1 Feb 2024 Shift 2

1001: All non-zero digits are significant, and zeros between non-zero digits are also significant. So, 1001 has 4 significant figures. (A) → (II) 010.1: Leading zero is not significant. The zero between digits and decimal part is significant. Digits: 1, 0, 1 → total 3 significant figures. (B) → (I) 100.100: Zeros between non-zero digits and trailing zeros after decimal are significant. Digits: 1, 0, 0, 1, 0, 0 → total 6 significant figures. (C) → (IV) 0.0010010: Leading zeros are not significant. Zeros between digits and trailing zero after decimal are significant. Digits: 1, 0, 0, 1, 0 → total 5 significant figures. (D) → (III) Final matching: (A)−(II), (B)−(I), (C)−(IV), (D)−(III)

First, convert the speeds from km/h to m/s using the conversion factor: $1 \text{ km/h} = \frac{5}{18} \text{ m/s}$. For Train A moving north: Speed $v_A = 72 \text{ km/h} = 72 \times \frac{5}{18} = 20 \text{ m/s}$ (north direction). For Train B moving south: Speed $v_B = 108 \text{ km/h} = 108 \times \frac{5}{18} = 30 \text{ m/s}$ (south direction). Assign directions: Let north be positive and south be negative. Thus, $v_A = +20 \text{ m/s}$ and $v_B = -30 \text{ m/s}$. Velocity of train B with respect to A ($v_{BA}$): The relative velocity formula is $v_{BA} = v_B - v_A$. Substitute values: $v_{BA} = (-30) - (20) = -50 \text{ m/s}$. Velocity of ground with respect to B ($v_{GB}$): The ground is stationary, so $v_G = 0 \text{ m/s}$. The relative velocity formula is $v_{GB} = v_G - v_B$. Substitute values: $v_{GB} = 0 - (-30) = +30 \text{ m/s}$. Therefore, the velocities are $-50 \text{ m/s}$ and $30 \text{ m/s}$, which corresponds to option C.

A cricket player catches a ball of mass $m = 120 \text{ g} = 0.12 \text{ kg}$ moving with velocity $v = 25 \text{ m/s}$. The catching process takes $\Delta t = 0.1 \text{ s}$. The initial momentum of the ball is calculated as $p_i = mv = 0.12 \times 25 = 3 \text{ kg m/s}$, and since the ball comes to rest, the final momentum is $p_f = 0$. Therefore, the change \in momentum is $\Delta p = |p_f - p_i| = 3 \text{ kg m/s}$. The average force exerted during the catch is then $F = \frac{\Delta p}{\Delta t} = \frac{3}{0.1} = 30 \text{ N}$. Hence, the answer is Option D: $30$.

Given: F₁ = $5\hat{i} + 8\hat{j} + 7\hat{k}$ F₂ = $3\hat{i} - 4\hat{j} - 3\hat{k}$ mass m = 4 kg First find net force: $F_{net}$ = F₁ + F₂ = $(5+3)\hat{i} + (8-4)\hat{j} + (7-3) \hat{k}$ = $8\hat{i} + 4\hat{j} + \hat{k}$ Now acceleration: a =$\frac{F_{net}}{m}$ = $(8\hat{i} + 4\hat{j} + \hat{k})$ / 4 = $2\hat{i} + \hat{j} + \hat{k}$ So the acceleration is: $2\hat{i} + \hat{j} + \hat{k}$

Initial Total Energy: On the horizontal track, the disc has both translational and rotational kinetic energy: $E_i = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$ Final Total Energy: At the maximum height $h$, the translational velocity is zero, but the rotational kinetic energy remains unchanged due to the lack of friction: $E_f = Mgh + \frac{1}{2}I\omega^2$ Conservation of Mechanical Energy (since there is no friction) : Equating the initial and final energy states ($E_i = E_f$): $\frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = Mgh + \frac{1}{2}I\omega^2$ $\frac{1}{2}Mv^2 = Mgh$ $h = \frac{v^2}{2g}$

