A rod swinging round in a magnetic field is a battery whose tip does most of the work. This file gets the ½Bωℓ² result, then applies it to pivots along the rod, bicycle spokes and spinning discs.
Rotating rod. Every piece of the rod cuts field lines, but the outer pieces move faster and cut more. Adding up all the tiny emfs gives ½Bωℓ².
Picture it
On a merry-go-round, a child near the centre barely moves while a child at the edge whizzes round. A rotating rod is the same: its tip moves fastest, its pivot does not move at all. So the tip end of the rod does most of the flux cutting.
In exam language
A small piece dx at distance x from the pivot moves with speed v = ωx, so its emf is de = B(ωx)dx. Adding along the rod:
e = ∫₀ˡ Bωx dx = ½Bωℓ²
The same answer comes from treating the whole rod as moving at its average speed ωℓ/2: e = Bℓ(ωℓ/2).
Trap
If the rotation is given in revolutions per second f, use ω = 2πf. In rpm, divide by 60 first: 120 rpm = 2 rev/s = 4π rad/s. For a rod turning in a horizontal plane in the earth's field, only the vertical component BV counts.
Pivot somewhere along the rod
Measure the emf from the pivot to each end separately. Both parts move the same way round, so the potential changes in the same sense from the pivot outwards to either end.
Arrangement
Pivot to one end
Pivot to other end
Between the two ends
Pivot at one end (Illus. 28, points A, B = ℓ/2, C)
A to B: Bωℓ²/8
B to C: 3Bωℓ²/8
A to C: Bωℓ²/2
Pivot at the middle
Bωℓ²/8
Bωℓ²/8
0
Pivot at ℓ/4 from one end (Illus. 29)
Bωℓ²/32
9Bωℓ²/32
Bωℓ²/4
Illustration 30. If the emf from pivot P to tip Q is 100 V, the emf from the midpoint M to Q is ½Bω(ℓ² − ℓ²/4) = ¾ × 100 = 75 V. Illustration 31. A rod of resistance r turning about one end, touching a ring, with R between centre and ring: I = (½Bωℓ²)/(R + r). All four checked: correct.
Spokes of a wheel
Spokes in parallel. Every spoke runs from the same hub to the same rim, so each is an identical cell connected between the same two points. Parallel identical cells give the emf of one.
e = ½BωR² (the same for 1 spoke or 100 spokes)
Trap
Do not multiply by the number of spokes. More spokes only lower the internal resistance, like adding identical cells in parallel.
A rotating disc
Faraday's disc. A solid disc is like a wheel with infinitely many spokes. The emf between centre and rim is steady, so this generator gives direct current.
Between
Emf
Centre and rim
½BωR²
Centre and a point at radius r
½Bωr²
Two points on the rim
0 (same radius)
Points at radii r₁ and r₂
½Bω(r₂² − r₁²)
Formula sheet
Formula
Meaning
When to use
Watch out
e = ½Bωℓ²
Rod rotating about one end
Pivot at an end, B ⊥ plane
ω in rad/s
e = ½Bω(r₂² − r₁²)
Between two points on a rotating rod or disc
Pivot inside, or annulus
Measure both from the pivot
e = Bωℓ²/8, ends equal
Rod pivoted at its middle
Centre to each end
End to end: 0
e = ½BωR²
Wheel with spokes
Any number of spokes
Not × N
e = ½BωR²
Disc, centre to rim
Faraday disc
Rim to rim: 0
I = ½Bωℓ²/(R + r)
Rod on a ring with a load R
Illustration 31
Include the rod's resistance
ω = 2πf = 2π × rpm/60
Converting rotation rates
Given rev/s or rpm
rpm ÷ 60 first
Allen pages 126–128 checked
Item
Check
Rotating rod derivation e = ½Bωℓ²
Correct
Illustration 28: Bωℓ²/8, 3Bωℓ²/8, Bωℓ²/2
Correct
Illustration 29: Bωℓ²/4
Correct
Spokes: parallel cells, independent of N
Correct
Disc: ½BωR² centre to rim, zero between rim points
Correct
Illustration 30: 75 V
Correct
Illustration 31: I = ½Bωℓ²/(R + r)
Correct
Beginner's Box 7 (continued) answer key
Q
Answer
Working
7 (iv)
3Bvℓ, top end L higher
Taking each marked length as ℓ: the vertical distance from L to N is ℓ + ℓ + ℓ, all perpendicular to v. The square in the middle adds nothing extra.
