9 Apr 2024 Shift 2

$\lambda = \frac{h}{\sqrt{2mE}}$, so $h = \lambda\sqrt{2mE}$. Dimensions: $[\lambda] = [L]$, $[m] = [M]$, $[E] = [ML^2T^{-2}]$. $[h] = [L] \cdot \$[M \cdot ML^2T^{-2}]^{1/2} = [L] \cdot \$[M^2L^2T^{-2}]^{1/2} = [L]\$[MLT^{-1}]\$ = [ML^2T^{-1}]$. The correct answer is Option 1: $[ML^2T^{-1}]$ .

Two cars travel towards each other at 20 m/s each, initially 300 m apart, both braking with retardation 2 m/s$^2$. We need the distance between them when they stop. Using the kinematic equation $v^2 = u^2 - 2as$ (where $v = 0$ at stop): $0 = (20)^2 - 2(2)s$ $0 = 400 - 4s$ $s = \frac{400}{4} = 100 \text{ m}$ Each car travels 100 m before stopping. Both cars move toward each other, so the total distance closed between them is: $s_{total} = 100 + 100 = 200 \text{ m}$ Initial separation was 300 m. After both cars stop: $d = 300 - 200 = 100 \text{ m}$ The distance between them when they come to rest is 100 m . The correct answer is Option (2): 100 m .

First lets write the Forces acting in Vertical and Horizontal direction In the Vertical direction, Forces are $F_g=mg\ due\ to\ gravity\ and\ T\sin\theta\ \ \left(\ tension\ of\ the\ rope\right)$ and \in Horizontal , The forces are $F\left(appplied\ force\right)\ and\ T\cos\theta\ \left(tension\ of\ the\ rope\right)$ First lets equate the Vertical Forces $mg\ =\ T\sin\ \theta\$ $10=\ T\times \ \frac{1}{\sqrt{\ 2}}$ Now substitute this and equal Horizontal forces $F=\ T\cos\theta$ $F=\ \sqrt{\ 2}\times \ 10\times \ \cos45^{\circ\ }\$ $F=\ \frac{10}{\sqrt{\ 2}}\times \ \sqrt{\ 2}$ $F=\ 10N$

Satellite mass = $10^3$ kg, initial orbit radius = $2R$. Total energy \in circular orbit: $E = -\frac{GMm}{2r}$. Initial energy: $E_1 = -\frac{GM \times 10^3}{2 \times 2R} = -\frac{GM \times 10^3}{4R}$. Using $GM = gR^2$: $E_1 = -\frac{gR \times 10^3}{4} = -\frac{10 \times R \times 10^3}{4} = -2500R$. Energy supplied: $\frac{10^4R}{6}$ J. New energy: $E_2 = E_1 + \frac{10^4R}{6} = -2500R + \frac{10000R}{6} = R(-2500 + \frac{5000}{3}) = R\frac{-7500+5000}{3} = \frac{-2500R}{3}$. $E_2 = -\frac{GMm}{2r_2} = -\frac{gR^2 \times 10^3}{2r_2} = -\frac{10^4R^2}{2r_2}$. $\frac{-2500R}{3} = \frac{-10^4R^2}{2r_2}$ $r_2 = \frac{10^4R^2 \times 3}{2 \times 2500R} = \frac{3 \times 10^4R}{5000} = 6R$. The correct answer is Option 4: $6R$ .

Excess pressure inside a soap bubble: $P = \frac{4T}{r}$, where $T$ is surface tension and $r$ is radius. Given $P_1 = 3P_2$: $\frac{4T}{r_1} = 3 \cdot \frac{4T}{r_2} \implies r_2 = 3r_1$. Volume ratio: $\frac{V_1}{V_2} = \frac{r_1^3}{r_2^3} = \frac{r_1^3}{27r_1^3} = \frac{1}{27}$. The correct answer is Option 3: $1:27$ .

