All the physics saved from the test, arranged by chapter and sub-topic. Each question carries what was given, what was asked, the concept and formula behind it, the steps in full, the fastest route through, and a freshly drawn diagram wherever the original had one.
29
Questions saved
17
Left blank
12
Correct
0
Wrong attempts
Current Electricity 23 · Magnetism and Matter 5 · Thermal 1. On NEET marking the seventeen blanks were worth 68 marks.
The headline is not the physics, it is the zero. Every physics question that was attempted was answered correctly. Nothing was marked wrong. All 68 marks lost here went to questions that were never started — and several of the blanks are shorter than the ones that were solved.
Chapters in this paper — red = wrong · amber = blank · green = correct
In an aluminium (Al) bar of square cross-section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7 × 10−8 Ωm and 1.0 × 10−7 Ωm respectively. The electrical resistance between the two faces P and Q of the composite bar is
2475/64 μΩ
KEY3750/128 μΩ
1875/49 μΩ
2475/132 μΩ
Given
Outer square side 7 mm, inner iron square side 2 mm, both running the full length.
Length between faces P and Q = 50 mm.
ρAl = 2.7 × 10−8 Ωm, ρFe = 1.0 × 10−7 Ωm.
Asked
Resistance between faces P and Q.
Concept to use
Both metals span the same two faces and therefore sit at the same pair of potentials — that makes them parallel, not series. Current divides between them. Each path has the same length (50 mm) but a different cross-sectional area, and the aluminium area is the outer square minus the hole, not the whole square.
Formula to use
R = ρL/A | 1/Req = 1/RAl + 1/RFe
Baby steps
Areas first. AFe = 2×2 = 4 mm2. AAl = 7×7 − 4 = 49 − 4 = 45 mm2. Subtracting the hole is the step that decides this question.
Convert once: 1 mm2 = 10−6 m2, L = 50 mm = 5 × 10−2 m.
Divide top and bottom by 10: 3750/128 μΩ ≈ 29.3 μΩ.
Answer
3750/128 μΩ
Shortcut
Two sanity checks let you pick the option without finishing the arithmetic. The parallel result must be smaller than the smaller branch (30 μΩ), so anything above 30 is out. And the denominator 49 in one option is a tell that the aluminium area was taken as the full 49 mm2 instead of 45 — a wrong-answer trap, not a right answer.
Iron is the poorer conductor here and has the smaller area, so it carries very little current: the composite behaves almost like the aluminium alone. Recognising that early tells you the answer sits just under 30 μΩ.
Q9Stretching a wire, then bending itNot attempted
A wire of resistance 100 Ω is stretched, so that its length increases by 20%. The stretched wire is then bent in the form of a rectangle whose length and breadth are in the ratio 3:2. The effective resistance between the ends of any diagonal of the rectangle is
KEY36 Ω
72 Ω
28.8 Ω
432 Ω
Given
Original wire: R = 100 Ω.
Stretched so length increases by 20%.
Bent into a rectangle with sides in the ratio 3 : 2.
Resistance wanted across a diagonal.
Asked
Effective resistance between two opposite corners.
Concept to use
Three separate ideas stack here. First, stretching conserves volume, so if length scales by k the area scales by 1/k and resistance by k2. Second, a uniform wire divides its resistance in proportion to length, so the rectangle’s perimeter of 10 equal parts (3+2+3+2) splits the total accordingly. Third, the two ways round the rectangle between opposite corners are two parallel paths.
Formula to use
Stretching: R′ = k2R | uniform wire: R ∝ length | two paths in parallel
Baby steps
Stretch factor k = 1.2, so R′ = (1.2)2 × 100 = 144 Ω. Note it is k2, not k — area shrinks as the length grows.
Bend into the rectangle. The perimeter is 3 + 2 + 3 + 2 = 10 parts, so one part carries 144/10 = 14.4 Ω.
Long side = 3 parts = 43.2 Ω; short side = 2 parts = 28.8 Ω.
Between opposite corners, each of the two routes is one long side plus one short side: 43.2 + 28.8 = 72 Ω each.
The two routes are in parallel: R = 72/2 = 36 Ω.
Answer
36 Ω
Shortcut
Between the ends of a diagonal, each path is always half the perimeter, so each path carries half the total resistance and the answer is simply total/4 = 144/4 = 36 Ω. The 3:2 ratio is a decoy — it changes nothing for a diagonal, though it would matter for adjacent corners.
The distractor 72 Ω is the half-perimeter value with the final parallel step forgotten; 28.8 Ω is a single short side. Both are stops along the correct route, which is why they appear.
Q10Drift velocity · total electron momentumNot attempted
If current of 80 A is passing through a straight conductor of length 10 m, then the total momentum of electrons in the conductor is (mass of electron = 9.1 × 10−31 kg and charge of electron = 1.6 × 10−19 C)
910 × 10−9 Ns
910 × 10−11 Ns
455 × 10−9 Ns
KEY455 × 10−11 Ns
Given
I = 80 A, L = 10 m.
me = 9.1 × 10−31 kg, e = 1.6 × 10−19 C.
Asked
Total momentum of all the conduction electrons in the wire.
Concept to use
Total momentum = (number of electrons) × (mass) × (drift velocity). Both the number and the drift velocity involve the unknowns n and A — but they cancel. That cancellation is the entire point of the question: you are not expected to know the electron density of the metal.
Formula to use
I = n A e vd | N = n A L → p = N m vd = mIL/e
Baby steps
Number of free electrons in the conductor: N = n A L.
Drift velocity from the current: vd = I / (n A e).
Total momentum p = N m vd = (nAL) · m · I/(nAe). The n and A cancel: p = mIL/e.
Substitute: p = (9.1×10−31 × 80 × 10) / (1.6×10−19).
Numerator = 9.1 × 800 × 10−31 = 7.28 × 10−28.
p = 7.28×10−28 / 1.6×10−19 = 4.55 × 10−9 = 455 × 10−11 Ns.
Answer
455 × 10−11 Ns
Shortcut
Memorise the collapsed result p = mIL/e. It says the total electron momentum depends only on the current and the length — not on the metal, the thickness, or the electron density. Once you have that, this is one substitution.
Both 455 options are the same number written differently: 455 × 10−11 = 4.55 × 10−9, whereas 455 × 10−9 is a hundred times too big. When two options share a mantissa, the exponent is the question — count the powers of ten deliberately.
Q11Temperature coefficient of resistance vs resistivityNot attempted
A metal wire has a coefficient of linear expansion α1 and a temperature coefficient of resistivity α2. Then the equivalent temperature coefficient of resistance of the wire is
(α1 + α2)
KEY(α2 − α1)
(2α1 − α2)
(2α2 − α1)
Given
Coefficient of linear expansion α1.
Temperature coefficient of resistivity α2.
Asked
The temperature coefficient of the wire’s resistance.
