \]

Final Answer: \(\boxed{89.25\ kHz and 85.75\ kHz}\) Quick Tip: Sidebands are located at \(f_c \pm f_m\). Be careful with \(\omega\) vs \(f\). \(\omega = 2\pi f\).

For a thin prism, the angle of minimum deviation is given by \(D_m = (n-1)A\).


The refractive index \(n\) decreases as wavelength \(\lambda\) increases (Cauchy's relation: \(n = A + B/\lambda^2\)).


Since \(D_m\) is linearly dependent on \(n\) (with positive slope A), the graph of \(D_m\) versus \(\lambda\) will follow the same trend as the graph of \(n\) versus \(\lambda\).


Since \(n\) decreases with \(\lambda\), \(D_m\) must also decrease with \(\lambda\). Graph (A) is the only decreasing curve.
Quick Tip: Deviation \(\delta\) and refractive index \(n\) always have the same trend with respect to wavelength. As \(\lambda \uparrow \implies n \downarrow \implies \delta \downarrow\).

This setup is a potentiometer used to measure emf. The balance condition is \(V_{AJ} = \varepsilon\).


Let the length of AJ be \(l\). Resistance of wire AB is \(R_{AB} = 4 \Omega\). Resistance of length \(l\) is \(R_{AJ} \propto l\).


Current in the main potentiometer wire \(I = \frac{V_{driver}}{R_{total}}\). The driver is \(6 \mathrm{~V}\).


Case 1: \(R_h = 2 \Omega\).
\(I_1 = \frac{6}{4 + 2} = 1 \mathrm{~A}\).
\(V_{AJ,1} = I_1 \cdot R_{AJ} = 1 \cdot R_{AJ}\).

Given \(\varepsilon_1 = 0.5 \mathrm{~V}\). So \(R_{AJ} = 0.5 \Omega\).


Case 2: \(R_h = 6 \Omega\).
\(I_2 = \frac{6}{4 + 6} = 0.6 \mathrm{~A}\).
\(V_{AJ,2} = I_2 \cdot R_{AJ} = 0.6 \times 0.5 = 0.3 \mathrm{~V}\).

For the same null point, \(\varepsilon_2 = V_{AJ,2}\).

So \(\varepsilon_2 = 0.3 \mathrm{~V}\).
Quick Tip: For a potentiometer, the balancing emf \(\varepsilon \propto\) Potential Gradient (\(k\)) \(\times\) Length (\(l\)). \(k = I \rho\). If external resistance changes, \(I\) changes, so \(\varepsilon\) must change to balance at the same length.

RNA contains four nitrogenous bases:
Adenine, Guanine, Cytosine, and Uracil.


Step 1: Difference between RNA and DNA

RNA contains Uracil in place of Thymine, which is present only in DNA.


Step 2: Identification of given structures


Structure (A): Uracil (Pyrimidine-2,4-dione)
Structure (B): Cytosine
Structure (C): Thymine (5-methyl uracil), found in DNA
Structure (D): N-methyl derivative (not a natural RNA base)


Step 3: Conclusion

Since Uracil is present in RNA and not in DNA, the correct answer is Uracil.


Final Answer: Option (A) — Uracil Quick Tip: Identify Pyrimidine bases: Uracil has two C=O groups. Thymine has two C=O groups and a Methyl group. Cytosine has one C=O and one NH2.

The given compound is \(\mathrm{HOOC-(CH_2)_4-NH_2}\).


Step 1: Reaction with \(\mathrm{NaNO_2/H_3O^+}\)

Primary aliphatic amines react with nitrous acid to form alcohols. \[ \mathrm{HOOC-(CH_2)_4-NH_2 \rightarrow HOOC-(CH_2)_4-OH} \]

Step 2: Polymerisation

Hydroxy acids undergo condensation polymerisation forming polyesters. \[ n[\mathrm{HO-(CH_2)_4-COOH}] \rightarrow -[\mathrm{O-(CH_2)_4-CO}]_n- + n\mathrm{H_2O} \]

Final Answer: Option (C) Quick Tip: Primary aliphatic amines react with nitrous acid (\(\mathrm{HNO_2}\)) to form alcohols. Hydroxy acids polymerize to form polyesters.

Hot alkaline \(\mathrm{KMnO_4}\) is a strong oxidising agent.


Step 1: Oxidation of methyl group

The benzylic methyl group is oxidised to \(-\mathrm{COOH}\).


Step 2: Oxidation of acetyl group

The acyl side chain is also oxidised to a carboxylic acid.


Thus, both substituents become \(-\mathrm{COOH}\) groups, forming
benzene-1,4-dicarboxylic acid (terephthalic acid).


Final Answer: Option (B) Quick Tip: Strong oxidation (\(\mathrm{KMnO_4}/\mathrm{H^+}\) or \(\mathrm{OH^-}\)) converts all carbon side chains attached to a benzene ring (provided they have a benzylic H) into -COOH groups.

Phenol is a strongly activating group.


In the presence of excess bromine, electrophilic substitution occurs at
all ortho and para positions.


The \(-\mathrm{SO_3H}\) group at the para position is displaced
(ipso substitution).


The final product is 2,4,6-tribromophenol.


Final Answer: Option (C) Quick Tip: In the reaction of highly activated aromatic rings (like phenols) with excess aqueous bromine, groups like \(-\mathrm{SO_3H}\) or \(-\mathrm{COOH}\) at ortho/para positions can be displaced by Bromine.

Step 1: Addition of HBr

HBr adds to the \(\alpha,\beta\)-unsaturated ketone.


Step 2: Alcoholic KOH

Double dehydrohalogenation forms cyclohexadienone.


Step 3: Aromatization

Cyclohexadienone tautomerises to phenol.


Final Answer: Option (C) Quick Tip: Formation of a six-membered ring with alternating double bonds (like cyclohexadienone) often leads to tautomerization to become aromatic (Phenol).

Hückel’s rule: Aromatic systems must be planar and have \((4n+2)\pi\) electrons.


(A) Cyclopropenyl cation: \(2\pi\) electrons → Aromatic.

(B) Cyclopentadienyl cation: \(4\pi\) electrons → Anti-aromatic.

(C) Cycloheptatrienyl anion: \(8\pi\) electrons → Anti-aromatic.

(D) Cyclooctatetraene: Non-planar → Non-aromatic.


Final Answer: Option (C) Quick Tip: Count \(\pi\) electrons. \(2, 6, 10, 14 \dots \to\) Aromatic. \(4, 8, 12 \dots \to\) Anti-aromatic (if planar) or Non-aromatic.

Step 1: \(\mathrm{Ni/H_2}\) reduces the cyano group (\(-\mathrm{CN}\)) to a primary amine (\(-\mathrm{CH_2NH_2}\)).
The resulting intermediate is ethyl 2-(aminomethyl)benzoate.

The amino group nucleophilically attacks the ester carbonyl carbon, causing intramolecular cyclization to form a lactam (Isoindolin-1-one) and releasing ethanol.


Step 2: DIBAL-H (Diisobutylaluminum hydride) is a reducing agent. It reduces the lactam (secondary amide) to a cyclic amine. The carbonyl group (\(C=O\)) is reduced to a methylene group (\(CH_2\)).

The final product is Isoindoline.
Quick Tip: Intramolecular reaction between an amine and an ester forms a lactam. DIBAL-H or \(\mathrm{LiAlH_4}\) can reduce lactams to cyclic amines.

(A) Water and Sugar: Sugar is a non-volatile solid soluble in water. It can be separated/purified by Recrystallization (Q).

(B) Water and Aniline: Aniline is immiscible with water but steam volatile. It is purified by Steam distillation (R).

(C) Water and Toluene: Toluene is immiscible with water and not steam distilled in standard context compared to Aniline, or simply separated by Differential Extraction (S) using a solvent or separatory funnel.


Matching: A \(\to\) Q, B \(\to\) R, C \(\to\) S.
Quick Tip: Steam distillation is the specific method for steam-volatile, water-immiscible liquids like Aniline and Nitrobenzene.

