5 Apr 2024 Shift 2

We are given the momentum vector at time t as $\vec{p}(t)=\hat{i}\cos(kt)-\hat{j}\sin(kt)$, and we wish to find the angle between the force $\vec{F}$ and the momentum $\vec{p}$. Since Newton's second law states that the force is the time derivative of the momentum, differentiating $\vec{p}(t)$ with respect to t yields $\vec{F}=\frac{d\vec{p}}{dt}=-k\hat{i}\sin(kt)-k\hat{j}\cos(kt)$. Next we compute the dot product of $\vec{F}$ and $\vec{p}$, which gives $\vec{F}\cdot\vec{p}=(-k\sin kt)(\cos kt)+(-k\cos kt)(-\sin kt)=-k\sin kt\cos kt+k\cos kt\sin kt=0$. Because the dot product is zero, the vectors are perpendicular and hence the angle between them is $\frac{\pi}{2}$. Therefore, the correct answer is Option (3): $\frac{\pi}{2}$.

$a/V^2$ has units of pressure: $[a]=PV^2=ML^{-1}T^{-2}\cdot L^6=ML^5T^{-2}$. $[b]=V=L^3$. $[a/b]=ML^5T^{-2}/L^3=ML^2T^{-2}$. The answer is Option (3): $ML^2T^{-2}$.

120 rev in 3 min = 40 rev/min = 2/3 rev/s. ω=2π·2/3=4π/3 rad/s. a=ω²r=(4π/3)²·9=16π²/9·9=16π² m/s². The answer is Option (4): 16π² m/s².

Find the kinetic friction force on a 50 kg box on a horizontal surface with $\mu_k=0.3$. The normal force on the horizontal surface is $N = mg = 50 \times 9.8 = 490$ N. From this, the kinetic friction force is $f_k = \mu_k N = 0.3 \times 490 = 147$ N. The correct answer is Option (2): 147 N .

We need to find how displacement depends on time for a body under constant power. The constant power relation is $P = Fv = mav = \text{constant}$ so $ma \cdot v = P \Rightarrow m\frac{dv}{dt} \cdot v = P$ which gives $mv \, dv = P \, dt$. Integrating: $\frac{mv^2}{2} = Pt$ (starting from rest) and thus $v = \sqrt{\frac{2Pt}{m}}$. Using $v = \frac{ds}{dt} = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$ we find $s = \sqrt{\frac{2P}{m}} \int_0^t t^{1/2} dt = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3}t^{3/2}$, so $s \propto t^{3/2}$. Displacement is proportional to $t^{3/2}$, which matches Option B. Therefore, the answer is Option B.

$T_p=6$h, $M_p=M_E/4$. For geostationary orbit: $T²\propto r³/M$. $r_p³=\frac{M_p}{M_E}\cdot\frac{T_p²}{T_E²}\cdot r_E³=\frac{1}{4}\cdot\frac{36}{576}\cdot r_E³=\frac{1}{4}\cdot\frac{1}{16}\cdot r_E³=\frac{r_E³}{64}$. $r_p=r_E/4=4.2\times10^4/4=1.05\times10^4$ km. The answer is Option (2): $1.05\times10^4$ km.

Match each definition carefully. (A) "A force that restores an elastic body of unit area to its original state" Force per unit area developed internally to restore original shape is definition of stress. So A→(III) (B) "Two equal and opposite forces parallel to opposite faces" Parallel tangential forces produce shear deformation, associated with shear modulus. So B→(IV) (C) "Forces perpendicular everywhere to the surface per unit area same everywhere" This describes uniform normal pressure acting in all directions, associated with bulk modulus. So C→(I) (D) "Two equal and opposite forces perpendicular to opposite faces" Normal tensile/compressive stress produces longitudinal strain, associated with Young's modulus. So D→(II)

For an adiabatic process, $PV^{\gamma}=\text{constant}$ Also ideal gas law gives PV=nRT or $V=\frac{nRT}{P}$ Substitute into adiabatic equation: $P\left(\frac{nRT}{P}\right)^{\gamma}=\text{constant}$ $P\cdot\frac{(nR)^{\gamma}T^{\gamma}}{P^{\gamma}}=\text{constant}$ $P^{1-\gamma}T^{\gamma}=\text{constant}$ Given pressure is proportional to cube of absolute temperature, $P\propto T^3$ So $P=kT^3$ Comparing with $P^{1-\gamma}T^{\gamma}=\text{constant}$ write $P\propto T^{\frac{\gamma}{\gamma-1}}$ Thus, $=3\frac{\gamma}{\gamma-1}$ Solving, $\gamma =3\gamma −3$ $2\gamma =3$ $\gamma =\frac{3}{2}$ But for an adiabatic process so, $\gamma=\frac{C_P}{C_V}$

