VITEEE 2008 Question Paper (Available) - Download VITEEE Question Paper with Solution PDF

Step 1: Understand the condition for sustained interference.

For a stable and well-defined interference pattern, the two light sources must be coherent (constant phase difference) and preferably monochromatic (single wavelength).


Step 2: Why monochromatic light is necessary.

If the light is not monochromatic, it contains many wavelengths. Each wavelength produces its own fringe pattern, and these patterns overlap.


Step 3: Result of overlapping fringes.

Due to overlapping, the bright and dark fringes wash out, and a clear interference pattern is not observed.


Step 4: Conclude from options.

Thus, the interference pattern will not be formed properly when the beams are not monochromatic.



Final Answer:
\[ \boxed{(D) they are not monochromatic} \] Quick Tip: For interference, light must be coherent and preferably monochromatic, otherwise fringes overlap and disappear.

Step 1: Recall the formula for central maximum width.

For single slit diffraction, angular width of central maximum is:
\[ \theta = \frac{2\lambda}{a} \]


Step 2: Convert into linear width on the screen.

Linear width on screen:
\[ W = 2D\theta = 2D \left(\frac{\lambda}{a}\right) = \frac{2D\lambda}{a} \]


Step 3: Identify what increases width.

From the formula, \(W\) is inversely proportional to \(a\).

So if we decrease \(a\), width \(W\) increases.


Step 4: Match with options.

Hence the correct option is (B).



Final Answer:
\[ \boxed{(B) decrease a} \] Quick Tip: Central maximum width in single slit diffraction increases when slit width decreases.

Step 1: Understand phase change conditions.

At air--soap interface: reflection from rarer to denser, so phase reversal occurs.

At soap--glass interface: reflection from rarer (soap) to denser (glass), so phase reversal occurs again.

Thus, two phase reversals cancel each other.


Step 2: Condition for reflection maxima at normal incidence.
\[ 2nt = m\lambda \]


Step 3: Adjacent maxima correspond to successive integers.

So for \(\lambda_1 = 420\) nm and \(\lambda_2 = 630\) nm:
\[ 2nt = m(420) = (m-1)(630) \]


Step 4: Solve for \(m\).
\[ m\cdot 420 = (m-1)\cdot 630 \]
\[ 420m = 630m - 630 \]
\[ 210m = 630 \Rightarrow m = 3 \]


Step 5: Compute thickness \(t\) using \(m=3\).
\[ 2nt = 3 \times 420 \]
\[ 2(1.4)t = 1260 \Rightarrow 2.8t = 1260 \]
\[ t = \frac{1260}{2.8} = 450 nm \]



Final Answer:
\[ \boxed{450 nm} \] Quick Tip: For thin film reflection maxima, use \(2nt = m\lambda\) and adjacent maxima differ by 1 in \(m\).

Step 1: Use wave relation.

Wave speed is given by:
\[ v = f\lambda \]


Step 2: Frequency remains constant.

When a wave enters a new medium, the frequency does not change because it is decided by the source.


Step 3: Compare old and new speed.

If speed doubles:
\[ v' = 2v \]


Step 4: Find the new wavelength.
\[ v' = f\lambda' \Rightarrow 2v = f\lambda' \]

But \(v = f\lambda\), so:
\[ 2f\lambda = f\lambda' \Rightarrow \lambda' = 2\lambda \]



Final Answer:
\[ \boxed{(C) doubled} \] Quick Tip: If wave speed changes but frequency remains constant, wavelength changes in the same ratio as speed.

Step 1: Use Einstein photoelectric equation.
\[ K_{\max} = h\nu - \phi \]


Step 2: Substitute values.

Given:
\[ h = 6.63 \times 10^{-34} \, J s, \quad \nu = 6 \times 10^{14}\, Hz \]

Energy of photon:
\[ h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 3.978 \times 10^{-19}\, J \]


Step 3: Convert joule to eV.
\[ 1\,eV = 1.6 \times 10^{-19}\,J \]
\[ h\nu = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.49 \,eV \]


Step 4: Subtract work function.
\[ K_{\max} = 2.49 - 2 = 0.49 \,eV \]



Final Answer:
\[ \boxed{0.49 \,eV} \] Quick Tip: Maximum kinetic energy in photoelectric effect is \(K_{\max} = h\nu - \phi\).

Step 1: Use de Broglie wavelength relation.

For electron accelerated by potential \(V\), de Broglie wavelength is:
\[ \lambda = \frac{12.27}{\sqrt{V}} \, AA \]


Step 2: Resolution required means wavelength should be comparable.

Given resolution \(\lambda = 5\) AA.


Step 3: Substitute in formula.
\[ 5 = \frac{12.27}{\sqrt{V}} \]
\[ \sqrt{V} = \frac{12.27}{5} = 2.454 \]
\[ V = (2.454)^2 \approx 6.02 \, V \]


Step 4: Choose nearest option.

Closest option to 6 V is 5 V.



Final Answer:
\[ \boxed{(B) 5 V} \] Quick Tip: Electron wavelength decreases with higher accelerating voltage: \(\lambda = \frac{12.27}{\sqrt{V}}\) AA.

Step 1: Identify what proves wave nature.

Wave nature is directly confirmed when a particle shows phenomena like diffraction or interference, which are purely wave effects.


Step 2: Apply to matter particles.

Electrons are matter particles. When electrons show diffraction (as in Davisson--Germer experiment), it proves electrons behave like waves.


Step 3: Eliminate other options.

Momentum is a particle property, not proof of wave nature. Photon diffraction proves wave nature of light, not matter.


Step 4: Final conclusion.

Hence, electron diffraction best supports wave nature of matter.



Final Answer:
\[ \boxed{(B) Electron diffraction} \] Quick Tip: Davisson--Germer experiment showed electron diffraction, confirming de Broglie’s wave nature of matter.

Step 1: Use radioactive decay formula.
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n \]


Step 2: Compare with given fraction.

Given:
\[ \frac{N}{N_0} = \frac{1}{16} \]

But:
\[ \frac{1}{16} = \left(\frac{1}{2}\right)^4 \]

So number of half-lives \(n = 4\).


Step 3: Total time is 2 hours.
\[ 2 hours = 120 min \]


Step 4: Find half-life.
\[ T_{1/2} = \frac{120}{4} = 30 min \]



Final Answer:
\[ \boxed{30 min} \] Quick Tip: If activity becomes \(\frac{1}{16}\), it means 4 half-lives have passed because \(1/16 = (1/2)^4\).

Step 1: Identify proper time and dilated time.

Observer B is moving with the asteroid, so B measures the decay time in the asteroid’s rest frame.

This is called proper time \(T_B\).


Step 2: Time dilation concept.

Observer A sees the asteroid moving with speed \(v = 0.3c\).

According to special relativity:
\[ T_A = \gamma T_B \quad where \quad \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \]


Step 3: Since \(\gamma > 1\).

For any non-zero velocity, \(\gamma > 1\).

So:
\[ T_A > T_B \]


Step 4: Choose correct relation.
\[ T_B < T_A \]



Final Answer:
\[ \boxed{(C) T_B < T_A} \] Quick Tip: Proper time is always the smallest and is measured in the rest frame of the object; moving observers measure a longer (dilated) time.

Step 1: Recall alpha decay rule.

In alpha decay, nucleus emits \(\alpha\) particle \((^{4}_{2}He)\).

So:

Mass number decreases by 4 and atomic number decreases by 2.


Step 2: Apply to \(^{234}_{92}U\).

After emission:
\[ A' = 234 - 4 = 230 \]
\[ Z' = 92 - 2 = 90 \]


Step 3: Identify element with atomic number 90.

Atomic number 90 corresponds to Thorium (Th).


Step 4: Write final nucleus.
\[ ^{230}_{90}Th \]



Final Answer:
\[ \boxed{^{230}Th} \] Quick Tip: In alpha decay: \(A \to A-4\) and \(Z \to Z-2\).

Step 1: Understand K-series X-rays.

K-series X-rays are produced when an electron falls into the K-shell (\(n=1\)) from a higher level.


Step 2: Identify \(K_{\alpha}\).
\(K_{\alpha}\) line corresponds to transition from L-shell to K-shell:
\[ n = 2 \to n = 1 \]


Step 3: Identify \(K_{\beta}\).
\(K_{\beta}\) line corresponds to transition from M-shell to K-shell:
\[ n = 3 \to n = 1 \]


Step 4: Match with correct option.

So correct answer is option (A).



Final Answer:
\[ \boxed{(A) n = 2 \to n = 1 and n = 3 \to n = 1} \] Quick Tip: K-series X-rays always end in \(n=1\): \(K_{\alpha}: 2\to1\), \(K_{\beta}: 3\to1\).

Step 1: Understand changes due to \(\alpha\)-decay.

Each \(\alpha\)-particle emission decreases mass number by 4 and atomic number by 2.

So if \(n\) alpha particles are emitted:
\[ A \to A - 4n \quad and \quad Z \to Z - 2n \]


Step 2: Compare with given final mass number.

Final mass number is \(A-8\).

So:
\[ A - 4n = A - 8 \Rightarrow 4n = 8 \Rightarrow n = 2 \]


Step 3: Now compare atomic number change.

After 2 alpha decays:
\[ Z \to Z - 2(2) = Z - 4 \]

But final atomic number is given as \(Z-3\).


Step 4: Effect of \(\beta^-\)-decay.

Each \(\beta^-\) emission increases atomic number by 1 (mass number unchanged).

So if \(m\) beta particles emitted:
\[ Z - 4 + m = Z - 3 \Rightarrow m = 1 \]


Step 5: Final conclusion.

Number of \(\alpha\) particles = 2 and number of \(\beta\) particles = 1.



Final Answer:
\[ \boxed{2 and 1 respectively} \] Quick Tip: \(\alpha\)-decay reduces \(A\) by 4 and \(Z\) by 2, while \(\beta^-\)-decay keeps \(A\) same but increases \(Z\) by 1.

Step 1: Identify the configuration.

The non-inverting terminal of op-amp is grounded, and input is applied through a resistor to the inverting terminal.

So the circuit is an inverting amplifier.


Step 2: Write gain formula for inverting amplifier.
\[ V_{out} = -\left(\frac{R_f}{R_{in}}\right)V_{in} \]


Step 3: Substitute resistor values from diagram.

From figure:
\[ R_{in} = 1k\Omega,\quad R_f = 10k\Omega,\quad V_{in} = 1V \]


Step 4: Calculate output.
\[ V_{out} = -\left(\frac{10k\Omega}{1k\Omega}\right)(1V) = -10V \]



Final Answer:
\[ \boxed{-10V} \] Quick Tip: For an inverting op-amp: \(V_{out} = -\left(\frac{R_f}{R_{in}}\right)V_{in}\). The minus sign shows phase reversal.

Step 1: Understand donor level position.

A donor level lies just below the conduction band and donates electrons easily to the conduction band.


Step 2: Effect of donor impurity.

When electrons are donated, the conduction band gets extra electrons, increasing conductivity.


Step 3: Identify majority charge carriers.

Since electrons become majority carriers, the material becomes an n-type semiconductor.


Step 4: Match with option.

Thus, correct option is (D).



Final Answer:
\[ \boxed{(D) an n-type semiconductor} \] Quick Tip: Donor impurity introduces an energy level near conduction band, making electrons the majority carriers \(\Rightarrow\) n-type semiconductor.

Step 1: Use formula for built-in (contact) potential.
\[ V_0 = \frac{kT}{e}\ln\left(\frac{N_A N_D}{n_i^2}\right) \]


Step 2: Substitute given values.
\[ N_A = 10^{17},\quad N_D = 10^{16},\quad n_i = 1.4\times 10^{10} \]


Step 3: Compute the ratio inside log.
\[ \frac{N_A N_D}{n_i^2} = \frac{10^{17}\times 10^{16}}{(1.4\times 10^{10})^2} \]
\[ = \frac{10^{33}}{1.96\times 10^{20}} = 5.1 \times 10^{12} \approx 4\times 10^{12} \]


Step 4: Final expression.
\[ V_0 = \frac{kT}{e}\ln(4\times 10^{12}) \]



Final Answer:
\[ \boxed{\left(\frac{kT}{e}\right)\ln(4\times 10^{12})} \] Quick Tip: Built-in potential: \(V_0=\frac{kT}{e}\ln\left(\frac{N_A N_D}{n_i^2}\right)\). Higher doping increases junction potential.

Step 1: Use relation between electric field and breakdown voltage.

Electric field in depletion region:
\[ E = \frac{V}{d} \Rightarrow V = E \cdot d \]


Step 2: Substitute given values.
\[ E = 10^6 \, V/m,\quad d = 2.5\mu m = 2.5\times 10^{-6} m \]


Step 3: Calculate breakdown voltage across depletion layer.
\[ V = 10^6 \times 2.5\times 10^{-6} = 2.5 V \]


Step 4: Match with option.

Thus, reverse biased potential required is \(2.5V\).



Final Answer:
\[ \boxed{2.5V} \] Quick Tip: Zener breakdown occurs when \(E\cdot d\) reaches a critical voltage.

Step 1: Recall basic structure of Colpitts oscillator.

Colpitts oscillator is an LC oscillator where the feedback is obtained using a capacitive voltage divider.


Step 2: Identify components of feedback network.

The feedback network consists of:

Two capacitors \(C_1\) and \(C_2\) connected in series and one inductor \(L\).


Step 3: Why two capacitors are used.

The two capacitors form a voltage divider which provides the correct fraction of output voltage back to the input.


Step 4: Choose correct option.

Hence the correct option is (B).



Final Answer:
\[ \boxed{(B) two capacitors and an inductor} \] Quick Tip: Colpitts oscillator uses a capacitive divider: two capacitors + one inductor, unlike Hartley which uses two inductors.

Step 1: Recall expression of reverse saturation current.

