COMEDK 2025 Evening Shift Question Paper with Solutions

Let the two intensities be

${𝐼}_1=9{𝐼}_0,{𝐼}_2=1{𝐼}_0$

so the amplitudes are

${𝑎}_1=\sqrt{\surd 9{𝐼}_0}=3\sqrt{\surd {𝐼}_0},{𝑎}_2=\sqrt{\surd {𝐼}_0}.$

Maximum intensity (in‐phase):

${𝐼}_{max}=({𝑎}_1+{𝑎}_2{)}^2=(3\sqrt{\surd {𝐼}_0}+\sqrt{\surd {𝐼}_0}{)}^2=16{𝐼}_0.$

Minimum intensity (out‐of‐phase):

${𝐼}_{min}=({𝑎}_1-{𝑎}_2{)}^2=(3\sqrt{\surd {𝐼}_0}-\sqrt{\surd {𝐼}_0}{)}^2=4{𝐼}_0.$

The desired ratio is

$\frac{{𝐼}_{max}+{𝐼}_{min}}{{𝐼}_{max}-{𝐼}_{min}}=\frac{16{𝐼}_0+4{𝐼}_0}{16{𝐼}_0-4{𝐼}_0}=\frac{20}{12}=\frac{5}{3}.$

So $𝛼=5$ and $𝛽=3$. (Option A)

Let’s denote the two nuclei by 1 and 2. Recall that

– Mass ∝ A (mass number)

– Volume ∝ R³ ∝ (Diameter)³

So

${𝜌}_1/{𝜌}_2=\frac{{𝐴}_1/{𝑉}_1}{{𝐴}_2/{𝑉}_2}=\frac{{𝐴}_1/{𝐴}_2}{({𝐷}_1/{𝐷}_2{)}^3}=\frac{\frac{5}{2}}{(\frac{2}{6}{)}^3}=\frac{5/2}{(1/3{)}^3}=(5/2)\times 27=\frac{135}{2}\approx 67.5:1$

None of the given choices match 135 : 2, but in the liquid‐drop model of the nucleus one shows that

$𝑅\propto {𝐴}^{1/3}⟹𝜌=\frac{𝐴}{{𝑅}^3}\approx \text{constant}$

for all nuclei. Hence to very good approximation

${𝜌}_1:{𝜌}_2=1:1$

Answer: Option A (1 : 1).

The only correct statement is B.

• A is false – Bohr’s model predicts discrete (line) spectra, not continuous ones.

• B is true – it postulates

$𝑚𝑣𝑟=𝑛\hslash ,𝑛=1,2,3,\dots$

i.e. the electron’s angular momentum is quantised.

• C and D are false – the Bohr model only applies to hydrogen-like (one-electron) atoms, so it doesn’t predict spectra for Ne⁺ (which has 9 electrons) or compare Ne⁺ with He⁺.

Given, ${𝑉}_{𝑆}=100\text{ V}$, ${𝑉}_{𝑍}=20\text{ V}$, ${𝑅}_{𝑆}=10\text{ k}\Omega$, and ${𝑅}_{𝐿}=5\text{ k}\Omega$.

Since the open-circuit voltage across the load,

${𝑉}_{\prime }^{𝐿}={𝑉}_{𝑆}\left(\frac{{𝑅}_{𝐿}}{{𝑅}_{𝑆}+{𝑅}_{𝐿}}\right)=100\times \frac{5}{15}\approx 33.33\text{ V}>{𝑉}_{𝑍},$

the Zener diode operates in the breakdown region and regulates the voltage across the load to ${𝑉}_{𝐿}={𝑉}_{𝑍}=20\text{ V}$.

The current through the series resistor ${𝑅}_{𝑆}$ is

${𝐼}_{𝑆}=\frac{{𝑉}_{𝑆}-{𝑉}_{𝑍}}{{𝑅}_{𝑆}}=\frac{100-20}{10\times {10}^3}=8\times {10}^{-3}\text{ A}=8\text{ mA}.$

The current through the load resistor ${𝑅}_{𝐿}$ is

${𝐼}_{𝐿}=\frac{{𝑉}_{𝑍}}{{𝑅}_{𝐿}}=\frac{20}{5\times {10}^3}=4\times {10}^{-3}\text{ A}=4\text{ mA}.$

Applying Kirchhoff's current law at the node,

${𝐼}_{𝑍}={𝐼}_{𝑆}-{𝐼}_{𝐿}=8-4=4\text{ mA}.$

Hence, the correct option is (C).

