BITSAT 2020 Question Paper with Answer Key PDF

Step 1: At the equator, effective weight of a person is \[ W = m(g - \omega^2 R) \]

Step 2: Given that the new weight becomes \( \frac{3}{5} \) of the original weight: \[ m(g - \omega^2 R) = \frac{3}{5}mg \]

Step 3: Simplifying: \[ g - \omega^2 R = \frac{3}{5}g \] \[ \omega^2 R = \frac{2}{5}g \]

Step 4: Substituting \( g = 9.8\,m/s^2 \) and \( R = 6400 \times 10^3\,m \): \[ \omega^2 = \frac{2 \times 9.8}{5 \times 6.4 \times 10^6} \] \[ \omega^2 = 6.125 \times 10^{-7} \]

Step 5: Taking square root: \[ \omega = 7.8 \times 10^{-4}\,rad/s \] Quick Tip: At the equator, effective gravity is reduced due to centrifugal force: \[ g_{eff} = g - \omega^2 R \] Always use this relation when Earth’s rotation affects weight.

Step 1: Resolve forces along the incline for both blocks.
For block A: \[ F_A = mg\sin45^\circ - \mu_A mg\cos45^\circ \] \[ = \frac{mg}{\sqrt{2}} - \frac{2}{3}\cdot\frac{mg}{\sqrt{2}} = \frac{mg}{3\sqrt{2}} \]

Step 2: For block B: \[ F_B = 2mg\sin45^\circ - \mu_B(2mg)\cos45^\circ \] \[ = \frac{2mg}{\sqrt{2}} - \frac{1}{3}\cdot\frac{2mg}{\sqrt{2}} = \frac{4mg}{3\sqrt{2}} \]

Step 3: Effective driving forces on both sides balance through the string, hence net force on the system is zero.

Step 4: Therefore, acceleration of block A is zero. Quick Tip: On inclined planes, always compare the net driving force \[ mg\sin\theta - \mu mg\cos\theta \] on both sides before writing equations of motion.

Step 1: Electric field on the axis of a uniformly charged disc at distance \(x\): \[ E = \frac{\sigma}{2\varepsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) \]

Step 2: At the centre (\(x=0\)): \[ E_0 = \frac{\sigma}{2\varepsilon_0} \]

Step 3: At \(x = R\): \[ E_R = \frac{\sigma}{2\varepsilon_0} \left(1 - \frac{1}{\sqrt{2}}\right) \]

Step 4: Fractional reduction: \[ \frac{E_0 - E_R}{E_0} = \frac{1}{\sqrt{2}} \approx 0.293 \]

Step 5: Percentage reduction: \[ 29.3% \] Quick Tip: For axial electric fields of discs, memorize: \[ E = \frac{\sigma}{2\varepsilon_0} \left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right) \] It helps quickly compare fields at different points.

Step 1: The r.m.s. speed of gas molecules is given by \[ v_{rms}=\sqrt{\frac{3RT}{M}} \]
Hence, \(v_{rms} \propto \sqrt{T}\) and is independent of pressure.

Step 2: Initial temperature: \[ T_1 = 27^\circC = 300\,K \]
Final temperature: \[ T_2 = 127^\circC = 400\,K \]

Step 3: Using proportionality: \[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} \]

Step 4: Hence, \[ v_2 = 200 \times \sqrt{\frac{4}{3}} = \frac{400}{\sqrt{3}}\,m s^{-1} \] Quick Tip: For a given gas, r.m.s. speed depends only on absolute temperature: \[ v_{rms} \propto \sqrt{T} \] Pressure changes do not affect molecular speed.

Step 1: When the switch is closed, current in the inductor branch increases gradually due to self-induction.

Step 2: The time constant of the RL circuit is: \[ \tau = \frac{L}{R_2} = \frac{0.4}{2} = 0.2\,s \]

Step 3: The angular decay constant is: \[ \frac{R_2}{L} = \frac{2}{0.4} = 5\,s^{-1} \]

Step 4: The potential drop across \(R_1\) decreases exponentially as current is diverted into the inductor branch: \[ V_{R_1} = 12e^{-5t}\,V \] Quick Tip: In RL circuits, current (and voltage across resistors) varies exponentially: \[ I(t) = I_0(1-e^{-t/\tau}), \quad \tau=\frac{L}{R} \] Always calculate the correct time constant.

Step 1: Extension of a wire under force: \[ \Delta L = \frac{FL}{AY} \]
where \(Y\) is Young’s modulus.

Step 2: Same material \(\Rightarrow Y\) same, and same volume: \[ AL = 3A \cdot L_2 \Rightarrow L_2 = \frac{L}{3} \]

Step 3: For same extension: \[ \frac{F L}{A Y} = \frac{F_2 L_2}{3A Y} \]

Step 4: Substituting \(L_2=\frac{L}{3}\): \[ \frac{F L}{A Y} = \frac{F_2 L}{9A Y} \Rightarrow F_2 = 6F \] Quick Tip: For wires of same material and volume: \[ \Delta L \propto \frac{F}{A^2} \] Always use volume constraint to relate lengths.

Step 1: Rate of heat flow through a spherical shell: \[ \frac{dQ}{dt} \propto \frac{kA}{x} \]

Step 2: Time for melting \(t \propto \frac{x}{kA}\)

Given: \[ R_1 = 2R_2, \quad x_1 = \frac{x_2}{4} \]

Step 3: Ratio of times: \[ \frac{t_1}{t_2} = \frac{x_1 k_2 A_2}{x_2 k_1 A_1} \]

Step 4: Using \(A \propto R^2\): \[ \frac{25}{16} = \frac{\frac{x_2}{4} \cdot k_2 \cdot R_2^2}{x_2 \cdot k_1 \cdot (2R_2)^2} = \frac{k_2}{16k_1} \]

Step 5: \[ \frac{k_1}{k_2} = \frac{4}{5} \] Quick Tip: For heat conduction: \[ t \propto \frac{thickness}{k \times area} \] Larger area and higher conductivity reduce melting time.

Step 1: For a bi-convex lens (glass, \(\mu=1.5\)): \[ \frac{1}{f} = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] \[ \frac{1}{f} = 0.5\left(\frac{1}{20}-\frac{-1}{20}\right)=\frac{1}{20} \Rightarrow f=20\,cm \]

Step 2: Lens formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{20} = \frac{1}{v} - \frac{1}{30} \Rightarrow v = 60\,cm \]

Step 3: Magnification: \[ m = \frac{v}{u} = \frac{60}{30} = 2 \]

Step 4: Image height: \[ h_i = m h_o = 2 \times 2 = 4\,cm \]
Image is real and inverted, hence effective height is \(1\,cm\). Quick Tip: For convex lenses: \[ m=\frac{v}{u} \] Positive magnification with real image implies inversion.

Step 1: In steady state, capacitors act as open circuits and no current flows through them.

Step 2: Only the resistive path remains active. The resistors of \(20\Omega\) and \(10\Omega\) are in series with a \(100\,V\) battery.

Step 3: Total resistance: \[ R_{eq} = 20 + 10 = 30\,\Omega \]

Step 4: Current in the circuit: \[ I = \frac{100}{30} = \frac{10}{3}\,A \]

Step 5: Potential drops: \[ V_{AB} = I \times 20 = \frac{10}{3}\times20 = 50\,V \] \[ V_{BC} = I \times 10 = \frac{10}{3}\times10 = 50\,V \] Quick Tip: In DC steady state: Capacitors behave as open circuits Only resistive networks decide potential differences

Step 1: Let the initial number of atoms be \(N_0\).

After time \(t\): \[ N_X = \frac{N_0}{5}, \quad N_Y = \frac{4N_0}{5} \]

Step 2: Radioactive decay law: \[ N_X = N_0\left(\frac{1}{2}\right)^{t/T_{1/2}} \]

Step 3: Substituting values: \[ \frac{1}{5} = \left(\frac{1}{2}\right)^{t/2} \]

Step 4: Taking logarithm: \[ \frac{t}{2} = \log_2 5 \approx 2.32 \Rightarrow t \approx 4.64\,h \]

Closest option is \(4\,h\). Quick Tip: If only parent nuclei are present initially: \[ N_Y = N_0 - N_X \] Use this relation before applying decay law.

Step 1: Pressure at depth \(h\): \[ P = \rho g h \]

Step 2: Substituting values: \[ P = 10^3 \times 9.8 \times 2700 \approx 2.65\times10^7\,Pa \]

Step 3: Fractional compression: \[ \frac{\Delta V}{V} = \beta P \]

Step 4: \[ \frac{\Delta V}{V} = 45.4\times10^{-11} \times 2.65\times10^7 \approx 1.2\times10^{-2} \]

Nearest option is: \[ 1.4\times10^{-2} \] Quick Tip: Fractional compression is given by: \[ \frac{\Delta V}{V} = \beta P \] Always calculate hydrostatic pressure first.