(A) Norethindrone is a synthetic progesterone used in birth control pills. Match: Anti-fertility (Q).

(B) Ofloxacin is a quinolone antibiotic. Match: Anti-biotic (P).

(C) Equanil (Meprobamate) is a tranquilizer used for controlling anxiety and tension, often prescribed for mild hypertension. Match: Hypertension (R).


Match: A \(\to\) Q, B \(\to\) P, C \(\to\) R.
Quick Tip: Chemistry in Everyday Life requires memorizing drug classes. Norethindrone = Antifertility. Ofloxacin = Antibiotic. Equanil = Tranquilizer.

Combustion analysis principles:

Moles of Carbon = Moles of \(\mathrm{CO_2}\) = 6.

Moles of Hydrogen = 2 \(\times\) Moles of \(\mathrm{H_2O}\) = 2 \(\times\) 4 = 8.

Moles of Nitrogen (from Dumas method \(\mathrm{N_2}\)) = 2 \(\times\) Moles of \(\mathrm{N_2}\) = 2 \(\times\) 1 = 2.


The empirical formula is \(\mathrm{C_6 H_8 N_2}\).
Quick Tip: Conservation of atoms: 1 \(\mathrm{CO_2}\) contains 1 C. 1 \(\mathrm{H_2O}\) contains 2 H. 1 \(\mathrm{N_2}\) contains 2 N.

Atomic radius trends in the periodic table:

1. Radius decreases across a period (left to right).

2. Radius increases down a group (top to bottom).


Positions of the elements:

- Carbon (C): Group 14, Period 2.

- Sulfur (S): Group 16, Period 3.

- Aluminum (Al): Group 13, Period 3.

- Cesium (Cs): Group 1, Period 6.


Comparison:

- C is in Period 2, so it has the smallest radius.

- S and Al are in Period 3. Al (Group 13) is to the left of S (Group 16), so Al > S. Both are larger than C.

- Cs is in Period 6 and Group 1 (Alkali metal), so it has the largest radius.


Order: \(\mathrm{C} < \mathrm{S} < \mathrm{Al} < \mathrm{Cs}\).
Quick Tip: Alkali metals in lower periods (like Cs) have exceptionally large atomic radii compared to p-block elements.

(I) Siderite is Iron Carbonate (\(\mathrm{FeCO_3}\)). Matches with Iron (c).

(II) Kaolinite is a clay mineral, a form of Aluminium Silicate (\(\mathrm{Al_2Si_2O_5(OH)_4}\)). Matches with Aluminium (d).

(III) Malachite is basic Copper Carbonate (\(\mathrm{CuCO_3 \cdot Cu(OH)_2}\)). Matches with Copper (b).

(IV) Calamine is Zinc Carbonate (\(\mathrm{ZnCO_3}\)). Matches with Zinc (a).


Correct match: (I)-(c); (II)-(d); (III)-(b); (IV)-(a).
Quick Tip: Memorize common ore formulas: Carbonates (Siderite, Calamine, Malachite) are frequent exam topics.

Binary hydrides of s-block elements (Group 1 and Group 2, except Be/Mg in some contexts) are ionic in nature.

These are also known as Saline (salt-like) hydrides because they are solid, non-volatile, and conduct electricity in the molten state, similar to salts.

NaH (Sodium Hydride) is formed by an alkali metal and hydrogen, making it a saline hydride.
Quick Tip: Hydrides Classification: Ionic/Saline (s-block), Covalent/Molecular (p-block), Metallic/Interstitial (d and f-block).

(a) True: Hydrogen combustion produces mainly water and nitrogen oxides (if air is used), but no carbon oxides or hydrocarbons, so it is cleaner than petrol.

(b) True: Hydrogen has very low density. Storing it as a gas requires heavy high-pressure cylinders. The weight of the cylinder is roughly 30 times that of a petrol tank for equivalent energy storage.

(c) True: Hydrogen can be stored in the form of metal hydrides (like \(\mathrm{NaNi_5}\), \(\mathrm{Ti-TiH_2}\), etc.).

(d) False: The calorific value of Hydrogen is \(\sim 142 \mathrm{~kJ/g}\), while LPG is \(\sim 50 \mathrm{~kJ/g}\). The statement swaps these values.


Therefore, statements (a), (b), and (c) are correct.
Quick Tip: Remember calorific values: Hydrogen (\(\sim 142 \mathrm{~kJ/g}\)) > LPG (\(\sim 50 \mathrm{~kJ/g}\)) > Petrol (\(\sim 47 \mathrm{~kJ/g}\)).

In Group 2, the basic character of hydroxides increases down the group.
\(\mathrm{Be(OH)_2}\) is amphoteric (reacts with both acids and bases).
\(\mathrm{Mg(OH)_2}\) is weakly basic.
\(\mathrm{Ca(OH)_2}\), \(\mathrm{Sr(OH)_2}\), and \(\mathrm{Ba(OH)_2}\) are strong bases.

Reaction with acid: \(\mathrm{Be(OH)_2 + 2HCl \to BeCl_2 + 2H_2O}\)

Reaction with base: \(\mathrm{Be(OH)_2 + 2NaOH \to Na_2[Be(OH)_4]}\) (Beryllate).
Quick Tip: Diagonal relationship: Beryllium shows properties similar to Aluminum. Both \(\mathrm{Be(OH)_2}\) and \(\mathrm{Al(OH)_3}\) are amphoteric.

Hydrolysis of halides occurs via the attack of water molecules (nucleophiles) on the central atom. This requires the central atom to have vacant orbitals (typically d-orbitals) to accept the lone pair from oxygen in water.


- Carbon (\(\mathrm{C}\)) in \(\mathrm{CCl_4}\) belongs to the 2nd period and has no d-orbitals. It cannot expand its coordination number beyond 4 to form the transition state needed for hydrolysis.


- Silicon (\(\mathrm{Si}\)), Tin (\(\mathrm{Sn}\)), and Lead (\(\mathrm{Pb}\)) are in lower periods (3, 5, 6) and possess vacant d-orbitals, allowing them to undergo hydrolysis.
Quick Tip: Absence of vacant d-orbitals in valence shell prevents hydrolysis of \(\mathrm{CCl_4}\) and \(\mathrm{NF_3}\).

Scandium (\(\mathrm{Sc}\)) has the electronic configuration \([\mathrm{Ar}] 3d^1 4s^2\).


By losing all 3 valence electrons, it achieves the stable noble gas configuration of Argon.


Thus, \(\mathrm{Sc}\) primarily exhibits only the \(+3\) oxidation state.


Other transition metals listed exhibit variable oxidation states due to the participation of \((n-1)d\) and \(ns\) electrons:


- \(\mathrm{Cu}\): \(+1, +2\)

- \(\mathrm{Ti}\): \(+2, +3, +4\)

- \(\mathrm{V}\): \(+2, +3, +4, +5\)
Quick Tip: Scandium (+3) and Zinc (+2) are the transition elements (by broad definition for Zn) in the 3d series that generally exhibit only one stable oxidation state.

(A) Cobalt (\(\mathrm{Co}\)) is the central metal ion in Vitamin \(\mathrm{B_{12}}\) (Cyanocobalamin). Match: (iii).

(B) Zinc (\(\mathrm{Zn}\)) is present in the enzyme Carbonic anhydrase. Match: (iv).

(C) Rhodium (\(\mathrm{Rh}\)) is used in Wilkinson's catalyst (\(\mathrm{[RhCl(PPh_3)_3]}\)) for hydrogenation of alkenes. Match: (i).

(D) Magnesium (\(\mathrm{Mg}\)) is the central metal in Chlorophyll, the green pigment in plants. Match: (ii).


Correct match: (A)-(iii), (B)-(iv), (C)-(i), (D)-(ii).
Quick Tip: Biological importance of metals: Fe (Hemoglobin), Mg (Chlorophyll), Co (Vitamin B12), Zn (Enzymes like Carbonic Anhydrase/Insulin).