7 (v) (a), (b)
Zero net emf around the closed loop; no current
In a uniform field the flux through a translating loop does not change. (Between the top and bottom points of the ring in (b) there is a potential difference 2BvR, but no current.)
8
e = μ₀Ia²v / 2πx(x + a), current clockwise
Near side PR (at x) and far side QS (at x + a) both cut field lines; their emfs oppose. e = Bₓav − Bₓ₊ₐav with B = μ₀I/2πx. The loop moves into weaker field, so its flux into the page falls: clockwise.
NEET practice: 29 questions
Watch the unit of rotation: rad/s, rev/s or rpm. Every number was recalculated in Python.
Q1Numerical
A 1 m rod rotates at 20 rad/s about one end, in a plane perpendicular to a uniform 0.5 T field. The emf between its ends is
(A)2.5 V
(B)10 V
(C)5 V
(D)20 V
Show the solution
Given
ℓ = 1 m, ω = 20 rad/s, B = 0.5 T
Asked
e
Concept
Rotating rod.
Formula
e = ½Bωℓ²
Baby steps
½ × 0.5 × 20 × 1 = 5 V.
Answer
(C) 5 V
Why not the others
10 V forgets the ½. 2.5 V halves twice. 20 V drops B and the ½.
Shortcut
Half B ω ℓ².
Where it went wrong
Using Bωℓ².
Q2Numerical
A 40 cm rod rotates at 5 revolutions per second about one end, in a plane perpendicular to a 0.2 T field. The emf between its ends is about
(A)0.16 V
(B)0.08 V
(C)1.01 V
(D)0.50 V
Show the solution
Given
ℓ = 0.4 m, f = 5 rev/s, B = 0.2 T
Asked
e
Concept
Convert to rad/s first.
Formula
e = ½B(2πf)ℓ²
Baby steps
ω = 10π rad/s.
e = ½ × 0.2 × 10π × 0.16 = 0.16π ≈ 0.50 V.
Answer
(D) 0.50 V
Why not the others
0.08 V uses ω = 5. 1.01 V forgets the ½. 0.16 V forgets π.
Shortcut
rev/s × 2π.
Where it went wrong
Using f instead of ω.
Q3Numerical
A 1 m metal rod rotates at 120 rpm in a horizontal plane about one end. The vertical component of the earth's field is 4 × 10⁻⁵ T. The emf between its ends is about
(A)2.5 × 10⁻⁴ V
(B)2.4 × 10⁻³ V
(C)8.0 × 10⁻⁵ V
(D)5.0 × 10⁻⁴ V
Show the solution
Given
ℓ = 1 m, 120 rpm, BV = 4 × 10⁻⁵ T
Asked
e
Concept
Horizontal rotation cuts the vertical component.
Formula
e = ½BVωℓ²
Baby steps
120 rpm = 2 rev/s → ω = 4π rad/s.
e = ½ × 4 × 10⁻⁵ × 4π × 1 = 8π × 10⁻⁵ ≈ 2.5 × 10⁻⁴ V.
Answer
(A) 2.5 × 10⁻⁴ V
Why not the others
2.4 × 10⁻³ V uses ω = 120. 8.0 × 10⁻⁵ V forgets π. 5.0 × 10⁻⁴ V forgets the ½.
Shortcut
rpm ÷ 60 × 2π.
Where it went wrong
Putting rpm straight in as ω.