Concept to use
Resistance depends on three things that all change with temperature: R = ρL/A. The resistivity rises (α2), the length rises (α1), and the cross-sectional area rises too — at twice the linear rate, since area is two-dimensional. Because the area is in the denominator, its growth reduces the resistance and partly cancels the other two effects.
Formula to use
R = ρL/A → αR = αρ + αL − αA = α2 + α1 − 2α1
Baby steps
Write how each factor scales for a small rise ΔT: ρ → ρ(1 + α2ΔT), L → L(1 + α1ΔT), A → A(1 + 2α1ΔT).
Area carries 2α1 because it is a two-dimensional quantity — this is the step the question is really testing.
For small ΔT, keep only first-order terms: R′ ≈ R[1 + (α2 + α1 − 2α1)ΔT].
So αR = α2 − α1.
Answer
(α2 − α1)
Shortcut
For any product of powers, coefficients simply add with the exponent as weight, and denominators come in with a minus sign. R = ρL1A−1 and A ∝ L2, so αR = α2 + α1 − 2α1 = α2 − α1. Writing it that way takes one line and never needs the binomial expansion.
Since α2 for metals (≈ 4 × 10−3) is far larger than α1 (≈ 2 × 10−5), the geometry correction is a small refinement — which is exactly why the everyday formula R = R0(1 + αΔT) ignores it.
Q42Resistance of wires of the same material and massNot attempted
The lengths of two copper wires A and B are 180 cm and 270 cm respectively. If the mass of wire A is twice the mass of wire B and the electrical resistance of wire A is 200 Ω, then the electrical resistance of wire B is
KEY900 Ω
400 Ω
600 Ω
300 Ω
Given
Same material (copper), so the same ρ and the same density d.
LA = 180 cm, LB = 270 cm.
mA = 2 mB.
RA = 200 Ω.
Asked
Resistance of wire B.
Concept to use
The areas are not given, so eliminate them using the mass. For a wire of density d, mass = d × A × L, so A = m/(dL). Substituting into R = ρL/A gives a formula in terms of the two things you do know — length and mass.
Formula to use
R = ρL/A and A = m/(dL) → R = ρd L2/m → R ∝ L2/m
Baby steps
Derive the working relation: R = ρL/A with A = m/(dL) gives R = ρdL2/m. For the same material, R ∝ L2/m.
Form the ratio: RB/RA = (LB/LA)2 × (mA/mB).
Lengths: LB/LA = 270/180 = 3/2, so the square is 9/4.
Keep the three proportionalities on one card and pick whichever matches the data given: R ∝ L/A (length and thickness known), R ∝ L2/m (length and mass known), R ∝ 1/A2 at fixed mass. Choosing the right one turns every wire question into a single ratio.
The L2 is easy to miss: B is longer and thinner (same material, less mass, more length), and both effects push its resistance up. If you use R ∝ L/m by mistake you get 300 Ω — which is exactly why 300 Ω is on the list.
Resistor networks and equivalent resistance
3 questions · 1 blank · 2 correct
Q31Choosing a combination to hit a target resistanceNot attempted
From the combination of resistors with resistance values R1 = R2 = R3 = 5 Ω and R4 = 10 Ω, which of the following combination is the best circuit to get an equivalent resistance of 6 Ω?
KEY(R1 + R2) in parallel with (R3 + R4)
(R1 + R2 + R4) in parallel with R3
R1 ∥ (R2 + R3) ∥ R4
fourth arrangement shown in the paper
Given
R1 = R2 = R3 = 5 Ω, R4 = 10 Ω.
Target equivalent resistance = 6 Ω.
Asked
Which circuit gives 6 Ω.
Concept to use
Test each option rather than trying to design the answer. Two quick bounds make most options fall without arithmetic: a series chain is always larger than its biggest member, and a parallel combination is always smaller than its smallest member.
Formula to use
Series: R = Ra + Rb | Parallel: R = RaRb/(Ra + Rb)
(R1 + R2) in parallel with (R3 + R4) — 10 ∥ 15 = 6 Ω
Shortcut
Use the parallel bound as a filter. The target is 6 Ω, so any option whose smallest parallel branch is below 6 Ω is impossible — that removes every arrangement in which a bare 5 Ω resistor sits alone across the terminals. Only the option where both branches exceed 6 Ω survives, and it is the one to compute.
The useful design fact behind this: to get 6 from 5s and a 10, you need the two parallel branches to be 10 and 15, since 1/6 = 1/10 + 1/15. Recognising 6 as the parallel pair (10, 15) is faster than testing four circuits.
Q37Ladder network · spotting a shared nodeAnswered correctly
Find the equivalent resistance across AB.
KEY1 Ω
2 Ω
3 Ω
4 Ω
Given
Five 2 Ω resistors arranged as in the figure, with a horizontal wire joining the two mid-points.
Asked
Equivalent resistance between A and B.
Concept to use
The plain wire joining the two mid-points is the key: a resistanceless wire forces both ends to the same potential, so those two points are electrically one node. Redraw with that single node C, and the tangle becomes three simple stages.
Formula to use
Points joined by a resistanceless wire are one node | then series and parallel as usual
Baby steps
Merge the two mid-points into a single node C, because the connecting wire has no resistance.
Between A and C there are now two 2 Ω resistors in parallel: 1 Ω.
Between C and B there are likewise two 2 Ω resistors in parallel: 1 Ω.
Path A–C–B is therefore 1 + 1 = 2 Ω.
The fifth resistor, 2 Ω, connects A to B directly and sits in parallel with that path.
RAB = (2 × 2)/(2 + 2) = 1 Ω.
Answer
1 Ω
Shortcut
Label nodes before doing any arithmetic. Give every point joined by bare wire the same letter, then list each resistor as a pair of letters. Once the network is written as A–C, A–C, C–B, C–B, A–B, the answer is immediate and the drawing cannot mislead you.
Node-labelling is the single most reliable technique for messy-looking networks — it also solves Q40 in this paper and would have solved Q5. If a circuit looks intimidating, relabel rather than redraw.
Q40Adding wires to a zigzag · balanced bridge againAnswered correctly
Five identical resistance wires of 1 Ω each are connected as shown in figure as clear lines. If two similar wires are added as shown by dashed lines, find the change in resistance between A and B.
KEY2 Ω
1 Ω
3 Ω
4 Ω
Given
Five 1 Ω wires in a zigzag chain from A to B through nodes 2, 3, 4, 5.
Two more 1 Ω wires added: one joining nodes 2 and 4, one joining nodes 3 and 5.
Asked
Change in resistance between A and B.
Concept to use
Before the additions the five wires are simply in series: 5 Ω. After the additions, look at the block between nodes 2 and 5 — it has four arms (2–3, 3–5, 2–4, 4–5) plus a cross-link 3–4. That is a Wheatstone bridge, and since all four arms are 1 Ω it is balanced, so the cross-link carries nothing.
