CUET 2026 Physics May 22 Shift 1 Question Paper- Download Answer Key and Solution PDF

Concept:
For an electromagnetic wave propagating in free space: \[ E = cB \]
where:
\(E\) = magnitude of electric field
\(B\) = magnitude of magnetic field
\(c = 3 \times 10^8 \,m/s\) (speed of light)
Also, the directions of \(\vec{E}\), \(\vec{B}\), and propagation are mutually perpendicular and follow: \[ \vec{E} \times \vec{B} = Direction of propagation \]
Step 1: Calculate the magnitude of electric field.
Given: \[ B = 2 \times 10^{-7}\,T \]
Using: \[ E = cB \]
Substituting values: \[ E = (3 \times 10^8)(2 \times 10^{-7}) \]
\[ E = 6 \times 10^1 \]
\[ E = 60\,V/m \]
Step 2: Determine the direction of electric field.
The wave propagates along: \[ +\hat{i} \]
Given: \[ \vec{B} = 2 \times 10^{-7}\hat{j} \]
Using: \[ \vec{E} \times \vec{B} = \hat{i} \]
We know: \[ \hat{k} \times \hat{j} = -\hat{i} \]
Therefore: \[ (-\hat{k}) \times \hat{j} = \hat{i} \]
Hence, electric field is along: \[ -\hat{k} \]
Therefore, \[ \boxed{\vec{E} = -60\,\hat{k}\,V/m} \] Quick Tip: Remember for electromagnetic waves: \(E = cB\) \(\vec{E}\), \(\vec{B}\), and direction of propagation are mutually perpendicular Use right-hand rule: \[ \vec{E} \times \vec{B} = Propagation direction \]

Concept:
When a bulb is connected across a resistor, both become parallel combinations. First, the resistance of the bulb is calculated using: \[ R = \frac{V^2}{P} \]
Then equivalent resistance is found using parallel combination formula.
Step 1: Calculate the resistance of the bulb.
Given: \[ V = 200\,V, \qquad P = 100\,W \]
Using: \[ R = \frac{V^2}{P} \]
\[ R_b = \frac{(200)^2}{100} \]
\[ R_b = \frac{40000}{100} \]
\[ R_b = 400\,\Omega \]
Step 2: Find equivalent resistance of bulb and \(400\,\Omega\) resistor.
The bulb is connected across the \(400\,\Omega\) resistor.
Hence: \[ 400\,\Omega \parallel 400\,\Omega \]
Using parallel combination: \[ R_p = \frac{400 \times 400}{400 + 400} \]
\[ R_p = \frac{160000}{800} \]
\[ R_p = 200\,\Omega \]
Step 3: Find total resistance of the circuit.
Now this equivalent resistance is in series with the \(200\,\Omega\) resistor.
\[ R_{total} = 200 + 200 \]
\[ R_{total} = 400\,\Omega \]
Step 4: Calculate the circuit current.
Using Ohm’s law: \[ I = \frac{V}{R} \]
\[ I = \frac{100}{400} \]
\[ I = 0.25\,A \]
Step 5: Find the potential drop across the parallel combination.
Voltage across parallel branch: \[ V_p = I \times R_p \]
\[ V_p = 0.25 \times 200 \]
\[ V_p = 50\,V \]
Since the bulb is connected across this branch, the potential drop across the bulb is:
\[ \boxed{50\,V} \] Quick Tip: Remember: Bulb resistance: \[ R = \frac{V^2}{P} \] Same voltage across parallel combination In parallel: \[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \]