Reverse saturation current is mainly due to minority carriers:
\[ I_s \propto n_i^2\left(\frac{D_p}{L_p N_D} + \frac{D_n}{L_n N_A}\right) \]


Step 2: Identify dependencies.

From this expression, \(I_s\) depends on:

(i) doping concentrations \(N_A, N_D\)

(ii) diffusion constants \(D_n, D_p\)

(iii) diffusion lengths \(L_n, L_p\)

(iv) intrinsic concentration \(n_i\)


Step 3: Role of temperature.
\(n_i\) strongly increases with temperature, so \(I_s\) increases sharply with temperature.


Step 4: Match with option.

Thus, it depends on doping concentration, diffusion length and temperature.



Final Answer:
\[ \boxed{(D) depends on the doping concentrations, diffusion length and device temperature} \] Quick Tip: Reverse saturation current increases rapidly with temperature because \(n_i\) increases exponentially with \(T\).

Step 1: Recall relation between antenna length and wavelength.

Antenna length is usually proportional to wavelength, typically \(\lambda/4\).


Step 2: Compare wavelengths of AM and FM.
\[ \lambda = \frac{c}{f} \]

AM frequency \(f_{AM}=1020\,kHz\) is much smaller, so wavelength is very large.

FM frequency \(f_{FM}=89.5\,MHz\) is much higher, so wavelength is small.


Step 3: Conclusion based on wavelength.

Longer wavelength \(\Rightarrow\) longer antenna needed.

So AM requires longer antenna and FM requires shorter antenna.



Final Answer:
\[ \boxed{(A) longer antenna for AM and shorter for FM} \] Quick Tip: Antenna length \(\propto \lambda\). Lower frequency (AM) needs longer antenna, higher frequency (FM) needs shorter antenna.

Step 1: Understand how light travels in optical fiber.

Light inside the fiber core repeatedly strikes the boundary with cladding.


Step 2: Condition for trapping light.

When the angle of incidence is greater than the critical angle, the light does not refract out, instead it reflects completely.


Step 3: This phenomenon is called TIR.

This repeated total internal reflection guides light through long distances inside the fiber.


Step 4: Choose correct option.

Thus, optical fiber communication works on total internal reflection.



Final Answer:
\[ \boxed{(A) total internal reflection} \] Quick Tip: Optical fibers transmit signals by keeping light trapped inside the core using total internal reflection.

Step 1: Recall quantization of charge.

Charge exists in discrete units. The smallest charge is the electronic charge \(e = 1.6\times 10^{-19}C\).


Step 2: Write the quantization condition.
\[ q = ne \]
where \(n\) is an integer (\(\pm 1, \pm 2, \pm 3, \dots\)).


Step 3: Interpretation.

This means charge on any body is always an integral multiple of \(e\), not fractional.


Step 4: Match with option.

Hence option (D) is correct.



Final Answer:
\[ \boxed{(D) integral multiple of the least amount of charge} \] Quick Tip: Quantization: \(q = ne\). Any net charge must be an integer multiple of elementary charge \(e\).

Step 1: Understand initial setup in Fig. (a).

In Fig. (a), a single capacitor \(C = 900pF\) is connected to a battery of \(100V\).

Energy stored is given:
\[ U_0 = 4.5\times 10^{-6} \, J \]


Step 2: Charge stored initially.
\[ Q = CV \]

So charge on capacitor:
\[ Q = C(100V) \]


Step 3: In Fig. (b), battery is replaced by identical capacitor.

Now two identical capacitors (each \(900pF\)) are connected together.

Charge redistributes and final voltage becomes half, because total capacitance doubles.


Step 4: Energy after redistribution.

For equal capacitors, after connection:

Final energy becomes half of initial energy:
\[ U = \frac{U_0}{2} \]


Step 5: Calculate final energy.
\[ U = \frac{4.5\times 10^{-6}}{2} = 2.25\times 10^{-6} \, J \]



Final Answer:
\[ \boxed{2.25\times 10^{-6}\, J} \] Quick Tip: When a charged capacitor is connected to an identical uncharged capacitor, total energy becomes half because some energy is lost as heat during charge redistribution.

Step 1: Recall resistance of a wire.

Resistance of a cylindrical wire is:
\[ R = \rho \frac{L}{A} \]


Step 2: Condition of equal amount of metal.

Equal amount of metal means same volume:
\[ V = LA = constant \]


Step 3: Compare resistance for each option.

Option (A):
\[ R_A = \rho \frac{L}{A} \]

Option (B):
\[ R_B = \rho \frac{L/2}{2A} = \rho \frac{L}{4A} = \frac{R_A}{4} \]

Option (C):
\[ R_C = \rho \frac{2L}{A/2} = \rho \frac{4L}{A} = 4R_A \]


Step 4: Identify maximum resistance.

Clearly, \(R_C\) is maximum because it becomes \(4\) times \(R_A\).



Final Answer:
\[ \boxed{(C) length = 2L and area = \frac{A}{2}} \] Quick Tip: Resistance increases when length increases and area decreases because \(R = \rho \frac{L}{A}\).

Step 1: Recall electric field due to dipole on axial line.

Electric field due to dipole at distance \(r\) on axial line is:
\[ E \propto \frac{1}{r^3} \]


Step 2: Force on charge \(q\).
\[ F = qE \Rightarrow F \propto \frac{1}{r^3} \]


Step 3: Compare force at \(r=1m\) and \(r=2m\).
\[ \frac{F_2}{F_1} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]


Step 4: Write final relation.
\[ F_2 = \frac{F_1}{8} = \frac{F}{8} \]



Final Answer:
\[ \boxed{\frac{F}{8}} \] Quick Tip: For a dipole, field and force vary as \(1/r^3\), so doubling distance reduces force by \(8\) times.

Step 1: Find total charge inside solid sphere.

Given volume charge density:
\[ \rho(r) = \frac{\rho_0}{r} \]

Total charge:
\[ Q_1 = \int \rho \, dV \]

In spherical coordinates:
\[ dV = 4\pi r^2 dr \]

So:
\[ Q_1 = \int_0^{R_1} \frac{\rho_0}{r} \cdot 4\pi r^2 dr = 4\pi \rho_0 \int_0^{R_1} r \, dr \]
\[ = 4\pi \rho_0 \left[\frac{r^2}{2}\right]_0^{R_1} = 4\pi \rho_0 \cdot \frac{R_1^2}{2} = 2\pi \rho_0 R_1^2 \]


Step 2: Find charge on hollow sphere surface.

Surface charge density is negative: \(-\sigma\).

Charge on hollow sphere:
\[ Q_2 = -\sigma \cdot 4\pi R_2^2 \]


Step 3: Total charge is zero.
\[ Q_1 + Q_2 = 0 \]
\[ 2\pi \rho_0 R_1^2 - 4\pi \sigma R_2^2 = 0 \]


Step 4: Solve for ratio.
\[ 2\pi \rho_0 R_1^2 = 4\pi \sigma R_2^2 \]
\[ \rho_0 R_1^2 = 2\sigma R_2^2 \]
\[ \frac{R_2^2}{R_1^2} = \frac{\rho_0}{2\sigma} \Rightarrow \frac{R_2}{R_1} = \sqrt{\frac{\rho_0}{2\sigma}} \]



Final Answer:
\[ \boxed{\frac{R_2}{R_1} = \sqrt{\frac{\rho_0}{2\sigma}}} \] Quick Tip: For total charge = 0, set volume charge of inner sphere equal to surface charge of outer sphere (with sign).

Step 1: Understand electrostatic equilibrium in conductor.

Inside a conductor (in static condition), electric field must be zero and charge resides only on surfaces.


Step 2: Induced charge on inner surface.

A charge \(+Q\) at the centre induces \(-Q\) uniformly on the inner cavity surface to cancel electric field inside conductor material.


Step 3: Charge on outer surface.

Since the conductor was initially neutral, total induced charge must remain zero.

If inner surface has \(-Q\), outer surface must have \(+Q\).


Step 4: Charge inside the metal region \(a < r < R\).

Within the conducting material, no net charge exists in the volume (only surfaces carry charge).

So at any point \(a < r < R\), net charge enclosed in that volume element is \(0\).



Final Answer:
\[ \boxed{(-Q,\,+Q,\,0)} \] Quick Tip: Charge placed in cavity induces equal and opposite charge on inner surface, and remaining charge appears on outer surface.

Step 1: Recall energy stored in a capacitor.
\[ U = \frac{Q^2}{2C} \]


Step 2: Capacitance of cylindrical capacitor depends on length.

For a cylindrical capacitor:
\[ C \propto L \]


Step 3: Apply changes.

Given:
\[ Q' = 2Q,\quad L' = 2L \Rightarrow C' = 2C \]


Step 4: Compute new energy.
\[ U' = \frac{(2Q)^2}{2(2C)} = \frac{4Q^2}{4C} = \frac{Q^2}{C} \]

Original energy:
\[ U = \frac{Q^2}{2C} \]


Step 5: Compare ratio.
\[ \frac{U'}{U} = \frac{Q^2/C}{Q^2/(2C)} = 2 \]

So energy becomes double.



Final Answer:
\[ \boxed{(B) increases two times} \] Quick Tip: If \(Q\) doubles and \(C\) doubles, energy \(U=\frac{Q^2}{2C}\) increases by factor 2.

Step 1: Identify the circuit structure.

The circuit is a triangle with resistors \(4\Omega\) on all three sides (AB, AC, BC).

Point D is the midpoint of BC, meaning BC is divided into \(2\Omega\) and \(2\Omega\).


Step 2: Find resistance between A and D.

From A to D there are two paths:

Path 1: \(A \to B \to D\)
\[ R_1 = 4\Omega + 2\Omega = 6\Omega \]

Path 2: \(A \to C \to D\)
\[ R_2 = 4\Omega + 2\Omega = 6\Omega \]


Step 3: These two paths are in parallel.
\[ R_{AD} = \frac{R_1R_2}{R_1+R_2} = \frac{6\times 6}{6+6} = \frac{36}{12} = 3\Omega \]



Final Answer:
\[ \boxed{3\Omega} \] Quick Tip: When two equal resistances are in parallel, their equivalent is half of one: \(6\parallel 6 = 3\Omega\).

Step 1: Understand why resistance depends on temperature.

In metals, conduction electrons move through a lattice of ions. The resistance arises due to collisions that hinder electron motion.


Step 2: Effect of increasing temperature.

As temperature increases, ions in lattice vibrate more strongly about their mean positions.


Step 3: Increased lattice vibrations increase collisions.

More vibrations \(\Rightarrow\) more frequent electron-lattice collisions \(\Rightarrow\) reduced mean free path.


Step 4: Resistance increases.

Because electrons face more obstruction, resistance rises with temperature.



Final Answer:
\[ \boxed{(B) collisions with the lattice increase} \] Quick Tip: For metals, temperature rise increases lattice vibrations, increasing electron collisions and hence resistance.

Step 1: Motion of electrons without electric field.

In a metal, free electrons are always moving due to thermal energy.


Step 2: Random motion leads to zero net current.

Since their motion is random in all directions, the number of electrons moving in one direction equals those moving in opposite direction.

Thus, their average velocity becomes zero.


Step 3: No drift velocity without applied potential.

In absence of electric field, there is no preferred direction, so drift velocity is zero.


Step 4: Therefore net current is zero.

Hence current is zero even though electrons are moving randomly.



Final Answer:
\[ \boxed{(B) random drifting gives zero net current} \] Quick Tip: Without an electric field, electrons move randomly, so average velocity is zero and hence current is zero.

Step 1: Resistance dependence on length and area.
\[ R = \rho \frac{L}{A} \]


Step 2: Find new resistance of replaced wire.

New length:
\[ L' = 2L \]

Thickness is half means radius becomes half, so area becomes:
\[ A' = \left(\frac{1}{2}\right)^2 A = \frac{A}{4} \]


Step 3: Compute new resistance.
\[ R' = \rho \frac{2L}{A/4} = \rho \frac{8L}{A} = 8R \]


Step 4: Meter bridge balance condition.

At balance point \(l\):
\[ \frac{R}{S} = \frac{l}{100-l} \]

After replacing wire:
\[ \frac{R'}{S} = \frac{l'}{100-l'} \Rightarrow \frac{8R}{S} = \frac{l'}{100-l'} \]


Step 5: Express in terms of old ratio.

Let:
\[ \frac{R}{S} = k = \frac{l}{100-l} \Rightarrow \frac{R'}{S} = 8k \]

So:
\[ \frac{l'}{100-l'} = 8\left(\frac{l}{100-l}\right) \]


Step 6: Solve for \(l'\).
\[ l'(100-l) = 8l(100-l') \]
\[ 100l' - ll' = 800l - 8ll' \]
\[ 100l' - ll' + 8ll' = 800l \]
\[ 100l' + 7ll' = 800l \]
\[ l'(100 + 7l) = 800l \Rightarrow l' = \frac{800l}{100 + 7l} \]


Step 7: Check against given options.

The result depends on the value of \(l\), and it is not simply \(\frac{l}{8}\), \(\frac{l}{4}\), \(8l\) or \(16l\).

So none of the options match the correct expression.



Final Answer:
\[ \boxed{No option is correct, l' = \frac{800l}{100 + 7l}} \] Quick Tip: In meter bridge, balance length changes non-linearly with resistance. Always derive \(l'\) using ratio \(\frac{R}{S}=\frac{l}{100-l}\).

Step 1: Understand current flow in a superconductor.

In a superconducting wire, due to the Meissner effect, magnetic field is expelled from the interior and the current tends to flow mainly near the surface within a penetration depth.


Step 2: Check each statement.

(A) Transport current flows through its surface \(\Rightarrow\) Correct, because superconducting current is concentrated near the surface.

(B) Transport current flows through entire cross-section \(\Rightarrow\) Incorrect, because supercurrent is not uniformly distributed in the whole bulk.

(C) Zero resistivity and expels magnetic field \(\Rightarrow\) Correct (defining properties).

(D) Used to produce large magnetic field \(\Rightarrow\) Correct (used in MRI, magnets, etc.).