The ball moves in two stages: (i) free fall through air for a height $h$ (ignore air drag), (ii) motion \in water with constant speed (terminal velocity). Stage (ii): terminal velocity \in water For a sphere of radius $r$ moving slowly through a viscous liquid, Stokes' law gives the viscous force $F_{\text{viscous}} = 6\pi \eta r v$. At terminal velocity all forces balance: weight $= \text{buoyancy} + \text{viscous force}$ Weight: $W = \rho_s \, V \, g$, Buoyant force: $B = \rho_w \, V \, g$, where $V = \frac{4}{3}\pi r^{3}$ is the volume. Hence $\rho_s V g = \rho_w V g + 6\pi \eta r v$ Simplify (take $V$ common): $(\rho_s - \rho_w)V g = 6\pi \eta r v$ Substitute $V = \frac{4}{3}\pi r^{3}$: $(\rho_s - \rho_w)\left(\frac{4}{3}\pi r^{3}\right)g = 6\pi \eta r v$ Cancel $\pi$ and rearrange for $v$: $v = \frac{2}{9}\,\frac{(\rho_s - \rho_w) g r^{2}}{\eta}$ $-(1)$ Insert the given data: $\rho_s = 1.0\times10^{5}\,\text{kg m}^{-3}$, $\rho_w = 1.0\times10^{3}\,\text{kg m}^{-3}$, $r = 1.0\times10^{-4}\,\text{m}$, $\eta = 9.8\times10^{-6}\,\text{N s m}^{-2}$, $g = 9.8\,\text{m s}^{-2}$. Compute the difference \in densities: $(\rho_s - \rho_w) = 1.0\times10^{5} - 1.0\times10^{3} = 9.9\times10^{4}\,\text{kg m}^{-3}$ Now evaluate $v$ from $(1)$: $v = \frac{2}{9}\,\frac{(9.9\times10^{4})(9.8)(1.0\times10^{-4})^{2}}{9.8\times10^{-6}}$ $r^{2} = (1.0\times10^{-4})^{2} = 1.0\times10^{-8}$, so $v = \frac{2}{9}\,\frac{(9.9\times10^{4})(9.8)(1.0\times10^{-8})}{9.8\times10^{-6}}$ The factor $9.8$ \in numerator and denominator cancels, giving $v = \frac{2}{9}\,\frac{9.9\times10^{4}\times1.0\times10^{-8}}{1.0\times10^{-6}}$ $9.9\times10^{4}\times1.0\times10^{-8}=9.9\times10^{-4}$; dividing by $1.0\times10^{-6}$ multiplies by $10^{6}$: $\frac{9.9\times10^{-4}}{1.0\times10^{-6}} = 9.9\times10^{2}=990$ Therefore $v = \frac{2}{9}\times 990 \approx 220\,\text{m s}^{-1}$ Stage (i): free fall \in air For free fall without air resistance $v^{2} = 2 g h \;\Rightarrow\; h = \frac{v^{2}}{2g}$ $-(2)$ Insert $v \approx 220\,\text{m s}^{-1}$ and $g = 9.8\,\text{m s}^{-2}$ into $(2)$: $h = \frac{(220)^{2}}{2\times9.8} = \frac{48400}{19.6} \approx 2.47\times10^{3}\,\text{m}$ The nearest value among the options is $\boxed{2518\ \text{m}}$ (Option B).

or a cyclic process, net work done is the sum of work in each path. Cycle is $C\to A\to B\to C$ For C\to A, volume is constant, so $W_{CA}=0$ For A\to B, given $PV^3=RT$ Temperature is $27^{\circ}C=300K$ and R=8 So $PV^3=8(300)=2400$ Thus $P=\frac{2400}{V^3}$ Work done from A to B is $W_{AB}=\int PdV$ $=\int_2^4\frac{2400}{V^3}dV$ $=2400\int_2^4V^{-3}dV$ $=2400\left(\frac{-1}{2V^2}\right)_2^4$ $2400\left(\frac{1}{8}-\frac{1}{32}\right)$ $=2400\cdot\frac{3}{32}$ =225J For B\to C, pressure is constant at P=10 So $W_{BC}=P(V_C-V_B)$ =10(2−4) =−20J Hence net work done in the complete cycle is W=0+225−20W =205J

Average translational KE: $K = \frac{3}{2}k_BT$. Initial: $T_1 = -78°C = 195$ K, $K_1 = K$. For $K_2 = 2K$: $T_2 = 2T_1 = 390$ K = $117°C$. The correct answer is Option 2: $117°C$ .