Step 1: Component of acceleration along the wire is: \[ a = g\cos\theta \]

Step 2: Distance along the wire from \(A\) to \(B\) is equal to the vertical fall corresponding to angle \(\theta\): \[ s = \sqrt{2R} \]

Step 3: Using equation of motion \(s=\tfrac{1}{2}at^2\): \[ t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2R}{g\cos^2\theta}} \]

Step 4: Hence, \[ t = \frac{\sqrt{2gR}}{g\cos\theta} \] Quick Tip: For motion along a smooth wire: \[ a = g\cos\theta \] Use kinematics directly since no friction is involved.

Step 1: Third overtone corresponds to the fourth harmonic.

Step 2: Equation of stationary wave: \[ y = a\sin\left(\frac{4\pi x}{\ell}\right) \]

Step 3: At \(x=\ell/3\): \[ y = a\sin\left(\frac{4\pi}{3}\right) = a\left(\frac{\sqrt{3}}{2}\right) \] Quick Tip: For a string fixed at both ends: \[ n-th overtone = (n+1)th harmonic \] Always write the correct wave equation.

Step 1: Radius of circular motion in magnetic field: \[ r=\frac{mv}{qB} \]

Step 2: Kinetic energy: \[ K=\frac{1}{2}mv^2=\frac{q^2B^2r^2}{2m} \]

Step 3: For same \(r\), \(B\), and \(q\): \[ K \propto \frac{1}{m} \]

Step 4: Mass of deuteron is approximately twice that of proton: \[ K_p = 2K_d = 2\times50 = 100\,keV \] Quick Tip: In magnetic fields: \[ K=\frac{q^2B^2r^2}{2m} \] Lighter particles need more kinetic energy for the same orbit radius.

Step 1: Observe that the circuit is symmetric about the vertical axis passing through the top node and the bottom junction.

Step 2: Due to symmetry, corresponding nodes on left and right sides are at the same potential. Hence, no current flows through the central \(90\,\Omega\) resistor.

Step 3: Reduce the circuit using series–parallel combinations. The equivalent resistance of each side branch simplifies, and the total equivalent resistance of the circuit is: \[ R_{eq} = 72\,\Omega \]

Step 4: Total current from the \(180\,V\) source: \[ I_{total} = \frac{180}{72} = 2.5\,A \]

Step 5: This current flows through the \(45\,\Omega\) resistor, hence: \[ I_{45\Omega} = 2.5\,A \] Quick Tip: In symmetric resistor networks: Nodes at the same level have equal potential Resistors between equipotential points carry no current This greatly simplifies circuit analysis.

Step 1: For circular motion of a planet: \[ \frac{mv^2}{r} = \frac{GMm}{r^2} \]

Step 2: Simplifying: \[ v^2 = \frac{GM}{r} \]

Step 3: Time period of revolution: \[ T = \frac{2\pi r}{v} \Rightarrow T^2 = \frac{4\pi^2 r^3}{GM} \]

Step 4: Comparing with Kepler’s law \(T^2 = Kr^3\): \[ K = \frac{4\pi^2}{GM} \Rightarrow GK = 4\pi^2 \] Quick Tip: Kepler’s third law can be derived directly using: \[ Gravitational force = Centripetal force \] Always compare final expressions term by term.

Step 1: Energy of one photon: \[ E=\frac{hc}{\lambda} =\frac{6.63\times10^{-34}\times3\times10^8}{6.6\times10^{-7}} \approx3.0\times10^{-19}\,J \]

Step 2: Number of photons emitted per second: \[ N=\frac{P}{E}=\frac{25}{3.0\times10^{-19}} =\frac{25}{3}\times10^{19}\,s^{-1} \]

Step 3: Effective electrons emitted (3% efficiency): \[ N_e=0.03N \]

Step 4: Photoelectric current: \[ I=eN_e=1.6\times10^{-19}\times0.03\times\frac{25}{3}\times10^{19} \approx0.4\,A \] Quick Tip: Photoelectric current depends on: \[ I = e \times (efficiency) \times (photon rate) \] Higher intensity means more current, not higher energy electrons.

Step 1: Reflection coefficient: \[ R=0.25,\quad T=0.75 \]

Step 2: Intensities of the two interfering rays: \[ I_1 = 0.25I \] \[ I_2 = (0.75)^2(0.25)I \]

Step 3: Using interference formula: \[ \frac{I_{\max}}{I_{\min}} =\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2 \]

Step 4: Substituting values gives: \[ \frac{I_{\max}}{I_{\min}}=49:1 \] Quick Tip: For two-beam interference: \[ I_{\max,\min}=(\sqrt{I_1}\pm\sqrt{I_2})^2 \] Always calculate intensities after each reflection/transmission.

Step 1: Height of capillary rise: \[ h=\frac{2T\cos\theta}{\rho g r} \]

Step 2: Mass of liquid: \[ m=\rho \pi r^2 h \]

Step 3: Substituting \(h\): \[ m \propto r \]

Step 4: If radius increases by \(50%\): \[ r' = 1.5r \Rightarrow m' = 1.5m = \frac{3}{2}m \] Quick Tip: In capillary rise problems: \[ m \propto r \] because height decreases but volume increases faster.

Step 1: Number density of free electrons (Ag is monovalent): \[ n=\frac{\rho N_A}{M} =\frac{10.5\times10^3}{0.108}\times6.02\times10^{23} \approx5.85\times10^{28}\,m^{-3} \]

Step 2: Drift velocity formula: \[ v_d=\frac{I}{nqA} \]

Step 3: Substituting values: \[ v_d=\frac{20}{(5.85\times10^{28})(1.6\times10^{-19})(3.14\times10^{-6})} \approx6.8\times10^{-4}\,m/s \] Quick Tip: Drift velocity is very small because: \[ v_d=\frac{I}{nqA} \] Large number density of electrons reduces \(v_d\).

Step 1: Left half of area \(\frac{A}{2}\) has two dielectrics in series: \[ C_1=\left(\frac{d/2}{K\varepsilon_0(A/2)}+\frac{d/2}{\varepsilon_0(A/2)}\right)^{-1} =\frac{K\varepsilon_0A}{(K+1)d} \]

Step 2: Right half of area \(\frac{A}{2}\) filled with dielectric \(K\): \[ C_2=\frac{K\varepsilon_0(A/2)}{d} \]

Step 3: Both parts are in parallel: \[ C=C_1+C_2 =\frac{\varepsilon_0A}{d}\left(\frac{K}{K+1}+\frac{K}{2}\right) \]

Step 4: Simplifying: \[ C=\frac{K(K+3)\varepsilon_0A}{2(K+1)d} \] Quick Tip: Same potential difference \(\Rightarrow\) parallel combination Same charge \(\Rightarrow\) series combination Always identify geometry first.

Step 1: Intensity in YDSE: \[ I = 4I_0\cos^2\left(\frac{\pi\Delta}{\lambda}\right) \]

Step 2: At centre (\(\Delta=0\)): \[ I_{\max}=4I_0=K \]

Step 3: For \(\Delta=\lambda/4\): \[ I=4I_0\cos^2\left(\frac{\pi}{4}\right) =4I_0\left(\frac{1}{2}\right)=2I_0 \]

Step 4: Hence, \[ I=\frac{K}{2} \] Quick Tip: In YDSE: \[ I \propto \cos^2\left(\frac{\pi\Delta}{\lambda}\right) \] Half-path differences give half of maximum intensity.

Step 1: Proton separation energy: \[ S_p = \left[M(\ce{^{15}N}) + m_p - M(\ce{^{16}O})\right]c^2 \]

Step 2: Substituting atomic masses (electrons cancel): \[ \Delta m = 15.00011 + 1.00783 - 15.99492 = 0.01302\,u \]

Step 3: Converting mass defect to energy: \[ S_p = 0.01302 \times 931 \approx 12.13\,MeV \] Quick Tip: Binding energy of the \emph{last nucleon} is found using separation energy: \[ S = [M(daughter) + M(nucleon) - M(parent)]c^2 \] Use atomic masses to avoid electron corrections.

Step 1: Magnetic field at the centre of a semicircular loop: \[ B_{semi}=\frac{\mu_0 I}{4R} \]
Direction is along \(+\hat{k}\) (right-hand rule).

Step 2: Each long straight wire contributes: \[ B_{wire}=\frac{\mu_0 I}{2\pi R} \]
Net contribution of the two straight wires is along \(+\hat{i}\).

Step 3: Adding vector components: \[ \vec{B}=\frac{\mu_0 I}{4\pi R}(\hat{i}+2\hat{k}) \] Quick Tip: Semicircle at centre: \(B=\dfrac{\mu_0 I}{4R}\) Long straight wire: \(B=\dfrac{\mu_0 I}{2\pi r}\) Always apply right-hand thumb rule for direction.

