Assertion (A) is true: The $\text{Fe}^{56}$ nucleus has the highest binding energy per nucleon (approx. 8.8 MeV), making it the most stable nucleus in nature compared to heavier elements like $\text{U}^{235}$.
Reason (R) is false: The total binding energy of a nucleus is the product of its binding energy per nucleon and its total number of nucleons (mass number A). Since $\text{U}^{235}$ has a much larger mass number than $\text{Fe}^{56}$ ($235 > 56$), the *total* binding energy of $\text{U}^{235}$ is much greater than that of $\text{Fe}^{56}$, even though its binding energy *per nucleon* is lower.
Dynamic or AC resistance is defined as the ratio of the change in applied voltage to the change in current.
$R = \frac{\Delta V}{\Delta I}$
Given $\Delta V = 5 \text{ V}$ and $\Delta I = 2 \mu\text{A} = 2 \times 10^{-6} \text{ A}$.
$R = \frac{5}{2 \times 10^{-6}} = 2.5 \times 10^6 \Omega$.
The most accurate measurement value is considered to be the arithmetic mean of all the observations.
$\text{Mean Length} = \frac{3.29 + 3.28 + 3.29 + 3.31 + 3.28}{5} \text{ cm}$
$= \frac{16.45}{5} \text{ cm} = 3.29 \text{ cm}$.
When $n$ identical drops of potential $V$ coalesce to form a single big drop, the potential of the big drop is given by $V_{big} = n^{2/3} V_{small}$.
Given $n = 1000$ and $V_{small} = 1 \text{ V}$.
$V_{big} = (1000)^{2/3} \times 1 = (10^3)^{2/3} = 10^2 = 100 \text{ V}$.
In a meter bridge, the balance condition is $\frac{R_1}{R_2} = \frac{l_1}{100 - l_1}$.
Initially, $l_1 = 40 \text{ cm}$. So, $\frac{R_1}{R_2} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3}$.
When $R_2$ is replaced by $2R_2$, the new balance condition is $\frac{R_1}{2R_2} = \frac{l_2}{100 - l_2}$.
Substituting $\frac{R_1}{R_2} = \frac{2}{3}$, we get $\frac{1}{2} \left( \frac{2}{3} \right) = \frac{l_2}{100 - l_2} \Rightarrow \frac{1}{3} = \frac{l_2}{100 - l_2}$.
$100 - l_2 = 3l_2 \Rightarrow 4l_2 = 100 \Rightarrow l_2 = 25 \text{ cm}$.
Potential difference between any two points is a relative physical quantity that does not depend on the arbitrary choice of the reference potential (or zero point).
If the reference potential at infinity is changed from 0 to 10 V, the absolute potential of both points A and B increases by 10 V.
New potential difference $V_A' - V_B' = (V_A + 10) - (V_B + 10) = V_A - V_B = 10 \text{ V}$.
In the given circuit, the 3 H and 6 H inductors are connected in parallel. Their equivalent inductance is:
$L_p = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 \text{ H}$.
This combination is in series with the 2 H inductor. Therefore, the total equivalent inductance between A and B is:
$L_{eq} = L_p + 2 \text{ H} = 2 + 2 = 4 \text{ H}$.
The standard equation for Simple Harmonic Motion is $x = A \sin(\omega t + \phi_0)$.
Given amplitude $A = 0.1 \text{ m}$, frequency $f = 25 \text{ Hz}$, and initial phase $\phi_0 = \frac{\pi}{4} \text{ radians}$.
Angular frequency $\omega = 2\pi f = 2\pi(25) = 50\pi \text{ rad/s}$.
Substituting the values, we get: $x = 0.1 \sin\left(50\pi t + \frac{\pi}{4}\right)$.
For a capacitor, the charge $Q$ stored is directly proportional to the potential difference $V$ across its plates, expressed as $Q = CV$.
Since capacitance $C$ is a constant for a given capacitor, the relationship between $Q$ and $V$ is linear ($y = mx$).
Therefore, the graph between $Q$ and $V$ is a straight line passing through the origin.
Two sources of light are said to be coherent if they emit light waves of the same frequency (or wavelength) and maintain a constant phase difference with respect to time.
Therefore, both conditions (1) and (2) must be satisfied.
