8 Apr 2024 Shift 1

The order of magnitude of a number expressed in scientific notation as $a \times 10^b$ (where $1 \leq a < 10$) is defined based on the value of $a$. According to the convention used \in this context: If $a \leq 5$, then the order of magnitude is $b$. If $a > 5$, then the order of magnitude is $b+1$. Therefore, $b$ is the order of magnitude when $a \leq 5$. Now, evaluating the options: Option A states that $b$ is the order of magnitude for $a \geq 5$. This is incorrect because when $a \geq 5$, the order of magnitude is $b$ only if $a \leq 5$, but for $a > 5$, it is $b+1$. Option B states that $b$ is the order of magnitude for $a \leq 5$. This is correct, as per the convention. Option C states that $a$ is the order of magnitude for $b \leq 5$. This is incorrect because the order of magnitude is an exponent (a power of 10), not the coefficient $a$. Option D states that $b$ is the order of magnitude for $5 < a \leq 10$. This is incorrect because when $a > 5$, the order of magnitude is $b+1$, not $b$. Thus, the correct option is B.

We need to find the percentage error in Young's modulus $Y = 49000\frac{M}{l}$. For $Y = k\frac{M}{l}$, taking the logarithm gives $\ln Y = \ln k + \ln M - \ln l$, and differentiating yields $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l}$. From the graph paper, the smallest scale division along the load axis is 5 g, so $\Delta M = 5$ g, and along the extension axis it is 0.02 cm, giving $\Delta l = 0.02$ cm. $\frac{\Delta Y}{Y} = \frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01 = 0.02 = 2\%$ The correct answer is Option (2): 2% .

Second hand: in 30 min = 30 revolutions. Distance = 30×2π×0.75 = 45π m. Minute hand: in 30 min = 0.5 revolution. Distance = 0.5×2π×0.60 = 0.6π m. x = 45π - 0.6π = 44.4π = 44.4×3.14 = 139.4 m. The correct answer is Option (3): 139.4.

A stationary particle breaks into two parts. We need the ratio $K_B : K_A$. By conservation of momentum, the initial momentum is zero since the particle is stationary, so $m_A v_A = m_B v_B$. The kinetic energies of the parts are given by $K_A = \frac{1}{2}m_A v_A^2, \quad K_B = \frac{1}{2}m_B v_B^2$. Thus, the ratio of the kinetic energies is $\frac{K_B}{K_A} = \frac{m_B v_B^2}{m_A v_A^2} = \frac{m_B v_B}{m_A v_A} \times \frac{v_B}{v_A}$. Since $m_B v_B = m_A v_A$ from conservation of momentum, the first factor equals 1, giving $\frac{K_B}{K_A} = 1 \times \frac{v_B}{v_A} = \frac{v_B}{v_A}$. Therefore $K_B : K_A = v_B : v_A$. The correct answer is Option (1): $v_B : v_A$ .

Three bodies A, B, C with equal kinetic energies and masses 400 g, 1.2 kg, 1.6 kg. Find the ratio of their momenta. Recall the relation between momentum and kinetic energy. $KE = \frac{p^2}{2m}$, so $p = \sqrt{2mKE}$. Since KE is the same for all three: $p_A : p_B : p_C = \sqrt{m_A} : \sqrt{m_B} : \sqrt{m_C}$ Substitute the masses. $m_A = 0.4$ kg, $m_B = 1.2$ kg, $m_C = 1.6$ kg. So $= \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6}$ Multiply all by $\sqrt{10}$ (equivalently, consider $4:12:16$ inside the roots): $= \sqrt{4} : \sqrt{12} : \sqrt{16} = 2 : 2\sqrt{3} : 4 = 1 : \sqrt{3} : 2$ The correct answer is Option (2): $1 : \sqrt{3} : 2$ .