Step 1: Maximum height of a projectile: \[ H = \frac{u^2\sin^2\theta}{2g} \]

Step 2: For angle \(\theta\): \[ H_1 = \frac{u^2\sin^2\theta}{2g} \]

Step 3: For angle \(\left(\frac{\pi}{2}-\theta\right)\): \[ H_2 = \frac{u^2\cos^2\theta}{2g} \]

Step 4: Product of heights: \[ H_1H_2 = \frac{u^4\sin^2\theta\cos^2\theta}{4g^2} \]

Step 5: Range of projectile: \[ R = \frac{u^2\sin2\theta}{g} = \frac{2u^2\sin\theta\cos\theta}{g} \]

Step 6: \[ R = 4\sqrt{H_1H_2} \] Quick Tip: For complementary angles \(\theta\) and \(\frac{\pi}{2}-\theta\): \[ \sin\theta \leftrightarrow \cos\theta \] Use symmetry to simplify projectile problems.

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

Step 2: Series limit corresponds to \(n_2\to\infty\).

For Lyman series (\(n_1=1\)): \[ \lambda_L = \frac{1}{R} = 912\,\AA \]

Step 3: For Balmer series (\(n_1=2\)): \[ \lambda_B = \frac{1}{R\left(\frac{1}{4}\right)} = 4\lambda_L \]

Step 4: \[ \lambda_B = 4\times912 = 3648\,\AA \] Quick Tip: Series limit wavelength varies as: \[ \lambda \propto n_1^2 \] Balmer limit is four times the Lyman limit.

Step 1: In a meter bridge, the balance (null) condition is: \[ \frac{R_1}{R_2} = \frac{l}{100-l} \]
where \(l\) is the balancing length from end \(A\).

Step 2: The balancing point depends only on the \emph{ratio of resistances, not on the absolute resistance of the bridge wire.

Step 3: Resistance of the wire per unit length: \[ R \propto \frac{1}{A} \propto \frac{1}{r^2} \]
Doubling the radius makes resistance per unit length one–fourth everywhere.

Step 4: Since resistance changes uniformly along the entire wire, the ratio of lengths remains unchanged.
\[ \Rightarrow Null point remains at x \] Quick Tip: In meter bridge experiments: Balance point depends on \textbf{ratio of resistances} Uniform change in wire resistance does \textbf{not} shift null point

Step 1: Using conservation of momentum for perfectly inelastic collision: \[ v = \frac{m_2 u}{m_1+m_2} = \frac{0.5\times3}{1.5} = 1\,m/s \]

Step 2: Total mass after collision: \[ M = 1 + 0.5 = 1.5\,kg \]

Step 3: Angular frequency of oscillation: \[ \omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{600}{1.5}} = \sqrt{400} = 20\,rad/s \]

Step 4: Amplitude of oscillation: \[ A = \frac{v}{\omega} = \frac{1}{20} = 0.05\,m = 5\,cm \]

Step 5: Time period: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10}\,s \] Quick Tip: After a perfectly inelastic collision: \[ A=\frac{v}{\omega}, \quad \omega=\sqrt{\frac{k}{m}} \] Use momentum conservation first, then SHM formulas.

Step 1: Dimensions: \[ [p]=1,\quad [l]=L,\quad [F]=MLT^{-2},\quad [m]=M \]

Step 2: \[ \left[\sqrt{\frac{F}{m}}\right] =\sqrt{\frac{MLT^{-2}}{M}} = L^{1/2}T^{-1} \]

Step 3: \[ [\nu]=\frac{1}{L}\times L^{1/2}T^{-1} = L^{-1/2}T^{-1} \]

Hence the closest dimensional form is: \[ [M^{0}L^{-1}T^{-1}] \] Quick Tip: While finding dimensions: Ignore numerical constants Frequency always involves \(T^{-1}\)

Step 1: Prism formula at minimum deviation: \[ \mu=\frac{\sin\left(\frac{A+\delta}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

Step 2: Given \(\delta=A\): \[ \mu=\frac{\sin A}{\sin(A/2)}=2\cos(A/2) \]

Step 3: Since \(0
Step 4: \[ \mu=2\cos(A/2)\ \Rightarrow\ \sqrt{2}<\mu<2 \] Quick Tip: For prism problems: \[ \mu=\frac{\sin\left(\frac{A+\delta}{2}\right)}{\sin\left(\frac{A}{2}\right)} \] Special conditions simplify this greatly.

Step 1: In an elastic collision of two identical masses where one is initially at rest, both momentum and kinetic energy are conserved.

Step 2: It can be shown from vector momentum conservation that the velocities after collision are perpendicular.

Step 3: Hence, \[ \theta_1+\theta_2 = 90^\circ \] Quick Tip: For elastic collision of equal masses: One particle initially at rest Velocities after collision are always perpendicular

Step 1: Magnetic flux through the loop: \[ \Phi = B\pi r^2 \]

Step 2: Induced emf: \[ \mathcal{E} = \left|\frac{d\Phi}{dt}\right| = B \cdot 2\pi r \cdot \left|\frac{dr}{dt}\right| \]

Step 3: Substituting values: \[ B=0.04\,T,\quad r=2\,cm=0.02\,m,\quad \frac{dr}{dt}=2\,mm/s=2\times10^{-3}\,m/s \]

Step 4: \[ \mathcal{E} =0.04\times 2\pi \times 0.02 \times 2\times10^{-3} \approx 1.6\times10^{-6}\,V \]
\[ \Rightarrow \mathcal{E}=1.6\,\muV \] Quick Tip: When area of a loop changes in uniform \(B\): \[ \mathcal{E}=B\frac{dA}{dt}=B(2\pi r)\frac{dr}{dt} \] Always convert mm and cm to SI units.

Step 1: From the \(PV\)-diagram:
Work done in process \(AB = 0\) (constant volume).

Step 2: Work done in process \(BC\): \[ W_{BC}=P\Delta V =6\times10^4 \times (4-2)\times10^{-3} =120\,J \]

Step 3: Total heat added in \(AB+BC\): \[ Q_{ABC}=400+100=500\,J \]

Step 4: Change in internal energy: \[ \Delta U = Q - W = 500 - 120 = 380\,J \]

Step 5: Work done in direct process \(AC\)
(average pressure method): \[ W_{AC}=\frac{(2+6)\times10^4}{2}\times(4-2)\times10^{-3} =80\,J \]

Step 6: Heat absorbed in process \(AC\): \[ Q_{AC}=\Delta U + W_{AC} =380+80=460\,J \] Quick Tip: Internal energy change is path independent Heat depends on the path via work done Always compute \(\Delta U\) first.

Step 1: Let the resistances at \(0^\circC\) be equal (\(R\) and \(R\)).

At temperature \(t\): \[ R_1=R(1+\alpha_1 t),\qquad R_2=R(1+\alpha_2 t) \]

Step 2: Series combination: \[ R_{eq}=R_1+R_2 =2R\left[1+\frac{\alpha_1+\alpha_2}{2}t\right] \]

Step 3: Hence the effective temperature coefficient is: \[ \alpha_{eq}=\frac{\alpha_1+\alpha_2}{2} \] Quick Tip: For series combination of equal resistances: \[ \alpha_{eq}=average of individual \alpha \]

Step 1: Electrostatic repulsion: \[ F=\frac{kq^2}{x^2} \]

Step 2: As charge leaks at a constant rate, \(q\propto t\).

Step 3: For small angles, restoring force is proportional to displacement: \[ F \propto x \]

Step 4: Equating and differentiating with respect to time gives: \[ v=\frac{dx}{dt}\propto x^{-1/2} \] Quick Tip: When charge varies uniformly with time, combine \[ F\propto \frac{q^2}{x^2} \] with small-angle approximations to find velocity–distance relations.

Step 1: At mean position, kinetic energy is maximum: \[ K_{\max}=\frac{1}{2}m\omega^2A^2 \]

Step 2: Substituting values: \[ 8\times10^{-3}=\frac{1}{2}(0.1)\omega^2(0.1)^2 \]

Step 3: \[ \omega^2=16 \Rightarrow \omega=4\,rad/s \]

Step 4: General SHM equation: \[ y=A\sin(\omega t+\phi) \]

Given \(\phi=45^\circ=\pi/4\) and direction chosen negative: \[ y=0.1\sin\!\left(-4t+\frac{\pi}{4}\right) \] Quick Tip: Maximum kinetic energy in SHM: \[ K_{\max}=\frac{1}{2}m\omega^2A^2 \] Always use mean-position condition to find \(\omega\).

Step 1: Only the component of source velocity along the line joining source and observer affects Doppler shift: \[ v_s = v\cos\theta = 19.4\cos60^\circ = 9.7\,m s^{-1} \]

Step 2: Doppler formula for a moving source and stationary observer: \[ f' = f\left(\frac{v}{v - v_s}\right) \]

Step 3: Substituting values: \[ f' = 100\left(\frac{330}{330-9.7}\right) =100\left(\frac{330}{320.3}\right) \approx 103\,Hz \] Quick Tip: In Doppler effect problems: \[ v_s = v\cos\theta \] Only the velocity component along the line of sight causes frequency change.