Mean free path is $\lambda =\frac{1}{\sqrt{\ 2n\sigma }}$ where n = number density \sigma \sigma\sigma = collision cross section For molecules of diameter d, $\sigma=\pi d^2$ So $\lambda =\frac{1}{\sqrt{\ 2n\pi\ d^2}}$

This question asks why vehicles carrying inflammable (flammable) fluids have metallic chains that touch the ground. When a vehicle moves along a road, friction between the tyres and the road surface, as well as friction between the vehicle body and the air, causes a build-up of static electric charge on the vehicle's metallic body. This is similar to how rubbing a balloon on cloth creates static charge. For ordinary vehicles, this small accumulation of charge is not a major concern. However, for vehicles carrying inflammable fluids (such as petroleum tankers), even a tiny spark can ignite the vapours and cause a catastrophic explosion. A spark occurs when the accumulated charge reaches a high enough potential difference to ionise the surrounding air and discharge suddenly. To prevent this, metallic chains are attached to the vehicle so that they drag along the ground. Since metals are good conductors of electricity, the chain provides a continuous conducting path from the vehicle body to the ground (earth). This allows the excess charge to flow harmlessly to the earth as it builds up, keeping the vehicle at ground potential and preventing any dangerous accumulation of charge. This is an application of the principle of earthing or grounding , which is widely used in electrical safety. Therefore, the correct answer is Option (4) : To conduct excess charge due to air friction to ground and prevent sparking.

We are given a galvanometer of resistance $G = 100 \, \Omega$ that, when connected \in series with a resistance of $400 \, \Omega$, measures a voltage of up to $10$ V. The full-scale deflection current of the galvanometer, $I_g$, can be found as follows. When the galvanometer is used as a voltmeter with a series resistance $R_s = 400 \, \Omega$, the total resistance \in the circuit is: $R_{total} = G + R_s = 100 + 400 = 500 \, \Omega$ Using Ohm's law, the full-scale deflection current is: $I_g = \frac{V}{R_{total}} = \frac{10}{500} = 0.02 \text{ A}$ To convert a galvanometer into an ammeter, a low resistance (called a shunt, $S$) is connected \in parallel with the galvanometer. The shunt diverts most of the current, allowing only $I_g$ to pass through the galvanometer. The formula for the shunt resistance is: $S = \frac{I_g \cdot G}{I - I_g}$ This comes from the condition that the voltage across the galvanometer equals the voltage across the shunt: $I_g \cdot G = (I - I_g) \cdot S$. Here, $I = 10$ A (the desired full-scale reading of the ammeter), $I_g = 0.02$ A, and $G = 100 \, \Omega$: $S = \frac{0.02 \times 100}{10 - 0.02} = \frac{2}{9.98} \approx 0.2004 \, \Omega$ Expressed \in the required form, $S \approx 0.2 \, \Omega = 20 \times 10^{-2} \, \Omega$. Therefore, $x = 20$. The correct answer is Option (3): 20 .

$P = \frac{V^2}{R}$ Let $H_1$ be the heat dissipated per second \in the $5\Omega$ resistor and $H_2$ be the heat dissipated \in the $10\Omega$ resistor, $\text{Ratio} = \frac{H_1}{H_2} = \frac{V^2 / 5}{V^2 / 10}$ (Voltage difference is same for both as they are \in parallel) $\frac{H_1}{H_2} = \frac{10}{5} = \frac{2}{1}$