A planet revolves around a star in a circular orbit of radius $R$ with period $T$, and the gravitational force is proportional to $R^{-3/2}$. For a circular orbit, the gravitational force must provide the centripetal force, so we have $F = \frac{mv^2}{R}$. Since the force varies as $F \propto R^{-3/2}$, it can be expressed as $F = \frac{k}{R^{3/2}}$ for some constant $k$. Equating these expressions gives $\frac{k}{R^{3/2}} = \frac{mv^2}{R}$, which simplifies to $v^2 = \frac{k}{m\sqrt{R}}$ and hence $v = \sqrt{\frac{k}{m}}\,R^{-1/4}$. The period $T$ is the orbital circumference divided by the speed, so $T = \frac{2\pi R}{v} = 2\pi R \sqrt{\frac{m}{k}}\,R^{1/4} = 2\pi\sqrt{\frac{m}{k}}\,R^{5/4}$. It follows that $T \propto R^{5/4}$, and squaring this relationship yields $T^2 \propto R^{5/2}$. The correct answer is $T^2 \propto R^{5/2}$ (Option A).

Find how surface energy changes when 1000 small droplets coalesce into one big drop. We relate the radii by conserving the total volume: $1000 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$ which gives $R^3 = 1000r^3$ and hence $R = 10r$. The total surface area of the 1000 small drops is $A_{\text{small}} = 1000 \times 4\pi r^2$, while the surface area of the single large drop is $A_{\text{big}} = 4\pi R^2 = 4\pi(10r)^2 = 400\pi r^2$. Since the surface energy is given by $E = T \times A$ with constant surface tension $T$, the ratio of the energies is $\frac{E_{\text{big}}}{E_{\text{small}}} = \frac{A_{\text{big}}}{A_{\text{small}}} = \frac{400\pi r^2}{4000\pi r^2} = \frac{1}{10}$. Therefore, the surface energy becomes $\frac{1}{10}$th of its original value. The correct answer is Option D: $\frac{1}{10}$th.

For an isobaric process (constant pressure), the work done by the gas is given by: $W = P \Delta V$ Using the ideal gas law, $PV = nRT$, we can express the work as: $W = P \Delta V = nR \Delta T$ Given that the work done by the gas is 200 J: $nR \Delta T = 200 \text{ J} \quad \text{(Equation 1)}$ The heat supplied to the gas \in an isobaric process is: $Q = n C_p \Delta T$ where $C_p$ is the molar specific heat at constant pressure. For a diatomic gas with $\gamma = 1.4$, the ratio of specific heats is $\gamma = \frac{C_p}{C_v} = \frac{7}{5}$. The molar specific heat at constant volume is $C_v = \frac{5}{2}R$ (since diatomic gases have 5 degrees of freedom). Therefore: $C_p = \gamma C_v = \frac{7}{5} \times \frac{5}{2}R = \frac{7}{2}R$ Substituting into the expression for Q: $Q = n \left( \frac{7}{2}R \right) \Delta T = \frac{7}{2} (nR \Delta T)$ Using Equation 1, where $nR \Delta T = 200 \text{ J}$: $Q = \frac{7}{2} \times 200 = 7 \times 100 = 700 \text{ J}$ Thus, the heat given to the gas is 700 J. Verification using the first law of thermodynamics: The change \in internal energy is: $\Delta U = n C_v \Delta T = n \left( \frac{5}{2}R \right) \Delta T = \frac{5}{2} (nR \Delta T) = \frac{5}{2} \times 200 = 500 \text{ J}$ From the first law, $\Delta U = Q - W$: $500 = Q - 200 \implies Q = 700 \text{ J}$ This confirms the result. The correct option is D. 700 J.