Q4Numerical
A bicycle wheel with 12 metal spokes of length 0.5 m turns at 2 rev/s in a plane perpendicular to a field of 5 × 10⁻⁵ T. The emf between the hub and the rim is about
(A)7.9 × 10⁻⁵ V
(B)9.4 × 10⁻⁴ V
(C)1.6 × 10⁻⁴ V
(D)2.5 × 10⁻⁵ V
Show the solution
Given
N = 12 spokes, R = 0.5 m, f = 2 rev/s, B = 5 × 10⁻⁵ T
Asked
e
Concept
Spokes are identical cells in parallel.
Formula
e = ½BωR²
Baby steps
ω = 4π rad/s.
e = ½ × 5 × 10⁻⁵ × 4π × 0.25 = 2.5π × 10⁻⁵ ≈ 7.9 × 10⁻⁵ V.
The number of spokes does not matter.
Answer
(A) 7.9 × 10⁻⁵ V
Why not the others
9.4 × 10⁻⁴ V multiplies by 12 spokes. 1.6 × 10⁻⁴ V forgets the ½. 2.5 × 10⁻⁵ V leaves out π.
Shortcut
One spoke's emf.
Where it went wrong
Multiplying by the number of spokes.
Q5Numerical
A metal disc of radius 0.2 m spins at 50 rad/s about its axis, which is parallel to a uniform 0.4 T field. The emf between its centre and rim is
(A)0.2 V
(B)0.8 V
(C)4 V
(D)0.4 V
Show the solution
Given
R = 0.2 m, ω = 50 rad/s, B = 0.4 T
Asked
e
Concept
Disc = many rotating rods in parallel.
Formula
e = ½BωR²
Baby steps
½ × 0.4 × 50 × 0.04 = 0.4 V.
Answer
(D) 0.4 V
Why not the others
0.8 V forgets the ½. 4 V uses R² = 0.4 instead of 0.04. 0.2 V halves again.
Shortcut
Same as a rod of length R.
Where it went wrong
Forgetting to square R.
Q6Concept
For the spinning disc of the previous question, the emf between two different points on its rim is
(A)0.8 V
(B)0.4 V
(C)zero
(D)depends on the angle between the points
Show the solution
Given
Two points at the same radius
Asked
e
Concept
Emf from the centre depends only on radius.
Formula
e = ½Bωr²
Baby steps
Both points are at r = R.
Each is ½BωR² from the centre.
Difference = 0.
Answer
(C) zero
Why not the others
0.4 V is centre to rim. 0.8 V doubles it. The angle does not change the radius.
Shortcut
Same radius, same potential.
Where it went wrong
Adding the two centre-to-rim emfs.
Q7Numerical
A 2 m rod rotates at 4 rad/s about its midpoint, in a plane perpendicular to a 0.5 T field. The emf between the midpoint and either end is
(A)4 V
(B)1 V
(C)2 V
(D)0.5 V
Show the solution
Given
ℓ = 2 m, pivot at midpoint, ω = 4 rad/s, B = 0.5 T
Asked
e (centre to end)
Concept
Each half is a rod of length ℓ/2 rotating about one end.
Formula
e = ½Bω(ℓ/2)² = Bωℓ²/8
Baby steps
ℓ/2 = 1 m.
e = ½ × 0.5 × 4 × 1 = 1 V.
Answer
(B) 1 V
Why not the others
4 V uses ½Bωℓ² for the whole rod. 2 V forgets the ½. 0.5 V halves again.
Shortcut
Half-rod of length 1 m.
Where it went wrong
Using the full length.
Q8Numerical
A 1 m rod rotates at 10 rad/s in a plane perpendicular to a 0.4 T field, about a point 25 cm from one end. The potential difference between its two ends is
(A)1.0 V
(B)2 V
(C)1.125 V
(D)1.25 V
Show the solution
Given
ℓ = 1 m, pivot at ℓ/4, ω = 10 rad/s, B = 0.4 T
Asked
e between ends
Concept
Measure from the pivot to each end and subtract (Allen Illustration 29).
Formula
e = ½Bω(r₂² − r₁²)
Baby steps
Short arm 0.25 m: ½ × 0.4 × 10 × 0.0625 = 0.125 V.
Long arm 0.75 m: ½ × 0.4 × 10 × 0.5625 = 1.125 V.