This question asks for the correct expressions for the electrostatic force and magnetic force acting on a charge $q$ moving with velocity $\vec{v}$. Recall that the electrostatic force on a charge $q$ \in an electric field $\vec{E}$ is given by: $\vec{F_1} = q\vec{E}$ This force acts \in the direction of $\vec{E}$ for a positive charge and opposite to $\vec{E}$ for a negative charge. Importantly, this force does not depend on the velocity of the charge. The magnetic force on a charge $q$ moving with velocity $\vec{v}$ \in a magnetic field $\vec{B}$ is given by: $\vec{F_2} = q(\vec{v} \times \vec{B})$ Key features of this force: - It involves the cross product of velocity and magnetic field, not a dot product. - The order matters: it is $\vec{v} \times \vec{B}$, not $\vec{B} \times \vec{v}$ (they differ by a negative sign). - This force is always perpendicular to both $\vec{v}$ and $\vec{B}$, so it does no work on the charge. - If the charge is stationary ($\vec{v} = 0$), the magnetic force is zero. Comparing with the given options shows that option (1) gives $\vec{F_1} = q\vec{E}$ and $\vec{F_2} = q(\vec{V} \times \vec{B})$, which matches both formulas correctly. Option (2) incorrectly uses $\vec{B}$ for the electrostatic force. Option (3) has the cross product \in the wrong order ($\vec{B} \times \vec{V}$). Option (4) incorrectly uses dot products instead of cross products. The correct answer is Option (1) : $\vec{F_1} = q\vec{E}$, $\vec{F_2} = q(\vec{V} \times \vec{B})$.

We are given a series LCR circuit with an AC signal of $200$ V and $50$ Hz. The inductor has inductance $L = 10$ mH, and the voltage across the inductor is $V_L = 31.4$ V. We need to find the current \in the circuit. Recall that the voltage across an inductor \in an AC circuit is related to the current by $V_L = I \times X_L$, where $X_L$ is the inductive reactance. The inductive reactance is given by the formula $X_L = \omega L = 2\pi f L$. Substituting the given values ($f = 50$ Hz, $L = 10 \text{ mH} = 10 \times 10^{-3}$ H) yields $X_L = 2\pi \times 50 \times 10 \times 10^{-3}$, so $X_L = 2\pi \times 0.5 = \pi \approx 3.14 \, \Omega$. Using the relationship $V_L = I \times X_L$ and solving for $I$ gives $I = \frac{V_L}{X_L} = \frac{31.4}{3.14} = 10 \text{ A}$. The correct answer is Option (3): 10 A .

The standard wavelength ranges for the given electromagnetic waves are:

1. Infra-red radiation: Extends from approximately $700 \text{ nm}$ to $1 \text{ mm}$. This matches (III) $1 \text{ mm to } 700 \text{ nm}$.
2. Ultraviolet radiation: Ranges from approximately $1 \text{ nm}$ to $400 \text{ nm}$. This matches (II) $400 \text{ nm to } 1 \text{ nm}$.
3. X-rays: Wavelengths typically fall between $0.01 \text{ nm}$ and $10 \text{ nm}$. The range (IV) $1 \text{ nm to } 10^{-3} \text{ nm}$ (i.e., $1 \text{ nm}$ to $0.001 \text{ nm}$) lies within the X-ray spectrum.
4. Gamma rays: Have the shortest wavelengths, generally less than $0.01 \text{ nm}$. This matches (I) $< 10^{-3} \text{ nm}$ (i.e., less than $0.001 \text{ nm}$).

Based on these matches, assuming the List-I items for matching are: (A) Infra-red, (B) Ultraviolet, (C) X-rays, (D) Gamma rays, we have:
(A) - (III)
(B) - (II)
(C) - (IV)
(D) - (I)

This corresponds to option A.

We need to evaluate two statements about the dispersion of white light through a prism. Analysis of Statement I: "When white light passes through a prism, the red light bends lesser than yellow and violet." This statement is true . When white light undergoes dispersion through a prism, different colours bend (refract) by different amounts. The degree of bending depends on the refractive index, which in turn depends on the wavelength. Red light has the longest wavelength in the visible spectrum and therefore has the smallest refractive index in a dispersive medium. Since a smaller refractive index means less bending (by Snell's law: $n_1 \sin \theta_1 = n_2 \sin \theta_2$), red light bends the least. The order of increasing bending is: red, orange, yellow, green, blue, indigo, violet (ROYGBIV). Analysis of Statement II: "The refractive indices are different for different wavelengths \in dispersive medium." This statement is also true . This is the fundamental property that defines a dispersive medium. In such a medium, the refractive index is a function of wavelength, described by Cauchy's equation: $n(\lambda) = A + \frac{B}{\lambda^2} + \frac{C}{\lambda^4} + \ldots$ where $A$, $B$, $C$ are material-dependent constants. Since $n$ decreases with increasing $\lambda$, violet light (shortest wavelength) has the highest refractive index and red light (longest wavelength) has the lowest. This variation of refractive index with wavelength is the cause of dispersion. Conclusion: Both statements are true, and Statement II provides the underlying reason for the phenomenon described in Statement I. The correct answer is Option (3) : Both Statement I and Statement II are true.

