5 Apr 2024 Shift 1

We need to find the angle between $\vec{Q}$ and the resultant of $(2\vec{Q} + 2\vec{P})$ and $(2\vec{Q} - 2\vec{P})$. Find the resultant. $\vec{R} = (2\vec{Q} + 2\vec{P}) + (2\vec{Q} - 2\vec{P}) = 4\vec{Q}$ The resultant is $4\vec{Q}$, which is \in the same direction as $\vec{Q}$. Therefore, the angle between $\vec{Q}$ and $\vec{R} = 4\vec{Q}$ is $0°$ . The correct answer is Option (2): $0°$.

Given readings: $4.62$ s, $4.632$ s, $4.6$ s, $4.64$ s. Calculate arithmetic mean. $\text{Mean} = \frac{4.62 + 4.632 + 4.6 + 4.64}{4} = \frac{18.492}{4} = 4.623 \text{ s}$ Apply significant figures rule. When adding/averaging numbers, the result should have the same number of decimal places as the measurement with the fewest decimal places. The readings have: 2, 3, 1, and 2 decimal places respectively. The least is 1 decimal place (from 4.6 s). So the mean should be rounded to 1 decimal place: $4.6$ s. The correct answer is Option (3): 4.6 s.

We need to find the dimensions of $\sqrt{uG}$ where $u$ is energy density and $G$ is the gravitational constant. Dimensions of energy density $u$. $[u] = \frac{\text{Energy}}{\text{Volume}} = \frac{[ML^2T^{-2}]\$}{[L^3]\$} = \$[ML^{-1}T^{-2}]\$$ Dimensions of gravitational constant $G$. From $F = \frac{Gm_1m_2}{r^2}$: $[G] = \frac{[F]\$[r^2]\$}{[m^2]\$} = \frac{[MLT^{-2}]\$\$[L^2]\$}{[M^2]\$} = \$[M^{-1}L^3T^{-2}]\$$ Dimensions of $\sqrt{uG}$. $[uG] = \$[ML^{-1}T^{-2}]\$ \times \$[M^{-1}L^3T^{-2}]\$ = \$[M^0L^2T^{-4}]\$ = \$[L^2T^{-4}]\$$ $[\sqrt{uG}] = \$[LT^{-2}]\$$ Identify the physical quantity. $[LT^{-2}]$ is the dimension of acceleration, which is force per unit mass . The correct answer is Option (4): Force per unit mass.

A wooden block (5 kg) and iron cylinder (25 kg) go down together with acceleration $a = 0.1$ m/s$^2$. The total mass is $m = 5 + 25 = 30$ kg. According to Newton's second law, the system accelerates downward at $0.1$ m/s$^2$. The forces acting on the system are weight (downward) and the normal reaction from the floor (upward). - Weight (downward): $mg = 30 \times 9.8 = 294$ N - Normal reaction from floor (upward): $N$ According to Newton's second law, $mg - N = ma$ Therefore, $N = mg - ma = m(g - a) = 30(9.8 - 0.1) = 30 \times 9.7 = 291 \text{ N}$ By Newton's third law, the action force of the system on the floor equals $N = 291$ N downward. The correct answer is Option (2): 291 N.

The work done against the force of gravity depends on the change in the object's gravitational potential energy, which is calculated as $W = \Delta PE = mgh$. This value is determined solely by the mass ($m$) and the vertical displacement ($h$), and it is independent of the path taken to lift the object.

In Case 1, the work done is $W_1 = mgh = (50 \text{ kg}) \cdot g \cdot (20 \text{ m})$.
In Case 2, despite moving along a ramp, the vertical height gained is the same. Thus, the work done against gravity is $W_2 = mgh = (50 \text{ kg}) \cdot g \cdot (20 \text{ m})$.
Since $W_1 = W_2$, the ratio of the work done is $W_1 : W_2 = 1 : 1$.