Ends differ by 1.125 − 0.125 = 1.0 V, which is Bωℓ²/4.
Answer
(A) 1.0 V
Why not the others
2 V is ½Bωℓ² for a pivot at an end. 1.125 V is only the long arm. 1.25 V adds the two arms.
Shortcut
Bωℓ²/4.
Where it went wrong
Adding the two arms instead of subtracting.
Q9Numerical
A rod PQ rotates about end P in a field perpendicular to its plane of rotation. The emf between P and Q is 80 V. The emf between the midpoint M and Q is
(A)60 V
(B)40 V
(C)20 V
(D)80 V
Show the solution
Given
ePQ = 80 V
Asked
eMQ
Concept
e ∝ (r₂² − r₁²) measured from the pivot.
Formula
eMQ = ½Bω(ℓ² − ℓ²/4) = ¾ ePQ
Baby steps
ePM = ¼ × 80 = 20 V.
eMQ = 80 − 20 = 60 V.
Answer
(A) 60 V
Why not the others
40 V assumes emf ∝ length. 20 V is P to M. 80 V is P to Q.
Shortcut
Outer half carries ¾ of the emf.
Where it went wrong
Halving because M is halfway.
Q10Numerical
For the same rod (ePQ = 80 V, rotating about P), the emf between P and the midpoint M is
(A)60 V
(B)40 V
(C)20 V
(D)10 V
Show the solution
Given
ePQ = 80 V
Asked
ePM
Concept
e ∝ ℓ² from the pivot.
Formula
ePM = ½Bω(ℓ/2)²
Baby steps
(½)² = ¼.
¼ × 80 = 20 V.
Answer
(C) 20 V
Why not the others
40 V treats e ∝ ℓ. 60 V is M to Q. 10 V uses ⅛.
Shortcut
Half the length, a quarter of the emf.
Where it went wrong
Treating emf as proportional to length.
Q11Numerical
A rod of length 0.5 m and resistance 1 Ω rotates at 40 rad/s about one end in a 0.5 T field perpendicular to its plane. Its other end slides on a conducting ring of negligible resistance, and a 1.5 Ω resistor joins the centre to the ring. The current is
(A)2.5 A
(B)1 A
(C)1.67 A
(D)0.5 A
Show the solution
Given
ℓ = 0.5 m, r = 1 Ω, ω = 40 rad/s, B = 0.5 T, R = 1.5 Ω
Asked
I
Concept
The rod is a cell with internal resistance r (Allen Illustration 31).
Formula
I = ½Bωℓ²/(R + r)
Baby steps
e = ½ × 0.5 × 40 × 0.25 = 2.5 V.
I = 2.5/(1.5 + 1) = 1 A.
Answer
(B) 1 A
Why not the others
2.5 A divides by the rod's 1 Ω only. 1.67 A ignores the rod's resistance. 0.5 A doubles the total resistance.
Shortcut
Emf over total resistance.
Where it went wrong
Leaving out the rod's own resistance.
Q12Numerical
A 0.5 m rod rotates about one end in a plane perpendicular to a 0.4 T field. For an emf of 1 V between its ends, the angular speed must be
(A)20 rad/s
(B)10 rad/s
(C)40 rad/s
(D)5 rad/s
Show the solution
Given
e = 1 V, ℓ = 0.5 m, B = 0.4 T
Asked
ω
Concept
Rearrange the rotating-rod formula.
Formula
ω = 2e/(Bℓ²)
Baby steps
Bℓ² = 0.4 × 0.25 = 0.1.
ω = 2 × 1/0.1 = 20 rad/s.
Answer
(A) 20 rad/s
Why not the others
10 rad/s forgets the 2. 40 rad/s doubles again. 5 rad/s uses ℓ instead of ℓ².
Shortcut
Undo the ½ by multiplying by 2.
Where it went wrong
Forgetting the factor 2.
Q13Concept
The length of a rod rotating about one end is doubled, with the same angular speed and field. The emf between its ends becomes
(A)eight times as large
(B)twice as large
(C)unchanged
(D)four times as large
Show the solution
Given
ℓ → 2ℓ
Asked
New e
Concept
e ∝ ℓ².