The solubility of gases in water decreases with an increase in temperature. In cold water, the solubility of oxygen is relatively high.

Standard environmental chemistry texts (like NCERT) state that the concentration of dissolved oxygen in cold water can reach up to \(10 \mathrm{~ppm}\) (parts per million). If DO falls below \(6 \mathrm{~ppm}\), it inhibits fish growth.
Quick Tip: Dissolved Oxygen (DO) vital range: \(\sim 10 \mathrm{~ppm}\) in cold water, \(< 6 \mathrm{~ppm}\) is polluted/harmful to aquatic life.

Photochemical smog is formed in warm, sunny climates by the action of sunlight on primary pollutants like nitrogen oxides and unsaturated hydrocarbons.

This process produces secondary pollutants such as Ozone (\(\mathrm{O_3}\)), Formaldehyde (\(\mathrm{HCHO}\)), Acrolein, and Peroxyacetyl nitrate (PAN).

PAN is a powerful lachrymator (eye irritant) and toxic to plants.
Quick Tip: Photochemical smog (Los Angeles smog) = Oxidizing (\(O_3\), PAN, \(NO_x\)). Classical smog (London smog) = Reducing (\(SO_2\), smoke, fog).

Reaction: \(\mathrm{2NaHCO_3 + H_2C_2O_4 \to Na_2C_2O_4 + 2H_2O + 2CO_2}\)

From stoichiometry, \(2\) moles of \(\mathrm{NaHCO_3}\) produce \(2\) moles of \(\mathrm{CO_2}\).

Thus, moles of \(\mathrm{NaHCO_3}\) reacted = moles of \(\mathrm{CO_2}\) produced.


Volume of \(\mathrm{CO_2}\) = \(0.25 \mathrm{~ml} = 0.25 \times 10^{-3} \mathrm{~L}\).

Molar Volume \(V_m = 25.0 \mathrm{~L/mol}\).

Moles of \(\mathrm{CO_2}\), \(n = \frac{0.25 \times 10^{-3}}{25.0} = 10^{-5} \mathrm{~mol}\).


Moles of \(\mathrm{NaHCO_3} = 10^{-5} \mathrm{~mol}\).

Mass of \(\mathrm{NaHCO_3} = n \times M = 10^{-5} \mathrm{~mol} \times 84 \mathrm{~g/mol} = 84 \times 10^{-5} \mathrm{~g}\).

Mass in mg = \(84 \times 10^{-5} \times 1000 \mathrm{~mg} = 0.84 \mathrm{~mg}\).


Mass of tablet = \(10 \mathrm{~mg}\).

Percentage of \(\mathrm{NaHCO_3} = \frac{0.84}{10} \times 100 = 8.4 %\).
Quick Tip: Always establish the stoichiometric relationship first. For Bicarbonate with acid, moles of Bicarbonate \(\approx\) moles of \(CO_2\).

The density of a crystalline solid is given by: \[ \rho = \frac{Z M}{N_A a^3} \]

where \(\rho\) is density, \(Z\) is number of atoms per unit cell, \(M\) is molar mass, \(N_A\) is Avogadro constant, and \(a\) is edge length.


Step 1: Given values \[ \rho = 9 \times 10^3\ \mathrm{kg\,m^{-3}}, \quad Z = 4 \ (FCC), \quad N_A = 6 \times 10^{23}\ \mathrm{mol^{-1}} \]
\[ a = 200\sqrt{2}\ \mathrm{pm} = 200\sqrt{2} \times 10^{-12}\ \mathrm{m} = 2\sqrt{2} \times 10^{-10}\ \mathrm{m} \]

Step 2: Volume of unit cell \[ a^3 = (2\sqrt{2} \times 10^{-10})^3 = 16\sqrt{2} \times 10^{-30}\ \mathrm{m^3} \]

Step 3: Calculate molar mass \[ M = \frac{\rho N_A a^3}{Z} \]
\[ M = \frac{(9 \times 10^3)(6 \times 10^{23})(16\sqrt{2} \times 10^{-30})}{4} \]
\[ M = 216\sqrt{2} \times 10^{-4} \]

Using \(\sqrt{2} \approx 1.414\): \[ M = 216 \times 1.414 \times 10^{-4} \approx 0.0305\ \mathrm{kg\,mol^{-1}} \]

Final Answer: \(\boxed{0.0305\ \mathrm{kg\,mol^{-1}}}\) Quick Tip: Be careful with unit conversions for edge length (pm to m) and density (\(kg/m^3\) vs \(g/cm^3\)). Consistent SI units yield M in kg/mol.

Heat therapy uses infrared radiation.
Given wavelength: \[ \lambda = 900\,nm = 900 \times 10^{-9}\,m = 9 \times 10^{-5}\,cm \]

Step 1: Rydberg formula \[ \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]

Given \(R_H = 1 \times 10^5\,\mathrm{cm^{-1}}\):
\[ \frac{1}{9 \times 10^{-5}} = 10^5\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]
\[ \frac{1}{9} = \frac{1}{n_1^2}-\frac{1}{n_2^2} \]

Step 2: Identify the series

Infrared lines correspond to the Paschen series, for which \(n_1 = 3\).
\[ \frac{1}{3^2}-\frac{1}{n_2^2}=\frac{1}{9} \]

This is satisfied when \(n_2=\infty\):
\[ \frac{1}{9}-0=\frac{1}{9} \]

Conclusion:

The required spectral line is Paschen series with transition \(\infty \to 3\).


Final Answer: Option (D) Quick Tip: Series limits (\(n_2 = \infty\)) represent the shortest wavelength/highest energy for that series. Lyman series limit: \(1/R \approx 91 nm\). Balmer limit: \(4/R \approx 365 nm\). Paschen limit: \(9/R \approx 820 nm\). 900nm is close to Paschen limit.

For the reaction \(\mathrm{X \rightleftharpoons Y}\), the temperature dependence of
standard Gibbs energy is: \[ \Delta_r G^\circ = 120 - \frac{3}{8}T \]

Step 1: Relation between \(\Delta_r G^\circ\) and equilibrium constant
\[ \Delta_r G^\circ = -RT \ln K \]

If \(\Delta_r G^\circ < 0\), then \(K > 1\) and products are favoured.

If \(\Delta_r G^\circ > 0\), then \(K < 1\) and reactants are favoured.


Step 2: Find temperature at which \(\Delta_r G^\circ = 0\) \[ 120 - \frac{3}{8}T = 0 \] \[ \frac{3}{8}T = 120 \] \[ T = \frac{120 \times 8}{3} = 320\,K \]

Step 3: Interpretation

For \(T < 320\,K\), \(\Delta_r G^\circ > 0\), so reactant X is favoured.

For \(T > 320\,K\), \(\Delta_r G^\circ < 0\), so product Y is favoured.


Step 4: Checking options

At \(T = 315\,K (<320\,K)\), reactant X is the major component.


Final Answer: Option (C) Quick Tip: Positive \(\Delta G^\circ\) implies non-spontaneous forward reaction at standard states, hence equilibrium lies to the left (Reactants dominate).

Two identical blocks of the same metal (same mass and same \(C_p\)) at temperatures \(T_1\) and \(T_2\) are brought into contact at constant pressure.


Step 1: Final equilibrium temperature

Since the blocks are identical and isolated, \[ T_f = \frac{T_1 + T_2}{2} \]

Step 2: Entropy change formula

For a process at constant pressure, \[ \Delta S = C_p \ln\left(\frac{T_f}{T_i}\right) \]

Step 3: Entropy change of each block
\[ \Delta S_1 = C_p \ln\left(\frac{T_f}{T_1}\right), \quad \Delta S_2 = C_p \ln\left(\frac{T_f}{T_2}\right) \]

Step 4: Total entropy change
\[ \Delta S_{total} = \Delta S_1 + \Delta S_2 \] \[ \Delta S_{total} = C_p \ln\left(\frac{T_f^2}{T_1 T_2}\right) \]

Step 5: Substitute \(T_f\)
\[ \Delta S_{total} = C_p \ln\left( \frac{\left(\frac{T_1 + T_2}{2}\right)^2}{T_1 T_2} \right) \]
\[ \Delta S_{total} = C_p \ln\left( \frac{(T_1 + T_2)^2}{4 T_1 T_2} \right) \]

Final Answer: Option (B) Quick Tip: Entropy of mixing bodies at different temperatures is always positive (irreversible process). The term inside the logarithm is \((Arithmetic Mean)^2 / (Geometric Mean)^2\), which is \(\ge 1\).