A cricket ball of mass 150 g moving at 20 m/s is caught in 0.1 s. Find the force on the player's hand. Recall Newton's second law in terms of impulse: $F = \frac{\Delta p}{\Delta t}$ where $\Delta p$ is the change \in momentum and $\Delta t$ is the time interval. The initial momentum is $p_i = mv = 0.15 \times 20 = 3$ kg m/s, and the final momentum is $p_f = 0$ (ball is brought to rest). Hence, $|\Delta p| = |p_f - p_i| = 3 \text{ kg m/s}$. Substituting into the formula gives $F = \frac{|\Delta p|}{\Delta t} = \frac{3}{0.1} = 30 \text{ N}$. By Newton's third law, the force exerted by the ball on the player's hand equals the force exerted by the hand on the ball. The correct answer is Option (4): 30 N .

For a planet of mass $m$ moving \in a circular orbit of radius $r$ around the Sun (mass $M_\odot$), the necessary centripetal force is provided by gravitation: $\frac{G M_\odot m}{r^{2}} = \frac{m v^{2}}{r}$ Solving for the orbital speed, $v = \sqrt{\frac{G M_\odot}{r}} \quad -(1)$ Angular momentum about the Sun is $L = m v r$ Substituting $v$ from $(1)$, $L = m r \sqrt{\frac{G M_\odot}{r}} = m \sqrt{G M_\odot\, r} \quad -(2)$ Applying $(2)$ to planets $A$ and $B$: $L_A = m_1 \sqrt{G M_\odot\, r_1}$ $L_B = m_2 \sqrt{G M_\odot\, r_2}$ Given $L_B = 3L_A$, $m_2 \sqrt{G M_\odot\, r_2} = 3 m_1 \sqrt{G M_\odot\, r_1}$ Canceling the common factor $\sqrt{G M_\odot}$, $m_2 \sqrt{r_2} = 3 m_1 \sqrt{r_1} \quad -(3)$ From $(3)$, $\frac{\sqrt{r_2}}{\sqrt{r_1}} = \frac{3 m_1}{m_2}$ Squaring both sides: $\frac{r_2}{r_1} = \frac{9 m_1^{2}}{m_2^{2}} \quad -(4)$ Kepler's third law for circular orbits gives the time period: $T = 2\pi \sqrt{\frac{r^{3}}{G M_\odot}} \propto r^{3/2} \quad -(5)$ Therefore, $\frac{T_A}{T_B} = \left(\frac{r_1}{r_2}\right)^{3/2} \quad -(6)$ Insert $\frac{r_1}{r_2}$ from $(4)$ into $(6)$: $\frac{T_A}{T_B} = \left(\frac{m_2^{2}}{9 m_1^{2}}\right)^{3/2}$ Compute the exponent: $\left(\frac{m_2^{2}}{9 m_1^{2}}\right)^{3/2} = \left(\frac{m_2}{m_1}\right)^{3} \left(\frac{1}{9}\right)^{3/2}$ Since $9 = 3^{2}$, we have $9^{3/2} = 3^{3} = 27$. Thus $\frac{T_A}{T_B} = \frac{1}{27}\left(\frac{m_2}{m_1}\right)^{3}$ Hence, the correct ratio is $\frac{1}{27}\left(\frac{m_2}{m_1}\right)^{3}$, which corresponds to Option B .

We need to identify the correct form of Bernoulli's equation. Bernoulli's equation applies to steady, incompressible, non-viscous fluid flow along a streamline. It is derived from the work-energy theorem (or equivalently, conservation of energy per unit volume). The three terms represent pressure energy, gravitational potential energy, and kinetic energy - all per unit volume: $P + \rho g h + \frac{1}{2}\rho v^2 = \text{constant}$ where $P$ is the fluid pressure, $\rho$ is the density, $g$ is gravitational acceleration, $h$ is the height, and $v$ is the fluid velocity. Option (1) uses mass $m$ instead of density $\rho$ - incorrect (that would give energy, not energy per unit volume). Option (2): $P + \rho gh + \frac{1}{2}\rho v^2 = \text{constant}$ - this is the correct form. Option (3) is missing the $\frac{1}{2}$ \in the kinetic term - incorrect. Option (4) has an extra $\frac{1}{2}$ in the gravitational term - incorrect. The correct answer is Option (2) .