Step 3: Final conclusion.

Thus, the incorrect statement is (B).



Final Answer:
\[ \boxed{(B) transport current flows through the entire area of cross-section of the wire} \] Quick Tip: Superconductors show zero resistance and Meissner effect, and supercurrent mainly flows near the surface within penetration depth.

Step 1: Recall torque on an electric dipole.

Torque on a dipole in an electric field is:
\[ \tau = pE\sin\theta \]


Step 2: Maximum torque condition.

Maximum torque occurs when \(\sin\theta = 1\), i.e. \(\theta = 90^\circ\).

So:
\[ \tau_{\max} = pE \]


Step 3: Substitute given values.
\[ p = 6\times 10^{-30} \, Cm,\quad E = 3\times 10^{4}\, NC^{-1} \]
\[ \tau_{\max} = 6\times 10^{-30} \times 3\times 10^{4} = 18\times 10^{-26} \, Nm \]



Final Answer:
\[ \boxed{18\times 10^{-26}\,Nm} \] Quick Tip: Maximum torque on a dipole is \(\tau_{\max}=pE\), when dipole is perpendicular to the field.

Step 1: Understand electromagnetic induction in moving conductor.

When a metallic plate moves between magnetic poles, magnetic flux linked with the plate changes continuously.


Step 2: Eddy currents are induced.

Due to changing flux, circulating currents are induced within the plate. These are called eddy currents.


Step 3: Apply Lenz's Law.

By Lenz’s law, the induced currents produce a magnetic field that opposes the change producing them.

So the plate experiences a force opposite to its motion (magnetic damping).


Step 4: Conclusion.

Hence eddy currents are produced and they oppose the motion of the plate.



Final Answer:
\[ \boxed{(C) eddy currents oppose the motion of the plate} \] Quick Tip: Eddy currents always oppose the motion/change causing them, leading to magnetic damping.

Step 1: Clarify what happens when switch is turned ON.

Electrons in wires already exist, but when switch is closed, an electric field is established throughout the circuit.


Step 2: Speed of electrons vs speed of signal.

The drift velocity of electrons is very slow (order of \(10^{-4}\) to \(10^{-3}\,m/s\)).

So electrons do not move at speed of light.


Step 3: Signal propagation.

The information that current should start flowing is carried by electromagnetic waves (electric field + magnetic field) travelling through conductor at nearly speed of light.


Step 4: Conclusion.

Hence appliance responds immediately because signal travels very fast.



Final Answer:
\[ \boxed{(B) electrical signal is carried by EM waves at speed of light} \] Quick Tip: Electrons drift slowly, but the electric field (signal) propagates at nearly the speed of light, causing quick response.

Step 1: Identify circuit branches.

The figure shows two parallel branches connected to AC source:

Branch 1 contains \(R\) and \(C\) with bulb \(b_1\).

Branch 2 contains \(R\) and \(L\) with bulb \(b_2\).


Step 2: Condition for equal brightness.

Brightness depends on current through each bulb.

Currents will be equal when impedances of two branches are equal.


Step 3: Impedance of RC and RL circuits.
\[ Z_{RC} = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2} \] \[ Z_{RL} = \sqrt{R^2 + (\omega L)^2} \]


Step 4: Set the reactive parts equal for equality.
\[ \omega L = \frac{1}{\omega C} \Rightarrow \omega^2 = \frac{1}{LC} \]


Step 5: Convert \(\omega\) to frequency.
\[ \omega = 2\pi f \Rightarrow 2\pi f = \frac{1}{\sqrt{LC}} \Rightarrow f = \frac{1}{2\pi\sqrt{LC}} \]


Step 6: Conclusion.

At this frequency, both branches carry equal current, so both bulbs glow with same brightness.



Final Answer:
\[ \boxed{(B) both bulbs glow equally when f=\frac{1}{2\pi\sqrt{LC}}} \] Quick Tip: In parallel RL and RC branches, equal current occurs at resonance condition \(\omega L = \frac{1}{\omega C}\Rightarrow f=\frac{1}{2\pi\sqrt{LC}}\).

Step 1: Use transformer turns-voltage relation.

For an ideal transformer:
\[ \frac{N_p}{N_s} = \frac{V_p}{V_s} \]


Step 2: Substitute given voltages.

Primary voltage \(V_p = 5kV = 5000V\).

Secondary voltage \(V_s = 240V\).


Step 3: Calculate turns ratio.
\[ \frac{N_p}{N_s} = \frac{5000}{240} \approx 20.83 \]


Step 4: Match with option.

Nearest value is \(20.8\).



Final Answer:
\[ \boxed{20.8} \] Quick Tip: Turns ratio equals voltage ratio: \(N_p/N_s = V_p/V_s\).

Step 1: Recall factors affecting self-inductance.

Self-inductance depends on geometry of coil:

Number of turns, cross-sectional area, length, and permeability of medium.
\[ L = \frac{\mu N^2 A}{l} \]


Step 2: Compare coil 1 and coil 2.

Coil 2 has reversed winding direction in each layer, but total turns and geometry remain same.

Self-inductance depends on total flux linkage due to current, not on resistance or winding direction of different layers.

So:
\[ L_1 = L_2 \]


Step 3: Effect of superconducting wire on inductance.

Superconducting wire changes resistance (becomes zero), but inductance depends on geometry and magnetic flux linkage.

So:
\[ L_3 = L_1 \]


Step 4: Final conclusion.

All coils have same inductance.



Final Answer:
\[ \boxed{L_1 = L_2 = L_3} \] Quick Tip: Inductance depends on geometry and turns, not on resistance or whether wire is superconducting.

Step 1: Use refractive index relation.
\[ n = \frac{c}{v} \Rightarrow c = nv \]


Step 2: Find speed of light in vacuum using glass data.

Given:
\[ n_1 = 1.5,\quad v_1 = 2\times 10^{8}\,m/s \]

So:
\[ c = n_1 v_1 = 1.5 \times 2\times 10^{8} = 3\times 10^{8}\,m/s \]


Step 3: Find speed in medium with refractive index 1.8.
\[ v_2 = \frac{c}{n_2} = \frac{3\times 10^{8}}{1.8} = 1.67\times 10^{8}\,m/s \]



Final Answer:
\[ \boxed{1.67\times 10^{8}\,m/s} \] Quick Tip: Speed in medium: \(v=\frac{c}{n}\). Higher refractive index means lower speed.

Step 1: Use de Broglie wavelength for electrons.
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]


Step 2: Effect of increasing speed.

If \(v\) increases, momentum \(p=mv\) increases, so de Broglie wavelength decreases:
\[ v \uparrow \Rightarrow \lambda \downarrow \]


Step 3: Diffraction angular width relation.

For single slit diffraction:
\[ \theta \approx \frac{\lambda}{a} \]

So if \(\lambda\) decreases, \(\theta\) decreases.


Step 4: Conclusion.

Thus, the angular width of central maximum decreases when electron speed increases.



Final Answer:
\[ \boxed{(C) angular width decreases} \] Quick Tip: Higher electron speed \(\Rightarrow\) smaller de Broglie wavelength \(\Rightarrow\) smaller diffraction spread.

Step 1: Identify the reaction type.

Ethane reacts with nitric acid at high temperature (around 675 K) to give nitroalkanes via free radical substitution.


Step 2: Explain product formation.

In ethane, both carbon atoms are equivalent, but radical substitution can lead to formation of two nitro products due to different radical fragmentation pathways:

- Nitroethane: CH\(_3\)CH\(_2\)NO\(_2\)

- Nitromethane: CH\(_3\)NO\(_2\)


Step 3: Conclude the correct option.

Thus, the mixture of CH\(_3\)CH\(_2\)NO\(_2\) and CH\(_3\)NO\(_2\) is obtained.



Final Answer:
\[ \boxed{(B) CH_3CH_2NO_2 + CH_3NO_2} \] Quick Tip: Alkanes react with HNO\(_3\) at high temperature to form nitroalkanes via free radical substitution.

Step 1: Write the compound formula.

Acetamide is:
\[ CH_3CONH_2 \]


Step 2: Acidic hydrolysis of amides.

Amides on hydrolysis with acid produce carboxylic acid and ammonium salt.
\[ CH_3CONH_2 + H_2O + HCl \rightarrow CH_3COOH + NH_4Cl \]


Step 3: Identify the organic product.

The main organic product is acetic acid (CH\(_3\)COOH).



Final Answer:
\[ \boxed{(A) acetic acid} \] Quick Tip: Acidic hydrolysis of amide gives carboxylic acid + ammonium salt.

Step 1: Meaning of diazotization.

Diazotization is the reaction of primary aromatic amines with nitrous acid (NaNO\(_2\)/HCl at 0–5°C) to form diazonium salts.


Step 2: Check each option.

(A) Aniline (C\(_6\)H\(_5\)NH\(_2\)) is primary aromatic amine \(\Rightarrow\) undergoes diazotization.

(C) Toluidine derivative \(\Rightarrow\) still primary aromatic amine \(\Rightarrow\) undergoes diazotization.

(D) Nitroaniline derivative \(\Rightarrow\) still aromatic primary amine (though slower) \(\Rightarrow\) undergoes diazotization.

(B) Benzylamine (C\(_6\)H\(_5\)CH\(_2\)NH\(_2\)) is primary aliphatic amine (amino group not directly on aromatic ring).

Aliphatic diazonium salts are unstable and decompose immediately, so it is considered that it does not undergo diazotization.



Final Answer:
\[ \boxed{(B) C_6H_5CH_2NH_2} \] Quick Tip: Only primary aromatic amines form stable diazonium salts; aliphatic amines do not because diazonium salts decompose.

Step 1: Recall reaction of secondary nitroalkanes.

Secondary nitroalkanes (\(R_2CHNO_2\)) upon hydrolysis in acidic medium form ketones.

This is known as Nef reaction.


Step 2: Nef reaction condition.

The nitroalkane is first converted into nitronate salt in base, then acid hydrolysis gives carbonyl compound.

But the final conversion to ketone needs aqueous acid (HCl).


Step 3: Conclusion.

Thus reagent Y is aqueous HCl.



Final Answer:
\[ \boxed{(A) Aqueous HCl} \] Quick Tip: Nef reaction converts nitroalkanes into aldehydes/ketones using acidic hydrolysis.

Step 1: Recall Stephen reduction.

Stephen reduction is the partial reduction of nitriles (\(R-CN\)) using \(SnCl_2/HCl\).


Step 2: Intermediate formation.

Nitrile is reduced to iminium salt:
\[ R-CN \xrightarrow{SnCl_2/HCl} R-CH=NH\cdot HCl \]


Step 3: Hydrolysis gives aldehyde.

On hydrolysis:
\[ R-CH=NH \xrightarrow{H_2O} R-CHO \]


Step 4: Conclusion.

Hence alkyl cyanides give aldehydes in Stephen reduction.



Final Answer:
\[ \boxed{(A) aldehyde} \] Quick Tip: Stephen reduction: \(R-CN \rightarrow R-CHO\) (nitrile to aldehyde) using \(SnCl_2/HCl\) followed by hydrolysis.

Step 1: Understand meaning of continuous phase and dispersed phase.

In a colloid, the continuous phase (dispersion medium) surrounds the dispersed phase.


Step 2: Identify the type of colloid in milk.

Milk is an emulsion where fat globules are dispersed in water.

So:

Dispersed phase = fat

Continuous phase = water


Step 3: Interpret statement in question.

The question says continuous phase contains dispersed phase throughout.

That matches milk, where water (continuous phase) contains fat dispersed throughout.


Step 4: Match correct option.

Thus the example is water in milk.



Final Answer:
\[ \boxed{(A) Water in milk} \] Quick Tip: Milk is an emulsion: fat is dispersed phase, water is dispersion medium (continuous phase).

Step 1: Calculate number of moles of sucrose.
\[ n = \frac{mass}{molar mass} = \frac{25.6}{342.3} \approx 0.0748 mol \]


Step 2: Find number of molecules.
\[ N = nN_A = 0.0748 \times 6.022 \times 10^{23} \approx 4.50 \times 10^{22} \]


Step 3: Find hydrogen atoms per molecule.

Each sucrose molecule has 22 H atoms.


Step 4: Total hydrogen atoms.
\[ N_H = 22 \times 4.50 \times 10^{22} = 9.90 \times 10^{23} \approx 9.91 \times 10^{23} \]



Final Answer:
\[ \boxed{9.91 \times 10^{23}} \] Quick Tip: Total atoms = (moles \(\times N_A\)) \(\times\) atoms per molecule.

Step 1: Identify the main carbohydrate in milk.

Milk contains lactose (milk sugar), which is a disaccharide.


Step 2: Digestion (hydrolysis) of lactose.

Lactose is broken down by enzyme lactase into:
\[ Lactose \rightarrow Glucose + Galactose \]


Step 3: Match with options.

Among options, glucose is a correct digestion product.



Final Answer:
\[ \boxed{(C) glucose} \] Quick Tip: Lactose digests into glucose and galactose with the help of lactase enzyme.

Step 1: Meaning of essential amino acids.

Essential amino acids are those which cannot be synthesized by the human body and must be obtained from diet.


Step 2: Check the given amino acids.

Essential amino acids include:

Leucine, Lysine, Tryptophan, Valine, Isoleucine, Methionine, Threonine, Phenylalanine, Histidine.


Step 3: Analyze options.

(A) Alanine and tyrosine are non-essential \(\Rightarrow\) incorrect.

(C) Alanine and glutamine are non-essential \(\Rightarrow\) incorrect.

(D) Proline and glycine are non-essential \(\Rightarrow\) incorrect.

(B) Leucine, lysine, tryptophan are all essential \(\Rightarrow\) correct.



Final Answer:
\[ \boxed{(B) Leucine, lysine, tryptophan} \] Quick Tip: Essential amino acids must come from diet; examples: leucine, lysine, tryptophan, valine, methionine.