For a hollow sphere of mass $M$ and radius $R$, the moment of inertia about its diameter is: $I_1 = \frac{2}{3}MR^2$ The radius of gyration is $k_1 = \sqrt{\frac{I_1}{M}} = \sqrt{\frac{2}{3}R^2}$ For the solid cylinder of mass $M$, radius $R$, and length $L = 4R$, the axis AB passes through one end. Using the Parallel Axis Theorem: $I_2 = I_{cm} + M\left(\frac{L}{2}\right)^2$ $I_2 = M\left(\frac{R^2}{4} + \frac{L^2}{12}\right) + M\frac{L^2}{4} = M\left(\frac{R^2}{4} + \frac{L^2}{3}\right)$ $I_2 = M\left(\frac{R^2}{4} + \frac{(4R)^2}{3}\right) = M\left(\frac{R^2}{4} + \frac{16R^2}{3}\right)$ $I_2 = MR^2 \left(\frac{3 + 64}{12}\right) = \frac{67}{12}MR^2$ The radius of gyration is $k_2 = \sqrt{\frac{I_2}{M}} = \sqrt{\frac{67}{12}R^2}$ $\text{Ratio} = \frac{k_1}{k_2} = \frac{\sqrt{\frac{2}{3}R^2}}{\sqrt{\frac{67}{12}R^2}}$ = $\sqrt{\frac{8}{67}}$ $x\ =\ 67$

A simple pendulum at height $R$ above Earth's surface has time period $T_1 = 4$ s. We need $T_2$ at height $2R$. The time period of a simple pendulum is $T = 2\pi\sqrt{\frac{l}{g'}}$ where $g'$ is the effective gravitational acceleration at height $h$: $g' = \frac{g}{(1 + h/R)^2}$ At height $h = R$: $g_1 = \frac{g}{(1+1)^2} = \frac{g}{4}$ $T_1 = 2\pi\sqrt{\frac{l}{g/4}} = 2\pi\sqrt{\frac{4l}{g}}$ At height $h = 2R$: $g_2 = \frac{g}{(1+2)^2} = \frac{g}{9}$ $T_2 = 2\pi\sqrt{\frac{l}{g/9}} = 2\pi\sqrt{\frac{9l}{g}}$ Dividing to find the relation between $T_2$ and $T_1$: $\frac{T_2}{T_1} = \frac{\sqrt{9}}{\sqrt{4}} = \frac{3}{2}$ Therefore, $T_2 = \frac{3}{2}T_1$ Multiplying both sides by 2: $2T_2 = 3T_1$ This corresponds to $3T_1 = 2T_2$. The correct answer is Option (4): $3T_1 = 2T_2$.

Coulomb force: $F_e=\frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}$ Gravitational force: $F_g=\frac{Gm_pm_e}{r^2}$ Ratio, $\frac{F_e}{F_g}=\frac{1}{4\pi\epsilon_0}\frac{e^2}{Gm_pm_e}$ Using approximate values, $\frac{1}{4\pi\epsilon_0}=9\times10^9$ $e=1.6\times10^{-19}$ $G=6.67\times10^{-11}$ $m_p=1.67\times10^{-27}$ $m_e=9.1\times10^{-31}$ Substituting gives order $\frac{F_e}{F_g}\sim10^{39}$

For a planet revolving around the sun in a circular orbit of radius aaa: The gravitational force provides the centripetal force. $\frac{GMm}{a^2}=\frac{mv^2}{a}$ From this, we get the orbital velocity: $v^2=\frac{GM}{a}$ Now kinetic energy of the planet is: $K=\frac{1}{2}mv^2=\frac{1}{2}m\cdot\frac{GM}{a}=\frac{GMm}{2a}$ So, (A) matches with (II) Gravitational potential energy of the sun-planet system is: $U=-\frac{GMm}{a}$ So, (B) matches with (I) Total mechanical energy is: $E=K+U=\frac{GMm}{2a}-\frac{GMm}{a}=-\frac{GMm}{2a}​$​ So, (C) matches with (IV) Escape energy per unit mass at the surface of a planet is: $\ \frac{\ 1}{2}​v_e^2​=\ \frac{\ GM}{r}$ $v_e=\sqrt{\frac{2GM}{r}}$ Energy per unit mass: $v_e^2=\frac{GM}{r}$ So, (D) matches with (III) Thus, the correct matching is: (A)−(II), (B)−(I), (C)−(IV), (D)−(III)

We need to evaluate both statements about capillarity and contact angle. Statement I: When a capillary tube is dipped into a liquid, the liquid neither rises nor falls. The contact angle may be $0°$. If the liquid neither rises nor falls, the contact angle must be $90°$ (not $0°$). When the contact angle is $0°$, the liquid completely wets the surface and rises in the capillary. Statement I is false . Statement II: The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well. This is correct. The contact angle depends on the nature of both the solid surface and the liquid (as well as the surrounding medium). It depends on the relative strengths of adhesive forces (solid-liquid) and cohesive forces (liquid-liquid). Statement II is true . The correct answer is Option (3): Statement I is false but Statement II is true.

