Every physics question lost on this paper, rebuilt in full. Each one carries what was given, what was asked, the concept behind it, the formula, the steps written out, a table justifying the right option and ruling out each of the others, the fastest route through, and an animated figure wherever seeing the thing settles the answer.
8Questions lost
4Attempted, wrong
4Left blank
36Marks at stake
The shape of this paper
Eight questions from a single chapter, split evenly: four attempted and missed, four left
blank. On NEET marking that is 32 marks not gained plus 4 in negatives.
The two halves are worth separating. The four blanks — Q107, Q118, Q131 and
Q134 — are all short numerical questions, and every one of them has a route under a
minute. Q118 in particular needs no algebra at all: adding each option and looking for 16 takes
about ten seconds.
The four wrong answers all turn on a sign, a factor or a direction, not on a missing
concept. In three of the four the chosen option is internally inconsistent with a formula already
known.
All eight come from one chapter, so they are grouped here by sub-topic. An even split — four attempted and missed, four left blank — and the two halves fail for quite different reasons.
Q 91Attempted · wrongField from a potential function
The potential V is varying with x as V = ½(y² − 4x) volt. The field at x = 1 m, y = 1 m, is
2î + ĵ Vm⁻¹
Her answer−2î + ĵ Vm⁻¹
Correct2î − ĵ Vm⁻¹
−2î + 2ĵ Vm⁻¹
Given
V = ½(y² − 4x) volt.
The field is wanted at the point x = 1 m, y = 1 m.
Asked
The electric field vector at that point.
Concept to use
The field is the negative gradient of the potential: each component of E is minus the rate at which V changes along that axis. The minus sign is the entire difficulty of this question — it says the field points downhill, from high potential towards low, and it flips the sign of every component you calculate.
DiagramAnimatedV falls as x grows, so the field points towards +x.
Formula to useEₓ = −∂V/∂x , Eₖ = −∂V/∂y
Baby steps
Expand: V = ½y² − 2x.
Differentiate with respect to x, holding y fixed: ∂V/∂x = −2.
So Eₓ = −(−2) = +2. Note the double negative — this is where the sign is usually lost.
Differentiate with respect to y: ∂V/∂y = y.
So Eₖ = −y, and at y = 1 that is −1.
E = 2î − ĵ V m⁻¹.
Answer
E = 2î − ĵ V m⁻¹
Why this option and not the others
Option
Verdict
Reason
2î + ĵ
rule out
The x component is right but the y sign is not. ∂V/∂y = +y, so Eₖ must be negative.
−2î + ĵ
rule out
Both signs are inverted — this is what you get by writing E = +∇V and forgetting the minus altogether.
2î − ĵ
keep
Eₓ = −(−2) = +2 and Eₖ = −(1) = −1. Both negatives applied correctly.
−2î + 2ĵ
rule out
Wrong on both counts: the x sign is inverted and the y magnitude is doubled, as though the ½ had been dropped before differentiating.
Shortcut
Do a physical check instead of trusting the algebra. V contains −2x, so V falls as x increases — and the field always points from high V to low V, which here means towards +x. So Eₓ must be positive, and two options die at once. Then V contains +½y², so V rises with y, so Eₖ must be negative. Two sign checks, no calculus needed to choose.
Where it went wrong
The x sign came out inverted, which means the field was taken as +∇V rather than −∇V somewhere in the working — note that the y component was signed correctly, so the minus was applied once and then dropped. The reliable fix is the physical check above: ask which way V is falling, because the field always points that way. Sign errors in a gradient are almost impossible to spot by re-reading the algebra and very easy to spot by asking which way is downhill.
Q 100Attempted · wrongIsolated capacitor · pulling the plates apart
A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are
constant, decreases, increases
Her answerconstant, decreases, decreases
Correctconstant, increases, decreases
increases, decreases, decreases
Given
A parallel plate capacitor, charged and then isolated.
The plate separation d is increased.
Asked
What happens to the charge, the potential difference and the capacitance, in that order.
Concept to use
Isolated is the key word: the capacitor is disconnected from any source, so no charge can flow in or out and Q is fixed. That makes Q the independent quantity, and everything else follows. C = ε₀A/d falls as d grows. And since V = Q/C with Q held constant, a falling C forces V to rise.
DiagramAnimatedIsolated means Q is nailed down; everything else follows from that.
Formula to useQ constant · C = ε₀A/d · V = Q/C
Baby steps
Isolated → no path for charge to move → Q stays constant.
C = ε₀A/d. Increasing d increases the denominator, so C decreases.
V = Q/C. Q is fixed and C has fallen, so V increases.
In the order asked: Q constant, V increases, C decreases.
Sanity check by energy: pulling the plates apart against their attraction means you do work, so the stored energy U = Q²/2C must rise — and with Q fixed, that requires C to fall. Consistent.
Answer
Constant, increases, decreases
Why this option and not the others
Option
Verdict
Reason
constant, decreases, increases
rule out
Capacitance cannot increase when d increases — d is in the denominator.
constant, decreases, decreases
rule out
C is right, but V cannot fall while Q is fixed and C is falling. V = Q/C forces V up.
constant, increases, decreases
keep
All three consistent with Q fixed and C = ε₀A/d.
increases, decreases, decreases
rule out
Charge cannot increase on an isolated conductor — there is nowhere for it to come from.
