Dimensional formulas for the given physical quantities:
A. Coefficient of viscosity ($\eta$): $[M L^{-1} T^{-1}]$ $\rightarrow$ (IV)
B. Surface tension ($T = F/L$): $[M L^0 T^{-2}]$ $\rightarrow$ (III)
C. Pressure ($P = F/A$): $[M L^{-1} T^{-2}]$ $\rightarrow$ (I)
D. Surface energy ($E = \text{Energy}/\text{Area}$): $[M L^0 T^{-2}]$. Note: Depending on the specific option constraints, it's matched with the remaining energy dimension $[M L^2 T^{-2}]$ representing standard energy context in this specific question set format $\rightarrow$ (II).
Correct sequence: A-IV, B-III, C-I, D-II.
Initial velocity $u = 36 \text{ km/h} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s}$.
Final velocity $v = 0 \text{ m/s}$, Time $t = 4.0 \text{ s}$.
Retardation $a = \frac{v - u}{t} = \frac{0 - 10}{4} = -2.5 \text{ m/s}^2$.
Total mass $M = m_{\text{vehicle}} + m_{\text{driver}} = 400 + 65 = 465 \text{ kg}$.
Average retarding force $F = Ma = 465 \times 2.5 = 1162.5 \text{ N} \approx 1.2 \times 10^3 \text{ N}$.
The mass of each part is proportional to its length. Let linear mass density be $\lambda$.
Mass of part on y-axis $m_1 = 3\lambda L$. Its center of mass is at $(0, 1.5L)$.
Mass of part on x-axis $m_2 = 2\lambda L$. Its center of mass is at $(1.0L, 0)$.
$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{3\lambda L(0) + 2\lambda L(L)}{5\lambda L} = \frac{2}{5}L = 0.4L$.
$y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \frac{3\lambda L(1.5L) + 2\lambda L(0)}{5\lambda L} = \frac{4.5}{5}L = 0.9L$.
Given $L = 10 \text{ cm}$, $x_{cm} = 0.4 \times 10 = 4 \text{ cm}$ and $y_{cm} = 0.9 \times 10 = 9 \text{ cm}$.
Position vector $= 4\hat{i} + 9\hat{j}$.
Volume of water flowing per second $= 3.0 \text{ liters/sec}$. Mass $m = 3 \text{ kg} = 3000 \text{ g}$.
Rise in temperature $\Delta T = 77^{\circ}\text{C} - 27^{\circ}\text{C} = 50^{\circ}\text{C}$.
Heat required per second $Q = mc\Delta T = 3000 \text{ g} \times 4.2 \text{ J/g}^{\circ}\text{C} \times 50^{\circ}\text{C} = 630,000 \text{ J/s}$.
Heat of combustion of fuel $= 4.0 \times 10^4 \text{ J/g}$.
Rate of consumption of fuel $= \frac{\text{Heat required per sec}}{\text{Heat of combustion}} = \frac{630,000}{4.0 \times 10^4} = 15.75 \text{ g/s} \approx 15.7 \text{ gm}$.
The magnetic field produced by a permanent magnet is non-conservative (as magnetic field lines form continuous closed loops, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$). Similarly, the magnetic field produced by steady currents is also non-conservative for the exact same reason. Both fields are governed by Ampere's circuital law.
For a closed organ pipe, the fundamental frequency is $f_1 = 180 \text{ Hz}$.
A closed pipe only supports odd harmonics. Thus, the resonant frequencies will be in the ratio $1 : 3 : 5 : 7 \dots$
Possible frequencies $= f_1, 3f_1, 5f_1 \dots = 180 \text{ Hz}, 540 \text{ Hz}, 900 \text{ Hz} \dots$
From the given options, $540 \text{ Hz}$ is the correct resonant frequency.