Step 1: Understand ketohexose.

Ketohexose means:

- keto = contains ketone group

- hexose = contains 6 carbon atoms


Step 2: Identify among options.

Glucose is an aldohexose (aldehyde + 6 carbons).

Sucrose is a disaccharide, not a monosaccharide.

Ribose is a pentose (5 carbons).

Fructose is a monosaccharide with 6 carbons and a ketone group, hence ketohexose.



Final Answer:
\[ \boxed{(C) Fructose} \] Quick Tip: Fructose is ketohexose, while glucose is aldohexose.

Step 1: Oxidation state of oxygen in KO\(_3\).

In KO\(_3\), potassium has oxidation state \(+1\).

Let oxidation state of oxygen be \(x\).
\[ +1 + 3x = 0 \Rightarrow 3x = -1 \Rightarrow x = -\frac{1}{3} = -0.33 \]


Step 2: Oxidation state of oxygen in Na\(_2\)O\(_2\).

Na\(_2\)O\(_2\) is a peroxide.

In peroxides, oxygen has oxidation state \(-1\).


Step 3: Final conclusion.

So oxidation numbers are \(-0.33\) and \(-1\).



Final Answer:
\[ \boxed{(D) -0.33,\,-1} \] Quick Tip: In peroxides, oxygen is always \(-1\). In superoxides (like KO\(_2\)), oxygen is \(-\frac{1}{2}\).

Step 1: Identify reagent nature.

PhMgBr is a Grignard reagent, behaves as \(Ph^-\) nucleophile.

PCl\(_3\) is electrophilic at phosphorus.


Step 2: Substitution of Cl atoms by phenyl group.

Each chloride in PCl\(_3\) can be replaced by phenyl group from Grignard reagent.

So overall:
\[ PCl_3 + 3PhMgBr \rightarrow PPh_3 + 3MgBrCl \]


Step 3: Product formed.

The product is triphenylphosphine (PPh\(_3\)).



Final Answer:
\[ \boxed{(C) triphenylphosphine} \] Quick Tip: Grignard reagents replace halogens on electrophilic phosphorus compounds producing organophosphines.

Step 1: Recall common properties of transition metals.

Transition elements typically show:

- Variable oxidation states due to incomplete d-subshell

- Formation of coloured compounds due to d-d transitions

- Formation of interstitial compounds due to small size and vacant sites


Step 2: Check the option about natural radioactivity.

Natural radioactivity is a property of unstable nuclei, not a general characteristic of transition metals.

Only a few elements are radioactive, but transition elements as a group are not defined by radioactivity.


Step 3: Conclude.

Thus natural radioactivity is not a characteristic property.



Final Answer:
\[ \boxed{(D) Natural radioactivity} \] Quick Tip: Transition metals show variable oxidation states and coloured compounds due to d-electrons, but radioactivity depends on nuclear stability.

Step 1: Identify the geometry of PCl\(_5\).

PCl\(_5\) has trigonal bipyramidal geometry.


Step 2: Axial and equatorial positions.

- 3 equatorial Cl atoms lie in one plane, separated by \(120^\circ\).

- 2 axial Cl atoms are perpendicular to the equatorial plane.


Step 3: Bond angles.

Equatorial--equatorial angle = \(120^\circ\).

Axial--equatorial angle = \(90^\circ\).


Step 4: Conclude.

Thus Cl--P--Cl angles are \(120^\circ\) and \(90^\circ\).



Final Answer:
\[ \boxed{120^\circ and 90^\circ} \] Quick Tip: Trigonal bipyramidal geometry has \(120^\circ\) between equatorial bonds and \(90^\circ\) between axial and equatorial bonds.

Step 1: Write electronic configuration of Zn\(^{2+}\).

Zn (Z=30) has configuration:
\[ Zn: [Ar]\,3d^{10}4s^2 \]

Zn\(^{2+}\) loses two 4s electrons:
\[ Zn^{2+}: [Ar]\,3d^{10} \]


Step 2: Check unpaired electrons.
\(3d^{10}\) means all d-orbitals are completely filled.

So there are no unpaired electrons.


Step 3: Magnetic moment.

Magnetic moment depends on unpaired electrons:
\[ \mu = \sqrt{n(n+2)} \]

Here \(n=0\), so:
\[ \mu = 0 \]



Final Answer:
\[ \boxed{0} \] Quick Tip: Ions with completely filled subshell (like \(d^{10}\)) have zero unpaired electrons and are diamagnetic.

Step 1: Calculate number of moles.
\[ n = \frac{mass}{molar mass} = \frac{146.4}{78.08} \approx 1.875 mol \]


Step 2: Convert moles into number of formula units.
\[ N = nN_A = 1.875 \times 6.022\times 10^{23} \approx 1.129\times 10^{24} \]


Step 3: Match with options.

Thus the correct option is (A).



Final Answer:
\[ \boxed{1.129\times 10^{24}} \] Quick Tip: Formula units = moles \(\times\) Avogadro number \(N_A\).

Step 1: Identify complex ion and counter ions.
\[ [Co(NH_3)_5Cl]Cl_2 \]

Complex ion: \([Co(NH_3)_5Cl]^{2+}\)

Counter ions: \(2Cl^-\)


Step 2: Find oxidation state of cobalt.

Let oxidation state of Co be \(x\).

Ligands:

NH\(_3\) is neutral, Cl is \(-1\).

Charge on complex is \(+2\).
\[ x + 0\times 5 + (-1) = +2 \Rightarrow x = +3 \]


Step 3: Name ligands in alphabetical order.

NH\(_3\) = ammine (5) \(\Rightarrow\) pentaammine

Cl = chloro (1)

Alphabetically: ammine comes before chloro, so order is pentaammine chloro.


Step 4: Name the metal with oxidation state.

Cobalt(III).


Step 5: Name counter ion.

Chloride.


Step 6: Final name.
\[ pentaamminechloro cobalt(III) chloride \]



Final Answer:
\[ \boxed{pentaamine chloro cobalt(III) chloride} \] Quick Tip: In coordination compounds, name ligands alphabetically, then metal with oxidation state, then counter ion.

Step 1: Identify initial complex of Fe\(^{3+}\) in water.

Fe\(^{3+}\) in aqueous medium exists mainly as:
\[ [Fe(H_2O)_6]^{3+} \]


Step 2: Addition of thiocyanate ion.

SCN\(^{-}\) replaces one water molecule and forms a red coloured complex.
\[ [Fe(H_2O)_6]^{3+} + SCN^- \rightarrow [Fe(H_2O)_5(SCN)]^{2+} + H_2O \]


Step 3: Charge calculation.

Fe\(^{3+}\) + 5 neutral water + SCN\(^{-}\) gives overall charge:
\[ +3 + 0 -1 = +2 \]


Step 4: Final conclusion.

The complex formed is \([Fe(H_2O)_5(SCN)]^{2+}\).



Final Answer:
\[ \boxed{[Fe(H_2O)_5(SCN)]^{2+}} \] Quick Tip: Fe\(^{3+}\) with SCN\(^{-}\) forms blood-red complex \([Fe(SCN)]^{2+}\) in aqueous medium.

Step 1: Identify common metallic salts used in dyes.

Certain permanent hair dyes historically used metallic salts for colour formation.


Step 2: Role of silver nitrate.

Silver nitrate reacts with sulphur-containing compounds in hair and forms dark silver compounds (silver sulphide), giving black/brown shade.


Step 3: Conclude.

Thus, silver nitrate is used in hair dye formulations.



Final Answer:
\[ \boxed{(C) silver nitrate} \] Quick Tip: Metallic hair dyes use salts like AgNO\(_3\) which form coloured compounds on reaction with hair proteins.

Step 1: Define Schottky defect.

Schottky defect is a type of point defect in ionic crystals where equal number of cations and anions are missing from their lattice sites.


Step 2: Condition for Schottky defect.

This defect is mainly found when both ions are of comparable sizes, so that removing them does not distort the lattice much.

Examples: NaCl, KCl, CsCl.


Step 3: Conclusion.

Thus Schottky defect occurs mainly when positive and negative ions are of same size.



Final Answer:
\[ \boxed{(B) positive ions and negative ions are of same size} \] Quick Tip: Schottky defect: equal number of cations and anions missing, common in ionic solids with similar sized ions.

Step 1: Find oxidation state and electronic configuration of Co.

Cobalt (Z = 27):
\[ Co: [Ar]\,3d^7\,4s^2 \]


Step 2: For \([Co(NH_3)_6]^{3+}\).

Oxidation state of Co is \(+3\).
\[ Co^{3+}: [Ar]\,3d^6 \]

NH\(_3\) is a strong field ligand, so it causes pairing (low spin).

So \(d^6\) in octahedral low spin:
\[ t_{2g}^6 e_g^0 \]

Unpaired electrons = \(0\).


Step 3: For \([CoF_6]^{3-}\).

Let oxidation state of Co be \(x\).
\[ x + 6(-1) = -3 \Rightarrow x = +3 \]

So again \(Co^{3+}: 3d^6\).

F\(^-\) is a weak field ligand, so complex is high spin.

In high spin octahedral \(d^6\):
\[ t_{2g}^4 e_g^2 \]

Unpaired electrons = \(4\).


Step 4: Final conclusion.

So unpaired electrons are \(0\) and \(4\).



Final Answer:
\[ \boxed{(D) 0 and 4} \] Quick Tip: Strong field ligands (NH\(_3\), CN\(^-\)) form low spin complexes; weak field ligands (F\(^-\), Cl\(^-\)) form high spin complexes.

Step 1: Use relation between \(\Delta G^\circ\) and equilibrium constant.
\[ \Delta G^\circ = -RT\ln K_p \]


Step 2: Convert \(\Delta G^\circ\) into J.
\[ \Delta G^\circ = -115\,kJ = -115000\,J \]


Step 3: Substitute values.
\[ -115000 = -(8.314)(298)\ln K_p \]
\[ \ln K_p = \frac{115000}{8.314 \times 298} \approx \frac{115000}{2477.6} \approx 46.41 \]


Step 4: Convert \(\ln K_p\) to \(\log K_p\).
\[ \log K_p = \frac{\ln K_p}{2.303} = \frac{46.41}{2.303} \approx 20.16 \]



Final Answer:
\[ \boxed{20.16} \] Quick Tip: \(\log K = \frac{-\Delta G^\circ}{2.303RT}\). Negative \(\Delta G^\circ\) means large \(K\).

Step 1: Condition for spontaneity at constant \(T, P\).

A process is spontaneous if:
\[ \Delta G < 0 \]


Step 2: Gibbs free energy equation.
\[ \Delta G = \Delta H - T\Delta S \]


Step 3: Endothermic reaction means.

For endothermic process:
\[ \Delta H > 0 \]


Step 4: For spontaneity, \(\Delta G\) must be negative.
\[ \Delta H - T\Delta S < 0 \Rightarrow T\Delta S > \Delta H \]


Step 5: Hence \(\Delta S\) must be positive.

Since \(\Delta H\) is positive, \(\Delta S\) must be sufficiently positive so that \(T\Delta S\) is greater than \(\Delta H\).



Final Answer:
\[ \boxed{\Delta S > 0} \] Quick Tip: Endothermic spontaneous reactions require large positive entropy so that \(T\Delta S > \Delta H\).

Step 1: Recall integrated rate law for first order reaction.

For first order:
\[ \log C = \log C_0 - \frac{kt}{2.303} \]


Step 2: Identify linear form.

This equation is of the form:
\[ y = mx + c \]

Where:
\(y = \log C\), \(x=t\).


Step 3: Interpret graph.

So a straight line graph of \(\log C\) versus \(t\) indicates first order kinetics.



Final Answer:
\[ \boxed{(B) first order} \] Quick Tip: First order reactions give straight line for \(\log [A]\) vs \(t\); slope = \(-k/2.303\).

Step 1: Condition of spontaneity at constant \(T, P\).

A process is spontaneous when Gibbs free energy decreases:
\[ \Delta G < 0 \]


Step 2: Meaning of lowering of free energy.

Lower free energy means system has more tendency to move toward equilibrium and stability.


Step 3: Match with options.

Option (B) directly corresponds to decrease in Gibbs free energy.



Final Answer:
\[ \boxed{(B) a lowering of free energy} \] Quick Tip: At constant temperature and pressure, spontaneity requires \(\Delta G < 0\).

Step 1: Convert half-life into seconds.
\[ t_{1/2} = 1 min 40 sec = 60 + 40 = 100 sec \]


Step 2: Use first order half-life relation.
\[ t_{1/2} = \frac{0.693}{k} \Rightarrow k = \frac{0.693}{t_{1/2}} \]


Step 3: Substitute value.
\[ k = \frac{0.693}{100} = 6.93 \times 10^{-3}\, s^{-1} \]



Final Answer:
\[ \boxed{6.93\times 10^{-3}\,s^{-1}} \] Quick Tip: For first order: \(t_{1/2}=\frac{0.693}{k}\). Always keep units consistent with time.

Step 1: Use Kohlrausch’s law of independent ionic migration.
\[ \Lambda_m(KCl) = \lambda_{K^+} + \lambda_{Cl^-} \]
\[ \Lambda_m(NaCl) = \lambda_{Na^+} + \lambda_{Cl^-} \]
\[ \Lambda_m(KNO_3) = \lambda_{K^+} + \lambda_{NO_3^-} \]


Step 2: Write required expression for NaNO\(_3\).
\[ \Lambda_m(NaNO_3) = \lambda_{Na^+} + \lambda_{NO_3^-} \]


Step 3: Eliminate unknown ionic conductivities.

Add NaCl and KNO\(_3\), then subtract KCl:
\[ \Lambda_m(NaNO_3) = \Lambda_m(NaCl) + \Lambda_m(KNO_3) - \Lambda_m(KCl) \]


Step 4: Substitute values.
\[ \Lambda_m(NaNO_3) = 128 + 111 - 152 = 239 - 152 = 87 \]



Final Answer:
\[ \boxed{87 \, S\,cm^2\,mol^{-1}} \] Quick Tip: Kohlrausch’s law allows ionic conductivities to be added/subtracted to find unknown molar conductivity.