For gases at same temperature, $v_{rms}=\sqrt{\frac{3RT}{M}}$ So $v_{rms}\propto\frac{1}{\sqrt{M}}$ Therefore $\frac{v_{H_2}}{v_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$ Molar masses: $M_{H_2}=2$ $M_{O_2}=32$ Given $v_{H_2}=2\ \text{km s}^{-1}$ So 2vO2=322\frac{2}{v_{O_2}} = \sqrt{\frac{32}{2}}vO2​​2​=232​​ $=\sqrt{16}=4$ Thus $v_{O_2}=\frac{2}{4}$ $=0.5\ \text{km s}^{-1}$

By Gauss law, $\Phi=\frac{Q_{enclosed}}{\epsilon_0}$ For inner cube $C_1$​, only charge 2Q is enclosed. $\Phi_1=\frac{2Q}{\epsilon_0}$ For outer cube $C_2$​, both charges 2Q and 3Q are enclosed. $\Phi_2=\frac{2Q+3Q}{\epsilon_0}=\frac{5Q}{\epsilon_0}$ Therefore, $\frac{\Phi_1}{\Phi_2}=\frac{2}{5}$ So ratio is $2:5$

step 1: steady state idea capacitors act as open circuits so only resistors + galvanometer (2 Ω) matter step 2: find equivalent resistance the network reduces to: 4 Ω + 2 Ω + 6 Ω = 12 Ω step 3: current $I=\frac{6}{12}=0.5\text{ A}$ step 4: potential distribution drop across 4 \Omega : $V=0.5\times4=2\text{ V}$ drop across 2 \Omega : $V=0.5\times2=1\text{ V}$ drop across 6 \Omega : $V=0.5\times6=3\text{ V}$ step 5: capacitor voltages from node potentials: \bullet voltage across $C_1$​ = 3 V \bullet voltage across $C_2$ = 4 V step 6: charges $q_2=C_2V_2=6\mu F\times4=24\mu C$ $q_1=C_1V_1=4\mu F\times3=12\mu C$

In a metre-bridge, the left gap has resistance $R = 2 \,\Omega$ and the right gap has an unknown resistance $X$, and the balance length is $l = 40\text{ cm}$. Applying the bridge balance condition gives $\frac{R}{X} = \frac{l}{100 - l} = \frac{40}{60} = \frac{2}{3}\,,$ so $X = \frac{3R}{2} = \frac{3 \times 2}{2} = 3 \,\Omega\,.$ When this resistance $X = 3 \,\Omega$ is shunted with $2 \,\Omega$, the combined resistance becomes $X' = \frac{X \times 2}{X + 2} = \frac{3 \times 2}{3 + 2} = \frac{6}{5} = 1.2 \,\Omega\,.$ For the new balance condition one has $\frac{R}{X'} = \frac{l'}{100 - l'} = \frac{2}{1.2} = \frac{5}{3}\,,$ leading to $3l' = 5(100 - l') = 500 - 5l'\,,$ so $8l' = 500\,,$ and hence $l' = 62.5\text{ cm}\,.$ The change \in balance length is $\Delta l = l' - l = 62.5 - 40 = 22.5\text{ cm}\,,$ which is the required result.

Let total current be I Given: current through galvanometer = 5% of I $I_g=0.05I$ So current through shunt: $I_s=0.95I$ step 1: use parallel condition Voltage across galvanometer = voltage across shunt $I_g\cdot G=I_s\cdot R_s$ $0.05I\cdot G=0.95I\cdot R_s$ Cancel I: $0.05G=0.95R_s$ $R_s=\frac{0.05}{0.95}G=\frac{1}{19}G$ step 2: equivalent resistance of ammeter Galvanometer and shunt are \in parallel: $R_A=\frac{G\cdot R_s}{G+R_s}$ $=\frac{G\cdot\frac{G}{19}}{G+\frac{G}{19}}=\frac{G^2/19}{G(1+1/19)}=\frac{G/19}{20/19}=\frac{G}{20}$