Freezing point depression \(\Delta T_f\) is proportional to molality \(m = \frac{moles of solute}{mass of solvent}\).

Since the solute (milk solids) amount is constant, \(\Delta T_f \propto \frac{1}{mass of water}\).


Let \(W_1\) be the initial mass of water in pure milk.

Let \(W_2\) be the final mass of water in diluted milk.

\(\frac{\Delta T_{f(pure)}}{\Delta T_{f(diluted)}} = \frac{W_2}{W_1}\).
\(\frac{0.5}{0.2} = \frac{W_2}{W_1} \implies \frac{5}{2} = \frac{W_2}{W_1}\).
\(W_2 = 2.5 W_1\).


The water added is \(W_{added} = W_2 - W_1 = 2.5 W_1 - W_1 = 1.5 W_1 = \frac{3}{2} W_1\).

This means 3 units of water were added for every 2 units of original water (or milk volume roughly).

Ratio: 3 cups of water to 2 cups of pure milk.
Quick Tip: \(\Delta T_f \times Mass of Solvent = Constant\) for a fixed amount of solute.

Given equilibrium: \[ \mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)} \]
with equilibrium constant \(K_p\).


For dissociation of ammonia: \[ \mathrm{2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)} \] \[ K_p'=\frac{1}{K_p} \]

Step 1: Assume equilibrium partial pressures

Let \(P_{\mathrm{N_2}}=x\). Then: \[ P_{\mathrm{H_2}}=3x \] \[ P_{products}=x+3x=4x \]

Given \(P_{\mathrm{NH_3}} \ll P_{total}\), \[ P_{total} \approx 4x = P \quad \Rightarrow \quad x=\frac{P}{4} \]

Step 2: Equilibrium constant expression \[ K_p'=\frac{P_{\mathrm{N_2}}(P_{\mathrm{H_2}})^3}{(P_{\mathrm{NH_3}})^2} =\frac{x(3x)^3}{(P_{\mathrm{NH_3}})^2} =\frac{27x^4}{(P_{\mathrm{NH_3}})^2} \]

Since \(K_p'=\frac{1}{K_p}\), \[ \frac{1}{K_p}=\frac{27x^4}{(P_{\mathrm{NH_3}})^2} \]

Step 3: Substitute \(x=\frac{P}{4}\) \[ \frac{1}{K_p} =\frac{27\left(\frac{P}{4}\right)^4}{(P_{\mathrm{NH_3}})^2} =\frac{27P^4}{256(P_{\mathrm{NH_3}})^2} \]
\[ (P_{\mathrm{NH_3}})^2 =K_p\frac{27P^4}{256} \]

Step 4: Take square root \[ P_{\mathrm{NH_3}} =\sqrt{K_p}\frac{\sqrt{27}P^2}{16} =\frac{3^{3/2}K_p^{1/2}P^2}{16} \]

Final Answer: Option (D) Quick Tip: For dissociation \(2A \to B + 3C\), mole fraction ratios are fixed. If \(P_{reactant}\) is small, total pressure is dominated by products, simplifying calculations.

The electrochemical cell is: \[ \mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,||\,M^{x+}(aq)\,|\,M(s)} \]

Zinc acts as the anode with: \[ E^\circ_{\mathrm{Zn^{2+}/Zn}} = -0.76\,V \]

Step 1: Standard cell potential
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
\[ E^\circ_{cell} = E^\circ_{cathode} - (-0.76) = E^\circ_{cathode} + 0.76 \]

Step 2: Important note

Electrode potential is an intensive property (energy per unit charge), hence
it is not divided by the number of electrons transferred.


Step 3: Calculate \(E^\circ_{cell}\) for each cathode


(A) \(\mathrm{Au^{3+}/Au}\): \[ E^\circ_{cell} = 1.40 + 0.76 = 2.16\,V \]

(B) \(\mathrm{Ag^+/Ag}\): \[ E^\circ_{cell} = 0.80 + 0.76 = 1.56\,V \]

(C) \(\mathrm{Fe^{3+}/Fe^{2+}}\): \[ E^\circ_{cell} = 0.77 + 0.76 = 1.53\,V \]

(D) \(\mathrm{Fe^{2+}/Fe}\): \[ E^\circ_{cell} = -0.44 + 0.76 = 0.32\,V \]

Step 4: Conclusion

The maximum \(E^\circ_{cell}\) is obtained with \(\mathrm{Au^{3+}/Au}\).


Final Answer: Option (A) Quick Tip: Standard Electrode Potential is an intensive property (Volt = Joule/Coulomb). It does not depend on the stoichiometric coefficients or number of electrons in the balanced equation.

Arrhenius equation: \(k = A e^{-E_a/RT}\).

Taking natural logarithm: \(\ln k = \ln A - \frac{E_a}{RT}\).

This can be written as: \(\ln k = \ln A - E_a \left( \frac{1}{RT} \right)\).

Comparing with linear equation \(Y = C + mX\):

Y-axis: \(\ln k\)

X-axis: \(\frac{1}{RT}\)

Slope (\(m\)) = \(-E_a\).


Given that the gradient (slope) is \((-y)\).

So, \(-E_a = -y \implies E_a = y\).

The energy required to activate the reactant is the Activation Energy \(E_a\), which is equal to \(y\).
Quick Tip: Pay attention to the X-axis variable. Usually, plot is vs \(1/T\) (slope \(-E_a/R\)). Here, plot is vs \(1/(RT)\) (slope \(-E_a\)).

Colloids are classified based on the physical state of the dispersed phase and dispersion medium.

- Solid Sol: Solid dispersed in Solid (e.g., Gem stones, coloured glass).

- Sol: Solid dispersed in Liquid (e.g., Paints).

- Gel: Liquid dispersed in Solid (e.g., Butter, Cheese).

- Emulsion: Liquid dispersed in Liquid (e.g., Hair cream, Milk).


Therefore, Gem stones are an example of solid sol.
Quick Tip: Memorize the colloid table. Solid in Solid = Solid Sol. Liquid in Solid = Gel. Liquid in Liquid = Emulsion.

Let \(y = \frac{x}{1+x^2}\).


Rearranging the terms, we get a quadratic in \(x\):

\(y(1+x^2) = x \implies y x^2 - x + y = 0\).


For \(x\) to be real, the discriminant \(D\) must be non-negative (\(D \ge 0\)).

\((-1)^2 - 4(y)(y) \ge 0\).

\(1 - 4y^2 \ge 0 \implies 4y^2 \le 1 \implies y^2 \le \frac{1}{4}\).


This implies \(-\frac{1}{2} \le y \le \frac{1}{2}\).


Since \(f(x)\) is continuous and defined for all real \(x\), the range is the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\).
Quick Tip: For rational functions of the form \(\frac{L(x)}{Q(x)}\) or \(\frac{Q(x)}{Q(x)}\), equating to \(y\) and using the discriminant condition (\(D \ge 0\)) is a standard method to find the range.

Let the roots be \(\alpha\) and \(\alpha^3\).


From the product of roots:
\(\alpha \cdot \alpha^3 = \frac{256}{81} \implies \alpha^4 = \left(\frac{4}{3}\right)^4\).


This gives the real roots \(\alpha = \frac{4}{3}\) or \(\alpha = -\frac{4}{3}\).


From the sum of roots:
\(\alpha + \alpha^3 = -\frac{k}{81}\).