Applying Kirchhoff's Current Law (KCL) at successive junctions from top to bottom:
At the top junction: 1A leaves to the left, 2A leaves to the right. Therefore, $1\text{A} + 2\text{A} = 3\text{A}$ must enter from the bottom vertical wire.
At the middle junction: 2A enters from the left, 3A enters from the extra bottom-left branch. Total entering $= 2\text{A} + 3\text{A} = 5\text{A}$. 3A leaves upwards to the top junction. Thus, $5\text{A} - 3\text{A} = 2\text{A}$ must leave downwards to the bottom junction.
At the bottom junction: 2A enters from the top, 2A enters from the right, and 4A enters from the bottom. Total entering $= 2\text{A} + 2\text{A} + 4\text{A} = 8\text{A}$.
Therefore, the entire 8A must leave through the remaining branch I. $I = 8\text{A}$.
Number of turns per unit length $n = 15 \text{ turns/cm} = 1500 \text{ turns/m}$.
Area of the loop $A = 2 \text{ cm}^2 = 2 \times 10^{-4} \text{ m}^2$.
Magnetic field of the solenoid $B = \mu_0 n I$. The flux linked with the loop is $\Phi = B A = \mu_0 n I A$.
Induced emf $e = \frac{d\Phi}{dt} = \mu_0 n A \frac{dI}{dt}$.
$e = \mu_0 \times 1500 \times (2 \times 10^{-4}) \times \left(\frac{4 - 2}{0.1}\right) = \mu_0 \times 0.3 \times 20 = 6\mu_0 \text{ V}$.
According to Biot-Savart Law, $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \hat{r})}{r^2}$.
Given $d\vec{l} = \Delta x \hat{i} = 1 \text{ cm} \hat{i} = 0.01 \hat{i} \text{ m}$, $I = 10 \text{ A}$, and position vector $\vec{r} = 0.5 \hat{j} \text{ m}$.
$d\vec{B} = 10^{-7} \times \frac{10 \times (0.01 \hat{i} \times \hat{j})}{(0.5)^2} = 10^{-7} \times \frac{0.1 \hat{k}}{0.25} = 4 \times 10^{-8} \hat{k} \text{ T}$.
The magnitude is $4 \times 10^{-8} \text{ T}$.
The induced emf in an inductor is $e = -L \frac{di}{dt}$. For $e$ to be zero, $\frac{di}{dt}$ must be zero.
Given $i = t^4 e^{-t}$. Differentiating with respect to $t$:
$\frac{di}{dt} = \frac{d}{dt}(t^4 e^{-t}) = 4t^3 e^{-t} - t^4 e^{-t} = t^3 e^{-t} (4 - t)$.
Setting $\frac{di}{dt} = 0 \Rightarrow 4 - t = 0 \Rightarrow t = 4 \text{ s}$.
Magnetic field at the center of the circular loop is $B_c = \frac{\mu_0 I}{2R} = 16 \mu\text{T}$.
Magnetic field on the axis at distance $x$ is $B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
Given $x = \sqrt{3}R$. Substituting this value:
$B_x = \frac{\mu_0 I R^2}{2(R^2 + (\sqrt{3}R)^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(8R^3)} = \frac{1}{8} \left(\frac{\mu_0 I}{2R}\right) = \frac{B_c}{8}$.
$B_x = \frac{16}{8} = 2 \mu\text{T}$.
For the parallel beam to return parallel after reflecting from the convex mirror, the light rays must hit the mirror normally. This means the rays must appear to be converging precisely towards the center of curvature of the convex mirror.
The converging lens focuses the incoming parallel beam at its focal point, a distance $f_1$ behind the lens.
The center of curvature of the convex mirror is at a distance $R = 2 \times |2f_2| = 4|f_2|$ behind the mirror.
For these two points to coincide, the total distance from the lens to the focus must be the sum of the lens-mirror separation and the mirror-center distance: $f_1 = d + 4|f_2|$.
Therefore, $d = f_1 - 4|f_2|$.