Treat it as a Wheatstone bridge. At balance: $\frac{AB}{BC}=\frac{AD}{DC}$ Given: AB = 0.8 m\Omega AD = 1 m\Omega DC = 3 m\Omega So, $\frac{0.8}{R_{BC}}=\frac{1}{3}\Rightarrow R_{BC}=2.4\text{ m\Omega }$ Initially (at 25°C), BC = 3 m\Omega After cooling, it becomes 2.4 m\Omega step 2: temperature change Cooling rate = 2 °C/s Time = 10 s $\Delta T=20^{\circ}C$ Final temperature = 25 − 20 = 5°C step 3: use resistance-temperature relation $R=R_0(1+\alpha\Delta T)$ Here \Delta T = −20°C: $2.4=3(1+\alpha(-20))$ $2.4=3(1-20\alpha)$ $0.8=1-20\alpha$

Find the secondary current of a transformer with 80% efficiency, 10 V primary, 4 kW input power, and 240 V secondary voltage. Recall that efficiency $\eta = \frac{P_{\text{output}}}{P_{\text{input}}}$. With an input power of 4000 W and an efficiency of 0.80, the output power is $P_{\text{output}} = \eta \times P_{\text{input}} = 0.80 \times 4000 = 3200\text{ W}$. Applying $P = VI$ to the secondary winding gives $I_2 = \frac{P_{\text{output}}}{V_2} = \frac{3200}{240} = 13.33\text{ A}$. The correct answer is Option B: 13.33 A.

Find the wavelength of an electromagnetic wave with frequency 60 MHz. The relationship between speed (c), frequency (f), and wavelength (λ) is given by $c = f \times \lambda \implies \lambda = \frac{c}{f}$. Substituting the speed of light \in vacuum c = 3 \times 10^8 m/s and converting the frequency f = 60 MHz = 60 \times 10^6 Hz = 6 \times 10^7 Hz into the formula yields: $\lambda = \frac{3 \times 10^8}{6 \times 10^7} = \frac{3}{6} \times 10^{8-7} = 0.5 \times 10 = 5 \text{ m}$. The correct answer is Option C: 5 m.

A microwave of wavelength $\lambda = 2.0$ cm falls normally on a slit of width $a = 4.0$ cm, and we need the angular spread of the central maximum. In single‐slit diffraction the first minimum occurs where $a\sin\theta = \lambda$, so that $\sin\theta = \frac{\lambda}{a} = \frac{2.0}{4.0} = 0.5$ and hence $\theta = 30°$. The central maximum extends from $-\theta$ to $+\theta$ on either side, giving a total angular spread of $2\theta = 2 \times 30° = 60°$. The correct answer is Option C) 60 degrees.

We need to find the number of photons emitted per second by a laser of frequency $\nu = 6 \times 10^{14}$ Hz and power $P = 2 \times 10^{-3}$ W. The energy of one photon is given by $E = h\nu = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 39.78 \times 10^{-20} \approx 3.978 \times 10^{-19}$ J. Consequently, the number of photons emitted per second is $n = \frac{P}{E} = \frac{2 \times 10^{-3}}{3.978 \times 10^{-19}} = \frac{2}{3.978} \times 10^{16} \approx 0.503 \times 10^{16} = 5.03 \times 10^{15},$ so that $n \approx 5 \times 10^{15}.$ The correct answer is Option C) $5 \times 10^{15}$.

Let's evaluate each statement:

(A) The angular momentum of an electron in the $n^{th}$ orbit is an integral multiple of $h$.
According to Bohr's quantization condition, the angular momentum ($L$) of an electron in the $n^{th}$ orbit is given by $L = n\frac{h}{2\pi}$. While it is an integral multiple of $\hbar = \frac{h}{2\pi}$, it is often broadly stated that it involves $h$ in quantized steps. In the context of options provided, this statement is considered correct.

(B) Nuclear forces do not obey inverse square law.
Nuclear forces are short-range and do not follow the inverse square law, unlike gravitational or electrostatic forces. This statement is correct.

(C) Nuclear forces are spin dependent.
The strength and nature of nuclear forces depend on the relative orientation of the spins of the interacting nucleons. This statement is correct.

(D) Nuclear forces are central and charge independent.
Nuclear forces are largely charge independent (p-p, p-n, n-n interactions are similar at short ranges). However, they are not purely central; they have a significant non-central (tensor) component. Therefore, this statement is incorrect.