Formula
e = ½Bωℓ²
Baby steps
(2)² = 4.
Answer
(D) four times as large
Why not the others
Twice treats e ∝ ℓ (true for a translating rod, not a rotating one). Unchanged and eight times have no basis.
Shortcut
Longer rod: more length and faster tip.
Where it went wrong
Using the Bℓv habit.
Q14Concept
A bicycle wheel's number of metal spokes is doubled, with the same radius, speed and field. The emf between hub and rim
(A)doubles
(B)stays the same
(C)halves
(D)becomes four times
Show the solution
Given
N → 2N
Asked
New e
Concept
Spokes are identical cells in parallel.
Formula
e = ½BωR²
Baby steps
Parallel identical cells give the emf of one.
So e is unchanged.
Answer
(B) stays the same
Why not the others
Doubling treats spokes as cells in series. Halving and four times have no basis.
Shortcut
Parallel cells: same emf.
Where it went wrong
Multiplying by the number of spokes.
Q15Concept
A rod rotates about one end in a plane that is parallel to a uniform magnetic field. The emf between its ends is
(A)½Bωℓ²
(B)zero
(C)Bωℓ²
(D)maximum
Show the solution
Given
B lies in the plane of rotation
Asked
e
Concept
Velocity is also in that plane, so v and B are both in the plane; v × B is perpendicular to the plane, not along the rod.
Formula
e = ∫(v × B) · dℓ
Baby steps
v × B points out of the plane of rotation.
The rod lies in the plane.
No component along the rod: e = 0.
Answer
(B) zero
Why not the others
The non-zero options need B perpendicular to the plane.
Shortcut
Field must be along the axis of rotation.
Where it went wrong
Assuming any rotation in a field gives ½Bωℓ².
Q16Direction
A rod rotates anticlockwise (as seen from the front) about one end, in a magnetic field pointing into the page. Which end is at higher potential?
(A)It changes every half turn
(B)The outer end
(C)Both ends are at the same potential
(D)The pivot (centre) end
Show the solution
Given
Anticlockwise rotation, B into page
Asked
Higher-potential end
Concept
Direction of q(v × B) along the rod.
Formula
F = q(v × B)
Baby steps
Take the rod along +x: its tip moves in +y (anticlockwise).
v × B = ĵ × (−k̂) = −î: towards the pivot.
Positive charge collects at the pivot: pivot is higher.
Answer
(D) The pivot (centre) end
Why not the others
The outer end is higher for clockwise rotation or a field out of the page. Equal potentials would need no emf. The answer does not change as the rod turns.
Shortcut
Work it out for one instant, it holds for all.
Where it went wrong
Cross-product sign slip.
Q17Direction
A rod rotates clockwise (as seen from the front) about one end, in a magnetic field pointing into the page. Which end is at higher potential?
(A)Both ends are at the same potential
(B)The pivot end
(C)The outer end
(D)The midpoint
Show the solution
Given
Clockwise rotation, B into page
Asked
Higher-potential end
Concept
Reversing ω reverses v, so reverses v × B.
Formula
F = q(v × B)
Baby steps
Rod along +x: tip moves in −y.
v × B = (−ĵ) × (−k̂) = +î: outward.
Outer end higher.
Answer
(C) The outer end
Why not the others
Pivot is the anticlockwise answer. Equal potentials need no emf. The midpoint is not an end.
Shortcut
Reverse rotation → swap ends.
Where it went wrong
Using the anticlockwise result.
Q18Direction
A rod rotates anticlockwise (as seen from the front) about one end, in a magnetic field pointing out of the page. Which end is at higher potential?
(A)The pivot end
(B)The outer end
(C)Both ends are at the same potential
(D)Neither, because no emf is induced
Show the solution
Given
Anticlockwise, B out of page
Asked
Higher-potential end
Concept
Reversing B reverses v × B.
Formula
F = q(v × B)
Baby steps
Rod along +x: tip moves in +y.
v × B = ĵ × k̂ = +î: outward.