Formula to use
Balance: R23/R35 = R24/R45 ⇒ bridge arm 3–4 carries no current
Baby steps
Before: A–2–3–4–5–B, five 1 Ω wires in series = 5 Ω.
After: node 2 now connects to both 3 and 4; node 5 connects to both 4 and 3. Take nodes 2 and 5 as the bridge terminals and nodes 3, 4 as the detector corners.
All four arms are 1 Ω, so 1/1 = 1/1 — balanced. The 3–4 wire carries zero current; delete it.
What remains between 2 and 5 is two paths of 2 Ω each (2–3–5 and 2–4–5), in parallel: 1 Ω.
Total: A–2 (1 Ω) + block (1 Ω) + 5–B (1 Ω) = 3 Ω.
Change = 5 − 3 = 2 Ω.
Answer
2 Ω (the resistance falls from 5 Ω to 3 Ω)
Shortcut
Adding wires can only ever reduce resistance, never raise it — extra paths mean more ways for current to flow. So the answer must be a drop, and since the original is 5 Ω the new value is 5 minus the option. Combined with the balanced-bridge collapse, the answer falls out in two lines.
This is the second balanced bridge in the paper — Q5 is the other one, and the same recognition solves both. If a network has four arms meeting at two mid-nodes with a link between them, test the ratio first.
Wheatstone bridge and meter bridge
3 questions · 1 blank · 2 correct
Q5Balanced Wheatstone bridge in disguiseNot attempted
Find the reading of the ammeter connected in the given circuit diagram (resistance of ammeter = 18 Ω).
The circuit is drawn flat, but it is a Wheatstone bridge: A and D are the supply corners, B and C are the detector corners, and the 3 Ω resistor is the bridge arm between them. If the ratio of the two arms on one side matches the ratio on the other, B and C sit at the same potential and the bridge arm carries no current at all — the network then collapses into two simple series chains.
Formula to use
Balance condition: RAB/RBD = RAC/RCD ⇒ no current in the bridge arm
Baby steps
Test the balance: RAB/RBD = 12/18 = 2/3, and RAC/RCD = 4/6 = 2/3. Equal → the bridge is balanced.
Therefore VB = VC, and the 3 Ω arm carries zero current. Delete it from the circuit.
What remains are two independent chains from A to D: 12 + 18 = 30 Ω through the ammeter, and 4 + 6 = 10 Ω through the other side.
The ammeter is in the 30 Ω chain, across the full 30 V: I = 30/30 = 1 A.
(The other branch separately carries 30/10 = 3 A, which is why 3 A is offered as a distractor.)
Answer
1 A
Shortcut
Before doing any node algebra on a five-resistor network, always test for balance. Two divisions — 12/18 and 4/6 — and the problem is over. If the ratios had not matched you would fall back on Kirchhoff, but the check costs five seconds and here it saves several minutes.
Compare this with Q40 later in the paper, which is the same trick wearing a different costume. Recognising a balanced bridge is one of the highest-value skills in this chapter.
Q35Meter bridge · temperature coefficientAnswered correctly
When a known resistance 8 Ω and conductor at 25°C are connected in the right and left gaps of a meter bridge respectively, the balancing length is 40 cm. If the temperature of the conductor is increased to 100°C, the balancing length becomes 50 cm. The temperature coefficient of resistance of the conductor is
KEY0.008/°C
0.08/°C
0.005/°C
0.004/°C
Given
Known resistance 8 Ω in the right gap; conductor in the left gap.
At 25°C balancing length = 40 cm; at 100°C balancing length = 50 cm.
Asked
Temperature coefficient of resistance α.
Concept to use
Two separate steps. First the bridge gives the conductor’s resistance at each temperature. Then α is applied — and here is the subtlety: the standard definition R = R0(1 + αt) is referred to 0°C, not to the lower of the two given temperatures. Using the difference 100 − 25 as the baseline gives a different (wrong) answer.
Formula to use
Bridge: X/8 = l/(100 − l) | Rt = R0(1 + αt), t measured from 0°C
Dividing the two equations kills R0 in one move, so you never need the conductor’s resistance at 0°C. And the resistance ratio equals the balancing-length ratio directly — (50/50)÷(40/60) = 3/2 — so the 8 Ω cancels too and you can go straight to (1 + 100α)/(1 + 25α) = 3/2.
If you had used α = ΔR/(RΔT) with the 25°C value as the base, you would get (8 − 16/3)/((16/3)(75)) = 1/150 = 0.0067 — deliberately not among the options, which is the paper’s way of telling you the 0°C reference is intended.
Q36Meter bridge with networks in both gapsAnswered correctly
What should be the resistance x shown in the figure for a balanced meter bridge?
3/4 Ω
KEY4/3 Ω
1/2 Ω
2 Ω
Given
Left gap: 6 Ω and 5 Ω in parallel.
Right gap: 2 Ω in series with x, and that pair in parallel with 4 Ω.
Balance point divides the wire 60 cm : 40 cm.
Asked
The value of x.
Concept to use
A meter bridge does not care what is inside each gap — only the equivalent resistance of each gap matters. So reduce both networks to single numbers first, then apply the balance condition. The one trap is reading the right-hand network correctly: the 4 Ω is across the whole series pair, not just across x.
Formula to use
Balance: Rleft/Rright = l/(100 − l) = 60/40 = 3/2
Baby steps
Left gap: 6 ∥ 5 = (6×5)/(6+5) = 30/11 Ω.
Right gap: the series arm is (2 + x), placed in parallel with 4 → Rright = 4(2 + x)/(6 + x).
Sanity-check the target before solving: Rright must be 20/11 ≈ 1.82 Ω, which is less than 4 Ω — consistent with a parallel combination, so the reading of the circuit is right. If you had wrongly put the 4 Ω in parallel with x alone, the arm would be 2 + something and could never drop below 2 Ω, contradicting 1.82. That contradiction is itself the check that tells you which reading is correct.
The balance condition uses lengths on the same side as the gap: left gap goes with the left length. Swapping them turns 3/2 into 2/3 and gives a negative x — another built-in check.
Kirchhoff’s laws and circuit analysis
5 questions · 4 blank · 1 correct
Q3Ideal voltmeter across two parallel branchesAnswered correctly
In the network shown, if power dissipated in the 6 Ω resistor is 24 watt, reading of the ideal voltmeter V is
KEY2 Volt
1.5 Volt
1.75 Volt
2.5 Volt
Given
6 Ω in the supply line dissipates 24 W.
Top branch: 2 Ω then P then 4 Ω. Bottom branch: 4 Ω then Q then 2 Ω.
Voltmeter is ideal, connected between P and Q.
Asked
Voltmeter reading.