Starting from Newton’s law of gravitation:

$𝐹=𝐺\frac{{𝑚}_1{𝑚}_2}{{𝑟}^2}$

Solving for $𝐺$:

$𝐺=𝐹\frac{{𝑟}^2}{{𝑚}_1{𝑚}_2}$

Now, in terms of units:

$\left[𝐹\right]=\text{N}=\frac{\text{kg}\text{m}}{{\text{s}}^2}$

$\left[𝑟\right]=\text{m}$

$\left[{𝑚}_1\right]=\left[{𝑚}_2\right]=\text{kg}$

So

$\left[𝐺\right]=\frac{(\frac{\text{kg}\text{m}}{{\text{s}}^2}){\text{m}}^2}{{\text{kg}}^2}=\text{N}{\text{m}}^2{\text{kg}}^{-2}$

That matches Option D.

We have in normal adjustment

• Tube length L = f₀ + fₑ = 36 cm

• Angular magnification M = f₀ / fₑ = 8

So

From M = f₀/fₑ = 8 ⇒ f₀ = 8 fₑ

Plug into f₀ + fₑ = 36 ⇒ 8 fₑ + fₑ = 36 ⇒ 9 fₑ = 36 ⇒ fₑ = 4 cm

Then f₀ = 8·4 = 32 cm

Answer: f₀ = 32 cm, fₑ = 4 cm (Option C).

Let’s collect the data and apply the standard formula for a rotating coil:

Given

• Number of turns: $𝑁=250$

• Coil dimensions: $20\text{ cm}\times 15\text{ cm}⟹𝐴=0.20\times 0.15=0.03m^2$

• Rotation rate: $𝑓=60$ Hz, so $𝜔=2𝜋𝑓=120𝜋\mathrm{rad}/s$

• Magnetic field: $𝐵=2\times {10}^{-2}T$

The induced emf varies as

$E\left(𝑡\right)=𝑁𝐵𝐴𝜔\sin \left(𝜔𝑡\right)$

so the peak (maximum) emf is

$\overline{)E_{max}=𝑁𝐵𝐴𝜔.}$

Substituting in:

$E_{max}=250\times (2\times {10}^{-2})\times 0.03\times \left(120𝜋\right)=18𝜋V.$

Answer: $18𝜋$ V, i.e. option C.

Let’s label:

• Diverging lens (L₁) at x=0, f₁=–20 cm

• Converging lens (L₂) at x=+10 cm, f₂=+15 cm

• Incoming rays are parallel ⇒ object for L₁ at infinity.

Image by L₁

Lens formula:

$\frac{1}{{𝑣}_1}+\frac{1}{{𝑢}_1}=\frac{1}{{𝑓}_1},{𝑢}_1=+\infty ,{𝑓}_1=-20$

⇒ $1/{𝑣}_1=1/(-20)$ ⇒ ${𝑣}_1=-20$cm.

So L₁ forms a virtual image 20 cm to its left.

That virtual image acts as an object for L₂.

Distance from that point (x=–20) to L₂ (x=+10) is

20+10 = 30 cm on the incoming side ⇒ ${𝑢}_2=+30$cm.

Image by L₂

$\frac{1}{{𝑣}_2}+\frac{1}{{𝑢}_2}=\frac{1}{{𝑓}_2}⟹\frac{1}{{𝑣}_2}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}⟹{𝑣}_2=+30\text{ cm.}$

${𝑣}_2>0$ means a real image 30 cm to the right of the converging lens.

Answer: Option C. The final image is real and 30 cm to the right of the converging lens.

First, find how many N₂ molecules are in 2 cm³ at NTP:

Volume per mole at NTP: 22.4 L = 22400 cm³

Number of moles in 2 cm³:

$𝑛=\frac{2{cm}^3}{22400{cm}^3/mol}=\frac{1}{11200}mol$

Number of molecules:

${𝑁}_{mol}=𝑁\times 𝑛=\frac{𝑁}{11200}$

At “normal” (room) temperature the vibrational mode of N₂ is essentially frozen, so each molecule has

• 3 translational + 2 rotational = 5 degrees of freedom.