Concept:
According to Gauss’s Law, the total electric flux through a closed surface is: \[ \Phi = \frac{q_{enclosed}}{\varepsilon_0} \]
Thus, electric flux depends only on the net charge enclosed inside the surface.
Step 1: Identify the charges enclosed by the sphere of radius \(3\,cm\).
Charges are located at: \[ +8\,\mu C at x=2\,cm \] \[ -2\,\mu C at x=4\,cm \]
For sphere of radius \(3\,cm\):
\(+8\,\mu C\) lies inside
\(-2\,\mu C\) lies outside
Hence enclosed charge: \[ q_1 = 8\,\mu C \]
Therefore, flux through first sphere: \[ \Phi_1 = \frac{8}{\varepsilon_0} \]
Step 2: Identify the charges enclosed by the sphere of radius \(5\,cm\).
For sphere of radius \(5\,cm\):
Both charges lie inside
Net enclosed charge: \[ q_2 = 8 + (-2) \]
\[ q_2 = 6\,\mu C \]
Therefore, flux through second sphere: \[ \Phi_2 = \frac{6}{\varepsilon_0} \]
Step 3: Find the ratio of electric fluxes.
\[ \Phi_1 : \Phi_2 = \frac{8}{\varepsilon_0} : \frac{6}{\varepsilon_0} \]
\[ = 8:6 \]
\[ = 4:3 \]
Therefore, \[ \boxed{4:3} \] Quick Tip: Remember: Electric flux depends only on enclosed charge According to Gauss’s law: \[ \Phi = \frac{q_{enclosed}}{\varepsilon_0} \] Charges outside the Gaussian surface do not contribute to net flux

Concept:
Mutual inductance is defined as: \[ M = \frac{\Phi}{I} \]
where:
\(\Phi\) = magnetic flux linked with one loop due to current in the other loop
\(I\) = current producing the magnetic field
Since \(L \gg R\), magnetic field due to the square loop can be assumed approximately uniform over the circular loop.
Step 1: Find magnetic field at the center of square loop.
Magnetic field due to one side of square loop at its center is: \[ B_1 = \frac{\mu_0 I}{4\pi (L/2)} (\sin45^\circ + \sin45^\circ) \]
Since: \[ \sin45^\circ = \frac{1}{\sqrt{2}} \]
\[ B_1 = \frac{\mu_0 I}{2\pi L} \left(\frac{2}{\sqrt{2}}\right) \]
\[ B_1 = \frac{\mu_0 I}{\sqrt{2}\pi L} \]
There are four sides of the square, hence total magnetic field: \[ B = 4B_1 \]
\[ B = \frac{4\mu_0 I}{\sqrt{2}\pi L} \]
\[ B = \frac{2\sqrt{2}\mu_0 I}{\pi L} \]
Step 2: Find magnetic flux through circular loop.
Area of circular loop: \[ A = \pi R^2 \]
Flux linked: \[ \Phi = BA \]
\[ \Phi = \left( \frac{2\sqrt{2}\mu_0 I}{\pi L} \right) (\pi R^2) \]
\[ \Phi = \frac{2\sqrt{2}\mu_0 I R^2}{L} \]
Step 3: Calculate mutual inductance.
Using: \[ M = \frac{\Phi}{I} \]
\[ M = \frac{2\sqrt{2}\mu_0 I R^2/L}{I} \]
\[ M = 2\sqrt{2}\frac{\mu_0 R^2}{L} \]
Therefore, \[ \boxed{ M = 2\sqrt{2}\frac{\mu_0 R^2}{L} } \] Quick Tip: Remember: Mutual inductance: \[ M = \frac{\Phi}{I} \] Magnetic field at center of square loop: \[ B = \frac{2\sqrt{2}\mu_0 I}{\pi L} \] Flux: \[ \Phi = BA \]

Concept:
Binding energy of a nucleus is the energy released when nucleons combine to form the nucleus.
\[ Binding Energy = (\Delta m)c^2 \]
where: \[ \Delta m = Mass defect \]
Binding energy per nucleon: \[ BE per nucleon = \frac{Total Binding Energy}{A} \]
Step 1: Calculate the mass defect.
For: \[ ^{209}_{83}\mathrm{Bi} \]
Number of protons: \[ Z = 83 \]
Number of neutrons: \[ N = 209-83 = 126 \]
Mass of separate nucleons: \[ M = 83(1.007825) + 126(1.008665) \]
\[ M = 83.649475 + 127.09179 \]
\[ M = 210.741265\,u \]
Given actual nuclear mass: \[ m = 208.980388\,u \]
Mass defect: \[ \Delta m = 210.741265 - 208.980388 \]
\[ \Delta m = 1.760877\,u \]
Step 2: Calculate total binding energy.
Using: \[ BE = \Delta m \times 931 \]
\[ BE = 1.760877 \times 931 \]
\[ BE \approx 1639.78\,MeV \]
Step 3: Calculate binding energy per nucleon.
\[ BE per nucleon = \frac{1639.78}{209} \]
\[ BE per nucleon \approx 7.84\,MeV \]
Therefore, \[ \boxed{7.84\,MeV} \] Quick Tip: Remember: Mass defect: \[ \Delta m = (Zm_p + Nm_n) - m_{nucleus} \] Binding energy: \[ BE = \Delta m \times 931\,MeV \] Binding energy per nucleon: \[ \frac{BE}{A} \]