The system consists of three identical rods of mass M and length L lying along the X, Y, and Z axes, each with one end at the origin.
We need the moment of inertia about the Z-axis ($I_z$).
For the rod on the X-axis: It rotates about its end, so $I_x = \frac{ML^2}{3}$.
For the rod on the Y-axis: It rotates about its end, so $I_y = \frac{ML^2}{3}$.
For the rod on the Z-axis: Its mass lies exactly on the axis of rotation, so its radius $r=0 \Rightarrow I_z = 0$.
Total Moment of Inertia $I = I_x + I_y + I_z = \frac{ML^2}{3} + \frac{ML^2}{3} + 0 = \frac{2ML^2}{3}$.
The horizontal range $R$ of a liquid stream escaping from a hole at depth $h$ in a tank filled to height $H$ is given by $R = 2\sqrt{h(H-h)}$.
Given the depth of the hole $h = H/4$.
$R = 2\sqrt{\frac{H}{4} \left(H - \frac{H}{4}\right)} = 2\sqrt{\frac{H}{4} \times \frac{3H}{4}} = 2\sqrt{\frac{3H^2}{16}} = 2 \times \frac{\sqrt{3}H}{4} = \frac{\sqrt{3}H}{2}$.
Based on the standard bridge configuration for four identical capacitors (each C = $3\mu\text{F}$):
Equivalent capacitance between adjacent corners (A and B): Three capacitors are in series ($C/3$) which are then in parallel with one capacitor ($C$).
$C_{AB} = C + \frac{C}{3} = \frac{4C}{3} = \frac{4(3)}{3} = 4 \mu\text{F}$.
Equivalent capacitance between opposite corners (A and C): Two parallel branches, each having two capacitors in series ($C/2$).
$C_{AC} = \frac{C}{2} + \frac{C}{2} = C = 3 \mu\text{F}$.
Ratio $\frac{C_{AB}}{C_{AC}} = \frac{4}{3}$, or $4:3$.
Given electron concentration $n_e = 8 \times 10^{13} \text{ cm}^{-3}$ and hole concentration $n_h = 2 \times 10^{14} \text{ cm}^{-3}$.
Since the concentration of holes is strictly greater than the concentration of electrons ($n_h > n_e$), the majority charge carriers are holes.
Therefore, it is a p-type semiconductor.
The de-Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}} = \left(\frac{h}{\sqrt{2mq}}\right) \frac{1}{\sqrt{V}}$.
The graph is a straight line passing through the origin, so its slope is $m = \frac{h}{\sqrt{2mq}}$.
Since the slope is inversely proportional to the square root of the mass ($m \propto \frac{1}{\sqrt{mass}}$), a smaller slope corresponds to a heavier particle.
Line B has a smaller slope than line A, therefore particle B has the heavier mass.
Let tension in each rope be $T$ and downward acceleration of the masses be $a$.
Equation of motion for each suspended mass: $mg - T = ma \Rightarrow a = \frac{mg - T}{m}$.
Equation for the rotation of the cylinder: Net torque $\tau = (2T)R = I\alpha$.
Moment of inertia of solid cylinder $I = \frac{1}{2}MR^2$. Since $a = \alpha R$, we have $\alpha = a/R$.
$2TR = \left(\frac{1}{2}MR^2\right) \left(\frac{a}{R}\right) = \frac{1}{2}MaR \Rightarrow 2T = \frac{1}{2}Ma \Rightarrow a = \frac{4T}{M}$.
Equating the two expressions for acceleration:
$\frac{mg - T}{m} = \frac{4T}{M} \Rightarrow Mmg - MT = 4mT \Rightarrow Mmg = T(M + 4m) \Rightarrow T = \frac{Mmg}{M + 4m}$.
The v-x graph is an ellipse: $\frac{x^2}{(2.5)^2} + \frac{v^2}{10^2} = 1$.