For an adiabatic process, $PV^{\gamma}=\text{constant}$ There are two adiabatic curves: For adiabatic through a and c: $P_aV_a^{\gamma}=P_cV_c^{\gamma}$ So, $\frac{P_a}{P_c}=\frac{V_c^{\gamma}}{V_a^{\gamma}}$ For adiabatic through b and d: $P_bV_b^{\gamma}=P_dV_d^{\gamma}$ So, $\frac{P_b}{P_d}=\frac{V_d^{\gamma}}{V_b^{\gamma}}$ Now a and b lie on same isotherm, so $P_aV_a=P_bV_b$ $\frac{P_a}{P_b}=\frac{V_b}{V_a}$ Also c and d lie on same isotherm, so $P_cV_c=P_dV_d$ $\frac{P_c}{P_d}=\frac{V_d}{V_c}$ Now divide the two adiabatic equations: $\frac{P_a/P_c}{P_b/P_d}=\left(\frac{\frac{V_C}{V_a}}{\frac{V_d}{V_b}}\right)^{\gamma}$ Using isothermal relations, $\frac{(V_c/V_a)}{(V_d/V_b)}=\left(\frac{V_bV_c}{V_aV_d}\right)^{\gamma}$ This simplifies only if $V_bV_c=V_aV_d$ Hence, $\frac{V_a}{V_d}=\frac{V_b}{V_c}$

We need to find the ratio of specific heats at constant volume for a monoatomic gas and a diatomic (rigid) gas. For an ideal gas, the molar specific heat at constant volume is $C_v = \frac{f}{2}R$, where $f$ is the number of degrees of freedom and $R$ is the universal gas constant. For a monoatomic gas, $f = 3$ (3 translational), so $C_{v_1} = \frac{3}{2}R$. For a diatomic gas (rigid, i.e., no vibrational modes), $f = 5$ (3 translational + 2 rotational), so $C_{v_2} = \frac{5}{2}R$. Therefore, $\frac{C_{v_1}}{C_{v_2}} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5}$. The correct answer is Option (2): $\frac{3}{5}$ .

Two conducting spheres of radii $a$ and $b$ are connected by a wire. We need the ratio of their charges. When two conductors are connected by a wire, charge flows until they reach the same electric potential. The potential of a conducting sphere with charge $Q$ and radius $r$ is: $V = \frac{Q}{4\pi\varepsilon_0 r}$ Equating the potentials of the two spheres gives $V_1 = V_2 \implies \frac{Q_1}{4\pi\varepsilon_0 a} = \frac{Q_2}{4\pi\varepsilon_0 b},$ which simplifies to $\frac{Q_1}{a} = \frac{Q_2}{b} \implies \frac{Q_1}{Q_2} = \frac{a}{b}.$ The correct answer is Option (1): $\frac{a}{b}$ .

The electromotive force (EMF) of the cell is $E = 2 \text{ V}$.
The terminal potential difference ($V$) of a cell is given by the formula $V = E - Ir$, where $I$ is the current flowing through the circuit and $r$ is the internal resistance.
When no current is drawn from the cell (i.e., in an open-circuit condition, $I=0$), the terminal potential difference is equal to its EMF.
Assuming the question refers to this open-circuit condition or the inherent EMF of the cell, then $V = E$.
Therefore, the terminal potential difference of the cell is $2 \text{ V}$.

Paramagnetic substances possess unpaired electrons, causing their atomic magnetic moments to align with an external magnetic field, thus statement A is true. They are weakly attracted to magnetic fields, not strongly, making statement B false. Their magnetic susceptibility, $\chi$, is small and positive ($\chi > 0$), so statement C is true. Consequently, they move from regions of weaker to stronger magnetic fields, not the reverse, rendering statement D false. Therefore, only statements A and C are correct.