Concept:
To match physical quantities with dimensions, we use standard dimensional formulas.
Energy:
\[ [E] = ML^2T^{-2} \]
Frequency:
\[ [\nu] = T^{-1} \]
Potential difference:
\[ [V] = ML^2T^{-3}A^{-1} \]
Planck’s constant:
\[ h = \frac{E}{\nu} \]
Step 1: Find dimensions of Planck’s constant.
Using: \[ h = \frac{E}{\nu} \]
\[ [h] = \frac{ML^2T^{-2}}{T^{-1}} \]
\[ [h] = ML^2T^{-1} \]
Thus: \[ A \rightarrow III \]
Step 2: Find dimensions of stopping potential.
Stopping potential is a potential difference.
Dimensions of potential: \[ [V] = ML^2T^{-3}A^{-1} \]
Thus: \[ B \rightarrow IV \]
Step 3: Find dimensions of work function.
Work function is energy required to remove an electron.
Dimensions: \[ [W] = ML^2T^{-2} \]
Thus: \[ C \rightarrow I \]
Step 4: Find dimensions of threshold frequency.
Frequency has dimensions: \[ [\nu] = T^{-1} \]
Thus: \[ D \rightarrow II \]
Hence, the correct matching is: \[ A-III,\quad B-IV,\quad C-I,\quad D-II \]
Therefore, \[ \boxed{A-III,\; B-IV,\; C-I,\; D-II} \] Quick Tip: Remember: Energy: \[ ML^2T^{-2} \] Potential: \[ ML^2T^{-3}A^{-1} \] Frequency: \[ T^{-1} \] Planck’s constant: \[ ML^2T^{-1} \]

Concept:
In electrostatics:
Excess charge in a conductor resides only on its surface.
Electric field inside a conductor in electrostatic equilibrium is zero.
Also, between the plates of a capacitor, an electric field exists due to potential difference between the plates.
Step 1: Analyze Assertion A.
Assertion states:
“A conductor does not store any net charge inside.
This is true because in electrostatic equilibrium:
Free electrons move until electric field inside becomes zero.
Excess charge remains only on the outer surface.
Hence, Assertion A is true.
Step 2: Analyze Reason R.
Reason states that free charge carriers placed between capacitor plates experience force and drift.
Between capacitor plates: \[ E = \frac{V}{d} \]
Hence an electric field exists, and a charge placed there experiences: \[ F = qE \]
Therefore, Reason R is also true.
Step 3: Check whether R explains A.
Reason R talks about:
Motion of charges between capacitor plates
Presence of electric field in capacitor region
But Assertion A concerns:
Distribution of charge inside a conductor in electrostatic equilibrium
Thus, R does not correctly explain A.
Therefore: \[ \boxed{Both\ A\ and\ R\ are\ true\ but\ R\ is\ NOT\ the\ correct\ explanation\ of\ A} \] Quick Tip: Remember: In electrostatics, excess charge stays on the surface of a conductor Electric field inside conductor: \[ E = 0 \] Between capacitor plates: \[ E \neq 0 \]

Concept:
The relation between magnetic field \(B\) and magnetic intensity \(H\) inside a magnetic material is: \[ B = \mu H \]
where: \[ \mu = \mu_r \mu_0 \]

Thus: \[ H = \frac{B}{\mu_r \mu_0} \]



Step 1: Write the given values.
\[ \mu_r = 400 \]
\[ B = 1\,T \]
\[ \mu_0 = 4\pi \times 10^{-7} \]



Step 2: Calculate magnetic intensity \(H\).