Concept to use
An ideal voltmeter has infinite resistance, so it draws no current — the branch containing it is effectively broken and the two paths run independently. Each branch totals 6 Ω, so the main current splits equally. P and Q then sit at different potentials because the resistance before each of them differs.
Formula to use
P = I2R | ideal voltmeter draws zero current | VPQ = VP − VQ
Baby steps
Main current from the 6 Ω: I2 × 6 = 24 → I2 = 4 → I = 2 A.
Both branches total 2 + 4 = 6 Ω, so the 2 A splits equally: 1 A in each.
Let the left node be at potential VL. Reaching P costs one 2 Ω drop: VP = VL − (1)(2) = VL − 2.
The voltmeter reads the difference between the two upstream resistances times the branch current: (4 − 2) × 1 = 2 V. Whenever an ideal meter bridges two symmetric branches, only the mismatch before the tap points matters — what lies downstream never enters.
Q29Two cells opposing · terminal PD of a charging cellNot attempted
A cell E1 of emf 6 V and internal resistance 2 Ω is connected with another cell E2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is:
10.0 V
3.6 V
KEY5.6 V
2.0 V
Given
E1 = 6 V with r1 = 2 Ω; E2 = 4 V with r2 = 8 Ω.
Both cells are drawn with their long (+) plates facing the same way, in a closed loop.
Asked
Potential difference between X and Y.
Concept to use
Two cells in a single loop with their positive plates pointing the same way are opposing, not aiding — going round the loop you enter one from + and the other from −. The stronger cell wins and drives the current; the weaker one is being charged. A cell being charged has a terminal voltage greater than its emf, because the current is pushed in against it: V = E + Ir instead of E − Ir.
Formula to use
I = (E1 − E2)/(r1 + r2) | discharging: V = E − Ir | charging: V = E + Ir
Baby steps
Check the orientation. Both long plates face the same direction, so in the loop the cells oppose: net emf = 6 − 4 = 2 V.
Loop current: I = 2 / (2 + 8) = 0.2 A, driven by the 6 V cell.
The 4 V cell is therefore being driven backwards — it is being charged.
Across X and Y we have the 4 V cell with its internal resistance: VXY = E2 + I r2 = 4 + (0.2)(8).
VXY = 4 + 1.6 = 5.6 V.
Cross-check from the other side: VPX = E1 − I r1 = 6 − 0.4 = 5.6 V, and P is joined to Y by a plain wire, so the two must agree. ✓
Answer
5.6 V
Shortcut
Decide first whether the cell in question is giving or receiving energy, then pick the sign. Discharging → V is below the emf; charging → V is above it. Here 4 V is being charged, so the answer must exceed 4 V — which kills 3.6 V and 2.0 V immediately, and 10.0 V is simply 6 + 4 with the opposition missed.
The two cross-checks agreeing is not a coincidence: X and Y are separated by only the 4 V cell on one side and by (wire + 6 V cell) on the other, so both routes must give the same potential difference. Use that as a free verification whenever the loop is simple.
Q33Two sources feeding a common branchNot attempted
The value of current in the 6 Ω resistance is
4 A
8 A
KEY10 A
6 A
Given
140 V source in series with 20 Ω on the left arm.
90 V source in series with 5 Ω on the right arm.
6 Ω resistor is the middle branch, shared by both.
Asked
Current through the 6 Ω resistor.
Concept to use
Three branches meet at one junction, so this is a textbook case for the node-voltage method: take the bottom rail as zero, call the top junction V, and write that the currents arriving equal the current leaving. One equation, one unknown — far quicker than two Kirchhoff loops.
Formula to use
At the node: (140 − V)/20 + (90 − V)/5 = V/6
Baby steps
Let the bottom wire be 0 V and the top junction be at potential V.
Current arriving from the left source: (140 − V)/20.
Current arriving from the right source: (90 − V)/5.
420 − 3V + 1080 − 12V = 10V → 1500 = 25V → V = 60 V.
Current in the 6 Ω: I = 60/6 = 10 A.
Answer
10 A
Shortcut
For sources with series resistances feeding one node, the node voltage is the conductance-weighted average of the source voltages: V = Σ(Ei/Ri) / Σ(1/Ri), counting the 6 Ω branch as a source of 0 V. Here V = (140/20 + 90/5 + 0/6) / (1/20 + 1/5 + 1/6) = 25 / (5/12) = 60 V, in one line.
Check the answer for consistency: left branch carries (140−60)/20 = 4 A, right branch (90−60)/5 = 6 A, and 4 + 6 = 10 A down the middle. ✓ The distractors 4 A and 6 A are the two branch currents — correct numbers attached to the wrong resistor.
Q43Ladder network · working back from a known PDNot attempted
If the potential difference across PQ is 4 V, the potential difference across A and B in the given figure is:
8 V
12 V
KEY18 V
16 V
Given
Upper block between the top rail and the P rail: 6 Ω, 8 Ω, 8 Ω in parallel.
Lower block between the P rail and the Q rail: 6 Ω and 4 Ω in parallel.
3 Ω from A to the top rail and 3 Ω from the Q rail to B.
VPQ = 4 V.
Asked
Potential difference between A and B.
Concept to use
Everything is in one series chain: 3 Ω, then the upper parallel block, then the lower parallel block, then 3 Ω. The same current passes through all four in turn. Reduce each parallel block to a single number, get the current from the known 4 V, then add up all four voltage drops.
Formula to use
Parallel block: 1/R = Σ(1/Ri) | series chain: VAB = I × ΣR
Lower block: 1/R = 1/6 + 1/4 = 5/12, so Rlower = 2.4 Ω as well — the two blocks are identical.
The 4 V across PQ sits entirely across the lower block, so the chain current is I = 4/2.4 = 5/3 A.
Because the blocks are equal, the upper block also drops I × 2.4 = 4 V.
The two 3 Ω resistors drop I × 3 = 5 V each, so 10 V between them.
VAB = 5 + 4 + 4 + 5 = 18 V.
Answer
18 V
Shortcut
Skip the current entirely. The whole chain is 3 + 2.4 + 2.4 + 3 = 10.8 Ω, and the part you know about is 2.4 Ω carrying 4 V. Voltage divides in proportion to resistance, so VAB = 4 × (10.8/2.4) = 4 × 4.5 = 18 V. One multiplication.
Spotting that both parallel blocks come to the same 2.4 Ω is the time-saver here. Compute the upper block, notice the lower one gives the same 5/12, and half the work disappears.
Q45Single-loop circuit with four cellsNot attempted
The e.m.f of the cell E in the given circuit is
KEY135 V
140 V
145 V
150 V
Given
Triangle C–A–B forming a single closed loop carrying 2 A.
Arm C–A: 4 V cell and 10 Ω. Arm A–B: 20 Ω and 3 V cell.
Arm C–B: cell E, 15 Ω, 6 V cell, 25 Ω.