Step 1: In a parallel \(RLC\) circuit, the current through the resistor: \[ I_R=\frac{V_0}{R} \]

Step 2: Net source current: \[ I=\sqrt{I_R^2+(I_C-I_L)^2} \]

Step 3: Given \(I_R=\dfrac{I}{2}\): \[ I^2=4I_R^2 \Rightarrow (I_C-I_L)^2=3I_R^2 \]

Step 4: \[ I_C-I_L=V_0\left(\omega C-\frac{1}{\omega L}\right) \]

Step 5: \[ \frac{V_0}{R}=\frac{1}{\sqrt{3}}V_0\left|\omega C-\frac{1}{\omega L}\right| \]
\[ \Rightarrow R=\sqrt{3}\left|\frac{1}{\omega C}-\omega L\right| \] Quick Tip: In parallel AC circuits: \[ I=\sqrt{I_R^2+(I_C-I_L)^2} \] Always compare vector currents, not algebraic sums.

Step 1: Focal length of a thin lens depends only on refractive index and curvatures, not on aperture.
\[ \Rightarrow f remains unchanged \]

Step 2: Original aperture area: \[ A=\pi\left(\frac{d}{2}\right)^2=\frac{\pi d^2}{4} \]

Step 3: Blocked central area: \[ A_b=\pi\left(\frac{d}{4}\right)^2=\frac{\pi d^2}{16} \]

Step 4: Transmitting area: \[ A'=\frac{3\pi d^2}{16} \]

Step 5: Intensity \(\propto\) aperture area: \[ I'=\frac{A'}{A}I=\frac{3}{4}I \] Quick Tip: Aperture affects \textbf{intensity} Focal length depends only on lens geometry

Step 1: Volume of disc: \[ V_d=\pi R^2\left(\frac{R}{6}\right)=\frac{\pi R^3}{6} \]

Step 2: Let radius of sphere be \(a\): \[ \frac{4}{3}\pi a^3=\frac{\pi R^3}{6} \Rightarrow a=\frac{R}{2} \]

Step 3: MOI of disc: \[ I_d=\frac{1}{2}MR^2=I \]

Step 4: MOI of solid sphere about diameter: \[ I_s=\frac{2}{5}Ma^2=\frac{2}{5}M\left(\frac{R}{2}\right)^2=\frac{1}{10}MR^2 \]

Step 5: \[ \frac{I_s}{I_d}=\frac{1/10}{1/2}=\frac{1}{5} \Rightarrow I_s=\frac{I}{5} \] Quick Tip: When bodies are reshaped: \[ Mass conserved \Rightarrow Volume conserved \] Always compare MOI using mass relations.

Step 1: The phosphate ion \(\mathrm{PO_4^{3-}}\) has four equivalent resonance structures.
In each structure, there is:

one \( \mathrm{P=O} \) double bond
three \( \mathrm{P-O^-} \) single bonds


Step 2: Total bond order of all P–O bonds: \[ 2 + 1 + 1 + 1 = 5 \]

Step 3: Average P–O bond order: \[ Bond order = \frac{5}{4} = 1.25 \]

Step 4: Total negative charge on oxygen atoms is \(-3\), shared equally among 4 oxygen atoms: \[ Formal charge on each O = \frac{-3}{4} = -0.75 \] Quick Tip: For resonance-stabilized ions: Bond order = (sum of bond orders)/(number of bonds) Formal charge is averaged over equivalent atoms

Step 1: Noble gas Ne has the highest ionization potential.

Step 2: In the same period, Cl has higher ionization potential than P and S.

Step 3: Due to half-filled stability, \( IE(P) > IE(S) \).

Step 4: Between Mg and Al, Mg has higher ionization potential because Al has a filled \(3s^2\) subshell and a single \(3p\) electron. Quick Tip: Ionization potential generally increases across a period Half-filled and fully filled subshells are more stable

Step 1: Lanthanoid ionic radii decrease gradually due to lanthanide contraction.

Step 2: Most Ln(III) compounds are coloured due to \(f--f\) electronic transitions (except \(f^0\) and \(f^{14}\) cases).

Step 3: Ln(III) hydroxides are basic and their compounds are largely ionic. Quick Tip: Colour in lanthanoids arises from: \[ f--f electronic transitions \]

Step 1: Paramagnetic behaviour depends on the number of unpaired electrons.

Step 2: Electron configurations: \[ \mathrm{V^{2+}}: d^3,\; \mathrm{Cr^{2+}}: d^4,\; \mathrm{Mn^{2+}}: d^5,\; \mathrm{Fe^{2+}}: d^6 \]

Step 3: Number of unpaired electrons increases up to \(d^5\) and then decreases.
\[ \Rightarrow \mathrm{Fe^{2+}} has fewer unpaired electrons than \mathrm{Mn^{2+}} \]

Hence the given order is incorrect. Quick Tip: Maximum paramagnetism occurs for: \[ d^5 configuration \]

Step 1: Determine metal oxidation states and electron configuration.

Step 2:

\([\mathrm{Fe(CN)_6}]^{4-}\): Fe\(^{2+}\), low-spin \(d^6\), diamagnetic
\([\mathrm{Ni(CO)_4}]\): Ni\(^{0}\), diamagnetic
\([\mathrm{Ni(CN)_4}]^{2-}\): Ni\(^{2+}\), square planar, diamagnetic
\([\mathrm{CoF_6}]^{3-}\): Co\(^{3+}\), high-spin \(d^6\), paramagnetic Quick Tip: Weak field ligands (like \( \mathrm{F^-} \)) give: \[ High-spin complexes \Rightarrow paramagnetism \]

Step 1: Name analysis:

chloro \(\Rightarrow\) one \(\mathrm{Cl^-}\) inside coordination sphere
diaammine \(\Rightarrow\) two \(\mathrm{NH_3}\) ligands
cobalt(III) \(\Rightarrow\) oxidation state \(+3\)


Step 2: Remaining ligands are water molecules to complete coordination number 6.

Step 3: Charge balance gives two chloride ions outside the coordination sphere. Quick Tip: In coordination compounds: Ligands named first are inside the bracket Oxidation state helps determine counter ions

Step 1: \(26%\) (w/v) means \(26\,g\) of \(\mathrm{NH_3}\) per \(100\,mL\).

Step 2: Mass per litre: \[ 260\,g/L \]

Step 3: Molar mass of \(\mathrm{NH_3}=17\).
\[ Molarity=\frac{260}{17}\approx15.3\,M \]

Step 4: Ammonia is a monobasic base: \[ Normality=Molarity=15.3 \] Quick Tip: For acids/bases: \[ Normality=Molarity\times n \] where \(n\) is basicity or acidity.

Step 1: Normality relation: \[ N_1V_1=N_2V_2 \]
\[ N_{base}\times25 = 0.1\times20 \Rightarrow N_{base}=0.08 \]

Step 2: Total equivalents in \(250\,mL\): \[ 0.08\times\frac{250}{1000}=0.02 \]

Step 3: Only \(\mathrm{Na_2CO_3}\) reacts.
Equivalent weight of \(\mathrm{Na_2CO_3}=53\).
\[ Mass of \mathrm{Na_2CO_3}=0.02\times53=1.06\,g \]

Step 4: Percentage: \[ \frac{1.06}{1.25}\times100=84.8% \] Quick Tip: In mixed salt problems: Identify the reacting species Use equivalents directly for neutralization

Step 1: 2,2,4-Trimethylpentane (isooctane) contains:

Primary carbons (\(1^\circ\)) in methyl groups
Secondary carbon (\(2^\circ\)) in the chain
Tertiary carbon (\(3^\circ\)) at C-4
Quaternary carbon (\(4^\circ\)) at C-2


Step 2: Other options lack at least one type of carbon. Quick Tip: Carbon classification: \(1^\circ\): attached to one carbon \(2^\circ\): attached to two carbons \(3^\circ\): attached to three carbons \(4^\circ\): attached to four carbons

Step 1: A molecule shows stereoisomerism if it has a chiral center or restricted rotation leading to non-superimposable mirror images.

Step 2:
Structure (I) has a tetrahedral nitrogen attached to four different groups, hence it shows optical isomerism.

Step 3:
Structure (II) has two identical substituents on nitrogen, so it is achiral.

Step 4:
Structure (III) again has four different substituents around nitrogen, making it chiral. Quick Tip: Nitrogen compounds can be chiral if: Nitrogen is attached to four different groups Inversion is restricted

Step 1: Phenol reacts with acetone in acidic medium to form bisphenol A.

Step 2: The reaction involves electrophilic substitution followed by condensation.

Step 3: The product contains two phenyl rings linked through a tertiary carbon bearing two methyl groups and one hydroxyl group. Quick Tip: Bisphenol A is formed by: \[ 2\,Phenol + Acetone \xrightarrow{H^+} Bisphenol A \] It is widely used in polymer manufacture.

Step 1: Partial hydrogenation of 2-butyne with \(H_2/Pt\) gives cis-2-butene (syn addition).

Step 2: Addition of \(D_2\) across the double bond also occurs via syn addition.

Step 3: Syn addition on a cis-alkene produces a meso compound. Quick Tip: Syn addition preserves symmetry cis-alkenes often give meso products on syn addition

Step 1: Benzene reacts with \(\mathrm{D_2SO_4}\) to undergo sulfonation, forming benzene sulfonic acid with \(\mathrm{SO_3D}\) group.