Inside surface SSS enclosed charges are: $+5q,\quad+q,\quad-2q$ Charges +3q and −4q are outside, so contribute zero net flux through closed surface. Net enclosed charge: $Q_{enc}=5q+q-2q=4q$ By Gauss law, $\Phi=\frac{Q_{enc}}{\epsilon_0}$ So, $\frac{4q}{\epsilon_0}$

step 1: observe symmetry The network is symmetric about the vertical axis (through the center). So the left and right junctions of the diamond are at the same potential. ⇒ no current flows through the horizontal central resistor ⇒ remove that resistor step 2: reduce the diamond Now between the left and right junctions, we have three independent parallel paths: • top path: two resistors → R + R = 2R • bottom path: two resistors → R + R = 2R • central vertical path: two resistors → R + R = 2R So: $R_{\text{diamond}}=2R\parallel2R\parallel2R$ $=\frac{2R}{3}$ step 3: include outer resistors From A to diamond: R From diamond to B: R So total: $R_{eq}=R+\frac{2R}{3}+R=2R+\frac{2R}{3}=\frac{8R}{3}$

A proton and a deuteron with the same kinetic energy enter a uniform magnetic field perpendicular to $\vec{B}$. We need the ratio of their radii of circular paths. When a charged particle moves perpendicular to a magnetic field, it follows a circular path with radius: $r = \frac{mv}{qB}$ Since $KE = \frac{1}{2}mv^2$, we get $mv = \sqrt{2m \cdot KE}$. Therefore: $r = \frac{\sqrt{2m \cdot KE}}{qB}$ For the proton: mass = $m_p$, charge = $e$, so $r_p = \frac{\sqrt{2m_p \cdot KE}}{eB}$ For the deuteron: mass = $2m_p$ (approximately twice the proton mass), charge = $e$ (same charge as proton), so: $r_d = \frac{\sqrt{2 \cdot 2m_p \cdot KE}}{eB} = \frac{\sqrt{2} \cdot \sqrt{2m_p \cdot KE}}{eB} = \sqrt{2} \cdot r_p$ The ratio is: $\frac{r_d}{r_p} = \sqrt{2} : 1$ The correct answer is Option (1): $\sqrt{2} : 1$ .

At $t=0$, the front edge of the square loop begins to enter the magnetic field.
The speed of the loop is $v = 2 \text{ cm/s}$.
At $t = 10 \text{ s}$, the distance moved by the loop is $d = v \times t = (2 \text{ cm/s}) \times (10 \text{ s}) = 20 \text{ cm}$.

The side length of the square loop is $L = 15 \text{ cm}$.
The magnetic field region is $50 \text{ cm}$ wide.

At $t = 10 \text{ s}$:
The front edge of the loop is at $x = 20 \text{ cm}$ from the left boundary of the magnetic field.
The back edge of the loop is at $x = 20 \text{ cm} - L = 20 \text{ cm} - 15 \text{ cm} = 5 \text{ cm}$ from the left boundary of the magnetic field.

Since $0 < 5 \text{ cm} < 20 \text{ cm} < 50 \text{ cm}$, both the front and back edges of the loop are entirely within the uniform magnetic field.
When the entire loop is moving within a uniform magnetic field, the magnetic flux ($\Phi_B = B \times A$) through the loop remains constant because the area $A$ exposed to the magnetic field does not change.
According to Faraday's Law of Induction, the induced emf is $\mathcal{E} = - \frac{d\Phi_B}{dt}$.
Since $\Phi_B$ is constant, $\frac{d\Phi_B}{dt} = 0$.
Therefore, the induced emf in the loop at $t = 10 \text{ s}$ is zero.

In an electromagnetic wave, the electric field $\vec{E}$, magnetic field $\vec{B}$, and direction of propagation $\vec{k}$ are mutually perpendicular. The given magnetic field is $B_y = B_0 \sin(kx + \omega t)$, so the wave propagates in the negative x-direction ($-\hat{i}$) and the magnetic field is in the y-direction ($\hat{j}$).
For $\vec{k} \propto \vec{E} \times \vec{B}$, if $\vec{k} = -\hat{i}$ and $\vec{B} = \hat{j}$, then $\vec{E}$ must be in the z-direction ($\hat{k}$) since $\hat{k} \times \hat{j} = -\hat{i}$.
The amplitude of the electric field $E_0$ is related to the amplitude of the magnetic field $B_0$ by $E_0 = c B_0$, where $c = 3 \times 10^8 \text{ m/s}$ is the speed of light.
Given $B_0 = 3.5 \times 10^{-7} \text{ T}$, then $E_0 = (3 \times 10^8 \text{ m/s}) \times (3.5 \times 10^{-7} \text{ T}) = 10.5 \times 10^1 \text{ V/m} = 105 \text{ V/m}$.
The electric field oscillates in phase with the magnetic field.
Therefore, the electric field equation is $E_z = 105 \sin(1.5 \times 10^3 x + 0.5 \times 10^{11} t) \text{ Vm}^{-1}$.