Case 1: If \(\alpha = \frac{4}{3}\), then \(\alpha + \alpha^3 = \frac{4}{3} + \frac{64}{27} = \frac{36+64}{27} = \frac{100}{27}\).
\(-\frac{k}{81} = \frac{100}{27} \implies -k = 300 \implies k = -300\).


Case 2: If \(\alpha = -\frac{4}{3}\), then \(\alpha + \alpha^3 = -\frac{100}{27}\).
\(-\frac{k}{81} = -\frac{100}{27} \implies k = 300\).


Since \(-300\) is in the options, the value is \(-300\).
Quick Tip: When roots are related (e.g., one is \(n\) times or power of other), assume roots as \(\alpha\) and \(f(\alpha)\), then use product and sum relations to solve for \(\alpha\).

\(\left(-2 - \frac{i}{3}\right)^3 = -\left(2 + \frac{i}{3}\right)^3\).


Using \((a+b)^3 = a^3 + b^3 + 3a^2b + 3ab^2\):

\(\left(2 + \frac{i}{3}\right)^3 = 8 + \left(\frac{i}{3}\right)^3 + 3(4)\left(\frac{i}{3}\right) + 3(2)\left(\frac{i}{3}\right)^2\).

\(= 8 - \frac{i}{27} + 4i - \frac{2}{3}\).

\(= \left(8 - \frac{2}{3}\right) + i\left(4 - \frac{1}{27}\right) = \frac{22}{3} + i\frac{107}{27} = \frac{198 + 107i}{27}\).


So, \(\left(-2 - \frac{i}{3}\right)^3 = -\frac{198 + 107i}{27} = \frac{-198 - 107i}{27}\).


Comparing with \(\frac{x+iy}{27}\), we get \(x = -198\) and \(y = -107\).

\(y - x = -107 - (-198) = 198 - 107 = 91\).
Quick Tip: Handle the negative sign carefully by factoring it out: \((-a-b)^3 = -(a+b)^3\). Simplify the complex arithmetic step-by-step.

Given \(A A^T = I\). This means \(A\) is an orthogonal matrix. The rows are orthonormal vectors.


Row 1 \(\cdot\) Row 1: \((0)^2 + (2q)^2 + r^2 = 1 \implies 4q^2 + r^2 = 1\). (i)


Row 2 \(\cdot\) Row 2: \(p^2 + q^2 + (-r)^2 = 1 \implies p^2 + q^2 + r^2 = 1\). (ii)


Row 1 \(\cdot\) Row 2: \(0(p) + 2q(q) + r(-r) = 0 \implies 2q^2 - r^2 = 0 \implies r^2 = 2q^2\).


Substitute \(r^2 = 2q^2\) into (i):
\(4q^2 + 2q^2 = 1 \implies 6q^2 = 1 \implies q^2 = \frac{1}{6}\).


Then \(r^2 = \frac{2}{6} = \frac{1}{3}\).


Substitute \(q^2\) and \(r^2\) into (ii):
\(p^2 + \frac{1}{6} + \frac{1}{3} = 1 \implies p^2 + \frac{1}{2} = 1 \implies p^2 = \frac{1}{2}\).


Therefore, \(|p| = \frac{1}{\sqrt{2}}\).
Quick Tip: For orthogonal matrices (\(AA^T = I\)), the sum of squares of elements in any row/column is 1, and the dot product of any two distinct rows/columns is 0.

The given system of equations is: \[ \begin{aligned} (1)\;& 2x + 2y + 3z = a
(2)\;& 3x - y + 5z = b
(3)\;& x - 3y + 2z = c \end{aligned} \]

Step 1: Determinant of coefficient matrix \[ \Delta = \begin{vmatrix} 2 & 2 & 3
3 & -1 & 5
1 & -3 & 2 \end{vmatrix} \]
\[ = 2(-2+15) - 2(6-5) + 3(-9+1) \]
\[ = 26 - 2 - 24 = 0 \]

Since \(\Delta = 0\), the system may have infinitely many solutions
provided it is consistent.


Step 2: Check linear dependence

Assume: \[ (1) + k(3) = (2) \]

Comparing coefficients of \(x\): \[ 2 + k = 3 \Rightarrow k = 1 \]

Checking \(y\) and \(z\) coefficients confirms the relation.

Hence, \[ LHS of (1) + LHS of (3) = LHS of (2) \]

Step 3: Condition on constants

For consistency: \[ a + c = b \]
\[ \boxed{b - c - a = 0} \]

Final Answer: Option (C) Quick Tip: If \(\Delta = 0\), finding the linear dependence between the rows of the coefficient matrix allows you to directly relate \(a, b, c\) without calculating \(\Delta_x, \Delta_y, \Delta_z\).

In the expansion of \((A+B)^n\) with \(n=8\) (even), the middle term is the \((\frac{n}{2} + 1)\)-th term, i.e., \(T_5\).

\(T_5 = {}^{8}C_4 \left(\frac{x^3}{3}\right)^4 \left(\frac{3}{x}\right)^4\).

\({}^{8}C_4 = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} = 70\).

\(T_5 = 70 \cdot \frac{x^{12}}{3^4} \cdot \frac{3^4}{x^4} = 70 x^8\).


Given \(T_5 = 5670\):
\(70 x^8 = 5670 \implies x^8 = \frac{5670}{70} = 81\).

\(x^8 - 81 = 0 \implies (x^4 - 9)(x^4 + 9) = 0\).


For real \(x\), \(x^4 = 9 \implies x^2 = 3 \implies x = \pm \sqrt{3}\).

(\(x^4 + 9 = 0\) yields complex roots).


The real values are \(\sqrt{3}\) and \(-\sqrt{3}\).

Sum of real values \(= \sqrt{3} + (-\sqrt{3}) = 0\).
Quick Tip: Middle term index for \((a+b)^n\) is \(n/2 + 1\) if \(n\) is even. Always check for "real values" constraint when solving polynomial equations.

Let the common ratio be \(r\).

\(\frac{a_3}{a_1} = \frac{a_1 r^2}{a_1} = r^2 = 25\).


We need to find \(\frac{a_9}{a_5}\).

\(\frac{a_9}{a_5} = \frac{a_1 r^8}{a_1 r^4} = r^4\).


Since \(r^2 = 25\), \(r^4 = (r^2)^2 = (25)^2 = 625 = 5^4\).
Quick Tip: In a G.P., ratio of terms depends only on the difference in their indices: \(\frac{a_m}{a_n} = r^{m-n}\).

Let the infinite geometric series be: \[ a,\; ar,\; ar^2,\; \dots \]
where \(a>0\) and \(0

Step 1: Sum of the infinite GP \[ \frac{a}{1-r} = 3 \Rightarrow a = 3(1-r) \]

Step 2: Series of cubes

The cubes form another GP: \[ a^3,\; a^3r^3,\; a^3r^6,\; \dots \]
with sum: \[ \frac{a^3}{1-r^3} = \frac{27}{19} \]

Step 3: Substitute \(a\) \[ a^3 = 27(1-r)^3 \]
\[ \frac{27(1-r)^3}{1-r^3} = \frac{27}{19} \]

Cancel 27: \[ \frac{(1-r)^3}{1-r^3} = \frac{1}{19} \]

Step 4: Factorisation \[ 1-r^3 = (1-r)(1+r+r^2) \]
\[ \frac{(1-r)^2}{1+r+r^2} = \frac{1}{19} \]

Step 5: Solve \[ 19(1-2r+r^2) = 1+r+r^2 \]
\[ 18r^2 - 39r + 18 = 0 \]
\[ 6r^2 - 13r + 6 = 0 \]
\[ (3r-2)(2r-3)=0 \]

Step 6: Valid solution \[ r=\frac{2}{3} \quad (since |r|<1) \]

Final Answer: Option (D) Quick Tip: For infinite geometric series, always apply the condition \(|r|<1\).

The given expression is the sum \(\sum_{k=0}^{r} {}^{20}C_{r-k} {}^{20}C_k\).


This represents the coefficient of \(x^r\) in the expansion of \((1+x)^{20} (1+x)^{20} = (1+x)^{40}\).