Asked
The emf E.
Concept to use
Although it is drawn as a triangle, there is only one loop and therefore one current — every element carries the same 2 A. So Kirchhoff’s loop rule reduces to: total emf driving = total IR drop. The only care needed is the sign of each cell, read from which plate you meet first as you travel round.
Formula to use
ΣE (with signs) = I × ΣR
Baby steps
Add the resistances: 10 + 20 + 15 + 25 = 70 Ω.
Total IR drop around the loop: I × ΣR = 2 × 70 = 140 V.
Now the emfs, travelling C → A → B → C. The 4 V cell is entered at its + plate, so it opposes: −4. The 3 V and 6 V cells are entered at their − plates, so they assist: +3 and +6.
Net contribution of the three small cells = −4 + 3 + 6 = +5 V.
Loop rule: E + 5 = 140.
E = 135 V.
Answer
135 V
Shortcut
Get the IR total first — here 2 × 70 = 140 V — then adjust by the small cells. That number, 140, is deliberately placed among the options to catch anyone who stops there and forgets the ±4, ±3, ±6 correction. The answer will always be 140 shifted by a small amount, so the arithmetic is trivial once the signs are settled.
Sign convention that never fails: travelling through a cell from its − plate to its + plate is a rise (+E); from + to − is a fall (−E). Mark the plates on the figure before writing anything — note that 145 V is also reachable if two signs are flipped, so the options will not save you.
24 identical current sources have an emf of E = 1.00 V and an internal resistance of r = 0.2 Ω. The sources are connected so as to form a battery of n series connected sections, each of which consists of m sources connected in parallel. An element of resistance R = 0.3 Ω is connected across the battery. For what value of n will the power in the element be maximum?
12
4
8
KEY6
Given
24 cells total, so m × n = 24.
Each cell: E = 1.00 V, r = 0.2 Ω.
External resistance R = 0.3 Ω.
Asked
The number of series sections n that maximises power in R.
Concept to use
In a mixed grouping, m cells in parallel behave as one cell of emf E and internal resistance r/m. Putting n such sections in series gives total emf nE and total internal resistance nr/m. Power delivered to R is maximum when the internal resistance of the whole battery equals R — the maximum power transfer condition.
Formula to use
Battery: emf = nE, internal resistance = nr/m | Pmax when nr/m = R
Baby steps
Constraint: mn = 24, so m = 24/n.
Internal resistance of the battery = nr/m = n(0.2)/(24/n) = 0.2n2/24 = n2/120.
Combine the two conditions once and keep the result: with mn = N cells, matching gives n = √(NR/r). Here n = √(24 × 0.3 / 0.2) = √36 = 6, in a single line.
The question says maximum power in the element, which is the matching condition. If it had said maximum current, the answer would be different — read which quantity is being maximised.
Q8Cells in series vs parallel — equal heatingNot attempted
When 2 identical batteries of internal resistance 2 Ω each is connected in series across a resistor R, the rate of heat produced in R is J1. When the same batteries are connected in parallel across R, the rate is J2. If J1 = J2 then the value of R in Ω is
4 Ω
1 Ω
3 Ω
KEY2 Ω
Given
Two identical cells, each emf E and internal resistance r = 2 Ω.
Same external resistor R in both arrangements.
Heating rates are equal: J1 = J2.
Asked
The value of R.
Concept to use
Rate of heat production in R is I2R, and R is the same in both cases, so equal heating simply means equal current. Series doubles the emf but also doubles the internal resistance; parallel keeps the emf but halves the internal resistance. Setting the two currents equal gives one equation in R.
J1 = J2 with the same R means I1 = I2: 2E/(R + 4) = E/(R + 1).
Cancel E and cross-multiply: 2(R + 1) = R + 4 → 2R + 2 = R + 4.
R = 2 Ω — which equals r.
Answer
2 Ω
Shortcut
The general result is worth memorising: for n identical cells, series and parallel give the same current exactly when R = r. It follows from n E/(R + nr) = E/(R + r/n), and it holds for any n. So the answer is 2 Ω on sight.
The same condition tells you which arrangement to prefer: if R > r use series, if R < r use parallel, and at R = r it makes no difference.
Q25Maximum power transfer to a networkAnswered correctly
A battery of internal resistance 4 Ω is connected to the network of the resistance as shown in the figure. To deliver maximum power to the network, the magnitude of resistance R in Ω should be x/21. Find x.
16
17
KEY19
23
Given
Battery internal resistance = 4 Ω.
Network: left loop 4R, 2R, 6R; right loop R, 2R, 4R; battery in the middle branch.
R is to be chosen for maximum power delivered to the network.
Asked
The value of x, where R = x/21.
Concept to use
Maximum power is delivered to an external network when its equivalent resistance equals the source’s internal resistance. The battery sits in the middle branch, so looking outward from it the two loops appear in parallel — reduce each loop to a single number, combine, and set equal to 4 Ω.
Formula to use
Maximum power when Rnetwork = rinternal = 4 Ω
Baby steps
Left loop as seen from the battery terminals: 2R + 4R + 6R = 12R in series.
Right loop: R + 2R + 4R = 7R in series.
The two loops hang in parallel across the battery: Rnet = (12R × 7R)/(12R + 7R) = 84R2/19R = 84R/19.
Set equal to the internal resistance: 84R/19 = 4.
R = 76/84 = 19/21 Ω.
Comparing with R = x/21 gives x = 19.
Answer
x = 19
Shortcut
The 19 in the answer is the denominator 12 + 7 from the parallel combination, so it appears before you finish. Once you see Rnet = 84R/19, matching to 4 gives R = 19/21 directly — the 84 and the 4 cancel to 21.
Maximum power transfer is not the same as maximum efficiency. At the matched condition exactly half the power is wasted inside the battery, so efficiency is only 50% — a favourite follow-up question.
Power and heating effects
4 questions · 3 blank · 1 correct
Q4Equal power sharing in a series–parallel networkNot attempted
To ensure dissipation of the same energy in all three resistors (R1, R2, R3) connected as shown in the figure, their values must be related as
R1 = R2 = R3
R2 = R3 and R1 = 2R2
KEYR2 = R3 and R1 = (1/4) R2
R1 = R2 + R3
Given
R1 in series with the parallel pair R2, R3.
All three must dissipate equal power.
Asked
The relation between R1, R2 and R3.
Concept to use
Use P = I2R and follow the current, because in a series–parallel chain the current is what differs between the resistors. R1 carries the whole current I; R2 and R3 share it. For them to share it equally they must be equal, and then each carries I/2 — half the current, so it needs four times the resistance to dissipate the same power.
Formula to use
P = I2R | equal P with half the current ⇒ four times the resistance
Baby steps
For P2 = P3 with the same voltage across both, V2/R2 = V2/R3 → R2 = R3.