The final answer is $\boxed{D}$

Concept: A ray passing through the focus of a convex lens emerges parallel . Parallel rays entering a convex lens converge at its focus . Condition for parallel output: For rays emerging from $L_1$​ to become parallel after $L_2$, the image formed by $L_1$​ must lie at the focus of L_2​ . Hence, separation between lenses: D = $f_1 + f_2$ = 10 + 15 = 25 Final Answer: 25 cm

UV light of energy 4.13 eV is incident on a metal surface with work function 3.13 eV. We need to find the maximum kinetic energy of ejected photoelectrons. Recall Einstein's photoelectric equation. $KE_{max} = h\nu - \phi$ where $h\nu$ is the energy of the incident photon and $\phi$ is the work function of the metal. Substitute the given values. $KE_{max} = 4.13 - 3.13 = 1.00 \text{ eV}$ Explanation: The work function $\phi = 3.13$ eV is the minimum energy required to eject an electron from the metal surface. Any photon energy in excess of the work function is converted into kinetic energy of the ejected electron. Since the incident photon has 4.13 eV and 3.13 eV is used to overcome the binding energy, the remaining 1.00 eV becomes the maximum kinetic energy of the photoelectron. The correct answer is Option (3): 1 eV .

Ground state energy of hydrogen: $E_1 = -13.6$ eV. Energy given = 10.2 eV. Total energy = $-13.6 + 10.2 = -3.4$ eV = $-\frac{13.6}{n^2}$. So $n^2 = 4$, $n = 2$. The electron goes to $n = 2$. Spectral lines from transitions: $n=2 \to n=1$. Number of spectral lines = $\frac{n(n-1)}{2} = \frac{2 \times 1}{2} = 1$. The correct answer is Option 4: 1 .

By conservation of momentum: $\vec{p}_1 + \vec{p}_2 = 0$ (nucleus initially at rest). So $m_1v_1 = m_2v_2$, and they move \in opposite directions. $\frac{v_1}{v_2} = \frac{m_2}{m_1} = \frac{1}{2}$. The heavier nucleus (mass ratio 2) moves slower, and the lighter one (mass ratio 1) moves faster, in opposite directions with speeds in ratio 1:2. The correct answer is Option 4: in opposite directions with speed in the ratio of 1:2 respectively .

Energy from fusion of 2 kg hydrogen: The fusion reaction: $4\,^1H + 2e^- \to \,^4_2He + 2\nu + 6\gamma + 26.7$ MeV This means 4 hydrogen atoms (combined mass $\approx 4 \times 1 = 4$ g/mol) produce 26.7 MeV per reaction. Moles of H \in 2 kg = $\frac{2000}{1} = 2000$ mol Number of fusion reactions = $\frac{2000}{4} = 500$ mol Total energy: $E_H = 500 \times N_A \times 26.7$ MeV Energy from fission of 2 kg $^{235}U$: Moles of $^{235}U$ = $\frac{2000}{235}$ mol Total energy: $E_U = \frac{2000}{235} \times N_A \times 200$ MeV Ratio: $\frac{E_H}{E_U} = \frac{500 \times 26.7}{\frac{2000}{235} \times 200} = \frac{13350}{\frac{400000}{235}} = \frac{13350 \times 235}{400000} = \frac{3137250}{400000} \approx 7.84$ The closest option is 7.62. The slight difference comes from the exact atomic masses used. The correct answer is Option A: 7.62.

The given I-V characteristic shows negligible current in forward bias initially, and then in reverse bias, the current remains low until a specific reverse voltage is reached. At this Zener voltage, the current increases sharply while the voltage across the device remains nearly constant. This sharp and controlled reverse breakdown, where the voltage is stabilized, is the defining characteristic of a Zener diode used for voltage regulation.

The circuit's output, $E$, is defined by the logic expression $E = \overline{(A \cdot B) + (\bar{B} \cdot \bar{A})}$.

1. For $A=0, B=1$ (to find $X$):
   The output of the top AND gate is $0 \cdot 1 = 0$.
   The output of the bottom AND gate is $\bar{1} \cdot \bar{0} = 0 \cdot 1 = 0$.
   The final NOR gate output is $E = \overline{0 + 0} = \overline{0} = 1$.
   Thus, $X=1$.

2. For $A=1, B=0$ (to find $Y$):
   The output of the top AND gate is $1 \cdot 0 = 0$.
   The output of the bottom AND gate is $\bar{0} \cdot \bar{1} = 1 \cdot 0 = 0$.
   The final NOR gate output is $E = \overline{0 + 0} = \overline{0} = 1$.
   Thus, $Y=1$.