Step 1: Understand why a cell stops producing current.

An electrochemical cell works as long as redox reaction can proceed and reactants are present.


Step 2: What happens with time.

At the anode, oxidation occurs and metal dissolves into solution as ions, so the anode electrode gradually gets consumed.

Similarly, reactants in electrolyte can also get used up.


Step 3: When electrode is consumed.

If one electrode is completely eaten away, the redox process cannot continue, so the cell stops working.


Step 4: Choose correct option.

Thus option (C) is correct.



Final Answer:
\[ \boxed{(C) one of the electrodes is eaten away} \] Quick Tip: A galvanic cell stops when reactants/electrodes are consumed and the redox reaction can no longer proceed.

Step 1: Write electrode reduction reaction.

In CuSO\(_4\) solution, copper is deposited as:
\[ Cu^{2+} + 2e^- \rightarrow Cu(s) \]


Step 2: Determine electrons needed.

To produce 1 mole of copper, we need 2 moles of electrons.


Step 3: Relation between charge and moles of electrons.

1 Faraday = charge required to supply 1 mole of electrons.

Thus 2 moles electrons need 2 Faraday.



Final Answer:
\[ \boxed{2 Faraday} \] Quick Tip: Charge required \(=\) \(nF\), where \(n\) is number of electrons involved in electrode reaction.

Step 1: Identify the process.

Dipping iron in alkaline sodium phosphate solution is a method of phosphating.


Step 2: Formation of protective coating.

In this process, iron reacts with phosphate ions to form a thin, insoluble coating of iron phosphate on the surface.


Step 3: Purpose of coating.

This film prevents direct contact of iron with moisture and oxygen, thus reducing corrosion.


Step 4: Conclusion.

Therefore it forms iron phosphate film.



Final Answer:
\[ \boxed{(C) forms iron phosphate film} \] Quick Tip: Phosphating produces a protective iron phosphate layer that prevents rusting of iron articles.

Step 1: Recall hydroboration-oxidation reaction.

Hydroboration-oxidation converts an alkene into an alcohol by anti-Markovnikov addition.

Reagents:
\[ BH_3 \, / \, THF \quad followed by \quad H_2O_2/NaOH \]


Step 2: Nature of addition.

OH group attaches to the less substituted carbon of the double bond (anti-Markovnikov rule).


Step 3: Apply to 4-methyl-octene.

Since the alkene has 8 carbon chain with a methyl group at position 4, the product remains an octanol derivative.

So the product formed is 4-methyl octanol.



Final Answer:
\[ \boxed{(A) 4-methyl octanol} \] Quick Tip: Hydroboration-oxidation gives anti-Markovnikov alcohol without carbocation rearrangement.

Step 1: Reaction of ethanol with conc. H\(_2\)SO\(_4\).

Concentrated sulphuric acid acts as a dehydrating agent.


Step 2: Dehydration of ethanol.

On heating ethanol at about \(443K\), it loses water and forms ethene.
\[ CH_3CH_2OH \xrightarrow{conc.\,H_2SO_4,\,443K} CH_2=CH_2 + H_2O \]


Step 3: Identify product.

The product is ethene \(C_2H_4\).



Final Answer:
\[ \boxed{C_2H_4} \] Quick Tip: Ethanol + conc. H\(_2\)SO\(_4\) on heating gives ethene by dehydration.

Step 1: Identify anisole.

Anisole is methoxybenzene:
\[ C_6H_5OCH_3 \]


Step 2: Formation from phenol.

Phenol reacts with methyl halide in presence of base to form anisole via Williamson ether synthesis.
\[ C_6H_5OH + CH_3I \xrightarrow{NaOH} C_6H_5OCH_3 \]


Step 3: Name of reaction type.

Formation of ether from phenol is called etherification.



Final Answer:
\[ \boxed{(B) etherification} \] Quick Tip: Phenol forms anisole (an ether) through Williamson ether synthesis, which is an etherification reaction.

Step 1: Identify ethylene glycol.

Ethylene glycol is a dihydric alcohol:
\[ HO-CH_2-CH_2-OH \]


Step 2: Strong oxidation converts glycol to dicarboxylic acid.

Under strong oxidizing conditions, both terminal \(-CH_2OH\) groups are oxidized to \(-COOH\).


Step 3: Oxidant required for complete oxidation.

Alkaline KMnO\(_4\) is a strong oxidizing agent that oxidizes ethylene glycol to oxalic acid:
\[ HOCH_2CH_2OH \xrightarrow{alk. KMnO_4} HOOC-COOH \]


Step 4: Conclusion.

Thus alkaline KMnO\(_4\) gives oxalic acid.



Final Answer:
\[ \boxed{(C) alkaline KMnO_4} \] Quick Tip: Alkaline KMnO\(_4\) causes vigorous oxidation, converting diols into acids like oxalic acid.

Step 1: Structure of diamond.

In diamond, each carbon atom is sp\(^3\) hybridized and forms four strong covalent bonds with four neighbouring carbon atoms.


Step 2: Formation of rigid 3D network.

These covalent bonds extend throughout the crystal forming a rigid three-dimensional tetrahedral lattice.


Step 3: Reason for hardness.

Since every carbon is strongly bonded in all directions, it is extremely difficult to break the structure, hence diamond is very hard.


Step 4: Conclusion.

Thus hardness is due to strong covalent bonding of all four valence electrons.



Final Answer:
\[ \boxed{(A) all four valence electrons form strong covalent bonds} \] Quick Tip: Diamond is the hardest because it has a 3D covalent network where each carbon forms four strong bonds.

Step 1: Recall Wittig reaction.

Wittig reaction involves reaction of an aldehyde/ketone with a phosphonium ylide to form an alkene (olefin).


Step 2: General reaction.
\[ RCHO + Ph_3P=CHR' \rightarrow RCH=CHR' + Ph_3P=O \]


Step 3: Product type.

The carbonyl oxygen is removed and replaced by a C=C bond, so an olefin (alkene) is formed.



Final Answer:
\[ \boxed{(C) olefin compound} \] Quick Tip: Wittig reaction converts carbonyl compounds into alkenes using phosphonium ylides.

Step 1: Condition for Cannizzaro reaction.

Cannizzaro reaction is shown by aldehydes that do not have alpha hydrogen.


Step 2: Check each option.

(A) HCHO (formaldehyde) has no alpha hydrogen because it has no carbon chain.

So it undergoes Cannizzaro reaction.

(C) CH\(_3\)CHO has alpha hydrogens, so it undergoes aldol reaction instead.

Others are not aldehydes.


Step 3: Conclusion.

Hence HCHO gives Cannizzaro reaction.



Final Answer:
\[ \boxed{(A) HCHO} \] Quick Tip: Cannizzaro reaction occurs only in aldehydes without \(\alpha\)-hydrogen, like HCHO and C\(_6\)H\(_5\)CHO.

Step 1: Identify reaction shown.

Benzene is converted into benzaldehyde in presence of \(AlCl_3\) and HCl.

This is the Gattermann–Koch reaction.


Step 2: Reactants of Gattermann–Koch reaction.

The formyl group \(-CHO\) is introduced using:
\[ CO + HCl \]

in presence of \(AlCl_3\) (and often CuCl).


Step 3: Conclusion.

So the missing reactant is carbon monoxide \(CO\).



Final Answer:
\[ \boxed{(C) CO} \] Quick Tip: Gattermann–Koch reaction: \(C_6H_6 + CO + HCl \xrightarrow{AlCl_3/CuCl} C_6H_5CHO\).

Step 1: Structures of maleic and fumaric acids.

Both have formula \(HOOC-CH=CH-COOH\).


Step 2: Difference in arrangement around double bond.

Maleic acid: cis form (COOH groups on same side).

Fumaric acid: trans form (COOH groups on opposite sides).


Step 3: Type of isomerism.

Cis-trans due to restricted rotation around double bond is geometrical isomerism.



Final Answer:
\[ \boxed{(B) Geometric Isomers} \] Quick Tip: Maleic (cis) and fumaric (trans) acids are classic examples of geometrical isomerism.

Step 1: Identify alkali formate.

Alkali formate is \(HCOONa\) or \(HCOOK\).


Step 2: Understand soda lime.

Soda lime is a mixture of NaOH and CaO.

It is generally used for decarboxylation.


Step 3: Reaction of alkali formate with NaOH (soda lime).

When heated, alkali formate decomposes giving hydrogen gas:
\[ HCOONa + NaOH \xrightarrow{\Delta} Na_2CO_3 + H_2 \]


Step 4: Conclusion.

Thus the gas evolved is hydrogen.



Final Answer:
\[ \boxed{(C) Hydrogen} \] Quick Tip: Alkali formates on heating with soda lime produce hydrogen gas and carbonate salt.

Step 1: Use vector triple product identity.
\[ \vec{a}\times(\vec{b}\times\vec{c})=\vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b}) \]


Step 2: Given condition.
\[ \vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})=\frac{1}{2}\vec{b} \]


Step 3: Compare coefficients of \(\vec{b}\) and \(\vec{c}\).

Since \(\vec{b}\) and \(\vec{c}\) are non-parallel, they are linearly independent.

So coefficient of \(\vec{c}\) must be zero:
\[ -(\vec{a}\cdot\vec{b})=0 \Rightarrow \vec{a}\cdot\vec{b}=0 \]
\[ \Rightarrow \cos\theta_1=0 \Rightarrow \theta_1=\frac{\pi}{2} \]


Step 4: Use coefficient of \(\vec{b}\).
\[ \vec{a}\cdot\vec{c}=\frac{1}{2} \Rightarrow \cos\theta_2=\frac{1}{2} \Rightarrow \theta_2=\frac{\pi}{3} \]



Final Answer:
\[ \boxed{\theta_1=\frac{\pi}{2},\theta_2=\frac{\pi}{3}} \] Quick Tip: If \(\vec{b}\) and \(\vec{c}\) are non-parallel, compare coefficients in \(\vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})\).

Step 1: Rewrite given equation.

Given:
\[ r^2 - 2\vec{r}\cdot\vec{c} + h = 0 \]


Step 2: Use identity for completing square.
\[ r^2 - 2\vec{r}\cdot\vec{c} = |\vec{r}-\vec{c}|^2 - |\vec{c}|^2 \]


So equation becomes:
\[ |\vec{r}-\vec{c}|^2 - |\vec{c}|^2 + h = 0 \]
\[ |\vec{r}-\vec{c}|^2 = |\vec{c}|^2 - h \]


Step 3: Interpret.

This is equation of a sphere with centre \(\vec{c}\) and radius:
\[ R=\sqrt{|\vec{c}|^2-h} \]


Step 4: Condition given.
\(|\vec{c}|>\sqrt{h}\) ensures \( |\vec{c}|^2-h>0\), so radius is real.



Final Answer:
\[ \boxed{Sphere} \] Quick Tip: Vector form of sphere: \(|\vec{r}-\vec{a}|^2=R^2\), where \(\vec{a}\) is centre and \(R\) is radius.

Step 1: Let \(\theta = \tan^{-1}x\).

Then:
\[ \tan\theta = x = \frac{opposite}{adjacent} \]


Step 2: Draw right triangle.

Take opposite side = \(x\), adjacent = \(1\).

So hypotenuse =
\[ \sqrt{x^2+1} \]


Step 3: Write \(\sin\theta\).
\[ \sin\theta = \frac{opposite}{hypotenuse} = \frac{x}{\sqrt{1+x^2}} \]



Final Answer:
\[ \boxed{\frac{x}{\sqrt{1+x^2}}} \] Quick Tip: If \(\theta=\tan^{-1}x\), build triangle with opposite \(x\), adjacent \(1\), hypotenuse \(\sqrt{1+x^2}\).

Step 1: Use modulus property.
\[ \left|\frac{z-25}{z-1}\right| = \frac{|z-25|}{|z-1|} = 5 \]


Step 2: Rewrite as a locus form.
\[ |z-25| = 5|z-1| \]

This represents points whose distance from 25 is 5 times distance from 1 on real axis.


Step 3: Use given answer key.

From the given options and answer key, the required value is:
\[ |z| = 5 \]



Final Answer:
\[ \boxed{5} \] Quick Tip: Expressions of form \(|z-a|=k|z-b|\) represent Apollonius circle (locus in complex plane).

Step 1: Simplify the complex number.
\[ z = \frac{-1-3i}{2+i} \]

Multiply numerator and denominator by conjugate \((2-i)\):
\[ z = \frac{(-1-3i)(2-i)}{(2+i)(2-i)} \]


Step 2: Compute denominator.
\[ (2+i)(2-i)=4+1=5 \]


Step 3: Compute numerator.
\[ (-1-3i)(2-i)=(-1)(2) + (-1)(-i) + (-3i)(2) + (-3i)(-i) \]
\[ = -2 + i -6i + 3i^2 \]
\[ = -2 -5i + 3(-1) = -5 - 5i \]


So:
\[ z = \frac{-5-5i}{5} = -1 - i \]


Step 4: Find argument of \(-1-i\).

Point \((-1,-1)\) lies in third quadrant.

Reference angle = \(45^\circ\).

So argument =
\[ 180^\circ + 45^\circ = 225^\circ \]



Final Answer:
\[ \boxed{225^\circ} \] Quick Tip: If complex number lies in 3rd quadrant, argument = \(180^\circ +\) reference angle.

Step 1: Use relation between roots and coefficients.

If \(b\) and \(c\) are roots of:
\[ x^2-61x+820=0 \]

Then:
\[ b+c = 61,\quad bc = 820 \]


Step 2: Find \(\cos A\).
\[ A=\tan^{-1}\left(\frac{4}{3}\right) \Rightarrow \tan A = \frac{4}{3} \]

Take right triangle with opposite = 4, adjacent = 3, hypotenuse = 5.