An LCR circuit at resonance has its resistance halved. Find the new current amplitude. At resonance in an LCR circuit, the inductive reactance equals the capacitive reactance: $X_L = X_C$. These cancel out, so the impedance equals just the resistance: $Z = R \quad \text{(at resonance)}$ $I = \frac{V}{Z} = \frac{V}{R}$ The resonance condition depends on $L$ and $C$ (not $R$), so the circuit remains at resonance. With $R' = R/2$: $I' = \frac{V}{R'} = \frac{V}{R/2} = \frac{2V}{R} = 2I$ The current amplitude doubles. The correct answer is Option (4): double .

The critical angle is $45°$. Find the ratio of refractive indices $n_1 : n_2$. Total internal reflection occurs when light travels from a denser to a rarer medium. The critical angle $C$ is given by: $\sin C = \frac{n_2}{n_1}$ where $n_1$ is the refractive index of the denser (first) medium and $n_2$ is that of the rarer (second) medium. $\sin 45° = \frac{n_2}{n_1} \implies \frac{1}{\sqrt{2}} = \frac{n_2}{n_1}$ $\frac{n_1}{n_2} = \sqrt{2}$ $n_1 : n_2 = \sqrt{2} : 1$ The correct answer is Option (2): $\sqrt{2} : 1$ .

A proton and electron have the same de Broglie wavelength. Find the ratio of their kinetic energies. $\lambda = \frac{h}{p} \implies p = \frac{h}{\lambda}$ Since both have the same $\lambda$, they have the same momentum $p$. $KE = \frac{p^2}{2m}$ Since $p$ is the same for both: $\frac{KE_p}{KE_e} = \frac{p^2/(2m_p)}{p^2/(2m_e)} = \frac{m_e}{m_p} = \frac{m_e}{1836 \, m_e} = \frac{1}{1836}$ $KE_p : KE_e = 1 : 1836$. The correct answer is Option (4): $1 : 1836$ .

Find the area of a non-reflecting surface given the radiation force and intensity. For a perfectly absorbing (non-reflecting) surface, the radiation pressure is: $P_{\text{rad}} = \frac{I}{c}$ where $I$ is the intensity (power per unit area) and $c = 3 \times 10^8$ m/s is the speed of light. The force on the surface is: $F = P_{\text{rad}} \times A = \frac{I \times A}{c}$. $I = 360$ W/cm$^2 = 360 \times 10^4$ W/m$^2 = 3.6 \times 10^6$ W/m$^2$. $F = \frac{I \times A}{c} \implies A = \frac{Fc}{I}$ $A = \frac{2.4 \times 10^{-4} \times 3 \times 10^8}{3.6 \times 10^6} = \frac{7.2 \times 10^4}{3.6 \times 10^6} = 0.02 \text{ m}^2$ The correct answer is Option (4): 0.02 m$^2$ .

Binding energy = $18 \times 10^8$ J. Find the mass difference between nucleons and the nucleus. $E = \Delta m \cdot c^2$ The binding energy represents the energy equivalent of the mass defect $\Delta m$. $\Delta m = \frac{E}{c^2} = \frac{18 \times 10^8}{(3 \times 10^8)^2} = \frac{18 \times 10^8}{9 \times 10^{16}} = 2 \times 10^{-8} \text{ kg}$ $1 \text{ kg} = 10^9 \; \mu\text{g}$, so: $\Delta m = 2 \times 10^{-8} \times 10^9 = 20 \; \mu\text{g}$ The correct answer is Option (2): 20 $\mu$g .

Let the output of the first AND gate be $X_1$ and the output of the NOR gate be $X_2$.

