CUET 2026 Physics May 22 Shift 2 Question Paper- Download Answer Key and Solution PDF

Concept:

Magnetic field inside a long solenoid is: \[ B=\mu_0 n I \]

where:

\(B\) = magnetic field
\(n\) = turns per unit length
\(I\) = current




Step 1: Write the formula.
\[ B=\mu_0 nI \]



Step 2: Substitute values.
\[ B= (4\pi \times 10^{-7}) \times1000 \times2 \]
\[ = 8\pi \times10^{-4}\,T \]



Step 3: Final answer.
\[ \boxed{ 8\pi \times10^{-4}\,T } \] Quick Tip: For long solenoids: \[ B\propto nI \] Increasing turns or current increases magnetic field strength.

Concept:

In an \(LCR\) circuit: \[ X_L=\omega L \]
and \[ X_C=\frac1{\omega C} \]

At resonance: \[ X_L=X_C \]

which makes impedance minimum and current maximum.



Step 1: Condition for resonance.

At resonance: \[ \omega L=\frac1{\omega C} \]

Thus: \[ X_L=X_C \]



Step 2: Effect at resonance.

Net reactance becomes zero: \[ X=X_L-X_C=0 \]

Therefore impedance: \[ Z=R \]

becomes minimum.



Step 3: Final conclusion.
\[ \boxed{ X_L=X_C } \] Quick Tip: At resonance: \[ Z=R \] and current becomes maximum in a series \(LCR\) circuit.

Concept:

In forward bias:

Positive terminal is connected to \(p\)-side.
Negative terminal is connected to \(n\)-side.


This reduces the barrier potential and narrows the depletion region.



Step 1: Understand depletion region.

The depletion layer contains immobile ions and acts as a barrier to charge flow.



Step 2: Effect of forward bias.

Forward bias opposes the built-in electric field.

Hence barrier potential decreases.

Therefore depletion layer thickness decreases.



Step 3: Conclusion.
\[ \boxed{ Depletion layer decreases } \] Quick Tip: Forward bias decreases barrier potential, while reverse bias increases it.

Concept:

Drift velocity is: \[ v_d=\frac{eE\tau}{m} \]

where:

\(E\) = electric field
\(\tau\) = relaxation time


Thus: \[ v_d\propto E \]



Step 1: Write relation.
\[ v_d=\frac{eE\tau}{m} \]



Step 2: Analyze dependence.

As electric field increases: \[ v_d \]
also increases.



Step 3: Final answer.
\[ \boxed{ Electric field increases } \] Quick Tip: Current density: \[ J=ne v_d \] Hence higher drift velocity produces larger current.

Concept:

Magnetic field at the center of a circular coil is: \[ B=\frac{\mu_0 NI}{2R} \]

where:

\(N\) = number of turns
\(I\) = current
\(R\) = radius




Step 1: Observe proportionality.

From: \[ B=\frac{\mu_0 NI}{2R} \]

we get: \[ B\propto I \]



Step 2: Analyze options.

Magnetic field increases linearly with current.

It decreases with radius.



Step 3: Final answer.
\[ \boxed{ Current through coil } \] Quick Tip: For circular loops: \[ B\propto \frac{I}{R} \] Smaller radius produces stronger magnetic field.

Concept:

When a charged particle moves perpendicular to a magnetic field, magnetic force acts as centripetal force.

Magnetic force: \[ F=qvB\sin\theta \]

For perpendicular motion: \[ \theta=90^\circ \]

thus: \[ F=qvB \]



Step 1: Identify direction of force.

Magnetic force is always perpendicular to velocity.

Therefore it changes direction of velocity but not speed.



Step 2: Compare with centripetal force.

For circular motion: \[ \frac{mv^2}{r}=qvB \]

Hence: \[ r=\frac{mv}{qB} \]

Thus particle moves in a circular path.



Step 3: Final answer.
\[ \boxed{ Circular path } \] Quick Tip: If velocity is perpendicular to magnetic field: \[ \vec v \perp \vec B \] the particle performs uniform circular motion.

Concept:

Alternating current changes direction periodically.

Positive half cycle and negative half cycle are equal and opposite.

Hence their net average over one full cycle becomes zero.



Step 1: Understand sinusoidal current.

AC current is generally: \[ i=I_0\sin\omega t \]

During first half cycle: \[ i>0 \]

During second half cycle: \[ i<0 \]



Step 2: Find average over complete cycle.

Average current: \[ I_{avg} = \frac1T\int_0^T I_0\sin\omega t\,dt \]

Since positive and negative areas cancel each other: \[ I_{avg}=0 \]



Step 3: Final conclusion.
\[ \boxed{ 0 } \] Quick Tip: Average value of AC over a complete cycle is zero, but RMS value is non-zero.

Concept:

In solids:

Conductors have overlapping valence and conduction bands.
Semiconductors have small band gap.
Insulators have large band gap.




Step 1: Understand conductors.

In conductors: \[ E_g \approx 0 \]

Valence band and conduction band overlap.

Hence electrons move freely.



Step 2: Compare with semiconductors and insulators.


Semiconductor:
\[ E_g\approx1\,eV \]

Insulator:
\[ E_g>5\,eV \]


Thus conductor has negligible forbidden gap.



Step 3: Final answer.
\[ \boxed{ 0\,eV } \] Quick Tip: Smaller band gap means easier movement of electrons and higher conductivity.

Concept:

Self inductance is: \[ L=\frac{\mu_0\mu_r N^2A}{l} \]

where:

\(N\) = number of turns
\(A\) = area
\(l\) = length
\(\mu_r\) = relative permeability




Step 1: Analyze dependence.

From formula: \[ L\propto \mu_r \]
and \[ L\propto \frac{N^2A}{l} \]

Thus inductance depends upon:

Shape and dimensions of coil
Nature of core material




Step 2: Check options.

Only option involving geometry and medium correctly represents all dependencies.



Step 3: Final answer.
\[ \boxed{ Geometry of coil and medium } \] Quick Tip: Greater number of turns and magnetic permeability increase self inductance.

Concept:

Efficiency of transformer: \[ \eta= \frac{Output Power} {Input Power} \times100 \]

An ideal transformer has:

No flux leakage
No copper loss
No hysteresis loss
No eddy current loss




Step 1: Understand ideal transformer.

In ideal transformer: \[ P_{in}=P_{out} \]

Thus: \[ \eta= \frac{P_{out}}{P_{in}}\times100 = 100% \]



Step 2: Final answer.
100% Quick Tip: Practical transformers always have efficiency less than \(100%\) due to energy losses.