Total degrees of freedom:

${\text{dof}}_{total}=5\times \frac{𝑁}{11200}=\frac{5𝑁}{11200}=\frac{𝑁}{2240}.$

Answer: Option C, $\frac{𝑁}{2240}$.

Let the pendulum bob of mass $𝑚=4$g $=0.004$kg carry a charge $𝑞=20\times {10}^{-6}$C in a horizontal field $𝐸=1000$V/m. At equilibrium the horizontal electric force balances the component of tension:

Electric force:

${𝐹}_{𝑒}=𝑞𝐸=(20\times {10}^{-6}\text{C})\left(1000\text{V/m}\right)=0.02\text{N}.$

Weight:

$𝑊=𝑚𝑔=\left(0.004\text{kg}\right)\left(9.8{\text{m/s}}^2\right)=0.0392\text{N}.$

At angle $𝜃$, equilibrium requires

$\tan 𝜃=\frac{{𝐹}_{𝑒}}{𝑊}=\frac{0.02}{0.0392}\approx 0.51\approx \frac{1}{2}.$

Hence

$𝜃={\tan }^{-1}\frac{1}{2},$

which is Option A.

From the given figure, the charges are located at the vertices of a right-angled isosceles triangle of side length $𝑟$. The distance between charges $𝑄$ and $2𝑞$ is $𝑟$, between $2𝑞$ and $-𝑞$ is $𝑟$, and the hypotenuse between $𝑄$ and $-𝑞$ has length $\sqrt{\surd {𝑟}^2+{𝑟}^2}=\sqrt{\surd 2}𝑟$.

The total electrostatic potential energy of the system is given by

$𝑈=\frac{1}{4𝜋{𝜀}_0}[\frac{𝑄\left(2𝑞\right)}{𝑟}+\frac{\left(2𝑞\right)(-𝑞)}{𝑟}+\frac{𝑄(-𝑞)}{\sqrt{\surd 2}𝑟}]$

For the net electrostatic energy to be zero,

$\frac{1}{4𝜋{𝜀}_0𝑟}[2𝑄𝑞-2{𝑞}^2-\frac{𝑄𝑞}{\sqrt{\surd 2}}]=0$

Since $𝑞\ne 0$, dividing both sides by $𝑞$,

$2𝑄-2𝑞-\frac{𝑄}{\sqrt{\surd 2}}=0$

$2𝑞=𝑄(2-\frac{1}{\sqrt{\surd 2}})$

$𝑞=𝑄(1-\frac{1}{2\sqrt{\surd 2}})$

Hence, the correct option is (B).

To emit a Paschen‐series photon (transition ending at $𝑛=3$), the atom must first be excited above $𝑛=3$. The lowest such level is $𝑛=4$.

Energy levels of hydrogen:

${𝐸}_{𝑛}=-\frac{13.6\text{ eV}}{{𝑛}^2}$

Ground‐state energy ${𝐸}_1=-13.6$ eV; fourth‐level energy

${𝐸}_4=-\frac{13.6}{4^2}=-0.85\text{ eV}.$

Minimum excitation energy from $𝑛=1$ to $𝑛=4$:

$\Delta 𝐸={𝐸}_4-{𝐸}_1=(-0.85)-(-13.6)=12.75\text{ eV}.$

Answer: Option B, 12.75 eV.

Option A

Reasoning:

In air, the thin lens equation is

$\frac1v+\frac1u=\frac1f\,,$

with the object at $u=1.5f$. That gives

$\frac1v=\frac1f-\frac1{1.5f}=\frac13\frac1f\quad\Rightarrow\;v=3f,$

a real, inverted image beyond $2f$.

Immersed in water ($n_w>1$), the lens’s focal length increases because its refractive‐power drops. In fact

$\frac1{f_{\rm water}}=\Big(\frac{n_{\rm lens}}{n_w}-1\Big)\biggl(\frac1{R_1}-\frac1{R_2}\biggr)$

is smaller than $\tfrac1f$ in air, so $f_{\rm water}>f$.