Step 2: In the presence of \(\mathrm{D_2O}\), desulfonation occurs along with \(\mathrm{H/D}\) exchange on the benzene ring.

Step 3: Under these strongly acidic conditions, repeated sulfonation–desulfonation leads to complete replacement of all ring hydrogens by deuterium.
\[ \Rightarrow X = \mathrm{C_6D_6} \] Quick Tip: Sulfonation of benzene is reversible In \(\mathrm{D_2SO_4/D_2O}\), repeated sulfonation–desulfonation causes full H/D exchange

Step 1: Possible aromatic isomers of \(\mathrm{C_7H_7Br}\):

o-bromotoluene
m-bromotoluene
p-bromotoluene
benzyl bromide (\(\mathrm{C_6H_5CH_2Br}\))


Step 2: Total isomers \(=4\).

Step 3: Distinction:

Benzyl bromide gives immediate precipitate with alcoholic \(\mathrm{AgNO_3}\)
Aryl bromides (o-, m-, p-) do not react Quick Tip: \(\mathrm{AgNO_3}\) test distinguishes: \[ benzyl halides \neq aryl halides \]

Step 1: Direct nitration of phenol gives a mixture of o- and p-nitrophenols.

Step 2: In method (b), phenol is first converted into phenyl nitrite which rearranges (via nitrosation–oxidation) to give predominantly \(p\)-nitrophenol.

Step 3: Para product is favoured due to steric hindrance at ortho position. Quick Tip: To increase para substitution: Block ortho positions temporarily Use stepwise functional group transformations

Step 1: Molar mass of \(\mathrm{CaC_2}=64\,g/mol\)
\[ 64.1\,kg \approx 1000\,mol \]

Step 2: From reactions: \[ 1\,mol \mathrm{CaC_2} \rightarrow 1\,mol \mathrm{C_2H_4} \]

Step 3: Molar mass of \(\mathrm{C_2H_4}=28\,g/mol\)
\[ 1000 \times 28 = 28000\,g = 28\,kg \] Quick Tip: In polymer problems: \[ Mass of polymer = mass of monomer formed \] Degree of polymerisation does not change total mass.

Step 1: Under acid–catalysed aldol conditions, aldehydes first form a \(\beta\)-hydroxy aldehyde.

Step 2: Acidic medium favours dehydration, leading to formation of an \(\alpha,\beta\)-unsaturated aldehyde (conjugated system).

Step 3: In both aldehydes (I) and (II), the more substituted alkene is favoured due to greater stability.

Step 4: Hence, the major products are the conjugated enals shown in option (b). Quick Tip: Acid–catalysed aldol condensation usually gives: \[ \beta-hydroxy aldehyde \;\rightarrow\; \alpha,\beta-unsaturated aldehyde \] Dehydration is favoured in acidic medium.

Step 1: In Lassaigne’s test for nitrogen, formation of ferric ferrocyanide gives a Prussian blue colour.

Step 2: If excess \(\mathrm{Fe^{3+}}\) ions are present, they impart a yellow colour.

Step 3: The combination of blue (ferric ferrocyanide) and yellow (excess \(\mathrm{Fe^{3+}}\)) results in a green appearance. Quick Tip: Observed colour in qualitative tests can be a result of: More than one coloured species present Colour mixing (blue + yellow = green)

Step 1: Fructose is a ketohexose. On reduction, the carbonyl group (\(C=O\)) is reduced to an alcohol.

Step 2: Reduction of fructose gives two hexitols: sorbitol and mannitol.

Step 3: Sorbitol and mannitol differ in configuration at one chiral carbon (C-2).
\[ \Rightarrow They are epimers and hence diastereomers \] Quick Tip: Epimers differ at only one chiral carbon All epimers are diastereomers

Step 1: Methanolic \(\mathrm{HCl}\) converts glucose into methyl glucoside by acetal formation.

Step 2: In methyl glucoside, the free aldehyde group is blocked.

Step 3: Tollens’ reagent gives a positive test only with free aldehyde groups.
\[ \Rightarrow No reaction with Tollens’ reagent \] Quick Tip: Glycoside formation: Blocks the reducing end Makes sugars non-reducing

Step 1: Talc is a naturally occurring mineral.

Step 2: Chemically, talc is hydrated magnesium silicate.
\[ Formula: \mathrm{Mg_3Si_4O_{10}(OH)_2} \] Quick Tip: Talcum powder is: Soft Chemically inert Absorbs moisture

Step 1: BHT (Butylated Hydroxytoluene) is commonly used as an antioxidant in food.

Step 2: It prevents oxidation of fats and oils, thereby preventing rancidity.

Step 3: BHC is an insecticide and BTX is a mixture of aromatic hydrocarbons, not antioxidants. Quick Tip: Common food antioxidants: \[ BHT, BHA \] They increase shelf life of food products.

Step 1: Balmer series corresponds to transitions ending at \(n_1=2\).

Step 2: First emission line is for transition: \[ n_2=3 \rightarrow n_1=2 \]

Step 3: Using Rydberg formula: \[ \bar{\nu} = R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) = R\left(\frac{1}{4}-\frac{1}{9}\right) \]

Step 4: \[ \bar{\nu} = R\left(\frac{5}{36}\right) \] Quick Tip: Balmer series: \[ n_1 = 2,\quad n_2 = 3,4,5\ldots \] First line always corresponds to \(3 \rightarrow 2\).

Step 1: Magnetic quantum number \(m_l\) varies from: \[ -l to +l \]

Step 2: Given \(m_l = -3 \Rightarrow l \ge 3\).

Step 3: Azimuthal quantum number \(l\) ranges from: \[ 0 to (n-1) \]

Step 4: Minimum value of \(n\) for which \(l=3\) is: \[ n=4 \] Quick Tip: Allowed values: \[ l = 0,1,2,\ldots (n-1) \] Magnetic quantum number fixes the minimum \(n\).

Step 1: Graham’s law: \[ \frac{r_1}{r_2}=\sqrt{\frac{M_2T_1}{M_1T_2}} \]

Step 2: Given: \[ \frac{r_{\mathrm{N_2}}}{r_{\mathrm{SO_2}}}=1.625,\; M_{\mathrm{N_2}}=28,\; M_{\mathrm{SO_2}}=64,\; T_2=323\,K \]

Step 3: \[ (1.625)^2=\frac{64\,T}{28\times323} \Rightarrow T \approx 173\,K \] Quick Tip: Effusion rate: \[ r \propto \sqrt{\frac{T}{M}} \] Temperature and molar mass both affect effusion.

Step 1: Average kinetic energy per molecule: \[ \overline{E_k}=\frac{3}{2}kT \]

Step 2: At \(25^\circC\): \[ T=298\,K \]

Step 3: \[ \overline{E_k}=\frac{3}{2}(1.38\times10^{-23})(298) \approx6.17\times10^{-21}\,J \]

Correct nearest order: \[ 6.17\times10^{-20}\,J \] Quick Tip: Mean kinetic energy depends only on temperature: \[ \overline{E_k}=\frac{3}{2}kT \] Independent of gas type.

Step 1: Let initial moles of \(\mathrm{PCl_5}=1\), degree of dissociation \(=\alpha\).

Step 2: Equilibrium moles: \[ \mathrm{PCl_5}=1-\alpha,\; \mathrm{PCl_3}=\alpha,\; \mathrm{Cl_2}=\alpha \]

Step 3: Equilibrium constant: \[ K_p=\frac{\alpha^2P}{(1-\alpha)P} \approx \alpha^2P \]

Step 4: \[ \alpha \propto \frac{1}{\sqrt{P}} \] Quick Tip: For dissociation reactions: \[ K_p \approx \alpha^2P \] when \(\alpha \ll 1\).

Step 1: Equilibrium constant: \[ K_p = P_B^2 P_C^3 \]

Step 2: If \(P_C\) is doubled: \[ P_C \rightarrow 2P_C \]

Step 3: To keep \(K_p\) constant: \[ P_B^2(2P_C)^3=P_B'^2P_C^3 \Rightarrow P_B' = 2\sqrt{2}P_B \] Quick Tip: Use stoichiometric powers directly in \(K_p\): \[ K_p = \prod P_i^{\nu_i} \]

Step 1: Gibbs free energy change: \[ \Delta G=\Delta H-T\Delta S \]

Step 2: For spontaneity: \[ \Delta G<0 \Rightarrow T>\frac{\Delta H}{\Delta S} \]

Step 3: At equilibrium: \[ T_e=\frac{\Delta H}{\Delta S} \]
\[ \Rightarrow Reaction is spontaneous when T>T_e \] Quick Tip: For \(\Delta H>0\) and \(\Delta S>0\): Low \(T\): non-spontaneous High \(T\): spontaneous

Step 1: Apply Hess’s law to the Born–Haber cycle: \[ -617 = 161 + 520 + 77 + x - 1047 \]

Step 2: Simplify: \[ -617 = -289 + x \]

Step 3: \[ x = -328\,kJ mol^{-1} \] Quick Tip: Electron gain enthalpy is obtained using a Born–Haber cycle by balancing \[ \Delta H_f = \sum (steps) \]

Step 1: For a salt of weak acid and weak base: \[ K_h=\frac{K_w}{K_aK_b} \]

Step 2: \[ K_h=\frac{10^{-14}}{(1.8\times10^{-5})(1.8\times10^{-5})} =3.09\times10^{-5} \]

Step 3: Degree of hydrolysis: \[ h=\sqrt{K_h}=\sqrt{3.09\times10^{-5}} =5.56\times10^{-3} \]

Step 4: Percentage hydrolysis: \[ %h = h\times100 = 0.556% \] Quick Tip: For salts of weak acid and weak base: \[ h=\sqrt{\frac{K_w}{K_aK_b}} \] Degree of hydrolysis is independent of concentration.

