CUET 2026 Physics May 29 Shift 2 Question Paper- Download Answer Key and Solution PDF

Step 1: Simplify the upper branch.
In the upper branch, one resistor \(R\) is in series with two resistors \(R\) connected in parallel.
The parallel combination is:
\[ R_p = \frac{R\cdot R}{R+R} = \frac{R}{2} \]
Therefore, the equivalent resistance of the upper branch is:
\[ R_{upper} = R+\frac{R}{2} = \frac{3R}{2} \]
Step 2: Simplify the lower branch.
The lower branch consists of two resistors \(R\) connected in parallel.
Thus,
\[ R_{lower} = \frac{R\cdot R}{R+R} = \frac{R}{2} \]
Step 3: Combine the two branches.
The upper and lower branches are connected in parallel between \(A\) and \(B\).
Hence,
\[ R_{eq} = \frac{R_{upper}\,R_{lower}} {R_{upper}+R_{lower}} \]
\[ = \frac{\left(\frac{3R}{2}\right)\left(\frac{R}{2}\right)} {\frac{3R}{2}+\frac{R}{2}} \]
\[ = \frac{\frac{3R^2}{4}} {2R} \]
\[ = \frac{3R}{8} \]
Therefore,
\[ \boxed{R_{eq}=\frac{3R}{8}} \]
Hence, the correct answer is:
\[ \boxed{(A)} \] Quick Tip: For two equal resistors \(R\) in parallel: \[ R\parallel R=\frac{R}{2} \] A useful strategy for complex resistor networks is: \[ \boxed{Simplify small parallel groups first, then combine series and parallel branches.} \]

Step 1: Identify the resistance values.
Each side of the triangle has resistance
\[ 4\,\Omega \]
Since \(D\) is the midpoint of the resistor \(BC\),
\[ BD=DC=2\,\Omega \]
Step 2: Find the two paths between \(A\) and \(D\).
Path 1:
\[ A \rightarrow B \rightarrow D \]
Resistance along this path:
\[ R_1=4+2=6\,\Omega \]
Path 2:
\[ A \rightarrow C \rightarrow D \]
Resistance along this path:
\[ R_2=4+2=6\,\Omega \]
Step 3: Combine the two paths.
The two \(6\,\Omega\) paths are connected in parallel between \(A\) and \(D\).
Therefore,
\[ R_{AD} = \frac{6\times6}{6+6} \]
\[ = \frac{36}{12} \]
\[ = 3\,\Omega \]
Therefore, the equivalent resistance between \(A\) and \(D\) is
\[ \boxed{3\,\Omega} \]
Hence, the correct answer is:
\[ \boxed{(C)} \] Quick Tip: Whenever a point divides a resistor into equal parts, first split the resistor accordingly. Here: \[ 4\,\Omega \rightarrow 2\,\Omega + 2\,\Omega \] Then identify all possible paths between the required terminals and reduce the circuit using series and parallel combinations.

Step 1: Analyze the upper triangular section.
Between the two upper junctions, there are two possible paths:
Direct path:
\[ 8\,\Omega \]
Upper path through the apex:
\[ 4+4=8\,\Omega \]
Therefore these two \(8\,\Omega\) resistances are in parallel.
\[ R_{upper} = 8 \parallel 8 = \frac{8\times8}{8+8} = 4\,\Omega \]
Step 2: Replace the upper section.
The circuit now becomes a rectangle with:
\[ 4\,\Omega \]
between the upper junctions and
\[ 12\,\Omega \]
between the lower junctions.
The side resistances are:
\[ 4\,\Omega \] on the left and \[ 4\,\Omega \] on the right.
Step 3: Find equivalent resistance between the lower junctions.
One path is directly:
\[ 12\,\Omega \]
The other path goes through the upper branch:
\[ 4+4+4 = 12\,\Omega \]
Hence,
\[ R = 12 \parallel 12 = 6\,\Omega \]
Step 4: Include the terminal resistors.
The \(6\,\Omega\) resistor from \(A\) to the left junction and the \(6\,\Omega\) resistor from the right junction to \(B\) are in series with the equivalent found above.
Thus,
\[ R_{AB} = 6+6+6 \]
\[ R_{AB} = 18\,\Omega \]
Therefore,
\[ \boxed{R_{AB}=18\,\Omega} \]
Hence, the correct answer is:
\[ \boxed{(B)} \] Quick Tip: Whenever a triangular resistor network contains: \[ R in parallel with (R_1+R_2), \] reduce that section first. Here: \[ 8\Omega \parallel (4\Omega+4\Omega) = 8\Omega \parallel 8\Omega = 4\Omega \] After simplification, the remaining network becomes a simple series-parallel combination.

