CUET 2026 Physics May 26 Shift 2 Question Paper- Download Answer Key and Solution PDF

Step 1: Recall Biot–Savart law.
Magnetic field due to a current element is: \[ dB \propto I \]
Hence magnetic field is directly proportional to current.
Step 2: Magnetic field at the center of circular loop.
For a circular loop: \[ B=\frac{\mu_0 I}{2R} \]
Thus: \[ B \propto I \]
Hence Assertion is true.
Step 3: Check the Reason.
Reason correctly explains why magnetic field depends on current.
Therefore: \[ \boxed{Both\ Assertion\ and\ Reason\ are\ true\ and\ Reason\ correctly\ explains\ Assertion} \]
Quick Tip: Remember: \[ B_{circular loop}=\frac{\mu_0 I}{2R} \] Magnetic field increases with current and decreases with radius.

Step 1: Recall electromagnetic spectrum order.
Energy order: \[ Gamma\ rays\ >\ X-rays\ >\ UV\ >\ Visible\ >\ IR\ >\ Microwaves\ >\ Radio\ waves \]
Step 2: Match each wave.
\[ Gamma rays \rightarrow Highest\ energy \]
\[ X-rays \rightarrow Medical\ imaging \]
\[ Microwaves \rightarrow Radar\ communication \]
\[ Radio waves \rightarrow Lowest\ frequency \]
Therefore: \[ \boxed{A-II,\ B-IV,\ C-III,\ D-I} \]
Quick Tip: Remember: \[ E=hf \] Higher frequency means higher energy.

Step 1: Recall de Broglie wavelength formula.
\[ \lambda=\frac{h}{\sqrt{2mK}} \]
where: \[ m=mass, \qquad K=kinetic energy \]
Step 2: Compare electron and proton wavelengths.
For same kinetic energy: \[ \lambda \propto \frac{1}{\sqrt{m}} \]
Thus: \[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \]
Therefore: \[ \boxed{\sqrt{\frac{m_p}{m_e}}} \]
Quick Tip: Remember: \[ \lambda \propto \frac{1}{\sqrt{m}} \] Lighter particle has greater de Broglie wavelength.

Step 1: Recall kinetic energy gained in electric field.
\[ K=qV \]
where: \[ q=charge, \qquad V=potential\ difference \]
Step 2: Calculate energies.
For proton: \[ K_p=eV \]
For alpha particle: \[ K_\alpha=2eV \]
Thus: \[ K_p:K_\alpha = 1:2 \]
Therefore: \[ \boxed{1:2} \]
Quick Tip: Remember: \[ K=qV \] Greater charge gains greater kinetic energy in same potential difference.

Step 1: Recall torque formula.
Torque on a magnetic dipole in a magnetic field is given by: \[ \tau = MB\sin\theta \]
Torque is maximum when: \[ \sin\theta = 1 \]
Thus: \[ \theta=90^\circ \]
Step 2: Relate angle with loop orientation.
Magnetic moment is perpendicular to plane of loop.
If magnetic moment is perpendicular to field: \[ \theta=90^\circ \] then plane of loop becomes parallel to magnetic field.
Thus Assertion is true.
Reason is also true and correctly explains Assertion.
Therefore: \[ \boxed{ Both\ Assertion\ and\ Reason\ are\ true\ and\ Reason\ is\ the\ correct\ explanation\ of\ Assertion.} \]
Quick Tip: Remember: \[ \tau_{\max}=MB \] Maximum torque occurs when magnetic moment is perpendicular to magnetic field.

Step 1: Use mirror formula.
\[ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \]
Given: \[ f=-20 cm, \qquad u=-30 cm \]
Step 2: Substitute values.
\[ \frac{1}{-20} = \frac{1}{v} + \frac{1}{-30} \]
\[ \frac{1}{v} = -\frac{1}{20} +\frac{1}{30} \]
\[ \frac{1}{v} = \frac{-3+2}{60} = -\frac{1}{60} \]
Thus: \[ v=-60 cm \]
Magnitude: \[ \boxed{60 cm} \]
Quick Tip: Remember sign convention: \[ f<0,\quad u<0 \] for concave mirrors.

Step 1: Recall magnifying power formula.
\[ M=1+\frac{D}{f} \]
where: \[ D=least\ distance\ of\ distinct\ vision \]
Step 2: Analyze relation.
\[ M \propto \frac{1}{f} \]
Hence smaller focal length gives larger magnifying power.
Therefore: \[ \boxed{Focal\ length\ decreases} \]
Quick Tip: Remember: \[ Smaller\ focal\ length \Rightarrow Greater\ magnification \]

\[ Magnetic flux \rightarrow Weber \]
\[ Magnetic field \rightarrow Tesla \]
\[ Capacitance \rightarrow Farad \]
\[ Resistance \rightarrow Ohm \]
Thus: \[ \boxed{A-II,\ B-I,\ C-III,\ D-IV} \]
Quick Tip: Remember: \[ 1 Tesla = \frac{Weber}{m^2} \]

Step 1: Use photon energy relation. \[ E=hf \]
Given: \[ E=6 eV \]
\[ 1 eV=1.6\times10^{-19} J \]
Thus: \[ E=9.6\times10^{-19} J \]
Step 2: Calculate frequency.
\[ f=\frac{E}{h} \]
\[ f= \frac{9.6\times10^{-19}} {6.63\times10^{-34}} \]
\[ f\approx1.45\times10^{15} Hz \]
Therefore: \[ \boxed{1.45\times10^{15} Hz} \]
Quick Tip: Remember: \[ 1 eV=1.6\times10^{-19} J \]

Step 1: Recall Einstein photoelectric equation.
\[ K_{\max}=hf-\phi \]
Thus: \[ K_{\max}\propto f \]
Step 2: Relation with stopping potential.
\[ eV_0=K_{\max} \]
Hence: \[ V_0\propto f \]
Thus stopping potential depends on frequency.
Both Assertion and Reason are true and Reason correctly explains Assertion.
Therefore: \[ \boxed{Both\ Assertion\ and\ Reason\ are\ true\ and\ Reason\ is\ the\ correct\ explanation\ of\ Assertion.} \]
Quick Tip: Remember: \[ V_0=\frac{hf-\phi}{e} \] Stopping potential increases with frequency.