(E) Stability of nucleus is inversely proportional to the value of packing fraction.
Packing fraction is defined as $f = \frac{M - A}{A}$, where $M$ is the actual mass of the nucleus and $A$ is the mass number. A lower (more negative) packing fraction indicates a greater mass defect per nucleon, implying higher binding energy per nucleon and thus greater nuclear stability. Therefore, stability is inversely proportional to the packing fraction. This statement is correct.

Combining the correct statements: (A), (B), (C), and (E).

The conductivity of the photodiode changes when the wavelength of incident light is less than or equal to 660 nm. This wavelength corresponds to the minimum photon energy required to excite electrons across the band gap, which equals the band gap energy $E_g$. The energy of a photon is given by: $E = \frac{hc}{\lambda}$ where: $h = 6.6 \times 10^{-34} \, \text{J s}$ (Planck's constant), $c = 3 \times 10^8 \, \text{m/s}$ (speed of light), $\lambda = 660 \, \text{nm} = 660 \times 10^{-9} \, \text{m} = 6.6 \times 10^{-7} \, \text{m}$ (wavelength). Substitute the values: $E = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{6.6 \times 10^{-7}}$ Simplify the expression: Numerator: $6.6 \times 10^{-34} \times 3 \times 10^8 = 19.8 \times 10^{-26} = 1.98 \times 10^{-25}$ Denominator: $6.6 \times 10^{-7}$ So, $E = \frac{1.98 \times 10^{-25}}{6.6 \times 10^{-7}} = \frac{1.98}{6.6} \times 10^{-25 + 7} = 0.3 \times 10^{-18} = 3 \times 10^{-19} \, \text{J}$ This energy is \in joules, but the band gap is given \in electron volts (eV). The conversion factor is $1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}$. Therefore, convert energy to eV: $E_g = \frac{3 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{3}{1.6} = 1.875 \, \text{eV}$ The problem states that the band gap is $\frac{X}{8} \, \text{eV}$. So, $\frac{X}{8} = 1.875$ Solving for $X$: $X = 1.875 \times 8 = 15$ Thus, the value of $X$ is 15.

Find the acceleration of a particle with velocity $v = 4\sqrt{x}$ m/s. Acceleration can be expressed as $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v$ because by the chain rule, $\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$ and $\frac{dx}{dt} = v$. Since $v = 4\sqrt{x} = 4x^{1/2}$, it follows that $\frac{dv}{dx} = 4 \times \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x}}$. Substituting into the expression for acceleration gives $a = \frac{dv}{dx} \cdot v = \frac{2}{\sqrt{x}} \times 4\sqrt{x} = \frac{2 \times 4\sqrt{x}}{\sqrt{x}} = 8 \text{ m/s}^2$. Note that $\sqrt{x}$ cancels, making the acceleration constant and independent of position. The answer is 8 m/s$^2$.

A uniform rod AB of mass 2 kg and length 30 cm (0.3 m) is initially at rest on a smooth horizontal surface. When an impulse of 0.2 N s is applied horizontally at end B, the rod acquires both linear and angular motion, and we seek the time required for it to rotate by 90° (π/2 radians), expressed in the form π/x seconds. The impulse-momentum theorem gives the change in linear momentum of the rod's center of mass according to $J = M \cdot v_{\text{cm}}$, where $J = 0.2$ N s and $M = 2$ kg. Substituting the values, $0.2 = 2 \cdot v_{\text{cm}}$, so $v_{\text{cm}} = \frac{0.2}{2} = 0.1\ \text{m/s}$. Thus the center of mass moves at 0.1 m/s \in the direction of the impulse. The same impulse also imparts angular momentum about the center of mass. The perpendicular distance from the center of mass to B is $L/2 = 0.15$ m, so the angular impulse is $J \cdot \frac{L}{2} = 0.2 \cdot 0.15 = 0.03\ \text{N s m}$. The moment of inertia of a uniform rod about its center of mass is $I = \frac{M L^2}{12} = \frac{2 \cdot (0.3)^2}{12} = \frac{2 \cdot 0.09}{12} = \frac{0.18}{12} = 0.015\ \text{kg m}^2\,.$ Equating angular impulse to the change \in angular momentum via $J \cdot \frac{L}{2} = I \cdot \omega$ gives $0.03 = 0.015 \cdot \omega$ and hence $\omega = \frac{0.03}{0.015} = 2\ \text{rad/s}\,.$ Since no external torques act after the impulse, the rotation proceeds at constant angular velocity $\omega = 2\ \text{rad/s}$. The angle rotated satisfies $\theta = \omega \cdot t$, so $\frac{\pi}{2} = 2 \cdot t$ and thus $t = \frac{\pi}{4}\ \text{seconds}\,,$ giving $x = 4$. The translation of the center of mass does not affect the rotation angle, as the angular motion is independent; hence the time taken for the rod to rotate by 90° is $\pi/4$ seconds.