For a cyclic process, ΔU=0 So from first law, Q=W Thus heat absorbed equals net work done, which is area enclosed by the cycle on the P-V diagram. The loop is a circle touching P=340, P=60 and V=340, V=60 So diameter is $340−60=280$ Hence radius is r=140 Area enclosed $W=\pi r^2$ $=\pi(140)^2$ $=19600\pi$ Now units: Pressure is \in cc and volume \in kPa (as labeled), so $1(\text{kPa})(\text{cc})=10^3\times10^{-6}=10^{-3}J$ Thus $Q=19600\pi\times10^{-3}$ $=19.6\pi\ \text{J}=61.6J$

Collision frequency of H$_2$ at 27°C is Z. Find collision frequency at 127°C. The collision frequency $Z \propto n^2 \sigma^2 \sqrt{T}$ where $n$ is number density and $\sigma$ is collision cross-section. In a closed chamber at constant volume, $n$ is constant and thus $Z \propto \sqrt{T}$. For $T_1 = 27 + 273 = 300$ K and $T_2 = 127 + 273 = 400$ K, $\frac{Z_2}{Z_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{400}{300}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}},$ so $Z_2 = \frac{2}{\sqrt{3}} Z.$ The correct answer is Option B : $\frac{2}{\sqrt{3}}Z$.

In the given circuit, resistors $R_2, R_3$, and $R_4$ are connected \in parallel as their terminals are connected to the same two points. The equivalent resistance of this parallel combination ($R_p$) is:
$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} = \frac{1}{8} + \frac{1}{4} + \frac{1}{8} = \frac{1+2+1}{8} = \frac{4}{8} = \frac{1}{2} \text{ }\Omega^{-1}$
This gives $R_p = 2\Omega$.

The resistor $R_1$ is \in series with this parallel combination. Thus, the total equivalent resistance of the circuit ($R_{eq}$) is:
$R_{eq} = R_1 + R_p = 10\Omega + 2\Omega = 12\Omega$

Using Ohm's law, the total current ($I$) supplied by the battery is:
$I = \frac{V}{R_{eq}} = \frac{12 \text{ V}}{12\Omega} = 1 \text{ A}$
The equivalent resistance is $12\Omega$ and the current is $1 \text{ A}$.

We use Ampere's Circuital Law, which states that the line integral of the magnetic field $\vec{B}$ around a closed loop is proportional to the enclosed current $I_{enc}$.
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$
To find the magnetic field outside the cable, we choose a circular Amperian loop concentric with the cable, with a radius $r$ greater than the radius of the outer conductor. This loop encloses both the central conductor (current $I$) and the outer conductor (current $-I$). The net current enclosed is $I_{enc} = I + (-I) = 0$. By symmetry, the integral becomes $B(2\pi r)$. Thus, Ampere's Law gives $B(2\pi r) = \mu_0(0) = 0$. Since $r \neq 0$, the magnetic field $B$ must be zero.

If two circular loops are concentric and lie in the same plane: Magnetic field at centre due to loop A: $B=\frac{(\mu ₀I)}{2b}$ Flux through loop B: $Φ=B\times areaofB$ $=\frac{\mu ₀I}{2b}\times (\pi a^2)$ So mutual inductance: $M=\frac{Φ}{I}$ $=\frac{(\mu ₀\pi a^2)}{2b}$

The given values are the amplitude $V_0 = 40$ V, the frequency $f = 4$ kHz = $4000$ Hz, and the capacitance $C = 12 \mu$F $= 12 \times 10^{-6}$ F. The displacement current through a capacitor equals the conduction current, so $i = C\frac{dV}{dt}$ For $V = V_0 \sin(\omega t)$, the maximum current is $i_{max} = C \omega V_0 = C \times 2\pi f \times V_0$ $i_{max} = 12 \times 10^{-6} \times 2\pi \times 4000 \times 40$ $i_{max} = 12 \times 10^{-6} \times 8\pi \times 10^3 \times 40$ $= 12 \times 8\pi \times 40 \times 10^{-3}$ $= 3840\pi \times 10^{-3}$ $= 3.84\pi \approx 3.84 \times 3.14 \approx 12.06 \text{ A}$ Thus the maximum displacement current is nearly $12$ A. The correct answer is Option (2): 12 A.