Outer end higher.
Answer
(B) The outer end
Why not the others
Pivot is the into-page answer. The emf is not zero.
Shortcut
Reverse B → swap ends.
Where it went wrong
Keeping the into-page answer.
Q19Concept
The emf of a rotating rod can be written as Bℓ × (ωℓ/2). The quantity ωℓ/2 is
(A)the speed of the pivot
(B)the speed of the rod's tip
(C)the angular speed of the rod
(D)the average speed of the points on the rod
Show the solution
Given
e = Bℓ(ωℓ/2)
Asked
Meaning of ωℓ/2
Concept
Speed grows linearly from 0 at the pivot to ωℓ at the tip.
Formula
vavg = (0 + ωℓ)/2
Baby steps
Pivot speed 0, tip speed ωℓ.
Linear variation → average = ωℓ/2.
Answer
(D) the average speed of the points on the rod
Why not the others
The tip moves at ωℓ. ω is angular speed. The pivot does not move.
Shortcut
Average of 0 and ωℓ.
Where it went wrong
Using the tip speed.
Q20Concept
Compared with the output of an ordinary rotating-coil generator, the emf between the centre and rim of a disc spinning at constant speed in a steady axial field is
(A)steady (direct)
(B)alternating at the rotation frequency
(C)alternating at twice the rotation frequency
(D)zero on average
Show the solution
Given
Faraday disc
Asked
Nature of output
Concept
Nothing about the geometry changes as the disc turns.
Formula
e = ½BωR² (constant)
Baby steps
Every radius always moves the same way through the same field.
So the emf is constant: DC.
Answer
(A) steady (direct)
Why not the others
Alternating outputs come from coils whose flux reverses. Zero average would need the emf to reverse.
Shortcut
Disc → DC; rotating coil → AC.
Where it went wrong
Assuming every rotating generator gives AC.
Q21Graph
Rods of different lengths rotate about one end at the same angular speed in the same field. Which graph shows the emf against length?
(A)
(B)
(C)
(D)
Show the solution
Given
ω, B fixed
Asked
e against ℓ
Concept
e ∝ ℓ².
Formula
e = ½Bωℓ²
Baby steps
Square law → parabola through the origin.
Answer
(B) the graph in option B
Why not the others
The straight line is a translating rod's Bℓv. The root curve and flat line have no basis.
Shortcut
ℓ² → parabola.
Where it went wrong
Using the translating-rod graph.
Q22Graph
A rod of fixed length rotates about one end in a fixed field. Which graph shows the emf against angular speed?
(A)
(B)
(C)
(D)
Show the solution
Given
ℓ, B fixed
Asked
e against ω
Concept
e ∝ ω.
Formula
e = ½Bωℓ²
Baby steps
Linear in ω → straight line through the origin.
Answer
(C) the graph in option C
Why not the others
The parabola squares ω. The flat line ignores ω. The hyperbola inverts it.
Shortcut
Only ℓ is squared.
Where it went wrong
Squaring ω.
Q23Graph
A rod rotates about one end O. Which graph shows the potential difference between O and a point at distance x along the rod, against x?
(A)
(B)
(C)
(D)
Show the solution
Given
Potential along a rotating rod
Asked
ΔV against x
Concept
ΔV(x) = ½Bωx².
Formula
ΔV ∝ x²
Baby steps
Near the pivot the rod moves slowly: ΔV grows slowly.
Farther out it grows faster: parabola opening upward.
Answer
(D) the graph in option D
Why not the others
The straight line assumes uniform speed. The curve that flattens has most change near the pivot, the opposite of reality. The flat line ignores the emf.
Shortcut
Slow near the pivot, fast near the tip.
Where it went wrong
Drawing the curve that bends the wrong way.
Q24Graph
Wheels identical except for the number of metal spokes are spun at the same speed in the same field. Which graph shows the hub-to-rim emf against the number of spokes?
(A)
(B)
(C)
(D)
Show the solution
Given
Spokes in parallel
Asked
e against N
Concept
Parallel identical cells.