1. The output of the first AND gate is $X_1 = A \cdot B$.
2. The output of the NOR gate is $X_2 = \overline{A+B}$.
3. The final output Y is the AND of $X_1$ and $X_2$:
   $Y = X_1 \cdot X_2 = (A \cdot B) \cdot (\overline{A+B})$
4. Applying De Morgan's theorem, $\overline{A+B} = \bar{A} \cdot \bar{B}$:
   $Y = (A \cdot B) \cdot (\bar{A} \cdot \bar{B})$
5. Rearranging the terms using commutativity and associativity:
   $Y = (A \cdot \bar{A}) \cdot (B \cdot \bar{B})$
6. Using the Boolean identity $X \cdot \bar{X} = 0$:
   $Y = 0 \cdot 0 = 0$

We need to find the density of a sphere measured with a vernier caliper. 9 main scale divisions = 10 vernier scale divisions. Smallest main scale division = 1 mm. $\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 - \frac{9}{10} = 0.1 \text{ mm} = 0.01 \text{ cm}$ Main scale reading = 2 cm. Vernier coincidence = 2nd division. $d = \text{MSR} + \text{VSR} \times \text{LC} = 2 + 2 \times 0.01 = 2.02 \text{ cm}$ Radius: $r = 1.01$ cm. $V = \frac{4}{3}\pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.01)^3$ $(1.01)^3 \approx 1.0303$ $V = \frac{4}{3} \times 3.1416 \times 1.0303 \approx 4.32 \text{ cm}^3$ $\rho = \frac{m}{V} = \frac{8.635}{4.32} \approx 2.0 \text{ g/cm}^3$ The correct answer is Option (1): 2.0 g/cm$^3$ .

Let's write the three vectors given in their deconstructed form: $\vec{OP} = A\hat{i}$ $\vec{OQ} = A\hat{j}$ $\vec{OR} = \dfrac{A}{\sqrt{2}}\hat{i} - \dfrac{A}{\sqrt{2}}\hat{j}$ Thus, adding the vectors, we get, $\vec{OP}+\vec{OQ}+\vec{OR} =\vec{R} = \left(A + \dfrac{A}{\sqrt{2}}\right)\hat{i} + \left(A - \dfrac{A}{\sqrt{2}}\right)\hat{j}$ And hence, the magnitude of the resultant vector, $|\vec{R}| = \sqrt{\left(A + \dfrac{A}{\sqrt{2}}\right)^2 + \left(A - \dfrac{A}{\sqrt{2}}\right)^2}$ $|\vec{R}| = \sqrt{2A^2 + 2\dfrac{A^2}{2}} = A\sqrt{3}$ Therefore, the value of $x$ is $3$.

Since the plate is uniform, we can find the COM by treating the object as a large rectangle with a smaller section removed from its top. We consider the full $3 \times 2$ rectangle and subtract the $1 \times 1$ empty square region \in the upper middle. Total Rectangle ($A_1$): The area is $3 \times 2 = 6$, with its geometric center at $(1.5, 1)$. Removed Square ($A_2$): The area is $1 \times 1 = 1$, with its geometric center at $(1.5, 1.5)$. Coordinates of the Center of Mass Because the figure is symmetric about the line $x = 1.5$, the $x$-coordinate is: $x_{cm} = \frac{A_1x_1 - A_2x_2}{A_1 - A_2} = \frac{6(1.5) - 1(1.5)}{5} = 1.5$ The $y$-coordinate is determined by the distribution of the remaining area along the vertical axis: $y_{cm} = \frac{A_1y_1 - A_2y_2}{A_1 - A_2} = \frac{6(1) - 1(1.5)}{6 - 1} = \frac{4.5}{5} = 0.9$ Taking the ratio of the $x$ and $y$ coordinates: $\text{Ratio} = \frac{x_{cm}}{y_{cm}} = \frac{1.5}{0.9}$ $\frac{15}{9} = \frac{n}{9}$ The value of $n$ is 15.