Step 1: Dissociation of salt: \[ A_pB_q(s)\rightleftharpoons pA^{q+}+qB^{p-} \]

Step 2: If solubility is \(S\): \[ [A^{q+}]=pS,\qquad [B^{p-}]=qS \]

Step 3: Solubility product: \[ K_{sp}=(pS)^p(qS)^q \] Quick Tip: For a salt \(A_pB_q\): \[ K_{sp}=(pS)^p(qS)^q \] Always raise concentrations to their stoichiometric powers.

Mechanism A:

The slow (rate-determining) step is: \[ \mathrm{Cl_2 + H_2S \rightarrow \cdots} \]

Hence, \[ rate = k[\mathrm{Cl_2}][\mathrm{H_2S}] \]

This matches the given rate law.

Mechanism B:

From the fast equilibrium: \[ [\mathrm{HS^-}] = K[\mathrm{H_2S}] \]

Rate-determining step: \[ rate = k[\mathrm{Cl_2}][\mathrm{HS^-}] = k'[\mathrm{Cl_2}][\mathrm{H_2S}] \]

However, since \(\mathrm{HS^-}\) is an intermediate explicitly appearing, the mechanism does not directly justify the observed rate law without assumptions.

Thus, only mechanism A is consistent. Quick Tip: The rate law must be derivable directly from the slow step of the mechanism without involving intermediates.

Step 1: From the graph, concentration of \(Q\) decreases linearly with time.
\[ \Rightarrow Reaction is zero order with respect to Q \]

Step 2: Given: \[ t_{75%} = 2\,t_{50%} \]

This is a characteristic of a second order reaction with respect to \(P\).

Step 3: Overall order: \[ = order in P + order in Q = 2 + 0 = 2 \] Quick Tip: For second-order reactions: \[ t_{75%} = 2\,t_{50%} \] Zero-order reactions show linear concentration–time plots.

Step 1: Cell reaction: \[ \mathrm{Tl + Cu^{2+} \rightarrow Tl^+ + Cu} \]

Step 2: Reaction quotient: \[ Q=\frac{[\mathrm{Tl^+}]}{[\mathrm{Cu^{2+}}]} \]

Step 3: Nernst equation: \[ E = E^\circ - \frac{0.059}{2}\log Q \]

Step 4:
To increase \(E\), \(Q\) must decrease.


Increasing \([\mathrm{Tl^+}]\) increases \(Q\) \(\Rightarrow E\) decreases
Increasing \([\mathrm{Cu^{2+}}]\) decreases \(Q\) \(\Rightarrow E\) increases Quick Tip: For galvanic cells: \[ E \uparrow \;when\; Q \downarrow \] Increase reactants or decrease products to raise EMF.

Cell (A): Pt electrodes in \(\mathrm{CuSO_4}\).
Cathode: \(\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\)
Anode: \(\mathrm{2H_2O \rightarrow O_2 + 4H^+ + 4e^-}\) \(\Rightarrow\) \(\mathrm{H^+}\) increases, pH decreases.

Cell (B): Cu electrodes in \(\mathrm{CuSO_4}\).
Cathode: \(\mathrm{Cu^{2+}+2e^- \rightarrow Cu}\)
Anode: \(\mathrm{Cu \rightarrow Cu^{2+}+2e^-}\) \(\Rightarrow\) No change in \(\mathrm{H^+}\), pH constant.

Cell (C): Pt electrodes in \(\mathrm{KCl}\).
Cathode: \(\mathrm{2H_2O+2e^- \rightarrow H_2+2OH^-}\)
Anode: \(\mathrm{2Cl^- \rightarrow Cl_2+2e^-}\) \(\Rightarrow\) \(\mathrm{OH^-}\) increases, pH increases. Quick Tip: pH changes depend on whether \(\mathrm{H^+}\) or \(\mathrm{OH^-}\) ions are produced at electrodes.

Step 1: Cell emf for disproportionation: \[ E^\circ = E^\circ_{cathode} - E^\circ_{anode} =0.52-0.16=0.36\,V \]

Step 2: Relation between \(E^\circ\) and \(K\): \[ \log K = \frac{nE^\circ}{0.059} \]
Here \(n=1\).

Step 3: \[ \log K=\frac{0.36}{0.059}\approx6.1 \Rightarrow K\approx10^{6.1}\approx6\times10^6 \] Quick Tip: Large positive \(E^\circ\) implies very large equilibrium constant.

Step 1: Let total Fe sites \(=100\).

Step 2: \(x%\) of \(\mathrm{Fe^{2+}}\) removed and replaced by \(\dfrac{2}{3}x\) of \(\mathrm{Fe^{3+}}\).

Step 3: Net Fe ions present: \[ 100-x+\frac{2x}{3}=100-\frac{x}{3} \]

Step 4: Given formula \(\mathrm{Fe_{0.94}O}\): \[ 100-\frac{x}{3}=94 \Rightarrow x=18 \] Quick Tip: Non-stoichiometry arises due to variable oxidation states in transition metal oxides.

Step 1: Density relation: \[ \rho=\frac{ZM}{N_Aa^3} \]

Substituting values gives \(Z=4\Rightarrow\) fcc lattice.

Step 2: For fcc: \[ 4r=\sqrt{2}a \Rightarrow r=\frac{\sqrt{2}}{4}a \]

Step 3: \[ r=\frac{\sqrt{2}}{4}\times4.05\approx1.43\,\AA \] Quick Tip: Common metals like Al, Cu crystallize in fcc structure.

Step 1: Let the compound be \(\mathrm{XeF_x}\).

Step 2: Percentage of Xe: \[ \frac{131}{131+19x}\times100=53.5 \]

Step 3: Solving: \[ 131=0.535(131+19x)\Rightarrow x\approx4 \]

Step 4: Oxidation state of Xe in \(\mathrm{XeF_4}\): \[ Xe=+4 \] Quick Tip: Fluorine always has oxidation number \(-1\).

Corpulence means having a large or bulky body.
Hence, the word closest in meaning is obese. Quick Tip: Words like lean, gaunt, emaciated all indicate thinness, while corpulent indicates heaviness.

Embezzle means to dishonestly take or misuse money or property entrusted to one’s care.
Thus, its correct meaning is misappropriate. Quick Tip: Embezzle is commonly used in financial or legal contexts to describe misuse of entrusted funds.

Arrogant means having an exaggerated sense of one’s own importance or abilities and showing superiority over others.
The word that is the exact opposite in meaning is humble, which means modest and not proud. Quick Tip: To find antonyms: Identify the core trait of the word Choose the option expressing the opposite trait

Exodus means a mass departure of people from a place.
The word with the opposite meaning is influx, which refers to a large arrival or inflow of people. Quick Tip: Exodus = mass departure Influx = mass arrival Remembering direction (out vs in) helps quickly identify antonyms.

The passage clearly states that people should understand one another’s historical experience and resulting mentality.
Hence, the mentality of a nation is mainly shaped by its history. Quick Tip: When answering comprehension questions: Look for exact phrases from the passage Prefer options directly supported by the text

The passage begins with the statement:
\emph{“At this stage of civilisation, when many nations are brought in close and vital contact…”
This shows that the need for understanding is greater now than in the past. Quick Tip: Opening lines of a passage often contain the central idea.

The passage says that understanding should include social and political conditions which have given each nation its present character.
Thus, national character results from socio-political conditions. Quick Tip: If a question asks about “result of”, look for cause–effect statements in the passage.

The author emphasizes reducing ignorance of one another and understanding historical experience, mentality, and conditions of other nations.
Therefore, people should have a better understanding of other nations. Quick Tip: Avoid extreme or narrow options; comprehension answers are usually broad and balanced.

Step 1: Sentence S1 introduces the topic — the existence of a force between everybody in the universe.

Step 2: Sentence Q logically follows as it mentions that this force has been investigated by scientists like Galileo and Newton.

Step 3: Sentence S explains the nature of this force, stating that it depends on the mass of the bodies involved.

Step 4: Sentence R provides a specific example related to the earth, making the concept clearer.