The resistivity of a metal increases with an increase in temperature due to the increased thermal agitation of lattice ions, which causes more frequent collisions of free electrons (decreasing relaxation time).
Conversely, the resistivity of a metal decreases when its temperature is decreased.
Statement A is correct: Dimensions define the physical nature of a quantity, represented as powers of fundamental base quantities.
Statement B is correct: Mean absolute error is a measure of the spread of values in the same units as the original measurement. Relative (or fractional) error is the ratio of mean absolute error to the mean value ($\frac{\Delta x}{x}$), so the units cancel out, making it dimensionless and unitless.
The particle is projected horizontally. Initial velocity $u_x = 30 \text{ m/s}$ and $u_y = 0$.
After $t = 3 \text{ s}$, the horizontal velocity remains constant: $v_x = 30 \text{ m/s}$.
The vertical velocity due to gravity is $v_y = gt = 10 \times 3 = 30 \text{ m/s}$ (downwards).
Let $\theta$ be the angle the velocity vector makes with the vertical (y-axis). Then $\tan\theta = \frac{v_x}{v_y}$.
$\tan\theta = \frac{30}{30} = 1 \Rightarrow \theta = 45^{\circ}$.
The given P-V curve shows $PV = \text{constant}$. This is the characteristic equation of an isothermal process, meaning the temperature $T$ remains constant throughout the process.
In a T-P diagram where T is on the y-axis, a constant temperature process is represented by a horizontal straight line.
From the P-V diagram, as the process goes from state 1 to state 2, the volume increases, which means the pressure P decreases.
Therefore, on the T-P graph, the horizontal line must have an arrow pointing in the direction of decreasing pressure (to the left). This perfectly matches graph (3).
Comparing the given wave equation $\vec{E} = 5 \cos(10^5 t + kx)\hat{k} \text{ V/m}$ with the standard form $E = E_0 \cos(\omega t + kx)$:
Amplitude $E_0 = 5 \text{ V/m}$.
Angular frequency $\omega = 10^5 \text{ rad/s}$.
Wave speed in free space $c = \frac{\omega}{k} \Rightarrow 3 \times 10^8 = \frac{10^5}{k} \Rightarrow k = \frac{1}{3 \times 10^3} = \frac{1}{3} \times 10^{-3} \text{ m}^{-1}$.
Wavelength $\lambda = \frac{2\pi}{k} = 2\pi \times (3 \times 10^3) = 6\pi \times 10^3 \text{ m}$.
Since all three individual statements match our derived values, all options are correct.
Coulomb's force between two charges is $F = \frac{k q_1 q_2}{d^2}$.
If both charges are doubled, the new charges are $q_1' = 2q_1$ and $q_2' = 2q_2$.
New force $F' = \frac{k (2q_1) (2q_2)}{d^2} = 4 \left( \frac{k q_1 q_2}{d^2} \right) = 4F$.
A. Torque ($\tau$) on a rigid body is given by $\tau = I\alpha$ (matches iii).
B. Rotational kinetic energy is $\frac{1}{2} I\omega^2$ (matches iv).
C. Moment of inertia of a uniform rod of mass $M$ and length $L$ about a perpendicular axis through its center is $\frac{1}{12}ML^2$ (matches ii).
D. Angular momentum ($L$ or $J$) of a rotating body is $I\omega$ (matches i).
Correct combination is a-iii, b-iv, c-ii, d-i.
In a moving coil galvanometer, the magnetic field exerts a deflecting torque on the coil. As the coil rotates, the suspension wire (usually made of phosphor-bronze) gets twisted.
The elasticity of this wire generates a mechanical restoring torque (couple) that opposes the magnetic torque and eventually brings the coil to equilibrium.
Given $d = 0.3 \text{ mm} = 3 \times 10^{-4} \text{ m}$ and $D = 1 \text{ m}$.
Distance of the $n^{\text{th}}$ dark fringe from the center is $y_{nD} = (2n-1)\frac{\lambda D}{2d}$.
For the 2nd dark fringe ($n=2$): $y_{2D} = \frac{3\lambda D}{2d} = 0.3 \text{ cm} = 3 \times 10^{-3} \text{ m}$.