Question: Which statement about stopping potential ($V_0$) is NOT true? In the photoelectric effect, the stopping potential $V_0$ is the minimum reverse potential needed to stop the most energetic photoelectrons. From Einstein's photoelectric equation: $eV_0 = h\nu - \phi$ where $h$ is Planck's constant, $\nu$ is the frequency of incident light, and $\phi$ is the work function of the metal. Option 1: "It is $\frac{1}{e}$ \times the maximum kinetic energy of electrons emitted." Since $V_0 = \frac{KE_{max}}{e}$, this is true by definition. Option 2: "It increases with increase \in intensity of the incident light." The stopping potential depends only on the frequency of light, not its intensity. Increasing intensity increases the number of photoelectrons (photocurrent) but does not change the maximum kinetic energy or stopping potential. This statement is NOT true. Option 3: "It depends on the nature of emitter material." From $V_0 = \frac{h\nu - \phi}{e}$, and since the work function $\phi$ depends on the material, $V_0$ does depend on the emitter material. This is true. Option 4: "It depends upon frequency of the incident light." From the equation, $V_0$ clearly depends on $\nu$. This is true. The statement that is NOT true is Option 2 .

We need to find how the angular momentum of an electron in a hydrogen atom depends on the radius of its orbit, using the Bohr model. In Bohr's model, the angular momentum of the electron in the $n$-th orbit is quantised: $L = mvr = n\hbar = \frac{nh}{2\pi}$, where $n$ is the principal quantum number ($n = 1, 2, 3, \ldots$), $m$ is the electron mass, $v$ is the orbital speed, and $r$ is the orbital radius. This tells us that $L \propto n$. In the Bohr model, the radius of the $n$-th orbit for hydrogen is given by $r_n = a_0 n^2$, where $a_0 = 0.529$ Angstrom is the Bohr radius, which shows that $r \propto n^2$. From $r \propto n^2$, we can write $n \propto \sqrt{r}$. Since $L \propto n$ and $n \propto \sqrt{r}$, it follows that $L \propto \sqrt{r}$. The correct answer is Option (2): $\sqrt{r}$ .

The logic circuit consists of two initial gates feeding into a final OR gate.

1. The top OR gate takes inputs A and B, producing $A+B$.
2. The bottom AND gate takes inputs B and 1, producing $B \cdot 1 = B$.
3. The final OR gate takes the outputs of the previous two gates, $(A+B)$ and $B$, as its inputs.

Thus, the final output Y is:
$Y = (A+B) + B$
Applying the associative and idempotent laws of Boolean algebra ($X+X=X$):
$Y = A + (B+B) = A+B$
The output Y is 0 only when both inputs to the OR gate are 0. Therefore, $A+B=0$ implies $A=0$ and $B=0$.

We are given a vernier callipers where 20 divisions on the vernier scale coincide with 19 divisions on the main scale, and the least count is 0.1 mm. We need to find the value of one main scale division (MSD). The given condition is $20 \text{ VSD} = 19 \text{ MSD}$ which implies $1 \text{ VSD} = \frac{19}{20} \text{ MSD}$. The least count (LC) is defined as the difference between one main scale division and one vernier scale division: $\text{LC} = 1 \text{ MSD} - 1 \text{ VSD}$. Substituting the expression for VSD from the previous relation gives $\text{LC} = 1 \text{ MSD} - \frac{19}{20} \text{ MSD} = \text{MSD} \left(1 - \frac{19}{20}\right) = \frac{1}{20} \text{ MSD}$. Since the least count is given as $\text{LC} = 0.1$ mm, one can write $0.1 = \frac{1}{20} \times \text{MSD}$ which yields $\text{MSD} = 0.1 \times 20 = 2 \text{ mm}$. Therefore, one main scale division is equal to 2 mm . The correct answer is Option (2): 2 .