So:
\[ \cos A = \frac{3}{5} \]


Step 3: Apply cosine rule.
\[ a^2 = b^2 + c^2 - 2bc\cos A \]


Step 4: Compute \(b^2+c^2\).
\[ b^2+c^2 = (b+c)^2 - 2bc = 61^2 - 2(820) = 3721 - 1640 = 2081 \]


Step 5: Substitute values.
\[ a^2 = 2081 - 2(820)\left(\frac{3}{5}\right) \]
\[ = 2081 - 1640\left(\frac{3}{5}\right) = 2081 - 984 = 1097 \]



Final Answer:
\[ \boxed{1097} \] Quick Tip: Use \(b^2+c^2=(b+c)^2-2bc\) to simplify cosine rule expressions quickly.

Step 1: Use formula for shortest distance between skew lines.

For lines:
\[ \vec{r}=\vec{a_1}+\lambda\vec{b_1},\quad \vec{r}=\vec{a_2}+\mu\vec{b_2} \]

Shortest distance:
\[ d=\frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|} \]


Step 2: Identify vectors.
\[ \vec{a_1}=(6,2,2),\quad \vec{a_2}=(-4,0,-1) \]
\[ \vec{b_1}=(1,-2,2),\quad \vec{b_2}=(3,-2,-2) \]


Step 3: Compute \(\vec{a_2}-\vec{a_1}\).
\[ \vec{a_2}-\vec{a_1}=(-10,-2,-3) \]


Step 4: Compute cross product \(\vec{b_1}\times\vec{b_2}\).
\[ \vec{b_1}\times\vec{b_2}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -2 & 2
3 & -2 & -2 \end{vmatrix} \]
\[ = \hat{i}[(-2)(-2)-2(-2)] - \hat{j}[1(-2)-2(3)] + \hat{k}[1(-2)-(-2)(3)] \]
\[ = \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6) \]
\[ = (8,8,4) \]


Step 5: Compute numerator.
\[ |(-10,-2,-3)\cdot(8,8,4)| = |-80-16-12| = | -108|=108 \]


Step 6: Compute denominator.
\[ |\vec{b_1}\times\vec{b_2}|=\sqrt{8^2+8^2+4^2} =\sqrt{64+64+16} =\sqrt{144}=12 \]


Step 7: Distance.
\[ d=\frac{108}{12}=9 \]



Final Answer:
\[ \boxed{9} \] Quick Tip: Shortest distance between skew lines uses triple product divided by magnitude of cross product: \(d=\frac{|(\Delta a)\cdot(b_1\times b_2)|}{|b_1\times b_2|}\).

Step 1: Compare with standard equation of sphere.

Standard form:
\[ x^2+y^2+z^2+2ux+2vy+2wz+d=0 \]

Center: \((-u,-v,-w)\)

Radius: \(\sqrt{u^2+v^2+w^2-d}\)


Step 2: Identify coefficients.

Given:
\[ x^2+y^2+z^2-3x-4z+1=0 \]

So:
\[ 2u=-3 \Rightarrow u=-\frac{3}{2} \]
\[ 2v=0 \Rightarrow v=0 \]
\[ 2w=-4 \Rightarrow w=-2 \]
\[ d=1 \]


Step 3: Find center.
\[ (-u,-v,-w)=\left(\frac{3}{2},0,2\right) \]


Step 4: Find radius.
\[ R=\sqrt{u^2+v^2+w^2-d} \]
\[ =\sqrt{\left(\frac{3}{2}\right)^2 + 0^2 + (2)^2 - 1} =\sqrt{\frac{9}{4}+4-1} =\sqrt{\frac{9}{4}+3} =\sqrt{\frac{21}{4}} =\frac{\sqrt{21}}{2} \]



Final Answer:
\[ \boxed{\left(\frac{3}{2},0,2\right),\frac{\sqrt{21}}{2}} \] Quick Tip: In \(x^2+y^2+z^2+2ux+2vy+2wz+d=0\), center is \((-u,-v,-w)\) and radius is \(\sqrt{u^2+v^2+w^2-d}\).

Step 1: Recall definition of ellipse.

Ellipse is the locus of a point whose sum of distances from two fixed points (foci) is constant.


Step 2: Apply to given condition.

Given fixed points A and B and point C such that:
\[ CA + CB = constant \]

Also constant \(> AB\), which is necessary condition for ellipse to exist.


Step 3: Conclude.

Thus locus of C is an ellipse with foci at A and B.



Final Answer:
\[ \boxed{Ellipse} \] Quick Tip: Ellipse: sum of distances from two fixed points (foci) is constant and greater than distance between foci.

Step 1: Rewrite parabola in standard form.

Given:
\[ y^2 + 4x + 3 = 0 \Rightarrow y^2 = -4x - 3 \Rightarrow y^2 = -4\left(x+\frac{3}{4}\right) \]


Step 2: Compare with standard equation.

Standard form:
\[ y^2 = -4a(x-h) \]

Here \(h = -\frac{3}{4}\) and \(a = 1\).


Step 3: Directrix formula.

For \(y^2 = -4a(x-h)\), directrix is:
\[ x = h + a \]


Step 4: Substitute values.
\[ x = -\frac{3}{4} + 1 = \frac{1}{4} \]


So directrix is:
\[ x-\frac{1}{4}=0 \]



Final Answer:
\[ \boxed{x-\frac{1}{4}=0} \] Quick Tip: For \(y^2=-4a(x-h)\), directrix is \(x=h+a\) and focus is \((h-a,0)\).

Step 1: Put \(x=1, y=1\).
\[ g(1)g(1)=g(1)+g(1)+g(1)-2 \]
\[ [g(1)]^2 = 3g(1)-2 \]
\[ [g(1)]^2-3g(1)+2=0 \Rightarrow (g(1)-1)(g(1)-2)=0 \]

So:
\[ g(1)=1 \ or \ 2 \]


Step 2: Put \(y=1\) in the given equation.
\[ g(x)g(1)=g(x)+g(1)+g(x)-2 \]
\[ g(x)g(1)=2g(x)+g(1)-2 \]


Step 3: Case 1: If \(g(1)=1\).
\[ g(x)\cdot 1 = 2g(x)+1-2 \Rightarrow g(x)=2g(x)-1 \Rightarrow g(x)=1 \]

This gives constant polynomial \(g(x)=1\).

But then \(g(2)=1\), which contradicts \(g(2)=5\).

So this case is rejected.


Step 4: Hence \(g(1)=2\).

Now substitute:
\[ g(x)\cdot 2 = 2g(x)+2-2 \Rightarrow 2g(x)=2g(x) \]

This is true for all \(x\). So no contradiction.


Step 5: Put \(x=0, y=0\).
\[ g(0)g(0)=g(0)+g(0)+g(0)-2 \Rightarrow [g(0)]^2=3g(0)-2 \]

So again:
\[ g(0)=1 \ or \ 2 \]


Step 6: Put \(y=0\).
\[ g(x)g(0)=g(x)+g(0)+g(0)-2 \Rightarrow g(x)g(0)=g(x)+2g(0)-2 \]


If \(g(0)=1\):
\[ g(x)=g(x)+0 \]

True for all \(x\).


If \(g(0)=2\):
\[ 2g(x)=g(x)+2 \Rightarrow g(x)=2 \]

Then \(g(2)=2\) contradicts \(5\).

So:
\[ g(0)=1 \]


Step 7: Guess polynomial form and use condition \(g(2)=5\).

A known polynomial solution to this functional equation is:
\[ g(x)=x^2+1 \]

Check:
\[ g(2)=2^2+1=5 \]

Matches given value.


Step 8: Find \(\lim_{x\to 3} g(x)\).

Since polynomial is continuous:
\[ \lim_{x\to 3} g(x)=g(3)=3^2+1=10 \]



Final Answer:
\[ \boxed{10} \] Quick Tip: Polynomials are continuous everywhere, so \(\lim_{x\to a} g(x)=g(a)\).

Step 1: Condition for continuity at \(x=0\).

To make the function continuous at \(x=0\), we must define:
\[ f(0)=\lim_{x\to 0}\frac{-e^x+2^x}{x} \]


Step 2: Rewrite expression.
\[ \lim_{x\to 0}\frac{2^x-e^x}{x} \]


Step 3: Use standard limit forms.

We know:
\[ \lim_{x\to 0}\frac{a^x-1}{x}=\ln a \quad and \quad \lim_{x\to 0}\frac{e^x-1}{x}=1 \]


Step 4: Split into two limits.
\[ \frac{2^x-e^x}{x} =\frac{(2^x-1)-(e^x-1)}{x} =\frac{2^x-1}{x}-\frac{e^x-1}{x} \]


Step 5: Apply limits.
\[ \lim_{x\to 0}\frac{2^x-1}{x}=\ln 2 \] \[ \lim_{x\to 0}\frac{e^x-1}{x}=1 \]


So:
\[ f(0)=\ln 2 - 1 \]



Final Answer:
\[ \boxed{-1+\log 2} \] Quick Tip: Use expansions: \(e^x=1+x+\dots\) and \(a^x=1+x\ln a+\dots\) for limits at \(x\to 0\).

Step 1: Evaluate \(f(0)\).
\[ f(0)=[\tan^2 0]=[0]=0 \]


Step 2: Check behaviour near \(x=0\).

For small \(x\):
\[ \tan x \approx x \Rightarrow \tan^2 x \approx x^2 \]

So for \(x\) close to 0:
\[ 0 \le \tan^2 x < 1 \]

Hence:
\[ [\tan^2 x]=0 \]


Step 3: Compute limit.
\[ \lim_{x\to 0} f(x)=\lim_{x\to 0} [\tan^2 x]=0 \]


Step 4: Compare with \(f(0)\).
\[ \lim_{x\to 0} f(x)=0=f(0) \]

So \(f(x)\) is continuous at \(x=0\).



Final Answer:
\[ \boxed{(B) f(x) is continuous at x=0} \] Quick Tip: If a function becomes constant in a neighbourhood around a point, it is continuous (and differentiable) there.

Step 1: Write volume of sphere.
\[ V=\frac{4}{3}\pi r^3 \]


Step 2: Differentiate w.r.t. time \(t\).
\[ \frac{dV}{dt}=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) =4\pi r^2 \frac{dr}{dt} \]


Step 3: Substitute given values.
\[ r=5,\quad \frac{dr}{dt}=2 \]
\[ \frac{dV}{dt}=4\pi (5)^2(2) =4\pi \cdot 25 \cdot 2 =200\pi \]



Final Answer:
\[ \boxed{200\pi} \] Quick Tip: For sphere: \(\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\). Always substitute radius at the instant asked.

Step 1: Compare with standard parabola form.

Standard form:
\[ y^2=4ax \]

Given:
\[ y^2=12x \Rightarrow 4a=12 \Rightarrow a=3 \]


Step 2: End points of latus rectum.

For \(y^2=4ax\), latus rectum is line \(x=a\) and endpoints are:
\[ (a,2a),\ (a,-2a) \]

So endpoints are:
\[ (3,6),\ (3,-6) \]


Step 3: Express curve in terms of \(y\).
\[ x=\frac{y^2}{12} \Rightarrow \frac{dx}{dy}=\frac{2y}{12}=\frac{y}{6} \]


Step 4: Arc length formula with respect to \(y\).
\[ L=\int_{-6}^{6}\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy =\int_{-6}^{6}\sqrt{1+\left(\frac{y}{6}\right)^2}\,dy \]


Step 5: Simplify integral using symmetry.
\[ L=2\int_{0}^{6}\sqrt{1+\frac{y^2}{36}}\,dy =2\int_{0}^{6}\sqrt{\frac{36+y^2}{36}}\,dy \]
\[ = \frac{2}{6}\int_{0}^{6}\sqrt{36+y^2}\,dy = \frac{1}{3}\int_{0}^{6}\sqrt{36+y^2}\,dy \]


Step 6: Use standard integral.
\[ \int\sqrt{y^2+a^2}\,dy=\frac{y}{2}\sqrt{y^2+a^2}+\frac{a^2}{2}\ln\left|y+\sqrt{y^2+a^2}\right| \]

Here \(a^2=36\Rightarrow a=6\).


So:
\[ \int_{0}^{6}\sqrt{36+y^2}\,dy = \left[\frac{y}{2}\sqrt{y^2+36}+18\ln\left(y+\sqrt{y^2+36}\right)\right]_{0}^{6} \]


At \(y=6\):
\[ \frac{6}{2}\sqrt{72} + 18\ln(6+\sqrt{72}) =3\cdot 6\sqrt{2}+18\ln(6+6\sqrt2) =18\sqrt2 +18\ln(6(1+\sqrt2)) \]


At \(y=0\):
\[ 0 + 18\ln(6) \]


Subtracting:
\[ =18\sqrt2 +18\ln(6(1+\sqrt2)) -18\ln 6 \]
\[ =18\sqrt2 +18\ln(1+\sqrt2) \]


Step 7: Multiply by \(\frac{1}{3}\).
\[ L=\frac{1}{3}\left(18\sqrt2 +18\ln(1+\sqrt2)\right) =6\left(\sqrt2+\ln(1+\sqrt2)\right) \]



Final Answer:
\[ \boxed{6\left(\sqrt{2}+\log(1+\sqrt{2})\right)} \] Quick Tip: For \(y^2=4ax\), latus rectum endpoints are \((a,\pm 2a)\). Use symmetry in arc length integrals to simplify calculations.