The object–lens distance remains $u=1.5f$, but now $u<f_{\rm water}$. For a converging lens with $u<f$, the image is

– virtual (on the same side as the object),

– erect,

– magnified.

That exactly matches Option A:

Magnified virtual erect image on the same side of the lens where the object is placed.

Compute the input power

${𝑃}_{in}={𝑉}_{𝑝}{𝐼}_{𝑝}=2200\text{V}\times 2\text{A}=4400\text{W}$

Account for the transformer’s efficiency

${𝑃}_{out}=𝜂{𝑃}_{in}=0.80\times 4400\text{W}=3520\text{W}$

Find the output current

${𝐼}_{out}=\frac{{𝑃}_{out}}{{𝑉}_{𝑠}}=\frac{3520\text{W}}{220\text{V}}=16\text{A}$

Answer: 16 A (Option C)

Stress and pressure both have the dimensions of force per unit area.

• A force F has dimensions

$\left[𝐹\right]=\left[𝑀𝐿{𝑇}^{-2}\right].$

• Dividing by area (dimensions ${𝐿}^2$) gives

$\frac{𝐹}{{𝐿}^2}=\left[𝑀𝐿{𝑇}^{-2}\right]\left[{𝐿}^{-2}\right]=\left[𝑀{𝐿}^{-1}{𝑇}^{-2}\right].$

Hence stress and pressure share the dimensional formula $\left[{𝑀}^1{𝐿}^{-1}{𝑇}^{-2}\right],$ so the correct answer is Option C.

Let $𝐴=\frac{ℎ𝑐}{𝑒𝜆}$ (in eV) and let ${𝜙}_{𝑒}$ be the work-function in eV. The photoelectric equations for stopping potentials ${𝑉}_1=15$V at $𝜆$ and ${𝑉}_2=3$V at $4𝜆$ are

${𝑉}_1=𝐴-{𝜙}_{𝑒}⟹15=𝐴-{𝜙}_{𝑒},$

${𝑉}_2=\frac{𝐴}{4}-{𝜙}_{𝑒}⟹3=\frac{𝐴}{4}-{𝜙}_{𝑒}.$

Subtracting (2) from (1):

$15-3=(𝐴-\frac{𝐴}{4})⟹12=\frac{3𝐴}{4}⟹𝐴=16\text{ eV}.$

Then from (1), ${𝜙}_{𝑒}=𝐴-15=1$eV.

The threshold wavelength ${𝜆}_0$ satisfies

${𝜙}_{𝑒}=\frac{ℎ𝑐}{𝑒{𝜆}_0}⟹{𝜆}_0=\frac{ℎ𝑐}{𝑒{𝜙}_{𝑒}}=\frac{𝐴𝜆}{{𝜙}_{𝑒}}=16𝜆.$

So the ratio ${𝜆}_0:𝜆$ is $\overline{)16:1}$ (Option C).

We model the wire as a spring with spring‐constant

$𝑘=\frac{𝑌𝐴}{𝐿}.$

The period of small vertical oscillations is

$𝑇=2𝜋\sqrt{\surd \frac{𝑚}{𝑘}}=2𝜋\sqrt{\surd \frac{𝑚𝐿}{𝑌𝐴}}.$

Squaring and solving for L gives

${𝑇}^2=\frac{4{𝜋}^2𝑚𝐿}{𝑌𝐴}⟹𝐿=\frac{{𝑇}^2𝑌𝐴}{4{𝜋}^2𝑚}.$

That’s Option D.

Let’s work in revolutions per second (rps) and use the fact that with constant torque (hence constant angular acceleration) the average speed times time gives total revolutions.

Convert speeds:

${𝑛}_{𝑖}=\frac{1800rpm}{60}=30rps,n_f=\frac{2400rpm}{60}=40rps.$

Average speed:

$\overline{𝑛}=\frac{{𝑛}_{𝑖}+{𝑛}_{𝑓}}{2}=\frac{30+40}{2}=35rps.$

Total revolutions in $𝑡=10$ s:

$𝑁=\overline{𝑛}\times 𝑡=35\times 10=350.$

Answer: 350 (Option A).