A liquid column of height 0.04 cm balances the excess pressure of a soap bubble. Find the diameter of the bubble. $\Delta P = \frac{4S}{R}$ where $S$ is the surface tension and $R$ is the radius. (Factor of 4 because a soap bubble has two surfaces.) $\Delta P = \rho g h = 8000 \times 10 \times 0.04 \times 10^{-2} = 8000 \times 10 \times 4 \times 10^{-4} = 32 \text{ Pa}$ $\frac{4S}{R} = \rho g h \implies R = \frac{4S}{\rho g h} = \frac{4 \times 0.28}{32} = \frac{1.12}{32} = 0.035 \text{ m}$ $D = 2R = 0.07 \text{ m} = 7 \text{ cm}$ The correct answer is 7 .

A closed and open pipe of same length. Find $a$ if the ratio of their 7th overtones is $\frac{a-1}{a}$. A closed pipe produces only odd harmonics: 1st, 3rd, 5th, 7th, ... The $n$th overtone corresponds to the $(2n+1)$th harmonic. The 7th overtone = 15th harmonic. $f_{\text{closed}} = \frac{15v}{4L}$ An open pipe produces all harmonics. The 7th overtone = 8th harmonic. $f_{\text{open}} = \frac{8v}{2L} = \frac{4v}{L}$ $\frac{f_{\text{closed}}}{f_{\text{open}}} = \frac{15v/(4L)}{4v/L} = \frac{15v}{4L} \times \frac{L}{4v} = \frac{15}{16}$ $\frac{15}{16} = \frac{a-1}{a} \implies a = 16$ The correct answer is 16 .

Find the electric flux through a surface of area 4 m$^2$. $\Phi = \vec{E} \cdot \vec{A} = \vec{E} \cdot (A\hat{n})$ $\vec{E} = \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}}$, $\hat{n} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$, $A = 4$ m$^2$. $\vec{E} \cdot \hat{n} = \frac{1}{\sqrt{6}} \times \frac{1}{\sqrt{6}}(2\times2 + 6\times1 + 8\times1) = \frac{4+6+8}{6} = \frac{18}{6} = 3$ $\Phi = (\vec{E}\cdot\hat{n}) \times A = 3 \times 4 = 12 \text{ Vm}$ The correct answer is 12 .

Find temperature $t$ (\in Kelvin) given resistance values at different temperatures. $R_t = R_0(1 + \alpha t)$ At $t = 100°C$: $R_{100} = R_0(1 + 100\alpha)$ $10.2 = 10(1 + 100\alpha) \implies 1.02 = 1 + 100\alpha \implies \alpha = \frac{0.02}{100} = 0.0002 \; °C^{-1}$ $10.95 = 10(1 + 0.0002t) \implies 1.095 = 1 + 0.0002t \implies t = \frac{0.095}{0.0002} = 475°C$ $T = 475 + 273 = 748 \text{ K}$ The correct answer is 748 .

An electron with KE = 5 eV enters a magnetic field of 3 $\mu$T. Find the electric field $E$ for the electron to move undeflected. $KE = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2 \times KE}{m}}$ Convert KE to Joules: $5 \text{ eV} = 5 \times 1.6 \times 10^{-19} = 8 \times 10^{-19}$ J. $v = \sqrt{\frac{2 \times 8 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{\frac{16 \times 10^{-19}}{9 \times 10^{-31}}} = \sqrt{1.778 \times 10^{12}}$ $v \approx 1.33 \times 10^6 \text{ m/s}$ $qE = qvB \implies E = vB$ $E = 1.33 \times 10^6 \times 3 \times 10^{-6} = 3.99 \approx 4 \text{ N/C}$ The correct answer is 4 NC$^{-1}$.

The work done in pulling the loop out of the magnetic field is entirely dissipated as heat in the resistance of the loop, assuming the pulling is done slowly and uniformly (constant velocity).