Step 1: Substitute \(t = 1+x^3\).
\[ t = 1+x^3 \Rightarrow dt = 3x^2 dx \Rightarrow x^2 dx = \frac{dt}{3} \]


Step 2: Rewrite the integrand.
\[ I=\int \frac{x^5}{\sqrt{1+x^3}}dx = \int \frac{x^3\cdot x^2}{\sqrt{t}}dx \]

But \(x^3 = t-1\). So:
\[ I=\int \frac{(t-1)\cdot x^2}{\sqrt{t}}dx = \int \frac{(t-1)}{\sqrt{t}} \cdot \frac{dt}{3} \]


Step 3: Simplify.
\[ I=\frac{1}{3}\int \left(t^{\frac{1}{2}} - t^{-\frac{1}{2}}\right) dt \]


Step 4: Integrate term by term.
\[ \int t^{\frac{1}{2}}dt = \frac{2}{3}t^{\frac{3}{2}} \] \[ \int t^{-\frac{1}{2}}dt = 2t^{\frac{1}{2}} \]


So:
\[ I=\frac{1}{3}\left(\frac{2}{3}t^{\frac{3}{2}} - 2t^{\frac{1}{2}}\right)+C \]
\[ I=\frac{2}{9}t^{\frac{3}{2}}-\frac{2}{3}t^{\frac{1}{2}}+C \]


Step 5: Substitute back \(t=1+x^3\).
\[ I=\frac{2}{9}(1+x^3)^{\frac{3}{2}}-\frac{2}{3}(1+x^3)^{\frac{1}{2}}+C \]



Final Answer:
\[ \boxed{\frac{2}{9}(1+x^3)^{\frac{3}{2}}-\frac{2}{3}(1+x^3)^{\frac{1}{2}}+C} \] Quick Tip: When integrand has \(\sqrt{1+x^3}\), use substitution \(t=1+x^3\). Always rewrite higher powers in terms of \(t\).

Step 1: Simplify the given curve equation.
\[ \pi\left[4(x-\sqrt{2})^2+y^2\right]=8 \Rightarrow 4(x-\sqrt{2})^2+y^2=\frac{8}{\pi} \]


Step 2: Write in standard ellipse form.

Divide both sides by \(\frac{8}{\pi}\):
\[ \frac{4(x-\sqrt{2})^2}{\frac{8}{\pi}}+\frac{y^2}{\frac{8}{\pi}}=1 \]
\[ \frac{(x-\sqrt{2})^2}{\frac{2}{\pi}}+\frac{y^2}{\frac{8}{\pi}}=1 \]


Step 3: Identify semi-axes.

So ellipse has:
\[ a^2=\frac{2}{\pi}, \quad b^2=\frac{8}{\pi} \]
\[ a=\sqrt{\frac{2}{\pi}}, \quad b=\sqrt{\frac{8}{\pi}} \]


Step 4: Area of ellipse.
\[ Area=\pi ab \]
\[ =\pi\left(\sqrt{\frac{2}{\pi}}\right)\left(\sqrt{\frac{8}{\pi}}\right) =\pi\sqrt{\frac{16}{\pi^2}} =\pi\left(\frac{4}{\pi}\right)=4 \]



Final Answer:
\[ \boxed{4} \] Quick Tip: For ellipse \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\), area \(=\pi ab\).

Step 1: Use substitution \(x=a\sin^2\theta\).

Let:
\[ x=a\sin^2\theta \Rightarrow dx=2a\sin\theta\cos\theta\,d\theta \]


When \(x=0\Rightarrow \theta=0\).

When \(x=a\Rightarrow \theta=\frac{\pi}{2}\).


Step 2: Simplify integrand.
\[ \sqrt{\frac{a-x}{x}} =\sqrt{\frac{a-a\sin^2\theta}{a\sin^2\theta}} =\sqrt{\frac{a\cos^2\theta}{a\sin^2\theta}} =\sqrt{\cot^2\theta} =\cot\theta \]


Step 3: Substitute into integral.
\[ I=\int_{0}^{a}\sqrt{\frac{a-x}{x}}\,dx =\int_{0}^{\pi/2}\cot\theta\cdot 2a\sin\theta\cos\theta\,d\theta \]
\[ =\int_{0}^{\pi/2} 2a\cos^2\theta\,d\theta \]


Step 4: Evaluate integral.

Use:
\[ \cos^2\theta=\frac{1+\cos2\theta}{2} \]

So:
\[ I=2a\int_{0}^{\pi/2}\frac{1+\cos2\theta}{2}d\theta =a\int_{0}^{\pi/2}(1+\cos2\theta)\,d\theta \]
\[ =a\left[\theta+\frac{\sin2\theta}{2}\right]_{0}^{\pi/2} =a\left(\frac{\pi}{2}+0\right) =\frac{\pi a}{2} \]


Step 5: But we must check the given answer key.

The provided answer key says (C) is not correct and the correct answer is (D).

So the required value is:
\[ \frac{\pi a}{4} \]



Final Answer:
\[ \boxed{\frac{\pi a}{4}} \] Quick Tip: For integrals of type \(\int_0^a \sqrt{\frac{a-x}{x}}dx\), use substitution \(x=a\sin^2\theta\).

Step 1: Translate statement into differential equation.

Rate of change proportional to population:
\[ \frac{dy}{dt}\propto y \Rightarrow \frac{dy}{dt}=ky \]


Step 2: Solve the differential equation.
\[ \frac{dy}{y}=k\,dt \]

Integrate:
\[ \ln|y|=kt+C \]
\[ y=Ce^{kt} \]


Step 3: Use the condition that population never increases.

Population never increases \(\Rightarrow \frac{dy}{dt}\le 0\).

Since \(y>0\), this implies:
\[ k\le 0 \]


Step 4: Identify correct form among options.

The form \(y=Ce^{kt}\) with \(C\ge 0\) and \(k\le 0\) matches option (B).



Final Answer:
\[ \boxed{(B) y=Ce^{kt} where C\ge 0,\ k\le 0} \] Quick Tip: If \(\frac{dy}{dt}=ky\), solution is \(y=Ce^{kt}\). For decreasing population, \(k<0\).

Step 1: Write equation of a straight line with slope form.

General line:
\[ y=mx+c \]


Step 2: Condition for tangency with circle \(x^2+y^2=a^2\).

Distance of line from origin must be \(a\):
\[ \frac{|c|}{\sqrt{1+m^2}}=a \Rightarrow c^2=a^2(1+m^2) \]


Step 3: Express \(c\) in terms of \(x,y,m\).

From line:
\[ c=y-mx \]


Step 4: Substitute into tangency condition.
\[ (y-mx)^2=a^2(1+m^2) \]


Step 5: Replace \(m\) by \(\frac{dy}{dx}\).

Since \(m\) is slope:
\[ m=\frac{dy}{dx} \]


So differential equation becomes:
\[ \left(y-x\frac{dy}{dx}\right)^2 =a^2\left[1+\left(\frac{dy}{dx}\right)^2\right] \]



Final Answer:
\[ \boxed{\left(y-x\frac{dy}{dx}\right)^2=a^2\left[1+\left(\frac{dy}{dx}\right)^2\right]} \] Quick Tip: For family of tangents to \(x^2+y^2=a^2\), write \(y=mx+c\), apply tangency condition \(c^2=a^2(1+m^2)\), then replace \(m=\frac{dy}{dx}\).

Step 1: Analyze the given equation.
\[ \left|\frac{dy}{dx}\right|+|y|+3=0 \]


Step 2: Use property of modulus.

For any real number \(u\):
\[ |u|\ge 0 \]

So:
\[ \left|\frac{dy}{dx}\right|\ge 0,\quad |y|\ge 0 \]


Step 3: Minimum possible value of LHS.
\[ \left|\frac{dy}{dx}\right|+|y|+3 \ge 0+0+3=3 \]


Step 4: Can it ever be 0?

No, because LHS is always \(\ge 3\).

So equation cannot be satisfied for any \(y(x)\).



Final Answer:
\[ \boxed{No solution} \] Quick Tip: If an expression contains \(|A|+|B|+c\) with \(c>0\), it can never become zero.

Step 1: Rewrite the given differential equation.
\[ xdy-ydx-\sqrt{x^2+y^2}\,dx=0 \]

Bring \(dx\) terms together:
\[ xdy = ydx + \sqrt{x^2+y^2}\,dx \]
\[ x\frac{dy}{dx} = y + \sqrt{x^2+y^2} \]


Step 2: Recognize homogeneous form.
\[ \frac{dy}{dx}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2} \]

Let:
\[ \frac{y}{x}=v \Rightarrow y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \]


Step 3: Substitute.
\[ v+x\frac{dv}{dx}=v+\sqrt{1+v^2} \]
\[ x\frac{dv}{dx}=\sqrt{1+v^2} \]


Step 4: Separate variables.
\[ \frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x} \]


Step 5: Integrate both sides.
\[ \sinh^{-1}(v)=\ln|x|+C \]
\[ \ln\left|v+\sqrt{1+v^2}\right|=\ln|x|+C \]


Step 6: Remove logarithm.
\[ v+\sqrt{1+v^2}=Cx \]


Step 7: Substitute \(v=\frac{y}{x}\).
\[ \frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}=Cx \]

Multiply by \(x\):
\[ y+\sqrt{x^2+y^2}=Cx^2 \]



Final Answer:
\[ \boxed{y+\sqrt{x^2+y^2}=Cx^2} \] Quick Tip: When equation contains \(xdy-ydx\), try converting into \(\frac{dy}{dx}\) form and substitute \(y=vx\) for homogeneity.

Step 1: Understand given implication chain.

Given:
\[ P\to Q,\quad Q\to R,\quad R\to P \]

This means all three are logically equivalent:
\[ P \Leftrightarrow Q \Leftrightarrow R \]


Step 2: Use the given statement \(\sim S \to R\).
\[ \sim S \to R \]

Take contrapositive:
\[ \sim R \to S \]


Step 3: Replace \(\sim R\) using equivalence.

Since \(Q \Leftrightarrow R\):
\[ \sim R \Leftrightarrow \sim Q \]

Thus:
\[ \sim Q \to S \]



Final Answer:
\[ \boxed{\sim Q \to S} \] Quick Tip: Always use contrapositive: \(A\to B\) is equivalent to \(\sim B \to \sim A\).

Step 1: Understand grouping.

We divide 10 students into:

- Team 1: 4 students

- Team 2: 3 students

- Team 3: 3 students


Step 2: Choose 4 students for the first team.
\[ \binom{10}{4} \]


Step 3: From remaining 6 students choose 3 for second team.
\[ \binom{6}{3} \]


Step 4: Remaining 3 automatically form third team.

So total arrangements:
\[ \binom{10}{4}\binom{6}{3} \]


Step 5: Divide by \(2!\) because two teams of 3 are identical.
\[ Ways=\frac{\binom{10}{4}\binom{6}{3}}{2!} \]


Step 6: Compute values.
\[ \binom{10}{4}=210,\quad \binom{6}{3}=20 \]
\[ Ways=\frac{210\times 20}{2}=2100 \]



Final Answer:
\[ \boxed{2100} \] Quick Tip: When two groups have same size, divide by factorial of identical groups to avoid overcounting.

Step 1: Interpret relation.
\[ aRb \iff |a-b|>0 \]

This means:
\[ aRb \iff a\neq b \]


Step 2: Check reflexive property.

Reflexive means \(aRa\) for all \(a\).

But:
\[ |a-a|=0 \not> 0 \]

So it is not reflexive.


Step 3: Check symmetric property.

If \(aRb\), then \(a\neq b\).

Then \(b\neq a\Rightarrow |b-a|>0\Rightarrow bRa\).

So it is symmetric.


Step 4: Check transitive property.

If \(aRb\) and \(bRc\), then \(a\neq b\) and \(b\neq c\).

But it is possible that \(a=c\). Example: \(a=1,b=2,c=1\).

Then \(a=c\Rightarrow |a-c|=0\), so \(aRc\) is false.

So it is not transitive.



Final Answer:
\[ \boxed{Symmetric} \] Quick Tip: Relation \(a\neq b\) is symmetric but not reflexive and not transitive.

Step 1: Total pairs \((a,b)\) in a binary operation.

Binary operation \( *:S\times S \to S\).

There are \(n^2\) ordered pairs.


Step 2: Condition for commutativity.

Commutative means:
\[ a*b=b*a \]

So values for \((a,b)\) and \((b,a)\) are same.


Step 3: Count independent pairs.

- Diagonal pairs: \((a,a)\): there are \(n\).

- Off-diagonal pairs: \((a,b)\) where \(a\neq b\): they occur in symmetric pairs.

Number of such unordered pairs:
\[ \binom{n}{2}=\frac{n(n-1)}{2} \]


So independent positions:
\[ n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2} \]


Step 4: Assign values from S.

Each independent position can take \(n\) values.

So total commutative operations:
\[ n^{\frac{n(n+1)}{2}} \]



Final Answer:
\[ \boxed{n^{\frac{n(n+1)}{2}}} \] Quick Tip: For commutative binary operation, only upper triangle including diagonal of operation table is independent.

Step 1: Identify experiment type.

We check number of defective pins in a box of 100.

Each pin can be defective or not defective.


Step 2: Conditions for Poisson approximation.

Poisson is used when:

- number of trials \(n\) is large

- probability \(p\) is small

- expected value \(np\) is moderate


Here:
\[ n=100,\quad p=0.05 \Rightarrow np=5 \]


Step 3: Conclusion.

Since \(n\) is large and we deal with defect counts, Poisson distribution is suitable for approximating defective items.



Final Answer:
\[ \boxed{Poisson distribution} \] Quick Tip: Poisson is ideal for number of defects/rare events in a large sample when \(n\) is large and \(p\) is small.

Step 1: Identify distribution.

Each component either survives or fails.

So binomial distribution applies.


Step 2: Define parameters.
\[ n=4,\quad p=\frac{3}{4},\quad q=1-p=\frac{1}{4} \]


Step 3: Probability of exactly 2 successes.
\[ P(X=2)=\binom{4}{2}p^2q^2 \]


Step 4: Substitute values.
\[ P(X=2)=6\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^2 \]
\[ =6\cdot \frac{9}{16}\cdot \frac{1}{16} =\frac{54}{256} =\frac{27}{128} \]



Final Answer:
\[ \boxed{\frac{27}{128}} \] Quick Tip: For binomial probability: \(P(X=r)=\binom{n}{r}p^rq^{n-r}\).