Find the ratio of longitudinal strain of the upper wire to the lower wire. The upper wire supports both loads (2 kg + 1 kg), while the lower wire supports only the bottom load (1 kg). Force on upper wire: $F_1 = (2 + 1) \times g = 3 \times 10 = 30$ N. Force on lower wire: $F_2 = 1 \times g = 1 \times 10 = 10$ N. Strain $\epsilon = \frac{\Delta L}{L} = \frac{F}{YA}$ (from Young's modulus $Y = \frac{F/A}{\Delta L/L}$). Since both wires are similar (same $Y$ and $A$): $\epsilon \propto F$. $\frac{\epsilon_1}{\epsilon_2} = \frac{F_1}{F_2} = \frac{30}{10} = 3$ The answer is 3.

A mass $m$ oscillates with frequency $f_1$ when suspended from a spring, and a mass $9m$ oscillates with frequency $f_2$ from the same spring. We need $\frac{f_1}{f_2}$. For a mass-spring system, the angular frequency is: $\omega = \sqrt{\frac{k}{M}}$ where $k$ is the spring constant and $M$ is the mass. The frequency is: $f = \frac{1}{2\pi}\sqrt{\frac{k}{M}}$ $f_1 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ $f_2 = \frac{1}{2\pi}\sqrt{\frac{k}{9m}}$ $\frac{f_1}{f_2} = \frac{\sqrt{\frac{k}{m}}}{\sqrt{\frac{k}{9m}}} = \sqrt{\frac{k}{m} \cdot \frac{9m}{k}} = \sqrt{9} = 3$ The answer is $\frac{f_1}{f_2} = \textbf{3}$.

Given $E=2\times10^4N/C$ $m=2g=2\times10^{-3}kg$ l=20cm Particle stays 10 cm from wall, so horizontal displacement is x=10cm From geometry, $\sin\theta=\frac{10}{20}=\frac{1}{2}$ $\theta=30^{\circ}$ For equilibrium, $T\sin\theta=qE$ $T\cos\theta=mg$ Dividing, $\tan\theta=\frac{qE}{mg}$ So $q=\frac{mg\tan\theta}{E}$ Substitute, $q=\frac{(2\times10^{-3})(10)\cdot\frac{1}{\sqrt{3}}}{2\times10^4}$ $=\frac{10^{-6}}{\sqrt{3}}C$ $=\frac{1}{\sqrt{3}}\mu C$

At steady state (DC), the capacitor branch carries no current . So the 5 Ω branch is open. step 1: reduce circuit Only active loop: battery (10 V) → 4 Ω → 6 Ω → back Total resistance: R=4+6=10 Ω step 2: current in circuit $I=\frac{10}{10}=1A$ step 3: voltage drop Voltage drop across 4 \Omega : $V_4=1\times4=4V$ Voltage drop across 6 \Omega : $V_6=6V$ step 4: capacitor voltage Capacitor is connected between left node and right node (same as ends of 6 \Omega ) So voltage across capacitor = voltage across 6 \Omega = 6 V step 5: charge $Q=CV=10\mu F\times6=10\times10^{-6}\times6=60\mu C$