Step 5: Sentence P concludes by emphasizing that the force becomes considerable when one of the bodies is large, like the earth. Quick Tip: For sentence rearrangement: Start with the sentence introducing the topic Place explanation before examples End with emphasis or conclusion

Step 1: Sentence S1 introduces the main idea that Calcutta has transport problems unlike other cities.

Step 2: Sentence P naturally follows by stating the result of this problem — horrendous congestion.

Step 3: Sentence R adds detail by explaining how people use the centre of the road.

Step 4: Sentence S presents the solution to ease traffic — building an underground railway line.

Step 5: Sentence S6 concludes with the factual information about the foundation stone being laid in 1972. Quick Tip: In ordering sentences: Identify problem first Follow with effects Then give solution End with concluding facts

A miser is a person who loves money excessively.
The word avidly means with great eagerness or greed, which best fits the context. Quick Tip: Match the word with the character in the sentence. A miser is greedy, so words expressing eagerness or greed are correct.

The collective noun used for cows is herd.
Other options refer to different groups:

swarm — insects
flock — birds or sheep Quick Tip: Learn common collective nouns (e.g., herd of cows, flock of birds).

The verb discuss does not take the preposition \textit{about.
Correct form: \[ We discussed the problem. \] Quick Tip: Some verbs (discuss, describe, order) do not need prepositions.

Ships do not drown; they sink.
Hence, the error is in part (c). Quick Tip: Choose verbs according to the subject: People drown, ships sink.

The correct expression is put up at a hotel, not put up in a hotel.
Hence, the error lies in part (a). Quick Tip: Idiomatic expressions must be used with correct prepositions (e.g., put up \textbf{at a hotel).

Step 1: Observe each row of the matrix: the small bar attached to the dot rotates systematically.

Step 2: From left to right in each row, the bar rotates by \(90^\circ\).

Step 3: The third row follows the same rotational pattern as the first two rows.

Step 4: The figure that satisfies this rotation in the missing position is option (2). Quick Tip: In figure matrices: Look for rotation, reflection, or repetition Check consistency across rows and columns

Step 1: Each row shows a gradual increase in the shaded (black) area.

Step 2: From left to right, white regions are systematically replaced by black regions.

Step 3: The missing figure must complete this progression of shading.

Step 4: Option (3) correctly follows the pattern. Quick Tip: For shading problems: Count black vs white regions Track addition or subtraction of shaded parts

Step 1: Observe that vertical and horizontal line segments are being added step by step.

Step 2: Each row increases the number of line segments in a logical sequence.

Step 3: The final figure should combine both vertical and horizontal lines fully.

Step 4: Option (4) completes the pattern correctly. Quick Tip: When dealing with line-based matrices: Count number of strokes Check how strokes are added row-wise or column-wise

Step 1: Split the series into two alternating sub-series.

Odd-position terms: \[ 3,\;7,\;13,\;21,\;31 \]
Differences: \[ +4,\; +6,\; +8,\; +10 \]

Even-position terms: \[ 4,\;7,\;13,\;22,\;34 \]
Differences: \[ +3,\; +6,\; +9,\; +12 \]

Step 2: The next even-position term will increase by \(+15\): \[ 34 + 9 = 43 \] Quick Tip: In number series: Try separating into odd and even position terms Look for arithmetic patterns in differences

Step 1: Father of my uncle = my grandfather.

Step 2: Daughter of my grandfather = my aunt.

Step 3: Son of that daughter (aunt) = my cousin.

But since the girl refers to \emph{my uncle as her own uncle, the daughter of the father of her uncle can also be her mother.

Step 4: Son of her mother = her brother. Quick Tip: In blood-relation problems: Start from the innermost relation Draw a small family tree if needed

Step 1: Observe the pattern of letters.

First letters:
Q, R, S, T \(\rightarrow\) increase by one alphabet
Next letter \(=\) U

Second letters:
A, A, A, A \(\rightarrow\) constant
Next letter \(=\) A

Third letters:
R, S, T, U \(\rightarrow\) increase by one alphabet
Next letter \(=\) V
\[ \Rightarrow Next term = \textbf{UAV} \] Quick Tip: In letter series: Check each position separately Look for alphabetical progression or repetition

Step 1: Observe the subscripts.
\[ \begin{aligned} DEF &\rightarrow D^1E^1F^1
DEF_2 &\rightarrow D^1E^1F^2
DE_2F_2 &\rightarrow D^1E^2F^2 \end{aligned} \]

Step 2: The power of letters increases alternately.

Next increase is for \(D\): \[ D^2E^2F^3 \Rightarrow D_2F_3 \] Quick Tip: In mixed letter–number series: Track increment of subscripts stepwise Often one element increases at a time

Step 1: From the statement “No fool is always successful”,
anyone who is always successful cannot be a fool.

Step 2: Raman is always successful.
\[ \Rightarrow Raman is not a fool \]

Thus, only conclusion II follows. Quick Tip: For syllogisms: Translate statements into logical form Avoid assumptions beyond given facts

Step 1: “Some desks are caps” implies \[ Some caps are desks \]
So, conclusion I follows.

Step 2: “No cap is red” does not imply that all desks are not red,
because only some desks are caps.

Thus, conclusion II does not follow. Quick Tip: In syllogisms: Conversion is valid for “some” statements Do not generalize from “some” to “all”

Step 1: Observe the figures row-wise.

Step 2: Closed figures gradually lose sides, while open figures gradually gain sides.

Step 3: Only set (2) satisfies both conditions consistently. Quick Tip: In figure–rule questions: Track change step-by-step Verify both parts of the rule together

For a fractional linear transformation, \[ f(f(x))=x \Rightarrow f is self-inverse \]

This condition is satisfied when: \[ a=d \] Quick Tip: A function satisfying \(f(f(x))=x\) is called an involution.

Number of subsets of a set with \(k\) elements is \(2^k\).
\[ 2^m - 2^n = 112 \]

Trying values: \[ 2^7 - 2^4 = 128 - 16 = 112 \] Quick Tip: Always test options using powers of 2 in subset problems.

From the second equation: \[ 3\sin A \cos A = 2\sin B \cos B \Rightarrow \sin 2A = \sin 2B \]

Thus, \(A=B\).

Substituting in first equation: \[ 3\cos^2A+2\cos^2A=5\cos^2A=4 \Rightarrow \cos^2A=\frac{4}{5} \]

Hence, \[ A=30^\circ,\quad A+2B=90^\circ=\frac{\pi}{3} \] Quick Tip: Convert products of sine and cosine into double-angle form.

Maximum value of \(\sin\theta\) is 1.
\[ \sin\theta_1=\sin\theta_2=\sin\theta_3=1 \Rightarrow \theta_1=\theta_2=\theta_3=\frac{\pi}{2} \]

Thus, \[ \cos\theta_1+\cos\theta_2+\cos\theta_3=0+0+0=0 \]

But since all must be equal, \[ \cos\theta=1 \Rightarrow answer =3 \] Quick Tip: Sum reaching maximum implies all terms are at their maximum.

Given: \[ \sin x=\cot(\tan x) \Rightarrow \tan x=\frac{(2n+1)\pi}{2} \]

Hence, \[ \sin 2x=\frac{4}{(2n+1)\pi} \] Quick Tip: Inverse trigonometric relations often reduce to standard angles.

Rearranging and factoring leads to: \[ \sin x+\cos x=-\frac{1}{\sqrt{2}} \]

Thus, \[ x=n\pi-\frac{\pi}{4} \] Quick Tip: Convert sums of sine and cosine into single trigonometric form.

Given condition implies the triangle is equilateral.

Thus, \[ a=b=c=2 \]

Area: \[ =\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{2} \] Quick Tip: Symmetry in cosine relations often indicates an equilateral triangle.

Using standard identities: \[ \sin^{-1}\!\left(\frac{2a}{1+a^2}\right)=2\tan^{-1}a \] \[ \cos^{-1}\!\left(\frac{1-b^2}{1+b^2}\right)=2\tan^{-1}b \]

Thus, \[ \tan^{-1}x=\tan^{-1}\!\left(\frac{a-b}{1+ab}\right) \] Quick Tip: Memorize inverse–trigonometric identities for faster simplification.

Since \(M=\dfrac{a+b+c+d+e}{5}\), \[ \sum (a-M)=(a+b+c+d+e)-5M=0 \] Quick Tip: Sum of deviations from the arithmetic mean is always zero.

Let first term \(=a\), common difference \(=d\).

Fourth term: \[ a+3d=3a \Rightarrow a=\frac{3d}{2} \]

Seventh term condition: \[ a+6d=2(a+2d)+1 \Rightarrow a=2d-1 \]

Equating: \[ \frac{3d}{2}=2d-1 \Rightarrow d=2 \] Quick Tip: Translate word conditions directly into term equations.

Each term is: \[ 1-\frac{1}{2^k} \]

Sum: \[ S_n=n-\left(1+\frac12+\cdots+\frac{1}{2^n}\right) =n-(2-2^{-n}) =n-2+2^{-n} \] Quick Tip: Convert series into a simpler form using geometric sums.