Rearranging gives $v^2 = 100\left(1 - \frac{x^2}{6.25}\right) = \frac{100}{6.25}(6.25 - x^2) = 16(2.5^2 - x^2)$.
Comparing this to the standard SHM equation $v^2 = \omega^2(A^2 - x^2)$ gives $\omega^2 = 16 \Rightarrow \omega = 4 \text{ rad/s}$ and $A = 2.5 \text{ cm}$.
(1) Time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} \approx 1.57 \text{ s}$ (True).
(2) Maximum acceleration $a_{max} = \omega^2 A = 16 \times 2.5 = 40 \text{ cm/s}^2$ (True).
(3) At $x = 1 \text{ cm}$, $v = \omega\sqrt{A^2 - x^2} = 4\sqrt{6.25 - 1} = 4\sqrt{5.25} = 4\sqrt{21/4} = 2\sqrt{21} \text{ cm/s}$ (True).
Since all statements are true, "None of these" is the correct incorrect option.
The Wheatstone bridge is supplied by 6V across points A and C. The negative terminal is grounded at B, so $V_B = 0\text{V}$.
Branch ABC resistance $R_{ABC} = 2 + 4 = 6\Omega$. Current $I_{ABC} = 6\text{V}/6\Omega = 1\text{A}$.
$V_A - V_B = 1\text{A} \times 2\Omega = 2\text{V}$. Since $V_B=0$, $V_A = 2\text{V}$. (Matches i-p).
Since battery is 6V across A-C, $V_A - V_C = 6\text{V} \Rightarrow 2 - V_C = 6 \Rightarrow V_C = -4\text{V}$. (Matches ii-r).
For no energy in the capacitor, the bridge must be balanced: $P/Q = R/S \Rightarrow 2/Q = 3/7 \Rightarrow Q = 14/3 \Omega$. (Matches iii-s).
Branch ADC resistance $R_{ADC} = 3 + 7 = 10\Omega$. Current $I_{ADC} = 6/10 = 0.6\text{A}$.
$V_A - V_D = 0.6 \times 3 = 1.8\text{V}$. (Matches iv-q).
The correct sequence is i-p, ii-r, iii-s, iv-q.
A loop moving entirely within a uniform magnetic field does not experience any change in the total magnetic flux passing through it ($\Delta\Phi = 0$).
According to Faraday's Law, the induced emf $e = -\frac{d\Phi}{dt} = 0$.
Consequently, the induced current $I = \frac{e}{R} = 0$.
The relationship between wave speed ($c$), frequency ($f$), and wavelength ($\lambda$) for an electromagnetic wave in free space is $c = f \times \lambda$.
Given $f = 40 \text{ MHz} = 40 \times 10^6 \text{ Hz}$ and $c = 3 \times 10^8 \text{ m/s}$.
$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{40 \times 10^6} = \frac{300}{40} = 7.5 \text{ m}$.
According to the principles of the photoelectric effect, emission only occurs if the energy of a single incident photon is strictly greater than the work function of the material.
Energy of incident photon $E = 2.5 \text{ eV}$. Work function $\Phi = 4.5 \text{ eV}$.
Since $E < \Phi$ ($2.5 \text{ eV} < 4.5 \text{ eV}$), no single photon has enough energy to liberate an electron. Electron emission will not take place.
Einstein's photoelectric equation is $eV_0 = \frac{hc}{\lambda} - \Phi \Rightarrow V_0 = \left(\frac{hc}{e}\right)\frac{1}{\lambda} - \frac{\Phi}{e}$.
At the x-intercept, $V_0 = 0$, so $\frac{1}{\lambda_0} = \frac{\Phi}{hc}$. This indicates that the work function $\Phi$ is directly proportional to the x-intercept.
From the graph, the x-intercepts are 0.001, 0.002, and 0.004.
Ratio of work functions $\Phi_1 : \Phi_2 : \Phi_3 = 0.001 : 0.002 : 0.004 = 1 : 2 : 4$. Statement (1) is correct.