When a point source of light is placed at the focus of a convex lens, the light rays after passing through the lens become parallel (collimated). Parallel rays have plane wavefronts - the wavefront is perpendicular to the direction of propagation, and since all rays are parallel, the wavefront is a flat plane. The correct answer is Option (2): plane.

According to Einstein's photoelectric equation, the maximum kinetic energy of an emitted photoelectron is given by $K_{max} = h\nu - \phi_0$ where $h$ is Planck's constant, $\nu$ is the frequency of incident radiation, and $\phi_0$ is the work function of the material.

The stopping potential $V_0$ is related to the maximum kinetic energy by $K_{max} = eV_0$. Combining these equations, we get $eV_0 = h\nu - \phi_0$, or $V_0 = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}$.

This is the equation of a straight line for $V_0$ vs $\nu$. The slope of this line is $\frac{h}{e}$. Since $h$ and $e$ are universal constants, the slope is the same for all materials, which is consistent with the parallel lines shown. Thus, Statement I is correct.

From the graph, the threshold frequency of $M_2$ is greater than that of $M_1$ ($\nu_{02} > \nu_{01}$). This implies the work function of $M_2$ is greater than that of $M_1$ ($\phi_{02} > \phi_{01}$). For the same incident frequency $\nu$, the kinetic energy is $K_{max} = h\nu - \phi_0$. Since $M_2$ has a larger work function, it will emit photoelectrons with less kinetic energy compared to $M_1$. Thus, Statement II is incorrect.

For an electron orbiting a nucleus with charge $Ze$: The potential energy of the electron is given by $U = -\frac{kZe^2}{r}$. Using the centripetal force condition $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$, we find the kinetic energy as $KE = \frac{1}{2}mv^2 = \frac{kZe^2}{2r} = -\frac{U}{2}$ Hence the total energy is $E = KE + U = -\frac{U}{2} + U = \frac{U}{2}$ It follows that $2E = U$ The correct answer is Option (2): \boxed{2E = U}.

The circuit consists of four NOR gates. Let the two terminals connected to the bulb be $P$ and $Q$. For the bulb to glow (ON), there must be a potential difference across it, which means the logic levels at $P$ and $Q$ must be different ($P \neq Q$).

The top part of the circuit determines the logic level at $P$. The NOR gate with input A acts as a NOT gate ($\overline{A}$). The NOR gate with inputs B and C gives $\overline{B+C}$. These two outputs feed into a final NOR gate.
$P = \overline{(\overline{A} + \overline{B+C})}$
Using De Morgan's theorem, we simplify this expression:
$P = \overline{\overline{A}} \cdot \overline{\overline{B+C}} = A \cdot (B+C)$

The bottom part of the circuit determines the logic level at $Q$. The NOR gate with input D acts as a NOT gate.
$Q = \overline{D}$

The bulb glows if $P \neq Q$. Let's test the combination from option C: $A = 1, B = 0, C = 0, D = 0$.
For this input, we calculate P:
$P = 1 \cdot (0+0) = 1 \cdot 0 = 0$
And we calculate Q:
$Q = \overline{0} = 1$
Since $P = 0$ and $Q = 1$, we have $P \neq Q$. Therefore, the potential difference exists, and the bulb will glow.