Let the magnetic moment of the bar‐magnet be μ. From the torque

$𝜏=𝜇𝐵\sin 𝜃$

we have

$𝜇=\frac{𝜏}{𝐵\sin 𝜃}=\frac{0.024}{(400\times {10}^{-4})(\sin {45}^{\circ })}=\frac{0.024}{0.04\frac{\sqrt{\surd 2}}{2}}=0.8485A\cdot m^2.$

When replaced by a solenoid of $𝑁=500$ turns and area $𝐴={10}^{-4}{m}^2$, its magnetic moment is

$𝜇=𝑁𝐼𝐴⟹𝐼=\frac{𝜇}{𝑁𝐴}=\frac{0.8485}{500\times {10}^{-4}}=16.97A=12\sqrt{\surd 2}A.$

Answer: Option C, $12\sqrt{\surd 2}A$.

According to Kepler’s third law, the square of the orbital period is proportional to the cube of the semi-major axis:

${𝑇}^2\propto {𝑎}^3.$

If the distance (i.e. $𝑎$) is doubled, ${𝑎}^{\prime }=2𝑎$, then

Compute the new period ${𝑇}^{\prime }$:

$({𝑇}^{\prime }{)}^2\propto (2𝑎{)}^3=2^3{𝑎}^3=8{𝑎}^3.$

Compare to the original:

$\frac{({𝑇}^{\prime }{)}^2}{{𝑇}^2}=\frac{8{𝑎}^3}{{𝑎}^3}=8⟹{𝑇}^{\prime }=𝑇\sqrt{\surd 8}=2\sqrt{\surd 2}𝑇.$

So the new year would last

$\overline{)2\sqrt{\surd 2}𝑇}$

(Option A).

First, convert temperatures to kelvin:

T₁ = 327 °C + 273 = 600 K

T₂ = 27 °C + 273 = 300 K

For any gas,

${𝑣}_{rms}=\sqrt{\surd \frac{3𝑅𝑇}{𝑀}},$

where M is the molar mass.

Since the rms speeds are equal,

$\sqrt{\surd \frac{3𝑅{𝑇}_1}{{𝑀}_{gas}}}=\sqrt{\surd \frac{3𝑅{𝑇}_2}{{𝑀}_{O_2}}}$

⇒

$\frac{{𝑇}_1}{{𝑀}_{gas}}=\frac{{𝑇}_2}{{𝑀}_{O_2}}$

⇒

${𝑀}_{gas}={𝑀}_{O_2}\frac{{𝑇}_1}{{𝑇}_2}=32\frac{600}{300}=64.$

Answer: 64.

The correct choice is Option D.

Explanation:

By definition the centre of mass of a system of point masses ${𝑚}_{𝑖}$ at positions ${𝐫}_{𝑖}$ is

${𝐑}_{cm}=\frac{1}{𝑀}\sum _{𝑖}{𝑚}_{𝑖}{𝐫}_{𝑖},𝑀=\sum _{𝑖}{𝑚}_{𝑖}.$

This formula shows that ${𝐑}_{cm}$ depends directly on how the mass is distributed (the values of each ${𝑚}_{𝑖}$ and their positions ${𝐫}_{𝑖}$).

Shape and size only matter insofar as they affect that mass distribution.

Although the numerical coordinates of ${𝐑}_{cm}$ change if you pick a different origin (Option B), the physical location of the centre of mass for a given distribution is determined solely by the masses and their arrangement.

Hence, “Depends on the distribution of its mass” is the fundamental property.

The kinetic energy gained is equal to the work done by the field:

$𝐾=𝑞𝐸𝑑$

Here

• $𝑞=60\times {10}^{-6}$ C

• $𝐸={10}^5$ N/C

• $𝑑=0.02$ m

So

$𝐾=(60\times {10}^{-6})\left({10}^5\right)\left(0.02\right)=0.12J=12\times {10}^{-2}J$

Hence $𝑚=12$. (Option A)

From the given circuit diagram, capacitors ${𝐶}_1$, ${𝐶}_2$, and ${𝐶}_3$ are connected in series across the supply voltage $𝑉=100\text{ V}$, while capacitor ${𝐶}_4$ is connected in parallel with this series combination.