Formula
e = ½BωR²
Baby steps
N does not appear.
Horizontal line.
Answer
(C) the graph in option C
Why not the others
The rising line treats spokes in series. The hyperbola makes more spokes weaker. The parabola has no basis.
Shortcut
N-independent.
Where it went wrong
Thinking more spokes means more emf.
Q25Assertion–reason
Assertion (A): The emf between hub and rim of a rotating wheel does not depend on the number of metal spokes. Reason (R): The spokes act as identical cells connected in parallel between hub and rim.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: N-independent. R: parallel identical cells.
Asked
Truth and link
Concept
Parallel identical emfs.
Formula
—
Baby steps
A is true.
R is true.
R is exactly why A holds.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
Circuit picture explains the result.
Where it went wrong
Choosing (B) by habit.
Q26Assertion–reason
Assertion (A): The emf between the two ends of a rod rotating about one end is Bωℓ². Reason (R): The speed of points on the rod increases linearly with distance from the pivot.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: Bωℓ². R: speed ∝ distance.
Asked
Truth and link
Concept
Linear speed gives ½Bωℓ² after integration.
Formula
e = ½Bωℓ²
Baby steps
R is true.
A is false: the emf is ½Bωℓ².
A false, R true.
Answer
(D) A is false, but R is true.
Why not the others
(A), (B) and (C) need A true.
Shortcut
Average speed is half the tip speed.
Where it went wrong
Using the tip speed for the whole rod.
Q27Assertion–reason
Assertion (A): For a rod rotating about its midpoint, the potential difference between its two ends is zero. Reason (R): Measured from the midpoint, the emf to each end is Bωℓ²/8 in the same sense.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: ends at equal potential. R: equal emfs from centre, same sense.
Asked
Truth and link
Concept
Both halves push charge the same way relative to the centre.
Formula
e = Bωℓ²/8 each
Baby steps
A is true.
R is true.
Equal changes from the same point → equal end potentials. R explains A.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
Symmetric halves cancel between the ends.
Where it went wrong
Adding the two halves.
Q28Assertion–reason
Assertion (A): Two points on the rim of a rotating metal disc in an axial field are at the same potential. Reason (R): The disc is made of a conducting material.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: rim points equal. R: disc conducts.
Asked
Truth and link
Concept
The reason is equal radius, not conductivity.
Formula
e = ½Bωr²
Baby steps
A is true.
R is true.
Conductivity alone does not make the potentials equal: the centre and rim of the same conductor differ. R does not explain A.
Answer
(B) Both A and R are true, but R is not the correct explanation of A.
Why not the others
(A) claims R explains A. (C) and (D) need a false statement.
Shortcut
Equal radius is the real reason.
Where it went wrong
Linking any true statement as the reason.
Q29Two statements
Statement I: The emf between the centre and rim of a rotating disc is ½BωR². Statement II: This emf doubles if the radius of the disc is doubled.
(A)Both Statement I and Statement II are true.
(B)Both Statement I and Statement II are false.
(C)Statement I is true, but Statement II is false.
(D)Statement I is false, but Statement II is true.
Show the solution
Given
Disc emf and scaling
Asked
Which are true
Concept
e ∝ R².
Formula
e = ½BωR²
Baby steps
Statement I is true.
Statement II is false: it becomes four times.
Answer
(C) Statement I is true, but Statement II is false.
Why not the others
Both-true fails on II. Both-false fails on I. I-false-II-true reverses them.
Shortcut
Square the radius factor.
Where it went wrong
Treating e ∝ R.
Answer key
1 C
2 D
3 A
4 A
5 D
6 C
7 B
8 A
9 A
10 C
11 B
12 A
13 D
14 B
15 B
16 D
17 C
18 B
19 D
20 A
21 B
22 C
23 D
24 C
25 A
26 D
27 A
28 B
29 C
Spread across letters: A 8, B 7, C 7, D 7. No letter repeats more than twice in a row. Question mix: Numerical 11, Concept 6, Direction 3, Graph 4, Assertion–reason 4, Two statements 1. Balancing seed 1002.