$\frac{3 \times \lambda \times 1}{2 \times 3 \times 10^{-4}} = 3 \times 10^{-3} \Rightarrow \frac{\lambda}{2 \times 10^{-4}} = 3 \times 10^{-3} \Rightarrow \lambda = 6 \times 10^{-7} \text{ m}$.
Distance of the $n^{\text{th}}$ bright fringe is $y_{nB} = \frac{n\lambda D}{d}$.
For 4th bright fringe ($n=4$): $y_{4B} = \frac{4 \times (6 \times 10^{-7}) \times 1}{3 \times 10^{-4}} = 8 \times 10^{-3} \text{ m} = 0.8 \text{ cm}$.
When thin lenses are placed in contact, their equivalent power is the algebraic sum of their individual powers.
$P_{eq} = P_1 + P_2 = 10\text{D} + 5\text{D} = +15\text{D}$.
The focal length in meters is $F = \frac{1}{P_{eq}}$. In centimeters, $F = \frac{100}{P_{eq}}$.
$F = \frac{100}{15} \text{ cm} = \frac{20}{3} \text{ cm}$.
When the Atwood machine is inside a lift accelerating upwards with $a = g$, the effective acceleration due to gravity acting on the masses is $g_{eff} = g + a = g + g = 2g$.
The tension in the string of an Atwood machine is $T = \frac{2 m_1 m_2}{m_1 + m_2} g_{eff}$.
$T = \frac{2(2)(3)}{2 + 3} (2g) = \frac{12}{5} (2 \times 10) = 2.4 \times 20 = 48 \text{ N}$.
Let $|\vec{A}| = |\vec{B}| = x$. The angle between them is $\theta$.
Given: $|\vec{A} + \vec{B}| = 2|\vec{A} - \vec{B}|$.
$\sqrt{x^2 + x^2 + 2x^2\cos\theta} = 2 \sqrt{x^2 + x^2 - 2x^2\cos\theta}$.
Squaring both sides: $2x^2(1 + \cos\theta) = 4 [2x^2(1 - \cos\theta)]$.
$1 + \cos\theta = 4(1 - \cos\theta) \Rightarrow 1 + \cos\theta = 4 - 4\cos\theta$.
$5\cos\theta = 3 \Rightarrow \cos\theta = \frac{3}{5}$. Thus, $\theta = \cos^{-1}\left(\frac{3}{5}\right)$.
The particle moves in a circle with constant speed $v$. The velocity vectors at points A and B have the same magnitude but different directions. The angle between the velocity vectors at A and B is equal to the angle subtended at the center, $\theta = 40^{\circ}$.
The magnitude of the change in velocity is $|\Delta \vec{v}| = |\vec{v}_B - \vec{v}_A| = \sqrt{v^2 + v^2 - 2v^2\cos\theta} = 2v \sin\left(\frac{\theta}{2}\right)$.
$|\Delta \vec{v}| = 2v \sin\left(\frac{40^{\circ}}{2}\right) = 2v \sin 20^{\circ}$.
The input voltage to the parallel circuit is $10\text{V}$. The Zener diode has a breakdown voltage of $V_z = 5\text{V}$.
Since $10\text{V} > 5\text{V}$, the Zener diode operates in the breakdown region and maintains a constant voltage of $5\text{V}$ across its parallel branches.
Therefore, the potential difference across the $1\text{k}\Omega$ resistor is $5\text{V}$.
Current through the $1\text{k}\Omega$ resistor $I = \frac{V}{R} = \frac{5\text{V}}{1000\Omega} = 5 \times 10^{-3} \text{ A} = 5 \text{ mA}$.
By the Principle of Continuity, the volume flow rate must be constant. Rate of flow in the main tube = Total rate of flow out of all holes.
$A_{tube} \times V_{tube} = N \times A_{hole} \times v_{hole}$
$(\pi R^2) \times V = (2n) \times \left( \pi \left(\frac{r}{2}\right)^2 \right) \times v_{hole}$
$R^2 V = 2n \left(\frac{r^2}{4}\right) v_{hole} = \frac{n r^2}{2} v_{hole}$
$v_{hole} = \frac{2 V R^2}{n r^2}$.
The de-Broglie wavelength of a charged particle accelerated through a potential difference $V$ is $\lambda = \frac{h}{\sqrt{2mqV}}$.