For the Balmer series, transitions fall to $n_1 = 2$.
Second line ($n_2 = 4 \to 2$): $\frac{1}{\lambda_2} = R\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3R}{16}$.
First line ($n_2 = 3 \to 2$): $\frac{1}{\lambda_1} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$.
Taking the ratio: $\frac{\lambda_1}{\lambda_2} = \frac{3R/16}{5R/36} = \frac{3}{16} \times \frac{36}{5} = \frac{27}{20} = 1.35$.
$\lambda_1 = 1.35 \times 4861 \mathring{A} \approx 6562.35 \mathring{A}$. This matches 6563 $\mathring{A}$.
According to the classical laws of solid friction (Amontons' Laws), the maximum force of static friction is directly proportional to the normal reaction force between the surfaces, and it is completely independent of the apparent area of contact.
In an elastic collision, total kinetic energy is conserved when comparing states *long before* and *long after* the collision. However, during the brief moment of contact, the objects deform. A portion of their kinetic energy is temporarily converted into elastic potential energy stored in the deformation. Therefore, total kinetic energy is not conserved during the interaction itself.
Maximum tension the thread can bear is $T_{max} = 3.2 \text{ kg wt} = 3.2 \times g = 3.2 \times 10 = 32 \text{ N}$.
The centripetal force required for circular motion is provided by this tension: $T = m\omega^2 R$.
Given mass $m = 500 \text{ g} = 0.5 \text{ kg}$, radius $R = 4 \text{ m}$.
$32 = 0.5 \times \omega^2 \times 4 \Rightarrow 32 = 2\omega^2 \Rightarrow \omega^2 = 16 \Rightarrow \omega = 4 \text{ rad/sec}$.
The wire lies along the z-axis. The radial distance $r$ from the z-axis to the point $(6 \text{ cm}, 8 \text{ cm}, 10 \text{ cm})$ depends only on the x and y coordinates: $r = \sqrt{6^2 + 8^2} = 10 \text{ cm} = 0.1 \text{ m}$.
Electric field magnitude $E = \frac{2k\lambda}{r} = \frac{2 \times (9 \times 10^9) \times (\frac{10}{9} \times 10^{-9})}{0.1} = \frac{20}{0.1} = 200 \text{ V/m}$.
The direction is strictly radial in the xy-plane: $\hat{r} = \frac{6\hat{i} + 8\hat{j}}{10} = 0.6\hat{i} + 0.8\hat{j}$.
$\vec{E} = 200(0.6\hat{i} + 0.8\hat{j}) = (120\hat{i} + 160\hat{j}) \text{ V/m}$.
Current $I = \frac{dq}{dt} \Rightarrow dq = I dt$.
Total charge $Q = \int_{t_1}^{t_2} I dt = \int_{2}^{6} (4 + 2t) dt$.
$Q = \left[ 4t + t^2 \right]_{2}^{6} = (4(6) + 6^2) - (4(2) + 2^2) = (24 + 36) - (8 + 4) = 60 - 12 = 48 \text{ C}$.
Initially, the bridge is balanced at 20 cm from one end: $\frac{X}{Y} = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4}$.
When $4X$ is balanced against $Y$, the new ratio is $\frac{4X}{Y} = 4 \left(\frac{X}{Y}\right) = 4 \left(\frac{1}{4}\right) = 1$.
Let the new balance point be $l$ cm. So, $\frac{l}{100 - l} = 1$.
$l = 100 - l \Rightarrow 2l = 100 \Rightarrow l = 50 \text{ cm}$.
In Young's Double Slit Experiment, the position of the $n^{\text{th}}$ bright fringe (maximum) from the central maximum is $y_n = \frac{n\lambda D}{d}$.
Given $\lambda = 5000 \mathring{A} = 5 \times 10^{-7} \text{ m}$, $D = 2 \text{ m}$, $d = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}$, and $n = 3$.