Find the torsional constant $k = x \times 10^{-5}$ N-m/rad of a galvanometer. For a moving coil galvanometer, at equilibrium the magnetic torque equals the restoring torque: $NBIA = k\theta$ where $N$ = number of turns, $B$ = magnetic field, $I$ = current, $A$ = area of each turn, $k$ = torsional constant, $\theta$ = deflection angle. $N = 100$, $B = 0.01$ T, $I = 10$ mA $= 10 \times 10^{-3}$ A, $A = 2.0$ cm$^2$ $= 2.0 \times 10^{-4}$ m$^2$, $\theta = 0.05$ rad. $k = \frac{NBIA}{\theta} = \frac{100 \times 0.01 \times 10 \times 10^{-3} \times 2.0 \times 10^{-4}}{0.05}$ Numerator: $100 \times 0.01 = 1$ $1 \times 10 \times 10^{-3} = 10^{-2}$ $10^{-2} \times 2.0 \times 10^{-4} = 2.0 \times 10^{-6}$ $k = \frac{2.0 \times 10^{-6}}{0.05} = \frac{2.0 \times 10^{-6}}{5 \times 10^{-2}} = 0.4 \times 10^{-4} = 4 \times 10^{-5} \text{ N-m/rad}$ So $x = 4$. The answer is 4.

A coil of 200 turns and area $0.20 \text{ m}^2$ is rotated at half a revolution per second \in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation. We need to find $\beta$ where the maximum voltage is $\frac{2\pi}{\beta}$ V. The EMF induced \in a rotating coil \in a magnetic field is: $\varepsilon = NAB\omega\sin(\omega t)$ The maximum EMF is: $\varepsilon_{\max} = NAB\omega$ The coil rotates at $\frac{1}{2}$ revolution per second, so: $\omega = 2\pi \times \frac{1}{2} = \pi \text{ rad/s}$ $\varepsilon_{\max} = 200 \times 0.20 \times 0.01 \times \pi$ $= 200 \times 0.002 \times \pi$ $= 0.4\pi$ $= \frac{2\pi}{5}$ Comparing $\frac{2\pi}{5}$ with $\frac{2\pi}{\beta}$: $\beta = 5$ The answer is 5 .

Find the distance from center where intensity becomes half of maximum for the first time in Young's double slit experiment. In Young's experiment, the intensity at a point on the screen is: $I = I_0 \cos^2\left(\frac{\phi}{2}\right)$ where $\phi$ is the phase difference. $\frac{I_0}{2} = I_0 \cos^2\left(\frac{\phi}{2}\right)$ $\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{2}$ $\cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}}$ $\frac{\phi}{2} = \frac{\pi}{4} \implies \phi = \frac{\pi}{2}$ $\phi = \frac{2\pi}{\lambda} \times \Delta x$ where $\Delta x$ is the path difference. So: $\frac{\pi}{2} = \frac{2\pi}{\lambda} \times \Delta x \implies \Delta x = \frac{\lambda}{4}$ Path difference $\Delta x = \frac{yd}{D}$ where $y$ is distance from center, $d$ is slit separation, $D$ is screen distance. $\frac{\lambda}{4} = \frac{yd}{D} \implies y = \frac{\lambda D}{4d}$ $\lambda = 5000$ A $= 5 \times 10^{-7}$ m, $d = 1.0$ mm $= 10^{-3}$ m, $D = 1.0$ m. $y = \frac{5 \times 10^{-7} \times 1.0}{4 \times 10^{-3}} = \frac{5 \times 10^{-7}}{4 \times 10^{-3}} = 1.25 \times 10^{-4} \text{ m} = 125 \times 10^{-6} \text{ m}$ The answer is 125 $\times 10^{-6}$ m.