Therefore, $X=1$ and $Y=1$.

Let angle between $\vec{A}$ and $\vec{B}$ be $\theta$. Resultant $\vec{R} = \vec{A} + \vec{B}$ is perpendicular to $\vec{A}$. $\vec{R} \cdot \vec{A} = 0$: $A^2 + AB\cos\theta = 0 \implies \cos\theta = -A/B$. $|\vec{R}| = B/2$: $A^2 + B^2 + 2AB\cos\theta = B^2/4$. $A^2 + B^2 - 2A^2 = B^2/4 \implies B^2 - A^2 = B^2/4 \implies A^2 = 3B^2/4$. $\cos\theta = -\frac{A}{B} = -\frac{\sqrt{3}}{2} \implies \theta = 150°$. The answer is $\boxed{150}$°.

Work = $\int_2^4 (3x^{2+}2x-5)dx = \$[x^{3+}x^{2-}5x]_2^4$ At x=4: 64+16-20 = 60. At x=2: 8+4-10 = 2. W = 60 - 2 = 58 J. The answer is $\boxed{58}$ J.

A circular disc rolls and slides down an inclined plane of length $l$. When it slides (no rolling), it takes $t$ seconds. When it rolls without slipping, it takes $\left(\frac{\alpha}{2}\right)^{1/2} t$ seconds. Find $\alpha$. When the disc slides without friction, the acceleration along the incline is: $a_{\text{slide}} = g\sin\theta$ Using $l = \frac{1}{2}a_{\text{slide}} t^2$: $t^2 = \frac{2l}{g\sin\theta} \quad \cdots (i)$ When a disc rolls without slipping, the moment of inertia about the centre is $I = \frac{1}{2}mR^2$, so $\frac{I}{mR^2} = \frac{1}{2}$. The translational acceleration along the incline is: $a_{\text{roll}} = \frac{g\sin\theta}{1 + I/(mR^2)} = \frac{g\sin\theta}{1 + 1/2} = \frac{2g\sin\theta}{3}$ Using $l = \frac{1}{2}a_{\text{roll}} t_{\text{roll}}^2$: $t_{\text{roll}}^2 = \frac{2l}{a_{\text{roll}}} = \frac{2l \cdot 3}{2g\sin\theta} = \frac{3l}{g\sin\theta} \quad \cdots (ii)$ $\frac{t_{\text{roll}}^2}{t^2} = \frac{3l/(g\sin\theta)}{2l/(g\sin\theta)} = \frac{3}{2}$ $t_{\text{roll}} = t\sqrt{\frac{3}{2}} = \left(\frac{3}{2}\right)^{1/2} t$ We are given $t_{\text{roll}} = \left(\frac{\alpha}{2}\right)^{1/2} t$. Comparing: $\frac{\alpha}{2} = \frac{3}{2} \implies \alpha = 3$. The answer is 3 .

$R = R_0(1+\alpha\Delta T)$. $62 = 50(1+2.4\times10^{-4}(T-27))$. $1.24 = 1 + 2.4\times10^{-4}(T-27)$. $0.24 = 2.4\times10^{-4}(T-27)$. $T-27 = 1000 \implies T = 1027$°C. The answer is $\boxed{1027}$.

A particle of mass $m = 0.50$ kg executes SHM under force $F = -50x$ N. We need to find the value of $x$ if the time period is $\frac{\pi}{x}$ s. Comparing with $F = -kx$, we get $k = 50$ N/m. $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{0.50}} = \sqrt{100} = 10 \text{ rad/s}$ $T = \frac{2\pi}{\omega} = \frac{2\pi}{10} = \frac{\pi}{5} \text{ s}$ Given $T = \frac{\pi}{x}$, comparing with $T = \frac{\pi}{5}$: $x = 5$ The answer is 5 .

Given $E⃗=(2x)x$ Cube side: a=2m From figure, cube extends from x=2 to x=4 Only two faces perpendicular to x-axis contribute flux, because field is along x-direction. At left face: x=2 $E_1=2(2)=4$ Area of face $A=2\times2=4$ Flux through left face (inward, negative): $\Phi_1=-E_1A$ $=-4(4)=-16$ At right face: x=4 $E_2=2(4)=8$ Flux through right face: $\Phi_2=E_2$ $=8(4)=32$ Net flux $\Phi=\Phi_1+\Phi_2$ $=32-16$ $=16$