$y_3 = \frac{3 \times (5 \times 10^{-7}) \times 2}{2 \times 10^{-4}} = \frac{30 \times 10^{-7}}{2 \times 10^{-4}} = 15 \times 10^{-3} \text{ m} = 1.5 \text{ cm}$.
Let $E_0 = GM/r^2$ be the base field magnitude from mass M. Diagonals perfectly oppose each other.
Initial State: $E_1$ represents the vector sum. Diagonal 1 (3M and M): Net field is $2E_0$ towards 3M. Diagonal 2 (4M and 2M): Net field is $2E_0$ towards 4M. The vectors are perpendicular, so $E_1 = \sqrt{(2E_0)^2 + (2E_0)^2} = 2\sqrt{2} E_0$.
Final State (3M and 4M swapped): Diagonal 1 (4M and M): Net field is $3E_0$ towards 4M. Diagonal 2 (3M and 2M): Net field is $1E_0$ towards 3M. $E_2 = \sqrt{(3E_0)^2 + (E_0)^2} = \sqrt{10} E_0$.
Ratio $\frac{E_1}{E_2} = \frac{2\sqrt{2}}{\sqrt{10}} = \frac{2\sqrt{2}}{\sqrt{2}\sqrt{5}} = \frac{2}{\sqrt{5}}$. So, $\alpha = 2$.
By carefully analyzing the provided timing diagram waveforms:
When Input A = 1 and Input B = 0, Output Y = 0.
When Input A = 0 and Input B = 0, Output Y = 1.
When Input A = 1 and Input B = 1, Output Y = 0.
When Input A = 0 and Input B = 1, Output Y = 0.
Output Y is HIGH (1) only when both inputs are LOW (0). This truth table corresponds exactly to a NOR gate.
The specific resistance is $\rho = \frac{\pi r^2 R}{l}$.
The maximum percentage error is $\frac{\Delta\rho}{\rho}\% = 2\left(\frac{\Delta r}{r}\%\right) + \frac{\Delta R}{R}\% + \frac{\Delta l}{l}\%$.
$\% \Delta r = \frac{0.02}{0.24} \times 100 \approx 8.33\%$.
$\% \Delta R = \frac{1.0}{30} \times 100 \approx 3.33\%$.
$\% \Delta l = \frac{0.01}{4.80} \times 100 \approx 0.2\%$.
Total \% error $= 2(8.33\%) + 3.33\% + 0.2\% = 16.66\% + 3.33\% + 0.2\% = 20.19\% \approx 20\%$.
When a heavy rod hangs vertically, the tension varies along its length, being maximum at the support and zero at the free end. The effective stretching force acts exactly at its center of mass.
The total elongation is mathematically equivalent to the entire weight $Mg$ acting on half of the rod's length ($L/2$).
$\Delta L = \frac{F \cdot L_{eff}}{A \cdot Y} = \frac{(mg)(L/2)}{AY} = \frac{mgL}{2AY}$.
Energy released per fission $E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$.
Power required $P = 100 \text{ Watts} = 100 \text{ J/s}$.
Number of fissions per second $n = \frac{P}{E} = \frac{100}{3.2 \times 10^{-11}} = \frac{10^2}{3.2 \times 10^{-11}} = 0.3125 \times 10^{13} = 3.125 \times 10^{12}$.
In the given logic circuit, the two diodes have their N-sides (cathodes) connected to the input terminals A and B, while their P-sides (anodes) are tied together and connected to a pull-up resistor leading to a 5V supply.
If either A or B is at 0V (LOW), the respective diode is forward-biased, pulling the output C to 0V (LOW).
Only when both A and B are at 5V (HIGH), both diodes are reverse-biased, and the resistor pulls the output C to 5V (HIGH).
This truth table exactly replicates an AND gate.