A body starts from rest on a frictionless plane. We are given the formula for the ratio $\frac{S_{n-1}}{S_n} = \left(1 - \frac{2}{x}\right)$ for $n = 10$, and we need to find $x$. We first recall the distance covered \in the $n$-th second under uniform acceleration. For a body starting from rest (initial velocity $u = 0$) with uniform acceleration $a$, the distance covered from $t = 0$ to $t = n$ is: $S(n) = \frac{1}{2}an^2$ The distance covered \in the $n$-th second (between $t = n-1$ and $t = n$) is: $S_n = S(n) - S(n-1) = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2$ Expanding $(n-1)^2 = n^2 - 2n + 1$: $S_n = \frac{a}{2}\left[n^2 - (n^2 - 2n + 1)\right]\$ = \frac{a}{2}(2n - 1)$ Similarly, replacing $n$ with $n-1$ \in the formula gives: $S_{n-1} = \frac{a}{2}(2(n-1) - 1) = \frac{a}{2}(2n - 3)$ Next, we compute the ratio $\frac{S_{n-1}}{S_n}$. $\frac{S_{n-1}}{S_n} = \frac{\frac{a}{2}(2n-3)}{\frac{a}{2}(2n-1)} = \frac{2n - 3}{2n - 1}$ This fraction can be rewritten as: $\frac{2n - 3}{2n - 1} = \frac{(2n-1) - 2}{2n - 1} = 1 - \frac{2}{2n - 1}$ Since $\frac{S_{n-1}}{S_n} = 1 - \frac{2}{x}$, equating this to $1 - \frac{2}{2n - 1}$ gives: $\frac{2}{x} = \frac{2}{2n - 1}$ $x = 2n - 1$ Finally, substituting $n = 10$ gives: $x = 2(10) - 1 = 20 - 1 = 19$ The value of $x$ is $\boxed{19}$.

All three masses move upward with acceleration a=$2 \text{ ms}^{-2}$ Total mass: $m_{total}=4+6+10=20kg$ Now treat the three blocks as a single body. Forces on the system: Upward: $T_1$ Downward: total weight $=20g$ Apply Newton's second law (upward positive): $T_1-20g=20\times2$ Substitute $g = 10 \text{ m/s}^2$ $T_1-200=40$ $T_1=240N$ Final answer: 240 N

The density is $\rho = 6 \times 10^4$ kg/m$^3$, the breaking stress is $\sigma = 1.2 \times 10^8$ N/m$^2$, and the effective acceleration is $g' = g/3 = 10/3$ m/s$^2$. The wire will break under its own weight when the stress at its top equals the breaking stress. $\sigma = \frac{F}{A} = \frac{\rho A L g'}{A} = \rho L g'$ At the breaking point, the length L is given by $L = \frac{\sigma}{\rho g'} = \frac{1.2 \times 10^8}{6 \times 10^4 \times \frac{10}{3}}$ $= \frac{1.2 \times 10^8}{2 \times 10^5} = 600 \text{ m}$ Therefore, the maximum length of the wire is 600 m.

Three capacitors are given as $C_1 = 25\mu$F, $C_2 = 30\mu$F, $C_3 = 45\mu$F, with a supply voltage $V = 100$ V. When these capacitors are connected \in parallel, the equivalent capacitance is $C_p = 25 + 30 + 45 = 100 \mu\text{F}$ and the energy stored \in this configuration is $E = \frac{1}{2}C_p V^2 = \frac{1}{2} \times 100 \times 10^{-6} \times 10000 = 0.5 \text{ J}$ In the case of series connection, the equivalent capacitance satisfies $\frac{1}{C_s} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45}$ LCM of 25, 30, 45 = 450: $\frac{1}{C_s} = \frac{18 + 15 + 10}{450} = \frac{43}{450}$ $C_s = \frac{450}{43} \mu\text{F}$ Thus the energy stored \in series is $E_s = \frac{1}{2}C_s V^2 = \frac{1}{2} \times \frac{450}{43} \times 10^{-6} \times 10000 = \frac{450}{43} \times 5 \times 10^{-3}$ Finally, the ratio of the energy stored \in series to that \in parallel is $\frac{E_s}{E} = \frac{C_s}{C_p} = \frac{450/43}{100} = \frac{450}{4300} = \frac{9}{86}$ Therefore, $E_s = \frac{9}{86}E$, which implies that $x = 86\,$.