The voltmeter (10 kΩ) is connected in parallel with the unknown resistance R, so the measured V-I relation corresponds to the equivalent resistance of the parallel combination. From the graph: At V = 8 V, I = 4 mA So equivalent resistance: $Req=\frac{V}{I}=\frac{8}{4\times10^{-3}}=2000\Omega =2k\Omega$ Now, Req=R∥10 k\Omega $\frac{1}{2000}=\frac{1}{R}+\frac{1}{10000}$ Solve: $\frac{1}{R}=\frac{1}{2000}-\frac{1}{10000}=\frac{5-1}{10000}=\frac{4}{10000}=\frac{1}{2500}$ R=2500 Ω=2.5 kΩ

We need to find the magnetic moment of a straight magnetic strip after it is bent into a semicircular shape. Original magnetic moment: $M = 44 \text{ Am}^2$ $\pi = \frac{22}{7}$ For a bar magnet (or magnetic strip), the magnetic moment is defined as: $M = m \times L$ where $m$ is the pole strength and $L$ is the length of the strip (distance between the poles). Here: $44 = m \times L \implies m = \frac{44}{L}$ When the strip of length $L$ is bent into a semicircular shape, the length of the strip equals the semicircular arc: $\pi R = L \implies R = \frac{L}{\pi}$ where $R$ is the radius of the semicircle. In a semicircle, the two ends of the strip (which carry the magnetic poles) are at diametrically opposite points. The distance between them is the diameter: $d = 2R = \frac{2L}{\pi}$ The pole strength $m$ remains unchanged (it is an intrinsic property of the magnet). The new magnetic moment is: $M' = m \times d = m \times \frac{2L}{\pi} = \frac{2mL}{\pi} = \frac{2M}{\pi}$ $M' = \frac{2 \times 44}{\pi} = \frac{88}{\frac{22}{7}} = \frac{88 \times 7}{22} = \frac{616}{22} = 28 \text{ Am}^2$ The answer is 28 Am$^2$.

We need to find the power dissipation in a series RC circuit connected to an AC source. Capacitive reactance: $X_C = 4\sqrt{3} \text{ } \Omega$ Resistance: $R = 4 \text{ } \Omega$ Peak voltage: $V_0 = 8\sqrt{2} \text{ V}$ For a series RC circuit, the impedance is given by: $Z = \sqrt{R^2 + X_C^2}$ Substituting the values: $Z = \sqrt{(4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \text{ } \Omega$ The relationship between peak and RMS voltage is: $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \text{ V}$ Using Ohm's law for AC circuits: $I_{rms} = \frac{V_{rms}}{Z} = \frac{8}{8} = 1 \text{ A}$ In an AC circuit, power is dissipated only \in the resistive component (capacitors and inductors do not dissipate power on average). The power dissipation formula is: $P = I_{rms}^2 \times R$ This can also be written as $P = V_{rms} \cdot I_{rms} \cdot \cos\phi$, where $\cos\phi = R/Z$ is the power factor. Both forms give the same result. $P = (1)^2 \times 4 = 4 \text{ W}$ Verification using power factor: $\cos\phi = \frac{R}{Z} = \frac{4}{8} = 0.5$ $P = V_{rms} \times I_{rms} \times \cos\phi = 8 \times 1 \times 0.5 = 4$ W. This confirms our answer. The answer is 4 W.

We need to find the thickness of a glass plate that shifts the central maximum to the position of the 4th bright fringe in Young's double slit experiment. Wavelength: $\lambda = 500 \text{ nm}$ Refractive index of glass plate: $\mu = 1.5$ Central maximum shifts to the position of the 4th bright fringe. When a glass plate of thickness $t$ and refractive index $\mu$ is placed \in front of one slit, it introduces an additional optical path. Inside the glass, the wavelength becomes $\lambda/\mu$, so the optical path through the glass is $\mu t$ instead of $t$. The extra path difference introduced is: $\Delta = \mu t - t = (\mu - 1)t$ The central maximum originally occurs where the path difference from both slits is zero. With the glass plate, the central maximum shifts to the point where the extra path difference due to the glass exactly compensates. For the central maximum to shift to the position of the $n^{th}$ bright fringe, the extra path difference must equal $n\lambda$: $(\mu - 1)t = n\lambda$ Here, $n = 4$ (shift to 4th bright fringe position): $(1.5 - 1) \times t = 4 \times 500 \text{ nm}$ $0.5 \times t = 2000 \text{ nm}$ $t = \frac{2000}{0.5} = 4000 \text{ nm} = 4 \text{ } \mu m$ The answer is 4 $\mu$m.