The coefficient is \({}^{40}C_r\).


The value of \({}^{n}C_r\) is maximum when \(r = n/2\) (if n is even).


Here \(n=40\), so maximum is at \(r = 40/2 = 20\).
Quick Tip: Sum of products of binomial coefficients often relates to Vandermonde's Identity: \(\sum \binom{n}{k} \binom{m}{r-k} = \binom{n+m}{r}\).

Consider the limit: \[ \lim_{x \to 0} \frac{\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2}{x^2} \]

Since the expression contains \([x]\) and \(|x|\), we evaluate RHL and LHL separately.


Right-Hand Limit (\(x \to 0^+\)):

For \(0 < x < 1\), \([x]=0\).
\[ |x| - \sin(x[x]) = x - \sin(0) = x \]
\[ \frac{\tan(\pi \sin^2 x) + x^2}{x^2} = \frac{\tan(\pi \sin^2 x)}{x^2} + 1 \]

Using \(\sin x \sim x\) and \(\tan y \sim y\) as \(x \to 0\): \[ \tan(\pi \sin^2 x) \sim \pi x^2 \]
\[ RHL = \pi + 1 \]

Left-Hand Limit (\(x \to 0^-\)):

Let \(x=-h\), where \(h>0\). Then \([x]=-1\).
\[ |x| - \sin(x[x]) = h - \sin h \]
\[ \frac{\tan(\pi \sin^2 h) + (h - \sin h)^2}{h^2} \]

As \(h \to 0\): \[ \tan(\pi \sin^2 h) \sim \pi h^2 \] \[ (h - \sin h)^2 \sim \frac{h^6}{36} \Rightarrow \frac{(h - \sin h)^2}{h^2} \to 0 \]
\[ LHL = \pi \]

Conclusion:

Since \(RHL \neq LHL\), the limit does not exist.


Final Answer: Option (D) Quick Tip: When \([x]\) or \(|x|\) is involved, always evaluate Left Hand Limit and Right Hand Limit separately.

Given: \[ x\ln(\ln x) - x^2 + y^2 = 4, \quad (y>0) \]

Step 1: Find \(y\) at \(x=e\)

Substitute \(x=e\): \[ e\ln(\ln e) - e^2 + y^2 = 4 \]
Since \(\ln e = 1\) and \(\ln 1 = 0\), \[ -e^2 + y^2 = 4 \] \[ y^2 = 4 + e^2 \] \[ y = \sqrt{4+e^2} \quad (y>0) \]

Step 2: Differentiate implicitly w.r.t. \(x\)
\[ \frac{d}{dx}[x\ln(\ln x)] - \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = 0 \]
\[ \ln(\ln x) + \frac{1}{\ln x} - 2x + 2y\frac{dy}{dx} = 0 \]

Step 3: Substitute \(x=e\) and \(y=\sqrt{4+e^2}\)
\[ 0 + 1 - 2e + 2\sqrt{4+e^2}\frac{dy}{dx} = 0 \]

Step 4: Solve for \(\dfrac{dy}{dx}\) \[ 2\sqrt{4+e^2}\frac{dy}{dx} = 2e - 1 \]
\[ \boxed{ \frac{dy}{dx} = \frac{2e-1}{2\sqrt{4+e^2}} } \]

Final Answer: Option (A) Quick Tip: Use Implicit Differentiation for equations mixing \(x\) and \(y\). Don't forget to find the specific \(y\) value for the given \(x\).

Given: \[ f(x)= \begin{cases} -1, & -2\le x<0
x^2-1, & 0\le x\le 2 \end{cases}, \quad g(x)=|f(x)|+f(|x|) \]

Case 1: \(x\in(-2,0)\)
\(|x|\in(0,2)\Rightarrow f(|x|)=x^2-1\), and \(|f(x)|=1\).
\[ g(x)=1+x^2-1=x^2 \]

Case 2: \(x\in[0,2)\)
\(f(|x|)=x^2-1\), \(|f(x)|=|x^2-1|\).

Subcase 2a: \(0\le x<1\)
\(|x^2-1|=1-x^2 \Rightarrow g(x)=0\)

Subcase 2b: \(1\le x<2\)
\(|x^2-1|=x^2-1 \Rightarrow g(x)=2x^2-2\)

Summary: \[ g(x)= \begin{cases} x^2, & -2 0, & 0\le x<1
2x^2-2, & 1\le x<2 \end{cases} \]

Check at \(x=0\): continuous and differentiable.

Check at \(x=1\): continuous but \[ LHD=0,\quad RHD=4\neq 0 \]

Hence, \(g\) is not differentiable at exactly one point (\(x=1\)). Quick Tip: Construct the piecewise function explicitly. Points where definition changes or modulus is zero are candidates for non-differentiability.

Given \(f(x)=3x^3-18x^2+27x-40\) and \[ x^2+30\le 11x \Rightarrow (x-5)(x-6)\le 0 \]
So \(S=[5,6]\).
\[ f'(x)=9(x-1)(x-3) \]
Critical points \(x=1,3\) lie outside \([5,6]\).

Since \(f'(x)>0\) on \([5,6]\), \(f\) is increasing.

Maximum at \(x=6\): \[ f(6)=648-648+162-40=122 \]

Maximum value =122. Quick Tip: If critical points lie outside the domain interval, the extrema occur at the endpoints. Check the sign of derivative to determine which endpoint is max/min.

Let \[ I=\int \frac{\sqrt{1-x^2}}{x^4}\,dx \]
Put \(x=\sin\theta\), \(dx=\cos\theta\,d\theta\).
\[ I=\int \frac{\cos^2\theta}{\sin^4\theta}\,d\theta =\int \cot^2\theta\,\csc^2\theta\,d\theta \]

Let \(u=\cot\theta\), \(du=-\csc^2\theta\,d\theta\): \[ I=-\int u^2\,du=-\frac{u^3}{3}+C \]

Back-substitute: \[ I=-\frac{1}{3}\left(\frac{\sqrt{1-x^2}}{x}\right)^3 \]

Thus \(m=3\), \(A(x)=-\frac{1}{3x^3}\), and \[ (A(x))^m=-\frac{1}{27x^9} \] Quick Tip: Trigonometric substitution (\(x=\sin\theta\)) is standard for \(\sqrt{a^2-x^2}\). Alternatively, factor out highest power of x from square root for algebraic integration.

For \(x\in[-2,0)\), \(\left[\frac{x}{\pi}\right]=-1\); for \(x\in[0,2]\), it is \(0\).
\[ I=\int_{-2}^0 \frac{\sin^2x}{-1/2}\,dx+\int_0^2 \frac{\sin^2x}{1/2}\,dx \] \[ =-2\int_{-2}^0\sin^2x\,dx+2\int_0^2\sin^2x\,dx \]

Using symmetry \(\int_{-2}^0\sin^2x\,dx=\int_0^2\sin^2x\,dx\), \[ I=0 \] Quick Tip: Split the integral at points where the greatest integer function changes value. Check for symmetry (odd/even functions) to simplify calculation.

Parabola: \(y=\frac{x^2}{4}\),\quad Line: \(y=\frac{x+2}{4}\).

Intersections: \[ x^2=x+2 \Rightarrow (x-2)(x+1)=0 \]
So \(x=-1,2\).
\[ A=\int_{-1}^2\left(\frac{x+2}{4}-\frac{x^2}{4}\right)dx =\frac14\int_{-1}^2(x+2-x^2)dx \]
\[ A=\frac14\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^2 =\frac{9}{8} \] Quick Tip: Area between parabola \(x^2=4ay\) and line \(y=mx+c\) with roots \(\alpha, \beta\) is \(\frac{|a|}{6} |\alpha - \beta|^3\) is a specific formula, or simply \(\frac{1}{6}|a|(\beta-\alpha)^3\) for standard forms. Standard integration is safer.