Step 1: Determine the resistances of the two arcs.
The wire forms a complete circle of total resistance:
\[ R=36\,\Omega \]
The points \(A\) and \(B\) subtend an angle of
\[ 60^\circ \]
Therefore, resistance of the smaller arc is:
\[ R_1 = 36\left(\frac{60}{360}\right) = 6\,\Omega \]
The remaining arc subtends
\[ 360^\circ-60^\circ=300^\circ \]
Hence,
\[ R_2 = 36\left(\frac{300}{360}\right) = 30\,\Omega \]
Step 2: Identify the combination.
The two arcs connect the same points \(A\) and \(B\).
Therefore, they are in parallel.
\[ R_{AB} = \frac{R_1R_2}{R_1+R_2} \]
\[ = \frac{(6)(30)}{6+30} \]
\[ = \frac{180}{36} \]
\[ = 5\,\Omega \]
Step 3: Identify the correct option.
\[ \boxed{R_{AB}=5\,\Omega} \]
Hence, the correct answer is:
\[ \boxed{(B)} \] Quick Tip: For a circular wire of total resistance \(R\): \[ R_{arc} = R\left(\frac{\theta}{360^\circ}\right) \] If two arcs connect the same terminals, they are in parallel. For an angle \(\theta\): \[ R_1=\frac{\theta}{360^\circ}R, \qquad R_2=\frac{360^\circ-\theta}{360^\circ}R \] and \[ R_{eq} = \frac{R_1R_2}{R_1+R_2} \]

Step 1: Identify the three paths between \(A\) and \(C\).
Upper path:
\[ A \rightarrow D \rightarrow C \]
Resistance:
\[ R_1=2+1=3\,\Omega \]
Lower path:
\[ A \rightarrow B \rightarrow C \]
Resistance:
\[ R_2=2+1=3\,\Omega \]
Diagonal path:
\[ A \rightarrow C \]
Resistance:
\[ R_3=3\,\Omega \]
Step 2: Observe the parallel combination.
All three branches connect directly between the same terminals \(A\) and \(C\).
Hence the three \(3\,\Omega\) resistors are in parallel.
\[ \frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \]
\[ \frac{1}{R_{eq}} = 1 \]
Therefore,
\[ R_{eq} = 1\,\Omega \]
Step 3: Select the correct option.
\[ \boxed{R_{eq}=1\,\Omega} \]
Hence, the correct answer is:
\[ \boxed{(B)} \] Quick Tip: If \(n\) equal resistors of resistance \(R\) are connected in parallel: \[ R_{eq} = \frac{R}{n} \] Here, \[ 3\Omega \parallel 3\Omega \parallel 3\Omega = \frac{3}{3} = 1\Omega \] Always look for complete paths joining the same two terminals before simplifying a resistor network.

Step 1: Apply the condition for suspension.
For the wire to remain suspended in air, the magnetic force must balance its weight.
\[ F_B=mg \]
Step 2: Use magnetic force on a current-carrying conductor.
The magnetic force is given by:
\[ F_B=BIL\sin\theta \]
For maximum upward force,
\[ \theta=90^\circ \]
Therefore,
\[ F_B=BIL \]
Step 3: Substitute the given values.
Mass:
\[ m=200\,g=0.2\,kg \]
Weight:
\[ mg=0.2\times10=2\,N \]
Using
\[ BIL=mg \]
\[ B\times2\times1.5=2 \]
\[ 3B=2 \]
\[ B=\frac{2}{3}\,T \]
\[ B=0.67\,T \]
Therefore,
\[ \boxed{B=0.67\,T} \]
Hence, the correct answer is:
\[ \boxed{(D)} \] Quick Tip: For a current-carrying wire suspended by a magnetic field: \[ BIL=mg \] Hence, \[ \boxed{B=\frac{mg}{IL}} \] This formula is frequently asked in CUET, JEE Main and NEET Physics examinations.