For a proton: $\lambda_p = \frac{h}{\sqrt{2 m_p e V}} = \lambda$.
For an $\alpha$-particle: mass $m_\alpha = 4m_p$, charge $q_\alpha = 2e$, and accelerating potential $V_\alpha = V/8$.
$\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)(V/8)}} = \frac{h}{\sqrt{2m_p e V \times (4 \times 2 \times \frac{1}{8})}} = \frac{h}{\sqrt{2m_p e V (1)}} = \lambda_p = \lambda$.
The particle starts from rest ($u=0$) and accelerates uniformly. Let the acceleration be $a$. At time $t_1$, velocity is $v$, so $a = \frac{v}{t_1}$.
The velocity at any given time $t$ is $v(t) = at = \left(\frac{v}{t_1}\right)t$.
By the work-energy theorem, the work done on the particle up to time $t$ is equal to its kinetic energy at that time:
$W = \frac{1}{2} m (v(t))^2 = \frac{1}{2} m \left(\frac{v}{t_1}t\right)^2 = \frac{1}{2} \left[ \frac{m v^2}{t_1^2} \right] t^2$.
The acceleration due to gravity at depth $d$ inside the Earth is given by $g_d = g\left(1 - \frac{d}{R}\right)$.
We are given that $g_d = \frac{1}{n} g$.
Substituting this into the equation: $\frac{g}{n} = g\left(1 - \frac{d}{R}\right) \Rightarrow \frac{1}{n} = 1 - \frac{d}{R}$.
$\frac{d}{R} = 1 - \frac{1}{n} = \frac{n-1}{n} \Rightarrow d = \frac{R(n-1)}{n}$.
The given wave equation is $y = a \sin \pi\left(\frac{t}{2} - \frac{x}{4}\right) = a \sin\left(\frac{\pi}{2}t - \frac{\pi}{4}x\right)$.
Comparing with standard form $y = A \sin(\omega t - kx)$, we have $\omega = \frac{\pi}{2} \text{ rad/s}$ and $k = \frac{\pi}{4} \text{ m}^{-1}$.
The wave velocity is $v = \frac{\omega}{k} = \frac{\pi/2}{\pi/4} = 2 \text{ m/s}$.
Distance moved by the wave in time $t = 8 \text{ s}$ is $d = v \times t = 2 \times 8 = 16 \text{ m}$.
The expression $\epsilon_0 \frac{d\phi_E}{dt}$ defines the displacement current ($I_d$) according to Maxwell's addition to Ampere's Law.
Since it fundamentally represents an electric current, its dimensions are identical to those of an electric current.
Therefore, the dimensional formula is $[A]$.
The distance fallen from rest in the $n^{\text{th}}$ second is $S_n = u + \frac{g}{2}(2n - 1)$.
Given $u = 0$, $g = 10 \text{ m/s}^2$, and $S_n = 35 \text{ m}$.
$35 = 0 + \frac{10}{2}(2n - 1) = 5(2n - 1) \Rightarrow 2n - 1 = 7 \Rightarrow 2n = 8 \Rightarrow n = 4$.
This means the particle fell 35m in the $4^{\text{th}}$ second. The next second is the $5^{\text{th}}$ second.
$S_5 = \frac{10}{2} (2(5) - 1) = 5(10 - 1) = 5 \times 9 = 45 \text{ m}$.
For an adiabatic process, the governing equation is $P V^\gamma = \text{constant}$.
From the ideal gas law $PV = nRT \Rightarrow V \propto \frac{T}{P}$. Substituting V:
$P \left(\frac{T}{P}\right)^\gamma = \text{const} \Rightarrow P^{1-\gamma} T^\gamma = \text{const} \Rightarrow P \propto T^{\frac{\gamma}{\gamma - 1}}$.
For a monoatomic gas, $\gamma = \frac{5}{3}$.
$C = \frac{\gamma}{\gamma - 1} = \frac{5/3}{5/3 - 1} = \frac{5/3}{2/3} = \frac{5}{2}$.
The concave mirror has focal length $f = -10 \text{ cm}$.
The rod is 10 cm long. Its end closer to the pole is at $u_1 = -20 \text{ cm}$. This point happens to be the center of curvature ($C = 2f = -20 \text{ cm}$). The image of this end forms exactly at C, so $v_1 = -20 \text{ cm}$.