Given: \[ \frac{dy}{dx}+\left(2+\frac1x\right)y=e^{-2x} \]
IF \(=xe^{2x}\).
\[ y(xe^{2x})=\int x\,dx=\frac{x^2}{2}+C \]

Using \(y(1)=\frac12e^{-2}\Rightarrow C=0\), \[ y=\frac{x}{2}e^{-2x} \]
\[ y'=\frac{e^{-2x}}{2}(1-2x) \]
\(y'<0\) for \(x>\frac12\). Hence decreasing on \((\frac12,1)\). Quick Tip: For monotonicity, check the sign of \(dy/dx\). \(dy/dx < 0\) implies decreasing.

Centers lie on x-axis: \((h,0)\) and \((-h,0)\).

Radius: \[ R=\sqrt{h^2+1} \]

Slope of tangent at \((0,1)\) for first circle: \[ m=h \]

Slope of line joining \((0,1)\) to \((-h,0)\): \[ m=\frac{1}{h} \]

Equating: \[ h=\frac1h\Rightarrow h=1 \]

Distance between centers \(=2h=2\). Quick Tip: Use symmetry of the configuration to simplify coordinates of centers. Tangent property: Tangent is perpendicular to radius at point of contact.

Intercepts: \(A(1,0)\), \(B(0,\tfrac12)\).

Circle: \(x^2+y^2-x-\frac y2=0\).

Tangent at origin: \[ 2x+y=0 \]

Distances: \[ p_A=\frac{2}{\sqrt5},\quad p_B=\frac{1}{2\sqrt5} \]

Sum: \[ \frac{2}{\sqrt5}+\frac{1}{2\sqrt5} =\frac{\sqrt5}{2} \] Quick Tip: Equation of tangent at origin for curve passing through origin is found by equating the lowest degree terms to zero.

Circle: \[ x^2+y^2-6x+8y-103=0 \]
Center \((3,-4)\), radius \(8\sqrt2\).

Vertices: \[ (3\pm8,\,-4\pm8) \Rightarrow (-5,4),(11,4),(-5,-12),(11,-12) \]

Distances from origin: \[ \sqrt{41},\sqrt{137},13,\sqrt{265} \]

Nearest distance \(=\sqrt{41}\). Quick Tip: For a square inscribed in a circle with sides parallel to axes, vertices are displaced from center by \((\pm r/\sqrt{2}, \pm r/\sqrt{2})\).

The equation of a tangent to the parabola \(y^2 = 4ax\) with slope \(m\) is \(y = mx + \frac{a}{m}\).


Here, \(4a = 4 \implies a = 1\). So, the tangent is \(y = mx + \frac{1}{m}\).


This line is also a tangent to the hyperbola \(xy = 2\).


Substitute \(y\) in the hyperbola equation:
\(x\left(mx + \frac{1}{m}\right) = 2\).

\(mx^2 + \frac{x}{m} - 2 = 0\).

\(m^2x^2 + x - 2m = 0\).


For tangency, the quadratic equation must have equal roots, so the discriminant \(D = 0\).

\(D = (1)^2 - 4(m^2)(-2m) = 0\).

\(1 + 8m^3 = 0\).

\(m^3 = -\frac{1}{8} \implies m = -\frac{1}{2}\).


Substitute \(m = -1/2\) into the tangent equation:
\(y = -\frac{1}{2}x + \frac{1}{-1/2}\).

\(y = -\frac{x}{2} - 2\).

\(2y = -x - 4 \implies x + 2y + 4 = 0\).
Quick Tip: For common tangent problems, assume the tangent equation of one curve (usually in slope form) and apply the condition of tangency (Determinant = 0 or distance = radius) to the second curve.

The equation of the ellipse is \(\frac{x^2}{2} + \frac{y^2}{1} = 1\).


A general point on the ellipse is \(P(\sqrt{2}\cos\theta, \sin\theta)\).


The equation of the tangent at \(P\) is \(\frac{x(\sqrt{2}\cos\theta)}{2} + \frac{y(\sin\theta)}{1} = 1\), which simplifies to \(\frac{x}{\sqrt{2}\sec\theta} + \frac{y}{\csc\theta} = 1\).


The intercepts on the axes are \(A(\sqrt{2}\sec\theta, 0)\) and \(B(0, \csc\theta)\).


Let \(M(h, k)\) be the midpoint of the segment AB.

\(h = \frac{\sqrt{2}\sec\theta + 0}{2} = \frac{\sec\theta}{\sqrt{2}} \implies \cos\theta = \frac{1}{\sqrt{2}h}\).

\(k = \frac{0 + \csc\theta}{2} = \frac{\csc\theta}{2} \implies \sin\theta = \frac{1}{2k}\).


Eliminating \(\theta\) using \(\cos^2\theta + \sin^2\theta = 1\):

\(\left(\frac{1}{\sqrt{2}h}\right)^2 + \left(\frac{1}{2k}\right)^2 = 1\).

\(\frac{1}{2h^2} + \frac{1}{4k^2} = 1\).


Replacing \((h, k)\) with \((x, y)\), the locus is \(\frac{1}{2x^2} + \frac{1}{4y^2} = 1\).
Quick Tip: When finding a locus involving a parameter (like \(\theta\)), express the trigonometric functions in terms of the coordinates (\(h, k\)) and use the Pythagorean identity (\(\sin^2\theta + \cos^2\theta = 1\)) to eliminate the parameter.

Let the required plane be: \[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]

Step 1: Use given points


Through \((0,-1,0)\): \[ -\frac{1}{b} = 1 \Rightarrow b=-1 \]

Through \((0,0,1)\): \[ \frac{1}{c} = 1 \Rightarrow c=1 \]

Thus the plane becomes: \[ \frac{x}{a} - y + z = 1 \]

Let \(A=\frac{1}{a}\), then: \[ Ax - y + z - 1 = 0 \]

Step 2: Normal vectors


Normal to required plane: \[ \vec{n}_1 = (A,-1,1) \]

Given plane: \[ y - z + 5 = 0 \Rightarrow \vec{n}_2 = (0,1,-1) \]

Step 3: Angle condition

\[ \cos\frac{\pi}{4} = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|} \]
\[ \frac{1}{\sqrt{2}} = \frac{|A(0)+(-1)(1)+(1)(-1)|}{\sqrt{A^2+2}\sqrt{2}} = \frac{2}{\sqrt{A^2+2}\sqrt{2}} \]
\[ 1=\frac{2}{\sqrt{A^2+2}} \Rightarrow A^2+2=4 \Rightarrow A^2=2 \Rightarrow A=\pm\sqrt{2} \]

Step 4: Direction ratios
\[ (\pm\sqrt{2},-1,1)\sim(\sqrt{2},1,-1) \]

Final Answer: Option (A) Quick Tip: Use the intercept form of the plane equation when intercepts or axial points are given. Remember angle between planes is the angle between their normals.

Given line: \[ \frac{x-3}{2}=\frac{y+2}{-1}=\frac{z-1}{3} \]
Direction vector: \[ \vec{b}=(2,-1,3) \]

Given plane: \[ 2x+3y-z=5 \]
Normal vector: \[ \vec{n}_p=(2,3,-1) \]

Step 1: Normal to required plane

The plane containing the line and its projection on the given plane
must be perpendicular to the given plane.
Hence its normal is perpendicular to both \(\vec{b}\) and \(\vec{n}_p\).
\[ \vec{n}=\vec{b}\times\vec{n}_p =\begin{vmatrix} \hat i & \hat j & \hat k
2 & -1 & 3
2 & 3 & -1 \end{vmatrix} =(-8,8,8) \]

Direction ratios \(\sim(-1,1,1)\).


Step 2: Equation of the plane

A point on the given line is \((3,-2,1)\).

Using point–normal form: \[ -1(x-3)+1(y+2)+1(z-1)=0 \]
\[ -x+y+z+4=0 \quad\Rightarrow\quad x-y-z=4 \]

Step 3: Check options


For \((2,0,-2)\): \[ 2-0-(-2)=4 \]

Hence the point lies on the plane.


Final Answer: Option (B) Quick Tip: The "plane of projection" of a line onto a plane is always perpendicular to the given plane and contains the line itself. Normal vector is \(\vec{d_{line}} \times \vec{n_{plane}}\).