Step 1: Apply the condition for suspension.
Since the wire is suspended in air, the magnetic force balances its weight.
\[ F_B=mg \]
Step 2: Use the magnetic force formula.
Magnetic force on a current-carrying conductor is:
\[ F_B=BIL\sin\theta \]
Since the magnetic field is perpendicular to the wire,
\[ \theta=90^\circ \]
and therefore,
\[ F_B=BIL \]
Step 3: Substitute the given values.
\[ m=500\,g=0.5\,kg \]
Taking
\[ g=9.8\,m s^{-2} \]
\[ mg=0.5\times9.8=4.9\,N \]
Using
\[ BIL=mg \]
\[ B\times1.4\times7=4.9 \]
\[ 9.8B=4.9 \]
\[ B=\frac{4.9}{9.8} \]
\[ B=0.5\,T \]
Therefore,
\[ \boxed{B=0.5\,T} \]
Hence, the correct answer is:
\[ \boxed{(C)} \] Quick Tip: For a wire suspended by magnetic force: \[ BIL=mg \] Hence, \[ \boxed{B=\frac{mg}{IL}} \] When the magnetic field is perpendicular to the wire, \[ \sin 90^\circ =1 \] which gives the maximum magnetic force.

Step 1: Apply the condition for suspension.
For the wire to remain suspended,
\[ F_B=mg \]
Step 2: Use the magnetic force formula.
Magnetic force on a current-carrying conductor is:
\[ F_B=BIL\sin\theta \]
For maximum upward force,
\[ \theta=90^\circ \]
Thus,
\[ BIL=mg \]
Step 3: Substitute the given values.
\[ m=250\,g=0.25\,kg \]
\[ L=175\,cm=1.75\,m \]
\[ B=0.7\,T \]
Taking
\[ g=9.8\,m s^{-2} \]
\[ I=\frac{mg}{BL} \]
\[ I = \frac{0.25\times9.8} {0.7\times1.75} \]
\[ I = \frac{2.45}{1.225} \]
\[ I=2\,A \]
Therefore,
\[ \boxed{I=2\,A} \]
Hence, the correct answer is:
\[ \boxed{(C)} \] Quick Tip: For a wire suspended by magnetic force: \[ BIL=mg \] Hence, \[ \boxed{I=\frac{mg}{BL}} \] Always convert: \[ cm \rightarrow m \] and \[ g \rightarrow kg \] before substitution.

Step 1: Use the force formula for a wire in a uniform magnetic field.
For a wire carrying current in a uniform magnetic field,
\[ \vec F = I\int d\vec l \times \vec B \]
Since \(\vec B\) is uniform,
\[ \vec F = I(\vec L)\times\vec B \]
where \(\vec L\) is the displacement vector joining the initial and final points of the wire.
Step 2: Find the effective displacement.
The wire consists of:
Left straight segment of length \(R\)
Semicircular arc of radius \(R\)
Right straight segment of length \(R\)
The horizontal displacement between the two end points is
\[ R + 2R + R = 4R \]
Thus,
\[ L=4R \] directed horizontally.
Step 3: Calculate the magnitude of force.
Since \(\vec L\) lies in the plane of the paper and \(\vec B\) is perpendicular to the plane,
\[ \theta=90^\circ \]
Therefore,
\[ F=BIL\sin 90^\circ \]
\[ F=BI(4R) \]
\[ F=4BiR \]
Step 4: Determine the direction.
Current flows from left to right and magnetic field is out of the plane.
Using the right-hand rule:
\[ \vec F=\vec L\times\vec B \] the force acts vertically downward.
Therefore,
\[ \boxed{F=4BiR} \] directed vertically downward.
Hence, the correct answer is:
\[ \boxed{(A)} \] Quick Tip: For any shaped wire placed in a {uniform} magnetic field, \[ \boxed{\vec F=I\,(\vec r_{final}-\vec r_{initial})\times\vec B} \] The force depends only on the displacement between the endpoints, not on the actual shape of the wire. This is a very common CUET/JEE conceptual question.

Step 1: Recall the magnetic properties of superconductors.
A superconductor exhibits:
\[ \boxed{Perfect diamagnetism} \]
This phenomenon is known as the \[ \boxed{Meissner Effect} \] in which magnetic flux is completely expelled from the interior of the material.
Step 2: Find the magnetic susceptibility.
For a perfect diamagnetic substance,
\[ \chi=-1 \]
Step 3: Use the relation between permeability and susceptibility.
The relative permeability is related to susceptibility by
\[ \mu_r = 1+\chi \]
Substituting
\[ \chi=-1 \]
gives
\[ \mu_r = 1-1 \]
\[ \mu_r = 0 \]
Step 4: Identify the correct option.
Thus,
\[ \boxed{\chi=-1,\qquad \mu_r=0} \]
Hence, the correct answer is:
\[ \boxed{(B)} \] Quick Tip: For superconductors: \[ \boxed{\chi=-1} \] \[ \boxed{\mu_r=0} \] because \[ \mu_r = 1+\chi \] The complete expulsion of magnetic field from a superconductor is called the \[ \boxed{Meissner Effect} \] which is a favorite CUET/JEE conceptual question.