CUET 2026 Physics June 6 Shift 1 Question Paper- Download Answer Key and Solution PDF

Concept:
According to Ohm's Law,
\[ V=IR \]
where
\[ V=Potential Difference, \quad I=Current, \quad R=Resistance \]
Step 1: Write the given values.
\[ V=10\,V \]
\[ R=5\,\Omega \]
Step 2: Use Ohm's law to calculate current.
\[ I=\frac{V}{R} \]
\[ I=\frac{10}{5} \]
\[ I=2\,A \]
Therefore,
\[ \boxed{I=2\,A} \]
\[ \boxed{Answer = (C)} \] Quick Tip: For numerical questions from Current Electricity, first check whether Ohm's law can be directly applied.

Concept:
According to de Broglie's hypothesis,
\[ \lambda=\frac{h}{p} \]
where
\[ \lambda=de Broglie wavelength \]
\[ h=Planck's constant \]
\[ p=momentum \]
Step 1: Observe the relation.
\[ \lambda=\frac{h}{p} \]
Since \(h\) is constant,
\[ \lambda\propto \frac{1}{p} \]
Step 2: Identify the correct option.
The wavelength is inversely proportional to momentum.
Therefore,
\[ \boxed{\lambda\propto \frac{1}{p}} \]
\[ \boxed{Answer = (B)} \] Quick Tip: Higher momentum means shorter wavelength.

Concept:
The magnetic field at the centre of a circular coil is
\[ B=\frac{\mu_0 I}{2R} \]
Step 1: Observe the proportionality.
\[ B\propto I \]
Step 2: Double the current.
If
\[ I\rightarrow 2I \]
then
\[ B\rightarrow 2B \]
Hence the magnetic field also doubles.
Therefore,
\[ \boxed{Magnetic Field = Double} \]
\[ \boxed{Answer = (B)} \] Quick Tip: Magnetic field at the centre of a coil is directly proportional to current.

Concept:
Kinetic Energy is given by
\[ KE=\frac12 mv^2 \]
Step 1: Substitute the given values.
\[ KE=\frac12 \times 4 \times 10^2 \]
\[ KE=2\times100 \]
\[ KE=200\,J \]
Therefore,
\[ \boxed{KE=200\,J} \]
\[ \boxed{Answer = (C)} \] Quick Tip: Kinetic energy depends on the square of velocity. Doubling velocity makes kinetic energy four times.

Concept:
Lens Formula:
\[ \frac1f=\frac1v-\frac1u \]
Step 1: Write the given values.
\[ f=+15\,cm \]
\[ u=-30\,cm \]
Step 2: Apply lens formula.
\[ \frac1{15} = \frac1v+\frac1{30} \]
\[ \frac1v = \frac1{15}-\frac1{30} \]
\[ = \frac2{30}-\frac1{30} \]
\[ = \frac1{30} \]
\[ v=30\,cm \]
Therefore,
\[ \boxed{v=30\,cm} \]
\[ \boxed{Answer = (C)} \] Quick Tip: An object placed at \(2F\) of a convex lens forms its image at \(2F\) on the other side.

Concept:
The rms speed of gas molecules is given by
\[ v_{rms}=\sqrt{\frac{3RT}{M}} \]
Therefore,
\[ v_{rms}\propto \sqrt{T} \]
Step 1: Write the relation between initial and final rms speeds.
\[ \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} \]
Step 2: Substitute the given values.
\[ \frac{v_2}{500} = \sqrt{\frac{1200}{300}} \]
\[ \frac{v_2}{500} = \sqrt{4} \]
\[ \frac{v_2}{500} = 2 \]
\[ v_2=1000\,m/s \]
Therefore,
\[ \boxed{v_2=1000\,m/s} \]
\[ \boxed{Answer = (C)} \] Quick Tip: The rms speed is directly proportional to the square root of absolute temperature.

Concept:
According to Einstein's photoelectric equation,
\[ K_{max}=h\nu-\phi \]
where
\[ K_{max}=Maximum Kinetic Energy \]
\[ h\nu=Energy of incident photon \]
\[ \phi=Work Function \]
Step 1: Write the given values.
\[ h\nu=5\,eV \]
\[ \phi=2\,eV \]
Step 2: Apply Einstein's equation.
\[ K_{max}=5-2 \]
\[ K_{max}=3\,eV \]
Therefore,
\[ \boxed{K_{max}=3\,eV} \]
\[ \boxed{Answer = (B)} \] Quick Tip: Photoelectric emission occurs only when photon energy exceeds the work function.

Concept:
For capacitors connected in parallel,
\[ C_{eq}=C_1+C_2 \]
Step 1: Write the given values.
\[ C_1=4\,\mu F \]
\[ C_2=6\,\mu F \]
Step 2: Calculate equivalent capacitance.
\[ C_{eq}=4+6 \]
\[ C_{eq}=10\,\mu F \]
Therefore,
\[ \boxed{C_{eq}=10\,\mu F} \]
\[ \boxed{Answer = (C)} \] Quick Tip: Capacitances add directly in parallel and decrease in series combination.

Concept:
Radioactive decay law:
\[ N=N_0\left(\frac12\right)^n \]
where
\[ n=\frac{t}{T_{1/2}} \]
Step 1: Calculate the number of half-lives elapsed.
\[ n=\frac{12}{4} \]
\[ n=3 \]
Step 2: Find the remaining fraction.
\[ \frac{N}{N_0} = \left(\frac12\right)^3 \]
\[ = \frac18 \]
Therefore,
\[ \boxed{\frac{N}{N_0}=\frac18} \]
\[ \boxed{Answer = (C)} \] Quick Tip: After every half-life, the amount of radioactive substance becomes half of its previous value.

Concept:
According to Faraday's Law,
\[ e= \left| \frac{\Delta\phi}{\Delta t} \right| \]
where
\[ \Delta\phi=Change in Magnetic Flux \]
\[ \Delta t=Time Interval \]
Step 1: Calculate the change in magnetic flux.
\[ \Delta\phi = 0.8-0.2 \]
\[ \Delta\phi=0.6\,Wb \]
Step 2: Apply Faraday's law.
\[ e = \frac{0.6}{0.1} \]
\[ e=6\,V \]
Therefore,
\[ \boxed{e=6\,V} \]
\[ \boxed{Answer = (C)} \] Quick Tip: Induced emf depends on the rate of change of magnetic flux, not on the flux alone.