Given: \[ \vec a=\hat i+2\hat j+4\hat k,\quad \vec b=\hat i+\lambda\hat j+4\hat k,\quad \vec c=2\hat i+4\hat j+(\lambda^2-1)\hat k. \]

Step 1: Coplanarity condition

For coplanar vectors, the scalar triple product is zero: \[ [\vec a\,\vec b\,\vec c]= \begin{vmatrix} 1 & 2 & 4
1 & \lambda & 4
2 & 4 & \lambda^2-1 \end{vmatrix}=0. \]

Apply \(R_2\to R_2-R_1\): \[ \begin{vmatrix} 1 & 2 & 4
0 & \lambda-2 & 0
2 & 4 & \lambda^2-1 \end{vmatrix}=0. \]

Expanding along the second row: \[ (\lambda-2)\big[(\lambda^2-1)-8\big]=0 \Rightarrow (\lambda-2)(\lambda^2-9)=0. \]

Thus, \[ \lambda=2,\,3,\,-3. \]

Step 2: Exclude zero cross product

For \(\lambda=3\) or \(-3\), \(\vec c=2\vec a\), hence \(\vec a\times\vec c=\vec 0\) (not allowed).

Therefore, \(\lambda=2\).

Step 3: Compute \(\vec a\times\vec c\)

With \(\lambda=2\), \[ \vec c=2\hat i+4\hat j+3\hat k. \]
\[ \vec a\times\vec c= \begin{vmatrix} \hat i & \hat j & \hat k
1 & 2 & 4
2 & 4 & 3 \end{vmatrix} = -10\hat i+5\hat j. \]

Final Answer: Option (C) Quick Tip: Use row operations in determinants to simplify the calculation of coplanarity conditions. Check for trivial solutions (parallel vectors) when the cross product is required to be non-zero.

Total number of observations \(N=30\).


Step 1: Mean
\[ \bar{x} =\frac{10\left(\frac12-d\right)+10\left(\frac12\right)+10\left(\frac12+d\right)}{30} =\frac{15}{30} =\frac12 \]

Step 2: Deviations from mean



\begin{tabular{c c c c
Outcome & Deviation & Square & Frequency

\hline \(\frac12-d\) & \(-d\) & \(d^2\) & 10
\(\frac12\) & \(0\) & \(0\) & 10
\(\frac12+d\) & \(d\) & \(d^2\) & 10
\end{tabular


Step 3: Variance
\[ \sigma^2=\frac{1}{N}\sum (x_i-\bar{x})^2 =\frac{10d^2+0+10d^2}{30} =\frac{20d^2}{30} =\frac{2}{3}d^2 \]

Step 4: Use given variance
\[ \frac{2}{3}d^2=\frac{4}{3} \Rightarrow 2d^2=4 \Rightarrow d^2=2 \]
\[ |d|=\sqrt{2} \]

Final Answer: Option (A) Quick Tip: Standard deviation/Variance is independent of change of origin. Shift the data by subtracting the mean (1/2) to simplify the calculation to just data points \(-d, 0, d\).

Set \(S = \{1, 2, \dots, 11\}\).

Number of Even integers (\(E\)) = \(\{2, 4, 6, 8, 10\}\) (5 numbers).

Number of Odd integers (\(O\)) = \(\{1, 3, 5, 7, 9, 11\}\) (6 numbers).


Event A: Sum is even.

Sum is even if both are even OR both are odd.

Number of ways \(n(A) = \binom{5}{2} + \binom{6}{2} = 10 + 15 = 25\).


Event B: Both numbers are even.

Number of ways \(n(B) = \binom{5}{2} = 10\).


Note that \(B \subset A\). We require \(P(B|A)\).
\(P(B|A) = \frac{n(B \cap A)}{n(A)} = \frac{n(B)}{n(A)} = \frac{10}{25} = \frac{2}{5}\).
Quick Tip: Parity of Sum: Even + Even = Even, Odd + Odd = Even, Even + Odd = Odd.

Given: \[ f_k(x)=\frac{1}{k}(\sin^k x+\cos^k x). \]

Step 1: Evaluate \(f_4(x)\) \[ f_4(x)=\frac{1}{4}(\sin^4 x+\cos^4 x). \]

Using \(\sin^2 x+\cos^2 x=1\), \[ \sin^4 x+\cos^4 x =(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x =1-2\sin^2 x\cos^2 x. \]
\[ f_4(x)=\frac{1}{4}-\frac{1}{2}\sin^2 x\cos^2 x. \]

Step 2: Evaluate \(f_6(x)\) \[ f_6(x)=\frac{1}{6}(\sin^6 x+\cos^6 x). \]

Using \((a^3+b^3)=(a+b)^3-3ab(a+b)\) with \(a=\sin^2 x\), \(b=\cos^2 x\), \[ \sin^6 x+\cos^6 x =1-3\sin^2 x\cos^2 x. \]
\[ f_6(x)=\frac{1}{6}-\frac{1}{2}\sin^2 x\cos^2 x. \]

Step 3: Compute the difference \[ \begin{aligned} f_4(x)-f_6(x) &=\left(\frac{1}{4}-\frac{1}{2}\sin^2 x\cos^2 x\right) -\left(\frac{1}{6}-\frac{1}{2}\sin^2 x\cos^2 x\right)
&=\frac{1}{4}-\frac{1}{6} =\frac{1}{12}. \end{aligned} \]

Final Answer: Option (A) Quick Tip: The expression \(f_4(x) - f_6(x)\) is a constant independent of \(x\). You can put \(x=0\) to find the value quickly. \(f_4(0) - f_6(0) = \frac{1}{4}(0+1) - \frac{1}{6}(0+1) = \frac{1}{12}\).

Let the sides of the triangle be \(a\), \(b\), and \(c\), where \[ a+b=x,\qquad ab=y. \]

Given: \[ x^2-c^2=y \]

Substitute \(x=a+b\) and \(y=ab\): \[ (a+b)^2-c^2=ab \]
\[ a^2+b^2+2ab-c^2=ab \]
\[ c^2=a^2+b^2+ab \]

Step 1: Use cosine rule

From cosine rule: \[ c^2=a^2+b^2-2ab\cos C \]

Comparing: \[ a^2+b^2+ab=a^2+b^2-2ab\cos C \]
\[ ab=-2ab\cos C \Rightarrow \cos C=-\frac12 \]
\[ C=120^\circ \]

Step 2: Circumradius formula
\[ R=\frac{c}{2\sin C} \]
\[ R=\frac{c}{2\sin120^\circ} =\frac{c}{2\cdot\frac{\sqrt3}{2}} =\frac{c}{\sqrt3} \]

Final Answer: Option (B) Quick Tip: Recognize algebraic structures similar to the Cosine Rule (\(c^2 = a^2 + b^2 \pm k ab\)) to determine the angle of the triangle.

Given: \[ q = F \quad and \quad (p \wedge q)\leftrightarrow r = T \]

Step 1: Determine \(r\)

Since \(q=F\), we have: \[ p \wedge q = F \]

Given: \[ F \leftrightarrow r = T \Rightarrow r = F \]

Thus: \[ q = F,\quad r = F,\quad p is arbitrary \]

Step 2: Check options



(A) \(p \wedge r = p \wedge F = F\) (Not tautology)
(B) \(p \vee r = p \vee F = p\) (Depends on \(p\))
(C) \((p \wedge r)\to(p \vee r)\)
\[ (p \wedge F)\to(p \vee F)=F\to p=T \]
(Always true)
(D) \((p \vee r)\to(p \wedge r)=p\to F=\neg p\) (Depends on \(p\))


Conclusion:

Only option (C) is always true.

Final Answer: Option (C) Quick Tip: Implication \(X \to Y\) is only false when X is True and Y is False. Since Intersection (\(X \cap Y\)) is a subset of Union (\(X \cup Y\)), the implication \((X \wedge Y) \to (X \vee Y)\) is always valid.