We are told that the maximum height reached by a projectile is 64 m, and we need to find the new maximum height when the initial velocity is halved. Since the maximum height depends on the vertical component of motion, we recall that for a projectile launched with initial velocity $u$ at angle $\theta$ with the horizontal, the maximum height is given by $H = \frac{u^2 \sin^2\theta}{2g}$. This formula follows from the condition that at maximum height the vertical velocity becomes zero; applying $v^2 = u_y^2 - 2gH$ with $v = 0$ and $u_y = u\sin\theta$ yields $0 = u^2\sin^2\theta - 2gH \implies H = \frac{u^2\sin^2\theta}{2g}$. From this expression, it is clear that with $\theta$ and $g$ held constant, the maximum height is proportional to the square of the initial velocity, $H \propto u^2$. Therefore, if the initial velocity is halved ($u' = u/2$), the new maximum height becomes $H' = \frac{(u/2)^2 \sin^2\theta}{2g} = \frac{u^2 \sin^2\theta}{4 \times 2g} = \frac{H}{4}$. Substituting the known value $H = 64$ m gives $H' = \frac{64}{4} = 16 \text{ m}$. The answer is 16 m.

We need to find the ratio of rotational kinetic energy to total kinetic energy for a hollow sphere rolling on a plane surface about its axis of symmetry. Since the sphere is thin-walled, its moment of inertia about an axis through its centre is $I = \frac{2}{3}mr^2$, where $m$ is the mass and $r$ is the radius. For rolling without slipping with velocity $v$ and angular velocity $\omega = v/r$, the translational kinetic energy is $K_{trans} = \frac{1}{2}mv^2$ and the rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{2}{3}mr^2 \cdot \frac{v^2}{r^2} = \frac{1}{3}mv^2$. Next, the total kinetic energy is $K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = mv^2\bigl(\tfrac{1}{2} + \tfrac{1}{3}\bigr) = mv^2 \cdot \frac{5}{6} = \frac{5}{6}mv^2$. From this, the ratio of rotational to total kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3}mv^2}{\frac{5}{6}mv^2} = \frac{\frac{1}{3}}{\frac{5}{6}} = \frac{1}{3} \times \frac{6}{5} = \frac{2}{5}$. Since this ratio equals $\frac{x}{5}$, we obtain $x = 2$. The answer is 2 .

For the hydraulic press to be in equilibrium, the pressure in both arms must be equal (Pascal's Law) $P_1 = P_2 \implies \frac{F_1}{A_1} = \frac{F_2}{A_2}$ Since the cross-sectional area $A$ is proportional to the square of the diameter ($A = \pi \frac{d^2}{4}$), $F_2 = F_1 \times \left( \frac{d_2}{d_1} \right)^2$ $F_2 = 10 \times \left( \frac{14}{1.4} \right)^2$ $F_2 = 10 \times (10)^2$ $F_2 = 10 \times 100 = 1000\text{ N}$

We are given that a sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz under a certain tension. We need to find the resonating length when the fundamental frequency is 600 Hz under the same tension. Recall that the fundamental frequency of a string fixed at both ends is given by: $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ Here $L$ is the vibrating length, $T$ is the tension, and $\mu$ is the linear mass density of the wire. Since both $T$ and $\mu$ remain constant, it follows that $f \propto \frac{1}{L}$, which means $f \times L$ is constant, or $f_1 L_1 = f_2 L_2$. Substituting the given values $f_1 = 400$ Hz, $L_1 = 90$ cm, and $f_2 = 600$ Hz gives $400 \times 90 = 600 \times L_2$ From this, $L_2 = \frac{400 \times 90}{600} = \frac{36000}{600} = 60 \text{ cm}$ The answer is 60 cm.

Dipole moment magnitude: $p=q(2r)=2qr$ At point P on axial line, distance from center is $OP=r$ Axial field of dipole: $E_P=\frac{2kp}{r^3}$ Given this is E so $E=\frac{2kp}{r^3}$ Now point R lies on equatorial line at distance OR=2r Equatorial field of dipole: Then $E_R=\frac{kp}{(2r)^3}$ $=\frac{kp}{8r^3}$ Using $E=\frac{2kp}{r^3}$ we get $E_R=\frac{1}{16}E$ So $\frac{E}{x}=\frac{E}{16}$

Wire 20Ω divided into 10 parts: each 2Ω. Pairs in parallel: 2||2=1Ω. 5 such pairs in series: 5Ω. The answer is $\boxed{5}$.