The equivalent capacitance ${𝐶}_{𝑠}$ of the series combination of ${𝐶}_1$, ${𝐶}_2$, and ${𝐶}_3$ is given by

$\frac{1}{{𝐶}_{𝑠}}=\frac{1}{{𝐶}_1}+\frac{1}{{𝐶}_2}+\frac{1}{{𝐶}_3}$

$\frac{1}{{𝐶}_{𝑠}}=\frac{1}{3}+\frac{1}{6}+\frac{1}{3}=\frac{5}{6}𝜇{\text{F}}^{-1}⟹{𝐶}_{𝑠}=1.2𝜇\text{F}$

The charge flowing through this series branch is

$𝑄={𝐶}_{𝑠}𝑉=1.2𝜇\text{F}\times 100\text{ V}=120𝜇\text{C}$

Since capacitors in series carry equal charge, the charge on the $6𝜇\text{F}$ capacitor is $𝑄=120𝜇\text{C}$.

The energy stored in the ${𝐶}_2=6𝜇\text{F}$ capacitor is

$𝑈=\frac{{𝑄}^2}{2{𝐶}_2}=\frac{(120\times {10}^{-6}\text{ C}{)}^2}{2\times 6\times {10}^{-6}\text{ F}}=1.2\times {10}^{-3}\text{ J}=1.2\text{ mJ}$

Hence, the correct option is (A).

Forces along the plane

– Down‐slope component of weight:

$𝑚𝑔\sin 𝜃=10\cdot 10\cdot \sin {30}^{\circ }=100\cdot \frac{1}{2}=50\text{ N}$

– Normal reaction:

$𝑁=𝑚𝑔\cos 𝜃=10\cdot 10\cdot \cos {30}^{\circ }=100\cdot \frac{\sqrt{\surd 3}}{2}=50\sqrt{\surd 3}\text{ N}$

– Friction (opposes upward motion):

$𝑓=𝜇𝑁=\frac{\sqrt{\surd 3}}{6}\times 50\sqrt{\surd 3}=\frac{50\cdot 3}{6}=25\text{ N}$

Newton’s 2nd law up the plane

Let $𝑃$ be the applied force up the plane. Then

$𝑃-(𝑚𝑔\sin 𝜃+𝑓)=𝑚𝑎$

$𝑃=𝑚𝑎+𝑚𝑔\sin 𝜃+𝑓$

Substitute numbers

$𝑃=10\cdot 2+50+25=20+50+25=95\text{ N}$

Answer: 95 N (Option D).

Understand the setup: When a galvanometer is "shunted by a wire," it means the galvanometer and the shunt wire are connected in parallel. In a parallel circuit, the voltage across each branch is the same.

Identify the given values:

Resistance of the galvanometer (${𝑅}_{𝑔}$) = $50\Omega$

Resistance of the shunt wire (${𝑅}_{𝑠}$) = $10\Omega$

Total current (${𝐼}_{𝑡𝑜𝑡𝑎𝑙}$) = $2\text{ A}$

Apply the voltage equality: Let the current passing through the galvanometer be ${𝐼}_{𝑔}$ and the current passing through the shunt be ${𝐼}_{𝑠}$. Since the voltage across parallel components is the same, we have:

${𝑉}_{𝑔}={𝑉}_{𝑠}$

${𝐼}_{𝑔}{𝑅}_{𝑔}={𝐼}_{𝑠}{𝑅}_{𝑠}$

Apply Kirchhoff's Current Law: The total current entering the parallel combination splits into the current through the galvanometer and the current through the shunt:

${𝐼}_{𝑡𝑜𝑡𝑎𝑙}={𝐼}_{𝑔}+{𝐼}_{𝑠}$

From this, we can express ${𝐼}_{𝑔}$ as:

${𝐼}_{𝑔}={𝐼}_{𝑡𝑜𝑡𝑎𝑙}-{𝐼}_{𝑠}$

Substitute and solve for ${𝐼}_{𝑠}$: Now, let's substitute the expression for ${𝐼}_{𝑔}$ into our voltage equality equation:

$({𝐼}_{𝑡𝑜𝑡𝑎𝑙}-{𝐼}_{𝑠}){𝑅}_{𝑔}={𝐼}_{𝑠}{𝑅}_{𝑠}$

Expand the equation:

${𝐼}_{𝑡𝑜𝑡𝑎𝑙}{𝑅}_{𝑔}-{𝐼}_{𝑠}{𝑅}_{𝑔}={𝐼}_{𝑠}{𝑅}_{𝑠}$

Move all terms with ${𝐼}_{𝑠}$ to one side:

${𝐼}_{𝑡𝑜𝑡𝑎𝑙}{𝑅}_{𝑔}={𝐼}_{𝑠}{𝑅}_{𝑠}+{𝐼}_{𝑠}{𝑅}_{𝑔}$

Factor out ${𝐼}_{𝑠}$:

${𝐼}_{𝑡𝑜𝑡𝑎𝑙}{𝑅}_{𝑔}={𝐼}_{𝑠}({𝑅}_{𝑠}+{𝑅}_{𝑔})$

Finally, solve for ${𝐼}_{𝑠}$:

${𝐼}_{𝑠}=\frac{{𝐼}_{𝑡𝑜𝑡𝑎𝑙}{𝑅}_{𝑔}}{{𝑅}_{𝑠}+{𝑅}_{𝑔}}$

Plug in the numbers:

${𝐼}_{𝑠}=\frac{\left(2\text{ A}\right)\times \left(50\Omega \right)}{10\Omega +50\Omega }$

${𝐼}_{𝑠}=\frac{100\text{ A}\Omega }{60\Omega }$

${𝐼}_{𝑠}=\frac{10}{6}\text{ A}$

${𝐼}_{𝑠}=\frac{5}{3}\text{ A}$

So, the current passing through the shunt is $\frac{5}{3}\text{ A}$.

Comparing this with the given options, it matches Option D.

Let $𝑉$ be the potential difference across the parallel combination.

The power dissipated across the $16\Omega$ resistor in the upper branch is

${𝑃}_{16}=\frac{{𝑉}^2}{16}$

Given ${𝑃}_{16}=2\text{W}$,

${𝑉}^2=2\times 16=32{\text{V}}^2$

The equivalent resistance of the lower branch is

${𝑅}_{\text{lower}}=6\Omega +4\Omega =10\Omega$

The current flowing through the lower branch is

${𝐼}_{\text{lower}}=\frac{𝑉}{{𝑅}_{\text{lower}}}=\frac{𝑉}{10}$

Therefore, the power dissipated across the $4\Omega$ resistor is

${𝑃}_4={𝐼}_2^{\text{lower}}\times 4={\left(\frac{𝑉}{10}\right)}^2\times 4=\frac{{𝑉}^2}{100}\times 4=\frac{32\times 4}{100}=1.28\text{W}$

Hence, the correct option is (C).

The acceleration of a particle is given by the slope of the velocity--time graph:

$𝑎=\frac{d𝑣}{d𝑡}$

For motion with uniform acceleration, the acceleration is constant, which means the velocity varies linearly with time:

$𝑣=𝑢+𝑎𝑡$

This is represented by a straight line with a constant non-zero slope on a velocity--time graph, as shown in (D).

Graph (A) represents constant velocity (zero acceleration), graph (B) represents a particle at rest, and graph (C) represents uniform motion with constant velocity.

Hence, the correct option is (D).

Magnetic susceptibility χ relates to relative permeability μr by

${𝜇}_{𝑟}=1+𝜒=1+799=800.$

The absolute permeability μ is

$𝜇={𝜇}_0{𝜇}_{𝑟}=(4𝜋\times {10}^{-7})\times 800=3200𝜋\times {10}^{-7}=3.2𝜋\times {10}^{-4}Tm{A}^{-1}.$

That matches Option B.

Let’s compare the two final pressures, starting from the same initial state $({𝑝}_1,{𝑉}_1,{𝑇}_1)$ and expanding to ${𝑉}_2=6{𝑉}_1$.

Isothermal expansion

• Temperature stays at ${𝑇}_1$.