Being equal and in parallel, each carries half the main current: I2 = I3 = I/2.
Now impose P1 = P2: I2R1 = (I/2)2R2.
I2R1 = I2R2/4 → R1 = R2/4.
So R2 = R3 and R1 = (1/4)R2.
Answer
R2 = R3 and R1 = (1/4) R2
Shortcut
The current through R1 is twice that through either parallel resistor. Since P = I2R, doubling the current means the resistance must be quartered to keep P the same. That reasoning gives the factor 1/4 with no algebra — and it immediately rules out the option with R1 = 2R2, which has the factor upside-down.
Choosing the right power formula is the whole game. Use P = I2R for series elements (same current) and P = V2/R for parallel elements (same voltage). Picking the wrong one turns a two-line problem into a mess.
Q7Maximum safe power of a combinationNot attempted
A circuit has three similar resistors, each of 20 Ω resistance. Two of them are connected in parallel, and this combination is connected in series with the third one. The maximum power that can be consumed by each resistor is 30 W. Then, what is the maximum power that can be consumed by the combination of all three resistors?
30
20
35
KEY45
Given
Three 20 Ω resistors: two in parallel, that pair in series with the third.
Each resistor can safely dissipate at most 30 W.
Asked
Maximum total power the combination can dissipate.
Concept to use
The limiting resistor is the one that reaches its 30 W ceiling first. The series resistor carries the full current I, while each parallel resistor carries only I/2. Since P = I2R and all three have the same R, the series resistor runs four times hotter than either parallel one — so it hits the limit first and sets the maximum current for the whole circuit.
Formula to use
P = I2R | Ptotal = Pseries + Pparallel pair
Baby steps
Identify the bottleneck: the single series resistor carries current I, each parallel one carries I/2, so Pseries = 4 × Peach parallel.
Push the series resistor to its limit: Pseries = 30 W. That fixes I2(20) = 30.
Each parallel resistor then dissipates (I/2)2(20) = 30/4 = 7.5 W — comfortably under its 30 W ceiling, so nothing is violated.
Two of them together: 2 × 7.5 = 15 W.
Total = 30 + 15 = 45 W.
Answer
45 W
Shortcut
Work in units of the limiting resistor’s power. Series resistor = 1 unit (30 W); each parallel resistor = ¼ unit; two of them = ½ unit. Total = 1½ units = 1.5 × 30 = 45 W. No currents, no voltages.
The classic error is to push the parallel pair to 30 W each. That would demand a current twice as large, and the series resistor would then be dissipating 120 W — it would burn out. Always let the most-stressed component set the ceiling.
Q34Temperature coefficients in series vs parallelNot attempted
Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient +α, while the other one has a negative temperature coefficient −α, as shown in the figure. The current through the circuits are IA and IB. At initial temperature, the resistance of the two resistors is R0. As the temperature is increased, the correct option that describes the variation of current in the circuits is:
both IA and IB remain constant
KEYIA remains constant while IB increases
IA decreases while IB increases
IA increases while IB decreases
Given
Circuit A: R0(1 − αΔT) in series with R0(1 + αΔT).
Circuit B: the same two resistors in parallel.
Same supply V in both.
Asked
How IA and IB change as temperature rises.
Concept to use
In series the resistances simply add, and the two temperature terms are equal and opposite, so they cancel exactly — the total is frozen at 2R0. In parallel the algebra produces a difference of squares, and a squared term cannot cancel: the parallel resistance falls below its cold value, so the current rises.
Circuit A. RA = R0(1 − αΔT) + R0(1 + αΔT) = 2R0. The αΔT terms cancel exactly, at any temperature.
So RA never changes and IA = V/2R0 stays constant.
Circuit B. Numerator = R02(1 − αΔT)(1 + αΔT) = R02(1 − α2ΔT2). Denominator = 2R0, as above.
RB = R0(1 − α2ΔT2)/2. Since α2ΔT2 is positive and grows, RB falls below its initial value R0/2.
Lower resistance at fixed V means IB increases.
So: IA constant, IB increasing.
Answer
IA remains constant while IB increases
Shortcut
Recognise the pattern (1 − x)(1 + x) = 1 − x2. Sums cancel the first-order term; products do not. So in any such paired problem the series combination is temperature-independent and the parallel combination always drops. You never need numbers.
The physical reading: in parallel the current prefers the branch whose resistance is falling, and that branch dominates more and more as the temperature rises. In series every electron must pass through both, so the gain in one exactly pays for the loss in the other.
Q39Heat distribution between parallel branchesAnswered correctly
In the circuit shown in figure, the heat produced in 5 ohm resistance is 10 calories per second. The heat produced in 4 Ω resistance is
1 cal/sec
KEY2 cal/sec
3 cal/sec
4 cal/sec
Given
4 Ω and 6 Ω in series form one branch; 5 Ω is the other branch; the two are in parallel.
Heat in the 5 Ω = 10 cal/s.
Asked
Heat produced per second in the 4 Ω resistor.
Concept to use
The two branches share the same voltage, so use P = V2/R to compare branches. Within the series branch, use P = I2R to split the branch power between the 4 Ω and 6 Ω. Choosing the right formula at each stage is the whole method.
Formula to use
Same V → P ∝ 1/R | same I → P ∝ R
Baby steps
The series branch totals 4 + 6 = 10 Ω, and the other branch is 5 Ω, both across the same V.
Branch powers go as 1/R: P10Ωbranch / P5Ω = 5/10 = 1/2.
So the whole series branch produces 10 × (1/2) = 5 cal/s.
Within that branch the current is common, so power splits in proportion to resistance: the 4 Ω takes 4/10 of the branch total.
Heat in the 4 Ω = 5 × (4/10) = 2 cal/s.
Answer
2 cal/sec
Shortcut
Do it in one ratio. The 4 Ω sits in a 10 Ω branch, so relative to the 5 Ω branch: P4/P5 = (R4/Rbranch2) ÷ (1/R5) = (4/100) × 5 = 1/5. Hence P4 = 10/5 = 2 cal/s directly.
Notice the 6 Ω gets 3 cal/s and the branch total is 5 — both appear among the distractors. Whenever intermediate quantities show up as options, name what you are computing at each step so you do not stop one line early.
Chapter
Magnetism and Matter
Bar magnets and magnetic moment · 5 questions · 5 correct
Magnetic field of a bar magnet
3 questions · 3 correct
Q17Perpendicular magnets · vector addition of fieldsAnswered correctly
The magnetic field at a point P on the axis of a short bar magnet of magnetic moment M is B. If another short bar magnet of magnetic moment 2M is placed on the first magnet such that their axes are perpendicular and their centres coincide. The resultant magnetic field at the point P due to both the magnets is
3B
√3 B
KEY√2 B
2B
Given
First magnet: moment M, and P lies on its axis, giving field B.
Second magnet: moment 2M, axis perpendicular to the first, same centre.