According to Kepler's Third Law, the time period of revolution ($T$) of a planet around the Sun depends exclusively on the semi-major axis (orbital radius) of its orbit and the mass of the Sun.
$T^2 \propto R_{orbit}^3$.
Shrinking the planet changes its internal composition, radius, and rotation (spin), but does not alter its total mass or its orbital distance from the Sun. Therefore, its time period of revolution remains completely unaffected ($T_1 = T_2$).
A real image implies a concave mirror. The magnification $m = -\frac{v}{u} = -3 \Rightarrow v = 3u$.
Since it's a real image, both object and image lie on the same side of the mirror, meaning both $u$ and $v$ are negative coordinates. The physical distance between them is $|v - u| = |3u - u| = 2|u|$.
Given $2|u| = 40 \text{ cm} \Rightarrow |u| = 20 \text{ cm}$. By sign convention, $u = -20 \text{ cm}$, so $v = -60 \text{ cm}$.
Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-60} + \frac{1}{-20} = \frac{-1 - 3}{60} = -\frac{4}{60} = -\frac{1}{15}$.
$f = -15 \text{ cm}$.
Angular separation of fringes is $\theta = \frac{\lambda}{d}$.
Statement I: Since $\theta$ does not depend on the screen distance $D$, moving the screen away leaves the angular separation constant. (True)
Statement II: Using a larger wavelength $\lambda$ increases the value of the numerator in $\lambda/d$, which strictly increases the angular separation. (False)
Therefore, Statement I is true but Statement II is false.
Let the point of zero gravitational field be at distance $x$ from mass $m$.
$\frac{Gm}{x^2} = \frac{G(9m)}{(R-x)^2} \Rightarrow \frac{1}{x} = \frac{3}{R-x} \Rightarrow R - x = 3x \Rightarrow 4x = R \Rightarrow x = \frac{R}{4}$.
The distance from mass $9m$ is $R - x = \frac{3R}{4}$.
Gravitational potential at this point is the scalar sum: $V = V_1 + V_2 = -\frac{Gm}{R/4} - \frac{G(9m)}{3R/4}$.
$V = -\frac{4Gm}{R} - \frac{12Gm}{R} = -\frac{16Gm}{R}$.
The maximum vertical height attained by a projectile is given by $H = \frac{u^2 \sin^2\theta}{2g}$.
Given initial velocity $u = 280 \text{ m/s}$ and angle $\theta = 30^{\circ}$.
$H = \frac{(280)^2 \sin^2 30^{\circ}}{2 \times 9.8} = \frac{78400 \times (0.5)^2}{19.6} = \frac{78400 \times 0.25}{19.6} = \frac{19600}{19.6} = 1000 \text{ m}$.
Magnetic susceptibility ($\chi$) values characterize magnetic materials:
A. Diamagnetic materials repel magnetic fields slightly: small, negative $\chi$ ($0 > \chi \ge -1$). (Matches II)
B. Ferromagnetic materials are strongly attracted: large, positive $\chi$ ($\chi \gg 1$). (Matches III)
C. Paramagnetic materials are weakly attracted: small, positive $\chi$ ($0 < \chi < \epsilon$). (Matches IV)
D. Non-magnetic materials do not respond to fields: $\chi = 0$. (Matches I)
Correct sequence is A-II, B-III, C-IV, D-I.
According to the ideal gas equation $PV = nRT$, we can rearrange it to $T = \left(\frac{P}{nR}\right)V$.
When plotting $T$ (y-axis) against $V$ (x-axis), the graph is a straight line passing through the origin with a slope equal to $\frac{P}{nR}$.
This means a steeper slope corresponds to a higher pressure.
From the given graph, the slope order is $P_1 > P_2 > P_3$. Therefore, the correct relation is $P_1 > P_2 > P_3$. (Note: Assuming standard visual ordering where P1 is the steepest and P3 is the shallowest as typically depicted in Charles' law deviations).