• By Boyle’s law:

${𝑝}_{2,iso}=\frac{{𝑝}_1{𝑉}_1}{{𝑉}_2}=\frac{{𝑝}_1}{6}.$

Adiabatic expansion

• No heat exchange, so

${𝑝}_1{𝑉}_{𝛾}^1={𝑝}_{2,ad}{𝑉}_{𝛾}^2⟹{𝑝}_{2,ad}={𝑝}_1(\frac{{𝑉}_1}{{𝑉}_2}{)}^{𝛾}={𝑝}_1{6}^{-𝛾}.$

• With $𝛾=1.4$:

$6^{-1.4}\approx \frac{1}{6^{1.4}}\approx \frac{1}{12.3}\approx 0.08⟹{𝑝}_{2,ad}\approx 0.08{𝑝}_1.$

Since ${𝑝}_{2,ad}\approx 0.08{𝑝}_1$ is smaller than ${𝑝}_{2,iso}=0.167{𝑝}_1$, the final pressure after adiabatic expansion is lower than after isothermal expansion.

Answer: Option D.

Terminal A is connected directly to node E, and terminal B is connected to the central junction.

The circuit forms a Wheatstone bridge between terminals A and B, with nodes D and C forming the opposite conjugate vertices. The branch resistances are ${𝑅}_{\text{AD}}=4𝑅$, ${𝑅}_{\text{BD}}=4𝑅$, ${𝑅}_{\text{AC}}=2𝑅$, and ${𝑅}_{\text{BC}}=2𝑅$, with a resistor of $5𝑅$ connected between nodes D and C.

For this bridge,

$\frac{{𝑅}_{\text{AD}}}{{𝑅}_{\text{BD}}}=\frac{4𝑅}{4𝑅}=1$

and

$\frac{{𝑅}_{\text{AC}}}{{𝑅}_{\text{BC}}}=\frac{2𝑅}{2𝑅}=1$

Since $\frac{{𝑅}_{\text{AD}}}{{𝑅}_{\text{BD}}}=\frac{{𝑅}_{\text{AC}}}{{𝑅}_{\text{BC}}}$, the Wheatstone bridge is balanced. The potentials at nodes D and C are equal, so no current flows through the $5𝑅$ resistor.

Removing the $5𝑅$ resistor, the network simplifies to two parallel branches between terminals A and B: branch ADB of resistance ${𝑅}_1=4𝑅+4𝑅=8𝑅$, and branch ACB of resistance ${𝑅}_2=2𝑅+2𝑅=4𝑅$.

The equivalent resistance between points A and B is

${𝑅}_{\text{AB}}=\frac{{𝑅}_1{𝑅}_2}{{𝑅}_1+{𝑅}_2}=\frac{8𝑅\times 4𝑅}{8𝑅+4𝑅}=\frac{8}{3}𝑅$

Hence, the correct option is (A).

For a single slit, the first minimum occurs when

$𝑎\sin 𝜃=𝜆$

where

• $𝑎$ is the slit width,

• $𝜆=800\text{ nm}=8.00\times {10}^{-7}\text{ m}$,

• $𝜃={45}^{\circ }$ so $\sin {45}^{\circ }=\frac{\sqrt{\surd 2}}{2}\approx 0.7071$.

Plug in the numbers:

$𝑎=\frac{𝜆}{\sin 𝜃}=\frac{8.00\times {10}^{-7}}{0.7071}\approx 1.13\times {10}^{-6}\text{ m}$

Converting to micrometers:

$1.13\times {10}^{-6}\text{ m}=1.13𝜇\text{m}$

That matches Option A.

The electric and magnetic fields in a free-space plane wave are related by

$𝐸=𝑐𝐵$

where $𝑐=3\times {10}^{8}m/s$. Plugging in $𝐵=2\times {10}^{-8}T$,

$𝐸=(3\times {10}^8)(2\times {10}^{-8})=6V/m.$

So the correct choice is Option B: 6 V/m.

At resonance ${𝑋}_{𝐿}={𝑋}_{𝐶}$, so the net impedance is purely resistive:

$𝑍=𝑅=44\Omega .$

The rms current is

${𝐼}_{rms}=\frac{{𝑉}_{rms}}{𝑅}=\frac{220}{44}=5\text{A}.$

The average power (over one cycle) is

$𝑃={𝑉}_{rms}{𝐼}_{rms}\cos 𝜙={𝑉}_{rms}{𝐼}_{rms}=\frac{{𝑉}_2^{rms}}{𝑅}=\frac{{220}^2}{44}=1100\text{W}=1.1\text{kW}.$

Answer: Option C (1.1 kW).