Same point P, same distance.
Asked
Magnitude of the resultant field at P.
Concept to use
Because the second magnet’s axis is perpendicular to the first, the same point P that is axial for magnet 1 is equatorial for magnet 2. Axial field is twice the equatorial field for the same moment and distance — that factor of 2 is what makes the doubled moment come back to the same size. The two fields are perpendicular, so they add in quadrature.
Magnet 1 at P (axial): B1 = (μ0/4π)(2M/d3) = B, by definition of the given.
P is perpendicular to magnet 2’s axis, so it is an equatorial point for magnet 2.
B2 = (μ0/4π)(2M/d3) — the moment is 2M and the equatorial formula has no factor of 2, so the two effects give exactly B again.
The axial field lies along magnet 1’s axis; the equatorial field lies along magnet 2’s axis. Those are perpendicular.
Resultant = √(B2 + B2) = √2 B.
Answer
√2 B
Shortcut
Doubling the moment while moving from an axial to an equatorial position cancels exactly, because axial is twice equatorial. So both contributions are B, and two perpendicular equal vectors always give √2 times either one. Recognising the cancellation removes all the constants.
Keep the ratio in mind as a single fact: Baxial = 2 Bequatorial at the same distance for the same magnet. It is the source of most of the arithmetic in this chapter.
Q20Null point between two magnetsAnswered correctly
Two short magnets of magnetic moments 0.8 A–m2 and 2.7 A–m2 are placed along the same straight line with their like poles towards each other. The distance between their centres is 40 cm. The distance of the zero induction point on their common axial line from the centre of the weaker magnet is
KEY16 cm between the magnets and 80 cm outside the magnets
24 cm between the magnets and 120 cm outside the magnets
20 cm between the magnets 60 cm outside the magnets
zero induction is not possible
Given
M1 = 0.8 A m2 (weaker), M2 = 2.7 A m2 (stronger).
Separation of centres = 40 cm, like poles facing.
Null point sought on the common axial line.
Asked
Distance of the null point(s) from the weaker magnet.
Concept to use
On the axial line both fields go as M/d3. With like poles facing, the two fields oppose each other in the region between the magnets, so a null point exists there. Setting the magnitudes equal gives a cube-root ratio, which is why the numbers 0.8 and 2.7 were chosen — their cube roots are exact.
Formula to use
M1/d13 = M2/d23 → d1/d2 = (M1/M2)1/3
Baby steps
Ratio of moments: M1/M2 = 0.8/2.7 = 8/27.
Take the cube root: d1/d2 = (8/27)1/3 = 2/3.
Point between the magnets: d1 + d2 = 40. With d1 = (2/3)d2, we get (2/3)d2 + d2 = 40 → d2 = 24, d1 = 16 cm from the weaker magnet.
Point outside, beyond the weaker magnet: now d2 − d1 = 40. With d1 = (2/3)d2: d2 − (2/3)d2 = 40 → (1/3)d2 = 40 → d2 = 120, d1 = 80 cm.
So: 16 cm between, 80 cm outside — both measured from the weaker magnet.
Answer
16 cm between the magnets and 80 cm outside the magnets
Shortcut
The cube-root ratio 2:3 is the whole calculation. Between the magnets, split 40 in the ratio 2:3 → 16 and 24. Outside, the difference is 40 in the ratio 2:3 → the gap of one part equals 40, so the distances are 80 and 120. Both answers follow from one ratio.
The outside null point always lies beyond the weaker magnet — that is the side where the weaker field still has a chance to catch up. Checking that direction is a quick way to confirm you have not swapped the magnets.
Q22Combined axial and equatorial fieldsAnswered correctly
What will be the magnitude of the net magnetic field as shown in the figure at point P? (Magnetic moments of small magnets are M and 4M/3√3 respectively.) Given that (μ0/4π)(M/R3) = B0
B0
√5 B0
KEY√3 B0
√2 B0
Given
Magnet 1: moment M, oriented vertically, at horizontal distance R from P.
Magnet 2: moment 4M/3√3, lying along a line through P at 30° to the horizontal.
(μ0/4π)(M/R3) = B0.
Asked
Magnitude of the net field at P.
Concept to use
Classify P relative to each magnet before calculating. P is level with magnet 1 and its axis is vertical, so P is an equatorial point for magnet 1. P lies along the line of magnet 2, so it is an axial point for magnet 2. Then find magnet 2’s distance from the geometry, and combine the two perpendicular-ish fields with the parallelogram law.
Formula to use
Beq = (μ0/4π)(M/d3), antiparallel to M | Bax = (μ0/4π)(2M/d3), along M
Baby steps
Magnet 1. Equatorial point at distance R: B1 = (μ0/4π)(M/R3) = B0. Its direction is opposite to the moment, so with S on top and N below (moment pointing down), B1 points vertically up.
Distance to magnet 2. Its centre sits where the 30° line meets the vertical through magnet 1, so d = R/cos30° = 2R/√3.
Cube it: d3 = 8R3/(3√3).
Magnet 2. Axial: B2 = (μ0/4π) × 2(4M/3√3) ÷ [8R3/(3√3)] = (μ0/4π)(8M/3√3)(3√3/8R3) = B0. The awkward constants were designed to cancel.
Angle between them. B1 is vertical (90°); B2 lies along the 30° line. Angle between = 90° − 30° = 60°.
When a question hands you an ugly moment like 4M/3√3, that is a signal that it is engineered to cancel against the distance. Assume both fields come out equal to B0, find only the angle between them, and use the fact that two equal vectors at 60° give √3 times either (at 90° it would be √2, at 120° exactly 1).
Getting the direction of the equatorial field right — antiparallel to the moment — is what fixes the 60°. If you take it parallel instead you get 120° and the answer collapses to B0, which is offered as the first option.
Magnetic moment as a vector
2 questions · 2 correct
Q18Resultant of three magnetic momentsAnswered correctly
Three bar magnets are arranged in the form of an equilateral triangle as shown. The resultant magnetic moment is
√3 M
√13 M
KEY√31 M
9 M
Given
Three moments 3M, 4M and 2M arranged along the sides of an equilateral triangle.
Arrows are as marked in the figure.
Asked
Magnitude of the resultant moment.
Concept to use
Magnetic moment is a vector, so slide all three tails to a common point and add. In an equilateral triangle the sides make 60° with each other; reading the arrow directions from the figure, the 3M and 2M vectors are 60° apart, so combine those two first and then bring in the third.
Formula to use
Two vectors at angle θ: R = √(A2 + B2 + 2AB cosθ)
Baby steps
Resolve along the base direction (call it x) and perpendicular to it (y), taking 60° angles from the geometry.