The capacitor ($C = 1.5\mu$F) discharges through a wire of resistance $R$ and the electric field drops to $\frac{1}{3}$ of its initial value \in $6.6\mu$s. Because the electric field is proportional to the voltage across the capacitor, $E \propto V$, the voltage during discharge follows $V = V_0 e^{-t/RC}$ When $V = V_0/3$, we have $\frac{1}{3} = e^{-t/RC}$ $\ln 3 = \frac{t}{RC}$ $RC = \frac{t}{\ln 3}$ Substituting $t = 6.6 \mu$s and $\ln 3 = \log_e 3 = 1.1$ gives $RC = \frac{6.6 \times 10^{-6}}{1.1} = 6 \times 10^{-6} \text{ s}$ $R = \frac{6 \times 10^{-6}}{1.5 \times 10^{-6}} = 4 \text{ }\Omega$ Hence, the answer is 4.

In half-deflection method, Current through galvanometer: $I=\frac{E}{(R+G)}$ Deflection \theta \propto I \Rightarrow $\theta =kI$ \Rightarrow $\theta =\frac{kE}{(R+G)}$ So: $\frac{1}{\theta }=\frac{\left(R+G\right)}{(kE)}$ \Rightarrow $1/\theta =(1/kE)R+(G/kE)$ This is of the form: $y=mx+c$ So slope $=1/(kE)$ From graph: Take two points: $(R=2,1/\theta =1)$ $(R=6,1/\theta =2)$ Slope: $m=\frac{(2−1)}{(6−2)}=1/4$ So: $1/(kE)=1/4$ Look at the markings: the vertical values like $1,1\frac{1}{2},2$ are not raw values - they are compressed scale values Because of this, the slope you calculate is actually 4 \times smaller than the true slope So real slope $=4\times (1/4)=1$ Now: $1/(kE)=1$ $\Rightarrow kE=1$ $\Rightarrow k\times2=1$ $\Rightarrow k=\frac{1}{2}A/div$ Now convert to required form: k=0.5A/div=5\times10⁻^1A/div So answer = 5

Given: Current $I = 2$ A, resistance $R = 1\Omega$, resistivity $\rho = 2 \times 10^{-6} \Omega$m, area $A = 10$ mm$^2 = 10 \times 10^{-6}$ m$^2$, and mass $m = 500$ g = 0.5 kg. First, we find the length of the wire using the relation $R = \frac{\rho L}{A}$ which gives $L = \frac{RA}{\rho} = \frac{1 \times 10 \times 10^{-6}}{2 \times 10^{-6}} = 5 \text{ m}$. Next, the magnetic force must balance the weight of the wire, so $BIL = mg$ and hence $B = \frac{mg}{IL} = \frac{0.5 \times 10}{2 \times 5} = \frac{5}{10} = 0.5 \text{ T}$. Therefore, $B = 0.5$ T $= 5 \times 10^{-1}$ T. The answer is 5.

Given: L = 1 H C = 20 μF = 20 × 10⁻⁶ F R = 300 Ω v(t) = 50√2 sin(100t) So: $V_{rms}=50V$ \omega = 100 rad/s step 1: calculate reactances Inductive reactance: $X_L=\omega L=100\times1=100\Omega$ Capacitive reactance: $X_C=1/(\omega C)$ = 1 / (100 \times 20\times 10⁻⁶) = 1 / (2\times 10⁻³) = 500 \Omega step 2: impedance of circuit $Z=\sqrt{\ \left(R^{2+}(X_L−X_C)^2\right)}$ $=\sqrt{\ \left(300^{2+}(100−500)^2\right)}$ $=\sqrt{\ 250000}$ $=500\Omega$ step 3: current $I_{rms}=V_{rms}/Z=50/500=0.1A$ step 4: voltage across capacitor $V_C=I\times X_C=0.1\times500=50V$

We are given that $\lambda = 5000$ Å $= 5 \times 10^{-7}$ m, $d = 0.3$ mm $= 3 \times 10^{-4}$ m, and $D = 200$ cm $= 2$ m. The position of the $n$-th maximum \in YDSE is given by $x_n = \frac{n\lambda D}{d}$ For the third maximum ($n = 3$), we then have $x_3 = \frac{3 \times 5 \times 10^{-7} \times 2}{3 \times 10^{-4}} = \frac{30 \times 10^{-7}}{3 \times 10^{-4}} = 10 \times 10^{-3} = 10 \text{ mm}$ Hence, the third maximum is located at 10 mm from the central maximum.