The other end of the rod is at $u_2 = -(20 + 10) = -30 \text{ cm}$.
Using the mirror formula: $\frac{1}{v_2} + \frac{1}{-30} = \frac{1}{-10} \Rightarrow \frac{1}{v_2} = \frac{1}{30} - \frac{1}{10} = \frac{1 - 3}{30} = -\frac{2}{30} = -\frac{1}{15} \Rightarrow v_2 = -15 \text{ cm}$.
Length of the image $= |v_1 - v_2| = |-20 - (-15)| = |-5| = 5 \text{ cm}$.
The ideal gas inside initially has pressure P and temp T. The force expanding the wire is $F_1 = P \cdot A$.
Initial elongation due to internal pressure $\Delta L_1 = \frac{F_1 L}{AY} = \frac{(PA)L}{AY} = \frac{PL}{Y}$.
The gas temperature is slowly increased to 2T. Since the volume remains practically constant (expansion is negligible compared to total volume), the new pressure $P_2 = 2P$ (Gay-Lussac's law).
New expanding force $F_2 = (2P) \cdot A$.
New elongation $\Delta L_2 = \frac{F_2 L}{AY} = \frac{(2PA)L}{AY} = \frac{2PL}{Y}$.
The *increase* in length caused by the heating process is $\Delta L = \Delta L_2 - \Delta L_1 = \frac{2PL}{Y} - \frac{PL}{Y} = \frac{PL}{Y}$.
In the series LCR circuit, Inductive Reactance $X_L = 15 \Omega$, Capacitive Reactance $X_C = 11 \Omega$, and Resistance $R = 3 \Omega$.
Impedance $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{3^2 + (15 - 11)^2} = \sqrt{9 + 16} = 5 \Omega$.
RMS Current $I = \frac{V_{rms}}{Z} = \frac{10}{5} = 2 \text{ A}$.
The potential difference across the series combination of the inductor and capacitor is the product of current and net reactance of the LC part.
$V_{LC} = I \times |X_L - X_C| = 2 \times |15 - 11| = 2 \times 4 = 8 \text{ V}$.
For a uniform disc, mass $M = \text{Volume} \times \text{Density} = (\pi R^2 t) \times \rho$. Since M and thickness t are constant, $R^2 \propto \frac{1}{\rho}$.
The moment of inertia of a disc is $I = \frac{1}{2} M R^2$. Since M is the same for both discs, $I \propto R^2$.
Combining these relations gives $I \propto \frac{1}{\rho}$.
Therefore, $\frac{I_1}{I_2} = \frac{\rho_2}{\rho_1} = \frac{8.9}{7.2}$, or $8.9 : 7.2$.
The rated resistance of the bulb is $R_b = \frac{V^2}{P} = \frac{100^2}{500} = \frac{10000}{500} = 20 \Omega$.
The normal operating current of the bulb is $I = \frac{P}{V} = \frac{500}{100} = 5 \text{ A}$.
To operate perfectly on a 230 V supply, the total resistance of the series circuit must be $R_{total} = \frac{230\text{V}}{5\text{A}} = 46 \Omega$.
Since $R_{total} = R_b + R$, the required series resistance is $R = 46 - 20 = 26 \Omega$.
For a pipe closed at one end, the fundamental frequency is $f_1 = \frac{v}{4L}$.
$f_1 = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \text{ Hz}$.
The permitted harmonics in a closed pipe are odd multiples of the fundamental: $f_n = (2n - 1)f_1$.
Possible frequencies are 100, 300, 500, 700, 900, 1100, 1300, 1500... Hz.
The frequencies strictly below 1250 Hz are 100, 300, 500, 700, 900, and 1100 Hz. There are exactly 6 such natural oscillations.
The range of water from a hole at height $y$ from the ground and depth $h$ from the surface is $R = 2\sqrt{hy}$. The velocity of efflux $v = \sqrt{2gh}$ depends on the effective head.
To double the range (from R to 2R) while maintaining the same physical hole geometry, the velocity of efflux must be doubled: $v' = 2v$.
Since $v \propto \sqrt{h_{eff}}$, doubling $v$ requires making the effective pressure head four times its original value: $h_{eff} = 4h = 4 \times 10 = 40 \text{ m}$.