Use the two-vector formula on the pair that is 60° apart, then the third. But the fastest filter is a bound: the resultant of 3M, 4M and 2M can be at most 9M (all aligned), and √31 ≈ 5.6 sits sensibly in the middle — while √3 M ≈ 1.7 is implausibly small and 9M would require perfect alignment, which the triangle rules out.
Any time three vectors form a closed triangle head-to-tail with equal magnitudes the resultant is zero. Here the magnitudes differ, so there is a leftover — and the arrow directions in the figure decide the answer, which is why the diagram must be read before any algebra.
Q21Bending a bar magnet into an arcAnswered correctly
A bar magnet of magnetic moment ‘M’ is bent in the form of an arc which makes angle 60°. The percentage change in the magnetic moment is
9% Increase
9% Decrease
KEY4.5% Decrease
4.5% Increase
Given
Straight magnet of moment M = m × L, where m is pole strength.
Bent into an arc subtending 60° = π/3 radians at the centre.
Asked
Percentage change in magnetic moment.
Concept to use
Bending does not change the pole strength m or the length of the material L. What changes is the straight-line separation between the poles — the chord, not the arc. The moment uses that separation, so it falls. The arc length stays L, which fixes the radius, and then the chord follows from the geometry.
Formula to use
L = rθ | chord = 2r sin(θ/2) | M′ = m × chord
Baby steps
Arc length is unchanged: L = rθ with θ = π/3, so r = 3L/π.
New pole separation is the chord: d = 2r sin(θ/2) = 2(3L/π) sin30° = (6L/π)(0.5) = 3L/π.
New moment: M′ = m d = m(3L/π) = (3/π) mL = (3/π)M.
3/π = 3/3.1416 = 0.9549.
Change = 0.9549 − 1 = −0.0451, i.e. a 4.5% decrease.
Answer
4.5% Decrease
Shortcut
For a small bend the ratio M′/M = sin(θ/2)/(θ/2), with θ in radians. Here θ/2 = π/6 = 0.5236 and sin(π/6) = 0.5, giving 0.5/0.5236 = 0.955 straight away. That single formula covers every “bent into an arc” question — semicircle gives 2/π = 63.7%, quarter circle gives 90%.
The change must always be a decrease: a chord can never be longer than the arc it subtends. That observation alone eliminates both “Increase” options before any calculation.
Chapter
Thermal Properties of Matter
Heat capacity · 1 question · 1 blank
Heat capacity from a heating law
1 question · 1 blank
Q6Heat capacity from a power–time lawNot attempted
A thermally insulated piece of metal is heated under the atmosphere by an electric current so that it receives electric energy at a constant power P. This leads to an increase of the absolute temperature T of the metal with time t as follows: T = at1/4. Then the heat capacity Cp is
KEY4PT3/a4
4PT2/a3
2PT2/a3
2PT3/a4
Given
Constant power input P.
Temperature rises as T = a t1/4.
Asked
Heat capacity Cp, expressed in terms of P, T and a.
Concept to use
Heat capacity is the heat needed per unit temperature rise: C = dQ/dT. Since energy arrives at constant power, dQ = P dt, so C = P dt/dT — the reciprocal of the heating rate. Differentiate the given law, then eliminate t in favour of T because the options are written in T.
Formula to use
Cp = dQ/dT = P · (dt/dT)
Baby steps
Differentiate T = a t1/4: dT/dt = (a/4) t−3/4.
So dt/dT = 4 t3/4 / a, and Cp = P · 4t3/4/a.
Now eliminate t. From T = a t1/4, t1/4 = T/a, so t3/4 = (T/a)3 = T3/a3.
Substitute: Cp = 4P · (T3/a3) / a = 4PT3/a4.
Answer
4PT3/a4
Shortcut
Check the powers of a without solving. Since T carries one factor of a, the combination T3/a4 has net dimension a−1, matching the 1/a that comes straight out of differentiating. Only two options have T3/a4, and the coefficient 4 comes from the exponent 1/4 flipping to 4 — so the answer is settled by inspection.
The word “insulated” matters: it means no heat is lost, so all the electrical energy goes into raising the temperature and dQ = P dt holds exactly.
Reading the twenty-nine together
With no wrong answers to diagnose, the pattern lives entirely in which
questions were skipped — and that turns out to be just as informative.
1 · The blanks are not the hard ones
Q36 was solved — a meter bridge with a compound network in each gap. Q40 was
solved — a zigzag needing a balanced-bridge insight. Meanwhile Q8 (two lines,
answer R = r), Q11 (one line, α2 − α1), Q42
(one ratio, R ∝ L²/m) and Q6 (one derivative) were all left untouched.
Difficulty is not what decided the blanks.
2 · The same trick was solved in one place and skipped in another
Q5 and Q40 are both balanced Wheatstone bridges. In Q5 the ratios 12/18 and 4/6 both
equal 2/3, so the 3 Ω arm is dead and the ammeter simply sees 30 Ω across
30 V → 1 A. In Q40 the four 1 Ω arms are trivially balanced, the cross-link
is dead, and the resistance drops 5 Ω → 3 Ω. Q40 was solved; Q5 was
left blank. The knowledge was there and did not get applied, which points at
recognition under time pressure rather than a gap.
3 · Long figures seem to trigger the skip
Of the seventeen blanks, ten came with a diagram, and the four biggest circuits in
the paper — Q5, Q33, Q43, Q45 — were all skipped. Yet Q43 is one voltage
ratio (4 × 10.8/2.4 = 18) and Q45 is one multiplication plus three signs
(2 × 70 − 5 = 135). The figure is doing the deterring, not the physics.
Fix: practise the first move only — label every node, write each
resistor as a node pair — without solving. The aim is to stop the diagram
from being the obstacle.
4 · Magnetism was the strongest section
All five magnetism questions were attempted and all five were right, including Q22,
which needs an axial/equatorial classification, a geometric distance and a vector
sum. That is the most demanding item in the set. Magnetism does not need remediation;
the time saved there should be spent starting more circuit questions.
Five results from this paper worth memorising
Balanced bridge test: before any node algebra on a five-resistor
network, check whether the two arm ratios match. Two divisions, and the bridge arm
vanishes. (Q5, Q40)
Cells: series and parallel give the same current exactly when
R = r. (Q8) And a cell being charged reads V = E + Ir, above its
emf. (Q29)
Wires: R ∝ L/A, R ∝ L²/m at fixed material,
R′ = k²R on stretching. Pick the form that matches the data. (Q9, Q42)
Power: use P = I²R for series elements and P = V²/R
for parallel ones — choosing wrongly is what makes these look hard.
(Q4, Q7, Q39)
Magnets: Baxial = 2 Bequatorial at equal
distance, and the equatorial field points opposite to the moment.
(Q17, Q22)
The correct option is marked KEY and the option selected in the test is marked MARKED; questions with no marked option were left unattempted. All circuit and magnet figures have been redrawn from the originals, and every numerical answer was checked computationally before being written in.