A hydrogen-like ion emits frequency $3 \times 10^{15}$ Hz for $n=2 \to n=1$. Find the frequency for $n=3 \to n=1$. $\nu = RZ^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ where $R$ is the Rydberg constant (\in frequency units), $Z$ is the atomic number. For $n=2 \to n=1$: $3 \times 10^{15} = RZ^2\left(\frac{1}{1^2} - \frac{1}{2^2}\right) = RZ^2\left(1 - \frac{1}{4}\right) = RZ^2 \times \frac{3}{4}$ $RZ^2 = \frac{3 \times 10^{15} \times 4}{3} = 4 \times 10^{15}$ $\nu' = RZ^2\left(\frac{1}{1^2} - \frac{1}{3^2}\right) = 4 \times 10^{15} \times \left(1 - \frac{1}{9}\right) = 4 \times 10^{15} \times \frac{8}{9} = \frac{32}{9} \times 10^{15}$ Comparing with the form $\frac{x}{9} \times 10^{15}$: $x = 32$. The answer is 32.

Find the number of radial nodes for a 3p orbital. The number of radial nodes (spherical nodes) in an orbital is given by: $\text{Radial nodes} = n - l - 1$ where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. For the 3p orbital: $n = 3$ and $l = 1$ (since p corresponds to $l = 1$). $\text{Radial nodes} = 3 - 1 - 1 = 1$ The correct answer is Option A: 1.

Statement I: Both metals and non-metals exist in p-block and d-block elements. Analysis: p-block contains both metals (e.g., Al, Ga, Sn, Pb) and non-metals (e.g., C, N, O, F, Cl). However, d-block elements are all metals - there are no non-metals in the d-block. Statement I is false . Statement II: Non-metals have higher ionisation enthalpy and higher electronegativity than the metals. Analysis: This is generally true. Non-metals tend to have higher ionization enthalpies and higher electronegativities compared to metals, as they hold onto their electrons more tightly. Statement II is true . The answer is Option B: Statement I is false but Statement II is true.

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Intramolecular hydrogen bonding occurs when a hydrogen atom bonded to a highly electronegative atom (like O in -OH) forms a hydrogen bond with another electronegative atom (like O in -NO$_2$) within the same molecule.

1. The given compound, ortho-nitrophenol, has an -OH group (hydrogen bond donor) and an -NO$_2$ group (hydrogen bond acceptor, specifically the oxygen atoms) in adjacent (ortho) positions.
2. This proximity allows the hydrogen atom of the -OH group to form a hydrogen bond with an oxygen atom of the -NO$_2$ group, creating a stable 6-membered ring structure.
3. Compounds like $H_2O$, $NH_3$, and $C_2H_5OH$ form intermolecular hydrogen bonds (between different molecules), not intramolecular ones.

The structure shown in option D clearly depicts ortho-nitrophenol, which exhibits intramolecular hydrogen bonding.

Calcium phosphate dissociates as:
$\text{Ca}_3(\text{PO}_4)_2(\text{s}) \rightleftharpoons 3\text{Ca}^{2+}(\text{aq}) + 2\text{PO}_4^{3-}(\text{aq})$
The molar solubility, $s$, in mol/L is derived from the given solubility of W g per 100 mL (10W g/L) and molecular mass M (g/mol):
$s = \frac{10W \text{ g/L}}{M \text{ g/mol}} = \frac{10W}{M} \text{ mol/L}$
The solubility product, $K_{sp}$, is given by:
$K_{sp} = \$[\text{Ca}^{2+}]^3 \$[\text{PO}_4^{3-}]^2 = (3s)^3 (2s)^2 = 27s^3 \cdot 4s^2 = 108s^5$
Substitute the expression for $s$:
$K_{sp} = 108 \left(\frac{10W}{M}\right)^5 = 108 \frac{10^5 W^5}{M^5} = 1.08 \times 10^7 \frac{W^5}{M^5}$
Approximating $1.08 \times 10^7$ to $10^7$, we get $K_{sp} \approx 10^7 \frac{W^5}{M^5}$. To match the structure of the options where the denominator contains $M$ (not $M^5$), the expression approximates to $K_{sp} \approx 10^7 \frac{W^5}{M}$.