First, calculate the induced electromotive force (EMF) in the loop. The change in magnetic flux ($\Delta\Phi$) occurs as the entire area of the loop ($A$) exits the magnetic field ($B$) over a time ($t$). For $N$ turns, the change in flux is:
$\Delta\Phi = N B A$
Given $N=10$, $B=0.5 \text{ T}$, $A=3.6 \times 10^{-3} \text{ m}^2$:
$\Delta\Phi = 10 \times 0.5 \text{ T} \times 3.6 \times 10^{-3} \text{ m}^2 = 18 \times 10^{-3} \text{ Wb}$
The average induced EMF ($E$) is the rate of change of magnetic flux:
$E = \frac{\Delta\Phi}{t}$
Given $t=1.0 \text{ s}$:
$E = \frac{18 \times 10^{-3} \text{ Wb}}{1.0 \text{ s}} = 18 \times 10^{-3} \text{ V}$
The work done ($W$) in pulling the loop out is equal to the energy dissipated in the resistance ($R$) over time ($t$). This can be calculated using the formula $W = P \times t$, where $P = E^2/R$ is the power dissipated:
$W = \frac{E^2}{R} t$
Given $R=100 \Omega$ and $t=1.0 \text{ s}$:
$W = \frac{(18 \times 10^{-3} \text{ V})^2}{100 \Omega} \times 1.0 \text{ s}$
$W = \frac{324 \times 10^{-6} \text{ V}^2}{100 \Omega} \times 1.0 \text{ s}$
$W = 3.24 \times 10^{-6} \text{ J}$
Rounding to one significant figure due to the magnetic field value ($B=0.5 \text{ T}$), the work done is $3 \times 10^{-6} \text{ J}$.

The final answer is $\boxed{3}$.

Find the angular divergence for the second-order minima in single-slit diffraction. $a\sin\theta = n\lambda$ where $a$ is the slit width, $n$ is the order, and $\lambda$ is the wavelength. $\sin\theta = \frac{2\lambda}{a} = \frac{2 \times 600 \times 10^{-9}}{0.4 \times 10^{-3}} = \frac{1200 \times 10^{-9}}{4 \times 10^{-4}} = 3 \times 10^{-3}$ Since $\sin\theta$ is very small, $\theta \approx \sin\theta = 3 \times 10^{-3}$ rad. The angular divergence is the total angle between the 2nd order minima on both sides of the central maximum: $\text{Angular divergence} = 2\theta = 2 \times 3 \times 10^{-3} = 6 \times 10^{-3} \text{ rad}$ The correct answer is 6 .

We need to find the maximum velocity of an alpha particle in a scattering experiment. At the distance of closest approach, all kinetic energy is converted to electrostatic potential energy: $\frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0}\frac{(2e)(Ze)}{d}$ $Z = 80$, $d = 4.5 \times 10^{-14}$ m, $m = 6.72 \times 10^{-27}$ kg, $e = 1.6 \times 10^{-19}$ C, $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$. $\frac{1}{2}mv^2 = \frac{9 \times 10^9 \times 2 \times 80 \times (1.6 \times 10^{-19})^2}{4.5 \times 10^{-14}}$ $9 \times 10^9 \times 160 \times 2.56 \times 10^{-38} = 9 \times 160 \times 2.56 \times 10^{-29}$ $= 9 \times 409.6 \times 10^{-29} = 3686.4 \times 10^{-29} = 3.6864 \times 10^{-26}$ $\frac{1}{2}mv^2 = \frac{3.6864 \times 10^{-26}}{4.5 \times 10^{-14}} = 8.192 \times 10^{-13}$ J $v^2 = \frac{2 \times 8.192 \times 10^{-13}}{6.72 \times 10^{-27}} = \frac{16.384 \times 10^{-13}}{6.72 \times 10^{-27}} = 2.438 \times 10^{14}$ $v = \sqrt{2.438 \times 10^{14}} \approx 1.56 \times 10^7 \text{ m/s} = 156 \times 10^5$ m/s The maximum velocity is approximately $156 \times 10^5$ m/s. Therefore, the answer is 156.