We need to find the number of turns $m$ \in a solenoid, given its dimensions and the magnetic field it produces. The magnetic field inside a long solenoid is given by $B = \mu_0 n I = \mu_0 \frac{m}{L} I$ where $\mu_0 = 4\pi \times 10^{-7}$ T m/A is the permeability of free space, $n = m/L$ is the number of turns per unit length, $m$ is the total number of turns, $L$ is the length of the solenoid, and $I$ is the current. Substituting the given values $B = 6.28 \times 10^{-3}$ T, $L = 0.5$ m, and $I = 5$ A into this formula gives $6.28 \times 10^{-3} = 4\pi \times 10^{-7} \times \frac{m}{0.5} \times 5$. This simplifies to $6.28 \times 10^{-3} = 4\pi \times 10^{-7} \times 10m$ and hence $6.28 \times 10^{-3} = 4\pi \times 10^{-6} \times m$. Using $\pi \approx 3.14$, we have $6.28 \times 10^{-3} = 4 \times 3.14 \times 10^{-6} \times m = 12.56 \times 10^{-6} \times m$. From this, $m = \frac{6.28 \times 10^{-3}}{12.56 \times 10^{-6}} = \frac{6.28}{12.56} \times 10^{3} = 0.5 \times 10^{3} = 500$. The answer is 500 .

We are given that the current in an inductor is $I = (3t + 8)$ A (where $t$ is \in seconds) and the magnitude of the induced EMF is $12$ mV, and we need to find the self-inductance. Since the EMF induced \in an inductor is given by $|emf| = L \frac{dI}{dt}$ where $L$ is the self-inductance and $\frac{dI}{dt}$ is the rate of change of current, we first compute this rate. For $I = 3t + 8$, differentiating with respect to time gives $\frac{dI}{dt} = \frac{d}{dt}(3t + 8) = 3 \text{ A/s}$, so the rate of change of current is constant at 3 A/s. Substituting into the EMF formula yields $12 \times 10^{-3} = L \times 3$, which gives $L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \text{ H} = 4 \text{ mH}$. The answer is 4 mH.

We need to find the position of the first order minima in a single slit diffraction experiment. Recall the condition for minima in single slit diffraction For a single slit of width $d$, the condition for the $n$-th order minima is: $d \sin\theta = n\lambda$ where $\lambda$ is the wavelength and $n = 1, 2, 3, \ldots$ Use the small angle approximation For small angles (which is valid when the slit width is much larger than the wavelength), $\sin\theta \approx \tan\theta = \frac{y}{D}$, where $y$ is the distance of the minima from the central maximum and $D$ is the distance to the screen. Substituting into the minima condition: $d \cdot \frac{y}{D} = n\lambda$ $y = \frac{n\lambda D}{d}$ Substitute the given values for the first order minimum ($n = 1$) Given: $\lambda = 550$ nm $= 550 \times 10^{-9}$ m, $D = 100$ cm $= 1$ m, $d = 0.20$ mm $= 0.20 \times 10^{-3}$ m $= 2 \times 10^{-4}$ m. $y = \frac{1 \times 550 \times 10^{-9} \times 1}{2 \times 10^{-4}}$ $y = \frac{550 \times 10^{-9}}{2 \times 10^{-4}} = 275 \times 10^{-5} \text{ m}$ Therefore, $x = 275$. The answer is 275 .

We are given that the shortest wavelength of the Lyman series is 915 Angstrom, and we need to find the longest wavelength of the Balmer series. Recall the Rydberg formula for hydrogen spectral lines The wavelength of spectral lines in hydrogen is given by: $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ where $R$ is the Rydberg constant, $n_1$ is the lower energy level, and $n_2$ is the upper energy level. Find the Rydberg constant from the Lyman series data The Lyman series corresponds to transitions where $n_1 = 1$. The shortest wavelength (series limit) occurs when $n_2 \to \infty$: $\frac{1}{\lambda_{\infty }} = R\left(\frac{1}{1^2} - \frac{1}{\infty }\right) = R$ Given $\lambda_{\infty } = 915$ Angstrom: $R = \frac{1}{915} \text{ per Angstrom}$ Calculate the longest wavelength of the Balmer series The Balmer series corresponds to transitions where $n_1 = 2$. The longest wavelength (smallest energy transition) occurs when $n_2 = 3$ (the closest upper level): $\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right)$ Finding the common denominator: $\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}$ Therefore: $\frac{1}{\lambda} = R \times \frac{5}{36} = \frac{1}{915} \times \frac{5}{36}$ $\lambda = \frac{36}{5R} = \frac{36 \times 915}{5} = \frac{32940}{5} = 6588 \text{ Angstrom}$ The answer is 6588 Angstrom.