This means we need to supply an equivalent pressure of an additional 30 meters of water column.
Extra pressure $\Delta P = \rho g \Delta h = 10^3 \times 10 \times 30 = 3 \times 10^5 \text{ Pa} = 3 \text{ atm}$.
For an ideal transformer, input power equals output power: $P_s = P_p = 10 \text{ kW} = 10^4 \text{ W}$.
The secondary voltage is $V_s = \frac{P_s}{I_s} = \frac{10^4}{25} \text{ V}$.
The transformer relation states $\frac{V_p}{V_s} = \frac{N_p}{N_s}$. We are given the turns ratio $N_p/N_s = 8/1 = 8$.
Therefore, $V_p = 8 \times V_s = 8 \times \left(\frac{10^4}{25}\right) = \frac{(10^4) \times 8}{25} \text{ V}$.
Horizontal range $R = \frac{u^2 \sin 2\theta}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$. Maximum height $h = \frac{u^2 \sin^2\theta}{2g}$.
We know the maximum possible range for a given speed is $R_{max} = \frac{u^2}{g}$. We must find an expression equivalent to $\frac{u^2}{g}$ using R and h.
Consider $\frac{R^2}{8h} = \frac{4 u^4 \sin^2\theta \cos^2\theta / g^2}{8 (u^2 \sin^2\theta / 2g)} = \frac{4 u^4 \sin^2\theta \cos^2\theta / g^2}{4 u^2 \sin^2\theta / g} = \frac{u^2 \cos^2\theta}{g}$.
Also note that $2h = \frac{u^2 \sin^2\theta}{g}$.
Adding these two derived components: $\frac{R^2}{8h} + 2h = \frac{u^2 \cos^2\theta}{g} + \frac{u^2 \sin^2\theta}{g} = \frac{u^2}{g} (\cos^2\theta + \sin^2\theta) = \frac{u^2}{g} = R_{max}$.
The energy required for an electronic transition in a hydrogen-like ion is $\Delta E \propto \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)$.
For $n=2 \to n=3$: $E_{2\to 3} = K \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = K \left(\frac{1}{4} - \frac{1}{9}\right) = K \left(\frac{5}{36}\right) = E$.
For $n=1 \to n=2$: $E_{1\to 2} = K \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = K \left(1 - \frac{1}{4}\right) = K \left(\frac{3}{4}\right)$.
Ratio $\frac{E_{1\to 2}}{E_{2\to 3}} = \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5}$.
Therefore, $E_{1\to 2} = \frac{27E}{5}$.
The Curie-Weiss law states that the magnetic susceptibility of a ferromagnetic material above its Curie temperature is $\chi = \frac{C}{T - T_c}$.
At $T_1 = 400 \text{ K}$, $\chi_1 = 0.02$. So, $0.02 = \frac{C}{400 - 300} = \frac{C}{100} \Rightarrow C = 0.02 \times 100 = 2$.
At $T_2 = 350 \text{ K}$, the new susceptibility is $\chi_2 = \frac{C}{350 - 300} = \frac{2}{50} = 0.04$.
A. The attractive force between the plates of a charged capacitor is $F = \frac{Q^2}{2A\epsilon_0} = \frac{1}{2}\epsilon_0 A E^2$ (Matches ii).
B. By the equipartition theorem, the average internal energy of a diatomic gas molecule (having 5 degrees of freedom) at temperature T is $\frac{5}{2}kT$ (Matches iii).
C. According to Biot-Savart law for a moving charge, the magnetic field is $B = \frac{\mu_0}{4\pi} \frac{qv\sin\theta}{r^2}$. Therefore, $B \propto qv$ (Matches v).
Correct arrangement: A-ii, B-iii, C-v.
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{280 \times 10^{-9}} \text{ J} = \frac{19.86 \times 10^{-26}}{280 \times 10^{-9}} \text{ J} = 7.09 \times 10^{-19} \text{ J}$.
To convert this energy to electron-volts (eV): $E \approx \frac{7.09 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.43 \text{ eV}$.
By Einstein's photoelectric equation, the maximum kinetic energy is $K_{max} = E - W = 4.43 \text{ eV} - 2.5 \text{ eV} = 1.93 \text{ eV}$.
The closest matching option is 1.9 eV.