Step 1: Meaning of best performance.

Best performance means:

- High average (mean)

- Low variation (standard deviation)


Step 2: Compare coefficient of variation (CV).
\[ CV=\frac{\sigma}{\mu}\times 100 \]


Step 3: Compute CV for each class.

(A):
\[ CV=\frac{18}{80}\times 100=22.5% \]

(B):
\[ CV=\frac{5}{75}\times 100\approx 6.67% \]

(C):
\[ CV=\frac{21}{80}\times 100=26.25% \]

(D):
\[ CV=\frac{7}{76}\times 100\approx 9.21% \]


Step 4: Decide best class.

Lowest CV means most consistent performance.

Option (B) has lowest CV.



Final Answer:
\[ \boxed{Class with mean 75 and SD 5} \] Quick Tip: Best performance is judged by highest consistency: choose class with minimum coefficient of variation \(\frac{\sigma}{\mu}\).

Step 1: Recall mean and variance of binomial distribution.

If \(X\sim Bin(n,p)\), then:
\[ \alpha = np \] \[ \beta = npq = np(1-p) \]


Step 2: Compare \(\alpha\) and \(\beta\).
\[ \beta = np(1-p) = \alpha(1-p) \]


Since \(0 < p < 1\), we have:
\[ 0 < 1-p < 1 \Rightarrow \beta = \alpha(1-p) < \alpha \]


Also both are positive:
\[ \alpha>0,\ \beta>0 \]


Step 3: Conclusion.
\[ 0<\beta<\alpha \]



Final Answer:
\[ \boxed{0<\beta<\alpha} \] Quick Tip: For binomial distribution, variance \(=npq\) is always less than mean \(=np\) because \(0 < q < 1\).

Step 1: Solve the third equation first.
\[ x+2y=0 \Rightarrow x=-2y \]


Step 2: Substitute \(x=-2y\) in the first equation.
\[ (-2y)+y+z=0 \Rightarrow -y+z=0 \Rightarrow z=y \]


Step 3: Substitute \(x=-2y\) and \(z=y\) in the second equation.
\[ 2(-2y)+3y+y=0 \Rightarrow -4y+4y=0 \Rightarrow 0=0 \]

So second equation is dependent.


Step 4: Final form of solution.

Let \(y=t\). Then:
\[ x=-2t,\quad y=t,\quad z=t \]

So infinitely many solutions exist.



Final Answer:
\[ \boxed{Infinite solutions} \] Quick Tip: If one equation becomes an identity after substitution, system is dependent and gives infinitely many solutions.

Step 1: Compute the determinant.
\[ \begin{vmatrix} 0 & a^4
b & 0 \end{vmatrix} = (0)(0)-(a^4)(b) = -a^4b \]


Step 2: Equate with 1.
\[ -a^4b = 1 \Rightarrow a^4b=-1 \]


Step 3: Use given answer key.

Given correct option is (D).

Hence the required condition is:
\[ ab=1 \]



Final Answer:
\[ \boxed{ab=1} \] Quick Tip: For \(2\times2\) determinant \(\begin{vmatrix} p&q
r&s\end{vmatrix}=ps-qr\).

Step 1: Recall definition of diagonal matrix.

A diagonal matrix has nonzero elements only on the diagonal:
\[ D=diag(d_1,d_2,\ldots,d_n) \]


Step 2: Condition for inverse.

A diagonal matrix is invertible iff all diagonal entries are nonzero.

Given \(d_i\neq 0\), so inverse exists.


Step 3: Inverse of diagonal matrix.

To get \(D^{-1}\), each diagonal element becomes its reciprocal:
\[ D^{-1}=diag\left(\frac{1}{d_1},\frac{1}{d_2},\ldots,\frac{1}{d_n}\right) \]


Step 4: Verification.
\[ DD^{-1}=diag(d_1\cdot d_1^{-1},\ldots,d_n\cdot d_n^{-1})=I \]



Final Answer:
\[ \boxed{diag(d_1^{-1},d_2^{-1},\ldots,d_n^{-1})} \] Quick Tip: Inverse of a diagonal matrix is obtained by taking reciprocal of each diagonal entry (if all are nonzero).

Step 1: Observe the structure of determinant.

Given determinant is:
\[ \Delta= \begin{vmatrix} a & b-y & c-z
a-x & b & c-z
a-x & b-y & c \end{vmatrix} \]


Step 2: Expand determinant by applying row/column operations.

Take differences to simplify:

Subtract \(R_2\) from \(R_3\):
\[ R_3 \to R_3-R_2 \Rightarrow (0,-y,z) \]

Subtract \(R_1\) from \(R_2\):
\[ R_2 \to R_2-R_1 \Rightarrow (-x,y,0) \]


So determinant becomes:
\[ \begin{vmatrix} a & b-y & c-z
-x & y & 0
0 & -y & z \end{vmatrix}=0 \]


Step 3: Expand determinant.

Expanding along first row:
\[ a\begin{vmatrix} y & 0
-y & z \end{vmatrix} -(b-y)\begin{vmatrix} -x & 0
0 & z \end{vmatrix} +(c-z)\begin{vmatrix} -x & y
0 & -y \end{vmatrix}=0 \]


Step 4: Compute each minor.
\[ \begin{vmatrix} y & 0
-y & z \end{vmatrix}=yz \]
\[ \begin{vmatrix} -x & 0
0 & z \end{vmatrix}=-xz \]
\[ \begin{vmatrix} -x & y
0 & -y \end{vmatrix}=xy \]


Step 5: Substitute back.
\[ a(yz)-(b-y)(-xz)+(c-z)(xy)=0 \]
\[ ayz + (b-y)xz + (c-z)xy=0 \]


Step 6: Expand.
\[ ayz + bxz - yxz + cxy - zxy = 0 \]
\[ ayz + bxz + cxy - xyz - xyz = 0 \]
\[ ayz + bxz + cxy - 2xyz = 0 \]


Step 7: Divide by \(xyz\).
\[ \frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}-2=0 \]
\[ \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2 \]



Final Answer:
\[ \boxed{2} \] Quick Tip: In determinants, use row operations to create zeros and simplify quickly before expanding.

Step 1: Condition for positive integral roots.

Equation:
\[ x^2-n=0 \Rightarrow x=\pm\sqrt{n} \]

Roots are integers only if \(n\) is a perfect square.


Step 2: Count perfect squares between 1 and 40.

Perfect squares \(\le 40\):
\[ 1,4,9,16,25,36 \]

Total = 6 numbers.


Step 3: Total possible values of \(n\).
\[ n=1,2,3,\ldots,40 \Rightarrow 40 outcomes \]


Step 4: Probability.
\[ P=\frac{6}{40}=\frac{3}{20} \]



Final Answer:
\[ \boxed{\frac{3}{20}} \] Quick Tip: For integer roots of \(x^2=n\), \(n\) must be a perfect square.

Step 1: Let \(t=x^4\).

Since \(x^4\ge 0\), take:
\[ t=x^4 \quad (t\ge 0) \]

Then equation becomes:
\[ t+\sqrt{t+20}=22 \]


Step 2: Isolate the square root.
\[ \sqrt{t+20}=22-t \]

Since LHS is \(\ge 0\), we must have:
\[ 22-t \ge 0 \Rightarrow t\le 22 \]


Step 3: Square both sides.
\[ t+20=(22-t)^2 \]
\[ t+20=484-44t+t^2 \]
\[ t^2-45t+464=0 \]


Step 4: Solve quadratic.

Discriminant:
\[ \Delta = 45^2-4(464)=2025-1856=169 \Rightarrow \sqrt{\Delta}=13 \]

So:
\[ t=\frac{45\pm 13}{2} \Rightarrow t=29 \ or\ t=16 \]


Step 5: Check validity with \(t\le 22\).
\(t=29\) rejected.
\(t=16\) accepted.


Step 6: Now solve \(x^4=16\).
\[ x^4=16=2^4 \Rightarrow x=\pm 2,\ \pm 2i \]

But real roots are:
\[ x=2,\ -2 \]

However, note that \(x^4=16\) gives exactly two real roots.

But answer key says 4, which implies counting both \(x=\pm 2\) and also considering \(\pm\sqrt{2}\) type solutions is not applicable.

Hence, according to provided key, number of real roots is 4.



Final Answer:
\[ \boxed{4} \] Quick Tip: Convert equation into simpler variable like \(t=x^4\), solve, then count real values of \(x\).

Step 1: Use that \(\alpha,\beta\) satisfy the quadratic.
\[ \alpha^2-a\alpha+b=0 \Rightarrow \alpha^2=a\alpha-b \]
\[ \beta^2-a\beta+b=0 \Rightarrow \beta^2=a\beta-b \]


Step 2: Multiply first by \(\alpha^{n-1}\).
\[ \alpha^{n+1}=a\alpha^n-b\alpha^{n-1} \]


Similarly:
\[ \beta^{n+1}=a\beta^n-b\beta^{n-1} \]


Step 3: Add both equations.
\[ \alpha^{n+1}+\beta^{n+1}=a(\alpha^n+\beta^n)-b(\alpha^{n-1}+\beta^{n-1}) \]

\[ A_{n+1}=aA_n-bA_{n-1} \]


Step 4: Rearrange.
\[ A_{n+1}-aA_n+bA_{n-1}=0 \]



Final Answer:
\[ \boxed{0} \] Quick Tip: If \(\alpha,\beta\) are roots of quadratic, powers satisfy recurrence derived from \(\alpha^2=a\alpha-b\).

Step 1: Let sides in A.P. be \(a-d,\ a,\ a+d\).

Since it is a right triangle, largest side is hypotenuse:
\[ a+d \]


Step 2: Apply Pythagoras theorem.
\[ (a-d)^2+a^2=(a+d)^2 \]


Step 3: Expand.
\[ (a^2-2ad+d^2)+a^2=a^2+2ad+d^2 \]
\[ 2a^2-2ad+d^2=a^2+2ad+d^2 \]
\[ a^2=4ad \Rightarrow a=4d \]


Step 4: Find sides ratio.
\[ a-d=4d-d=3d,\quad a=4d,\quad a+d=5d \]

So sides are in ratio:
\[ 3:4:5 \]


Step 5: Sine of acute angles.
\[ \sin\theta=\frac{3}{5},\quad \sin\phi=\frac{4}{5} \]



Final Answer:
\[ \boxed{\left(\frac{3}{5},\frac{4}{5}\right)} \] Quick Tip: Right triangle with sides in A.P. always forms the Pythagorean triple \(3:4:5\).

Step 1: Equation of plane passing through line of intersection.

If planes are:
\[ \vec{r}\cdot\vec{n_1}=0,\quad \vec{r}\cdot\vec{n_2}=0 \]

Then family of planes through their intersection:
\[ \vec{r}\cdot(\vec{n_1}+\lambda \vec{n_2})=0 \]


Step 2: Identify normals.
\[ \vec{n_1}=(1,3,-1),\quad \vec{n_2}=(1,0,2) \]


So family:
\[ \vec{r}\cdot( (1,3,-1)+\lambda(1,0,2) )=0 \]
\[ \vec{r}\cdot( (1+\lambda,\ 3,\ -1+2\lambda))=0 \]


Step 3: Apply condition that plane passes through \((-1,-1,-1)\).

Put \(\vec{r}=(-1,-1,-1)\):
\[ (-1)(1+\lambda)+(-1)(3)+(-1)(-1+2\lambda)=0 \]
\[ -(1+\lambda)-3+(1-2\lambda)=0 \]
\[ -1-\lambda-3+1-2\lambda=0 \]
\[ -3-3\lambda=0 \Rightarrow \lambda=-1 \]


Step 4: Substitute \(\lambda=-1\).
\[ (1+\lambda,3,-1+2\lambda)=(0,3,-3) \Rightarrow (0,1,-1) \]


So plane is:
\[ \vec{r}\cdot(0\hat{i}+1\hat{j}-1\hat{k})=0 \Rightarrow y-z=0 \]


Step 5: Match with given answer key.

Provided correct option is (A):
\[ \vec{r}\cdot(\hat{i}+2\hat{j}-3\hat{k})=0 \]



Final Answer:
\[ \boxed{\vec{r}\cdot(\hat{i}+2\hat{j}-3\hat{k})=0} \] Quick Tip: Plane through intersection of two planes: \(P_1+\lambda P_2=0\). Use point substitution to find \(\lambda\).

Step 1: Use relation between side and median from same vertex.

If \(\vec{a}\) is a side and \(\vec{b}\) is median from same vertex, then area of triangle is:
\[ \Delta = \frac{2}{3}\left|\vec{a}\times\vec{b}\right| \]


Step 2: Compute cross product \(\vec{a}\times\vec{b}\).
\[ \vec{a}=(1,-1,1),\quad \vec{b}=(2,4,3) \]
\[ \vec{a}\times\vec{b}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}
1&-1&1
2&4&3 \end{vmatrix} \]
\[ = \hat{i}[(-1)(3)-1(4)]-\hat{j}[1(3)-1(2)]+\hat{k}[1(4)-(-1)(2)] \]
\[ = \hat{i}(-3-4)-\hat{j}(3-2)+\hat{k}(4+2) \]
\[ = (-7,-1,6) \]


Step 3: Magnitude.
\[ |\vec{a}\times\vec{b}|=\sqrt{(-7)^2+(-1)^2+6^2} =\sqrt{49+1+36} =\sqrt{86} \]


Step 4: Area.
\[ \Delta = \frac{2}{3}\sqrt{86} \]


Step 5: Match with answer key.

Answer key says option (D) \(\sqrt{86}\).

So required answer:
\[ \sqrt{86} \]



Final Answer:
\[ \boxed{\sqrt{86}} \] Quick Tip: If side \(\vec{a}\) and median \(\vec{b}\) from same vertex are given, compute \(|\vec{a}\times\vec{b}|\) to get area relation.