We need to find the number of moles of methane required to produce 11 g of $CO_2$ after complete combustion. Write the balanced equation for the complete combustion of methane $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ From the balanced equation, 1 mole of $CH_4$ produces 1 mole of $CO_2$. This is a 1:1 molar ratio. Calculate the moles of $CO_2$ produced The molar mass of $CO_2 = 12 + 2(16) = 44$ g/mol. $\text{Moles of } CO_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{11}{44} = 0.25 \text{ mol}$ Use the stoichiometric ratio to find moles of $CH_4$ Since the molar ratio of $CH_4$ to $CO_2$ is 1:1: $\text{Moles of } CH_4 = \text{Moles of } CO_2 = 0.25 \text{ mol}$ Verification: 0.25 mol of $CH_4 = 0.25 \times 16 = 4$ g of methane, which upon complete combustion produces 0.25 mol of $CO_2 = 0.25 \times 44 = 11$ g. This matches the given data. The correct answer is Option (4): 0.25 .

Statement I: The metallic radius of Na is 1.86 A and the ionic radius of $Na^+$ is less than 1.86 A. Statement II: Ions are always smaller \in size than the corresponding elements. Evaluation of Statement I: When sodium loses an electron to form $Na^+$, it loses the entire outermost shell (3s). The resulting ion has the electronic configuration of neon, with a much smaller radius. The ionic radius of $Na^+$ is approximately 0.95 A, which is indeed less than 1.86 A. Statement I is correct . Evaluation of Statement II: This statement claims ions are always smaller than their parent atoms. This is true for cations (positive ions lose electrons, reducing electron-electron repulsion and the outermost shell). However, for anions (negative ions), the addition of electrons increases electron-electron repulsion and the ionic radius is larger than the atomic radius. For example, $Cl^-$ (1.81 A) is larger than Cl (0.99 A). Therefore, Statement II is false . Statement I is correct but Statement II is false, which corresponds to Option 4 .

To determine the shape of each molecule, we use VSEPR theory by calculating the number of bond pairs (BP) and lone pairs (LP) around the central atom.

1. ICl (A): This is a diatomic molecule. All diatomic molecules have a linear shape. Thus, (A) matches with (IV).

2. ICl$_3$ (B): Iodine (I) is the central atom. It has 7 valence electrons. It forms 3 bonds with Cl atoms.
   Number of bond pairs = 3.
   Number of lone pairs = $(7 - 3)/2 = 2$.
   Total electron pairs = $3 + 2 = 5$. For 5 electron pairs (3 BP, 2 LP), the molecular geometry is T-shape. Thus, (B) matches with (I).

3. ClF$_5$ (C): Chlorine (Cl) is the central atom. It has 7 valence electrons. It forms 5 bonds with F atoms.
   Number of bond pairs = 5.
   Number of lone pairs = $(7 - 5)/2 = 1$.
   Total electron pairs = $5 + 1 = 6$. For 6 electron pairs (5 BP, 1 LP), the molecular geometry is square pyramidal. Among the given options, (II) is "pyramidal," which includes square pyramidal. Thus, (C) matches with (II).

4. IF$_7$ (D): Iodine (I) is the central atom. It has 7 valence electrons. It forms 7 bonds with F atoms.
   Number of bond pairs = 7.
   Number of lone pairs = $(7 - 7)/2 = 0$.
   Total electron pairs = $7 + 0 = 7$. For 7 electron pairs (7 BP, 0 LP), the molecular geometry is pentagonal bipyramidal. Thus, (D) matches with (III).

Matching these results:
(A) - (IV)
(B) - (I)
(C) - (II)
(D) - (III)

This corresponds to option A.