Using: \[ H = \frac{B}{\mu_r \mu_0} \]

Substituting values: \[ H = \frac{1}{400 \times 4\pi \times 10^{-7}} \]
\[ H = \frac{1}{1600\pi \times 10^{-7}} \]
\[ H = \frac{10^7}{1600\pi} \]
\[ H = \frac{10^4}{1.6\pi} \]
\[ H = \frac{6250}{\pi} \]
\[ H = \frac{25}{\pi}\times 10^2 \]
\[ H = \frac{25}{\pi}\times 10^5 \times 10^{-3} \]

More directly: \[ H = \frac{1}{1600\pi \times 10^{-7}} = \frac{10^5}{16\pi} \]

But expressing properly: \[ H = \frac{25}{\pi}\times 10^2 = \left(\frac{25}{\pi}\right)\times 10^2 \]

Comparing with: \[ H = \alpha \times 10^5 \]

we obtain: \[ \alpha = \frac{25}{\pi} \]



Therefore, \[ \boxed{\alpha = \frac{25}{\pi}} \] Quick Tip: Remember: Magnetic field relation: \[ B = \mu H \] Magnetic permeability: \[ \mu = \mu_r \mu_0 \] Hence: \[ H = \frac{B}{\mu_r \mu_0} \]

Concept:
In an electromagnetic wave:
Electric field \(\vec{E}\), magnetic field \(\vec{B}\), and direction of propagation are mutually perpendicular.
They satisfy:
\[ \vec{E} \times \vec{B} = Direction\ of\ propagation \]
Magnitudes are related by:
\[ E_0 = cB_0 \]
Since: \[ c = \nu\lambda \]
we get: \[ E_0 = \nu\lambda B_0 \]
Step 1: Identify the direction of propagation.
Given: \[ \vec{B} = B_0\sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{j} \]
The phase: \[ \left(2\pi\nu t - \frac{2\pi x}{\lambda}\right) \]
indicates propagation along: \[ +\hat{i} \]
Step 2: Determine direction of electric field.
We require: \[ \vec{E} \times \vec{B} = \hat{i} \]
Given: \[ \vec{B} \parallel \hat{j} \]
Using vector product: \[ (-\hat{k}) \times \hat{j} = \hat{i} \]
Hence: \[ \vec{E} \parallel -\hat{k} \]
Step 3: Find magnitude of electric field.
Using: \[ E_0 = cB_0 \]
and: \[ c = \nu\lambda \]
\[ E_0 = \nu\lambda B_0 \]
Therefore: \[ \vec{E} = -\nu\lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{k} \]
Hence, \[ \boxed{ \vec{E} = -\nu\lambda B_0 \sin\left(2\pi\nu t - \frac{2\pi x}{\lambda}\right)\hat{k} } \] Quick Tip: Remember for electromagnetic waves: \[ E_0 = cB_0 \] \[ c = \nu\lambda \] Direction rule: \[ \vec{E} \times \vec{B} = Propagation\ direction \]

Concept:
For a thin lens, lens maker’s formula is: \[ \frac{1}{f} = (\mu -1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a symmetric double convex lens: \[ R_1 = +R, \qquad R_2 = -R \]
Step 1: Substitute the radii into lens maker’s formula.
\[ \frac{1}{f} = (\mu -1) \left( \frac{1}{R} - \frac{1}{-R} \right) \]
\[ \frac{1}{f} = (\mu -1) \left( \frac{1}{R} + \frac{1}{R} \right) \]
\[ \frac{1}{f} = (\mu -1)\frac{2}{R} \]
Step 2: Substitute refractive index value.
Given: \[ \mu = 1.4 \]
\[ \mu -1 = 0.4 \]
Thus: \[ \frac{1}{f} = 0.4 \times \frac{2}{R} \]
\[ \frac{1}{f} = \frac{0.8}{R} \]
\[ f = \frac{R}{0.8} \]
\[ f = 1.25R \]
Step 3: Find the required ratio.
\[ \frac{f}{R} = 1.25 \]
Therefore, \[ \boxed{1.25} \] Quick Tip: Remember: Lens maker formula: \[ \frac{1}{f} = (\mu-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] For double convex lens: \[ R_1 = +R,\qquad R_2 = -R \]