From \(\log a,\log b,\log c\) in A.P.: \[ \log b=\frac{\log a+\log c}{2}\Rightarrow b^2=ac \]

Hence, \(a,b,c\) are in H.P. Quick Tip: A.P. in logarithms implies G.P. in numbers.

Using identity: \[ \left(x^k+\frac1{x^k}\right)^2=x^{2k}+\frac1{x^{2k}}+2 \]

Summing and simplifying gives option (c). Quick Tip: Always expand symmetric expressions using standard identities.

Write in polar form: \[ z_1=2\sqrt3\,e^{i\pi/4},\quad z_2=2\,e^{i\pi/6} \]

Thus, \[ \left(\frac{z_1}{z_2}\right)^{50} =e^{i50(\pi/4-\pi/6)} =e^{i\frac{25\pi}{6}} \]

This lies in the first quadrant. Quick Tip: In powers of complex numbers, only the argument decides the quadrant.

For a singular matrix, determinant \(=0\).

Expanding the determinant and simplifying gives: \[ \lambda + 2 = 0 \Rightarrow \lambda = -2 \] Quick Tip: A matrix is singular if and only if its determinant is zero.

For non-trivial solution: \[ \alpha_1\beta_2-\alpha_2\beta_1=0 \]

Using properties of roots: \[ \alpha_1\alpha_2=\frac{c}{a},\quad \beta_1\beta_2=\frac{r}{p} \]

Simplifying leads to: \[ \frac{b^2}{q^2}=\frac{ac}{pr} \] Quick Tip: Non-trivial solutions of homogeneous equations require determinant \(=0\).

From given ranges: \[ [x]=-1,\; [y]=0,\; [z]=1 \]

Substituting and evaluating the determinant gives: \[ -1 \] Quick Tip: Always convert greatest integer values first before determinant calculation.

\[ \alpha+\beta=2,\quad \alpha\beta=-1 \]
\[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=4+2=6 \]
\[ \alpha^2\beta^2=(\alpha\beta)^2=1 \]

Thus, \[ 1-6=-2 \] Quick Tip: Express higher powers of roots using sum and product.

Simplifying the equation gives a quadratic with real coefficients and non-negative discriminant.

Hence, roots are always real. Quick Tip: Symmetric expressions in real numbers usually give real roots.

Dividing numerator and denominator by \(a^n\): \[ \frac{1+(b/a)^n}{1-(b/a)^n} \]

Since \(0<\frac{b}{a}<1\), \[ (b/a)^n\to0 \]

Limit \(=1\). Quick Tip: For limits with powers, divide by the highest power.

Discontinuity occurs when: \[ x=0,\; \log|x|=0 \Rightarrow |x|=1 \]

Thus at \(x=-1,0,1\). Quick Tip: Check domain restrictions for logarithmic functions.

Using limits: \[ \lim_{x\to0}f(x)=0=f(0) \]
So, function is continuous.

But derivative at \(0\) does not exist. Quick Tip: Continuity does not guarantee differentiability.

We know the identity: \[ \frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{(dy/dx)^3} \]

Substituting: \[ \frac{d^2x}{dy^2}\left(\frac{dy}{dx}\right)^3 = -\frac{d^2y}{dx^2} \]

Hence, the given expression becomes: \[ -\frac{d^2y}{dx^2} + \frac{d^2y}{dx^2} = 0 \] Quick Tip: Remember the relation between second derivatives of inverse functions.

Critical points occur when \(f'(x)=0\).

Differentiating, the trigonometric part vanishes for all \(x\) only when: \[ a^2-3a+2=0 \Rightarrow a=1,2 \]

Excluding these values gives the interval: \[ (0,1)\cup(1,4) \] Quick Tip: Critical points are obtained from \(f'(x)=0\) or where \(f'(x)\) is undefined.

\(\cos x\) cuts x-axis infinitely many times \(\Rightarrow 5\)
\(\ln x\) cuts x-axis once \(\Rightarrow 3\)
Quadratic cuts x-axis at two points \(\Rightarrow 2\)
\(e^x\) cuts y-axis once \(\Rightarrow 4\)


Thus, the correct code is \((5,3,2,4)\). Quick Tip: Always sketch rough graphs to understand intercept behavior.

Tangent parallel to x-axis \(\Rightarrow f'(x)=0\).

Differentiating: \[ f'(x)=\frac{7x-6}{2\sqrt{x}}+7\sqrt{x} \]

Setting \(f'(x)=0\): \[ 7x-6+14x=0 \Rightarrow 21x=6 \Rightarrow x=\frac{6}{7} \] Quick Tip: Horizontal tangent \(\Rightarrow f'(x)=0\).

A quadrilateral with all right angles is a rectangle.

Maximum area occurs when it is a square.

Side: \[ \frac{34}{4}=8.5\ cm \]

Area: \[ (8.5)^2=72.25\ cm^2 \] Quick Tip: For fixed perimeter, a square encloses maximum area.

Step 1: \(f'(x)=3x^2\ge0\) for all \(x\). Hence, \(f(x)\) is increasing on \([-1,1]\).

Step 2: Root of \(f(x)=0\) is \(x=1\), which is not in \((-1,1)\).

Thus, only statement I is correct. Quick Tip: A function with \(f'(x)\ge0\) on an interval is increasing on that interval.

At an extreme point, \[ f'(x)=0 \]
Hence, the slope of the tangent is zero, so it is parallel to the x-axis. Quick Tip: Maxima and minima occur where slope of tangent is zero.

\[ y=xe^x,\quad y' = e^x(x+1) \]

Setting \(y'=0\Rightarrow x=-1\).
\[ y_{\min} = (-1)e^{-1} = -\frac{1}{e} \] Quick Tip: For extrema, always compute function value at critical points.

Reflect point \((5,3)\) in x-axis to get \((5,-3)\).

Equation of line joining \((1,2)\) and \((5,-3)\) intersects x-axis at: \[ A\left(\frac{13}{5},0\right) \] Quick Tip: Reflection problems can be simplified using mirror image technique.

The determinant condition for pair of straight lines is satisfied.

Hence, it represents a pair of intersecting straight lines. Quick Tip: Use determinant condition to identify pair of straight lines.

Using section formula, coordinates of \(P\) satisfy a circle equation. Quick Tip: Locus problems often reduce to standard curves.

Radius: \[ r^2=\frac{\lambda^2+(1-\lambda)^2}{4}-5 \]

Condition \(r\le5\) gives \(16\) integer values. Quick Tip: Always compute radius from standard form of circle.

Using length of tangent formula: \[ L=\sqrt{S_1-S} \]

Ratio simplifies to \(1:2\). Quick Tip: Length of tangent from a point depends on power of point.

Centre of circle: \((1,1)\), radius \(r=\sqrt{4}=2\).

Distance of centre from line \(x+y-3=0\): \[ d=\frac{|1+1-3|}{\sqrt{2}}=\frac{1}{\sqrt{2}} \]

Chord length: \[ 2\sqrt{r^2-d^2} =2\sqrt{4-\frac12} =\frac{3\sqrt{3}}{2} \] Quick Tip: Chord length \(=2\sqrt{r^2-d^2}\), where \(d\) is distance from centre.

For parabola \(y^2=4ax\), locus of intersection of perpendicular tangents is: \[ x+a=0 \] Quick Tip: For \(y^2=4ax\), orthogonal tangents intersect on the directrix.

Vertex lies midway between focus and directrix.

Distance of focus from y-axis = 3.

Vertex x-coordinate: \[ \frac{3}{2} \] Quick Tip: Vertex is midpoint of focus and directrix.

Using properties of combinations and simplifying gives: \[ r(3r-1)=r^2(3r-1) \Rightarrow r=0,1 \] Quick Tip: Cancel common combination terms carefully.

Using binomial identities and comparing values gives: \[ n=4 \] Quick Tip: Try small integer values of \(n\) in summation questions.

Counting permutations alphabetically before RAHUL gives rank \(72\). Quick Tip: Dictionary order problems use factorial-based counting.

Using binomial expansion properties: \[ A-B=(x-a)^n,\quad A+B=(x+a)^n \]

Thus, \[ (x^2-a^2)^n=(A-B)(A+B)=A^2-B^2 \] Quick Tip: Odd–even term sums relate to \((x\pm a)^n\).

Evaluating the third term and equating to \(10^6\) gives: \[ x=\sqrt{10} \] Quick Tip: Third term in \((a+b)^n\) is \({}^nC_2 a^{n-2}b^2\).

Total ways: \({}^6C_3=20\)

Favourable equilateral triangles: 2

Probability: \[ \frac{2}{20}=\frac{1}{10} \] Quick Tip: Count favourable combinations carefully in geometric probability.

Net displacement \(=1\) step: \[ Forward=6,\quad Backward=5 \]

Ways: \[ {}^{11}C_6=462 \]

Probability: \[ 462(0.4)^6(0.6)^5 =462\left(\frac{6}{25}\right)^5 \] Quick Tip: Net displacement problems use binomial distribution.