The final answer is $\boxed{A}$

Assertion (A): $NH_3$ and $NF_3$ both have pyramidal shapes with a lone pair on nitrogen. The dipole moment of $NH_3$ is greater than that of $NF_3$. Reason (R): In $NH_3$, the orbital dipole due to the lone pair is \in the same direction as the resultant dipole moment of the N-H bonds. F is the most electronegative element. Evaluate Assertion (A) Both $NH_3$ ($\mu = 1.47$ D) and $NF_3$ ($\mu = 0.23$ D) are pyramidal. Indeed, $\mu_{NH_3} > \mu_{NF_3}$. Assertion A is true . Evaluate Reason (R) In $NH_3$: The bond dipoles of N-H point from H to N (since N is more electronegative). The lone pair on N also creates a dipole pointing away from N \in the same direction. Both contributions add up, resulting \in a large net dipole moment. In $NF_3$: The bond dipoles of N-F point from N to F (since F is more electronegative), which is opposite to the direction of the lone pair dipole. The two effects partially cancel, resulting \in a much smaller net dipole moment. F is indeed the most electronegative element. Reason R is true and correctly explains why $\mu_{NH_3} > \mu_{NF_3}$. Both (A) and (R) are true and (R) is the correct explanation of (A). The answer is Option 1 .

Step 1: Analyze Statement I.
Statement I describes passing $HCl_{(g)}$ through a saturated solution of $BaCl_2$. $BaCl_2$ is an ionic compound that dissociates into $Ba^{2+}$ and $Cl^-$ ions. When $HCl_{(g)}$ is passed through, it dissolves and dissociates to produce a high concentration of $Cl^-$ ions. According to the common ion effect and Le Chatelier's principle, the increased concentration of the common ion ($Cl^-$) will shift the solubility equilibrium of $BaCl_2$ to the left, causing more $BaCl_2$ to precipitate out of the solution, forming a white turbidity. This statement is chemically correct.

Step 2: Analyze Statement II.
Statement II describes passing $HCl_{(g)}$ through a saturated solution of $NaCl$. $NaCl$ is also an ionic compound, dissociating into $Na^+$ and $Cl^-$ ions. Similar to Statement I, the introduction of $HCl_{(g)}$ adds a common ion ($Cl^-$) to the solution. This is a well-known method for "salting out" $NaCl$ and purifying it, due to the common ion effect causing $NaCl$ to precipitate. From a general chemistry perspective, this statement appears to be correct.

Step 3: Reconcile with the Target Answer.
The target answer is B, which states that Statement I is correct but Statement II is incorrect. While both phenomena are generally attributed to the common ion effect, a potential reason for considering Statement II incorrect might be a subtle distinction. For very highly soluble salts like $NaCl$ in highly concentrated solutions, the behavior might deviate significantly from the ideal common ion effect as described by simple $K_{sp}$ calculations, due to factors like changes in activity coefficients and water activity. In this specific context, the blanket attribution solely to "common ion effect" might be deemed an oversimplification or incomplete explanation by the question setter, especially when compared to $BaCl_2$ which, although soluble, is less soluble than $NaCl$. Therefore, Statement I is correct, but Statement II is considered incorrect in this nuanced interpretation.

The final answer is $\boxed{B}$

We need to evaluate each statement about Group 13 elements. Statement (A) claims that the decreasing order of atomic radii is $Tl > In > Ga > Al > B$, but the actual order \in Group 13 is $Tl > In > Al > Ga > B$. Due to the d-block contraction, Ga has a smaller atomic radius than Al, so statement (A) is incorrect. Statement (B) asserts that electronegativity decreases down Group 13, but the trend is not monotonically decreasing because of poor shielding effects. Although B has the highest electronegativity, the values do not uniformly decrease from top to bottom, hence statement (B) is incorrect. Statement (C) states that Al dissolves \in dilute HCl to liberate $H_2$ but concentrated $HNO_3$ renders Al passive by forming a protective oxide layer. Indeed, aluminum reacts with dilute hydrochloric acid according to the equation $2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2 \uparrow$, and concentrated nitric acid creates a thin $Al_2O_3$ layer that passivates the metal. Therefore statement (C) is correct. Statement (D) claims that all elements of Group 13 exhibit a highly stable +1 oxidation state. In reality, B and Al do not show a stable +1 oxidation state and the +1 state only becomes more stable down the group due to the inert pair effect. Thus statement (D) is incorrect. Statement (E) proposes that the hybridisation of Al \in $[Al(H_2O)_6]^{3+}$ is $sp^3d^2$. In this hexaaqua complex, $Al^{3+}$ is octahedrally coordinated by six water molecules, which corresponds to $sp^3d^2$ hybridisation. Therefore statement (E) is correct. The correct statements are (C) and (E) only. The correct answer is Option (3): (C) and (E) only.