Shortcut
The whole family of these questions turns on one word. ISOLATED means Q is fixed, so work with V = Q/C. CONNECTED TO A BATTERY means V is fixed, so work with Q = CV. Decide which quantity is nailed down before touching anything else, and the rest is one substitution. Getting this backwards is the single commonest error in capacitance.
Where it went wrong
Q and C were both identified correctly, so the isolated condition was recognised. The slip is in V: with Q pinned and C falling, V = Q/C must rise. It is worth noticing that the chosen option is internally inconsistent — if Q is constant and C decreases, V cannot also decrease, because that would violate V = Q/C. Testing an option against the defining equation before selecting it catches this in seconds.
Q 105Attempted · wrongTwo charged plates brought together
Two identical metal plates are given positive charges Q₁ and Q₂ (< Q₁) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is
(Q₁ − Q₂)/C
Correct(Q₁ − Q₂)/2C
2(Q₁ − Q₂)/C
Her answer(Q₁ + Q₂)/2C
Given
Two identical plates carrying charges Q₁ and Q₂, with Q₂ < Q₁.
Both charges are positive.
They are brought together to form a capacitor of capacitance C.
Asked
The potential difference between the plates.
Concept to use
When two charged plates face each other, the charge rearranges itself onto the four faces. The outer faces each take (Q₁+Q₂)/2, and the inner faces take +(Q₁−Q₂)/2 and −(Q₁−Q₂)/2. Only the inner faces produce field in the gap, so only they set the potential difference. The effective capacitor charge is therefore half the difference, not half the sum.
DiagramAnimatedCharge splits between inner and outer faces — only the inner ones count.
Formula to useQ_inner = (Q₁ − Q₂)/2 V = Q_inner / C
Baby steps
Total charge on the system is Q₁ + Q₂, and it must be conserved.
Inside the metal the field must be zero, which forces the outer faces to be equal: each carries (Q₁+Q₂)/2.
What is left for the inner faces is +(Q₁−Q₂)/2 on one and −(Q₁−Q₂)/2 on the other — equal and opposite, as a capacitor requires.
Only these inner charges create field in the gap.
V = Q_inner/C = (Q₁ − Q₂)/2C.
Answer
(Q₁ − Q₂)/2C
Why this option and not the others
Option
Verdict
Reason
(Q₁−Q₂)/C
rule out
The right combination of charges but the factor of 2 is missing — only half the difference ends up on the inner faces.
(Q₁−Q₂)/2C
keep
Half the difference sits on the inner faces, and that is the charge that makes the field in the gap.
2(Q₁−Q₂)/C
rule out
Off by a factor of four from the correct value.
(Q₁+Q₂)/2C
rule out
This is the charge on each outer face. Outer-face charge produces field outside the capacitor, not between the plates, so it contributes nothing to V.
Shortcut
Test the formula on a case you already know. If Q₂ = −Q₁ — an ordinary capacitor with equal and opposite charges — the answer must reduce to V = Q₁/C. Substituting into (Q₁−Q₂)/2C gives (Q₁+Q₁)/2C = Q₁/C. ✓ Substituting into the sum version gives zero, which is plainly wrong. One substitution eliminates three options.
Where it went wrong
The sum was used where the difference was needed — (Q₁+Q₂)/2 is the charge on each outer face, and outer faces produce no field between the plates. The physical picture to hold on to is that a capacitor works on equal and opposite charge, so the relevant quantity must vanish when the two plates carry the same charge. Q₁ = Q₂ should give V = 0, and only the difference formulae do that.
Q 106Attempted · wrongPotential energy and distance
Consider the following statements and choose the correct option. Statement I – If the distance between two charges is increased, the potential energy of the system always decreases. Statement II – Potential energy between two charges is inversely proportional to the distance between them.
Her answerBoth statements I and II are correct.
Statement I is correct, but statement II is incorrect.
CorrectStatement I is incorrect, but statement II is correct.
Both statements I and II are incorrect.
Given
Two point charges separated by a distance r.
U = kq₁q₂/r.
Asked
Which of the two statements are correct.
Concept to use
Statement II is just the formula and is correct: U goes as 1/r. Statement I adds the word always, and that is where it fails. The magnitude of U does fall as r grows — but the value of U depends on the sign of q₁q₂. For unlike charges U is negative, and moving them apart takes it from a large negative number up towards zero, which is an increase.
DiagramAnimatedFor unlike charges, U rises towards zero as the separation grows.
Formula to useU = kq₁q₂/r like charges: U > 0 and falls unlike: U < 0 and rises
Baby steps
Statement II first. U = kq₁q₂/r, so U is inversely proportional to r. Correct.
Statement I. Test it on like charges: U is positive and falls towards zero as r grows. The statement holds here.
Now test it on unlike charges: U is negative, say −10 J at close range, and −2 J further away. −2 is greater than −10, so U has increased.
The word always is therefore false. Statement I is incorrect.
Answer: I incorrect, II correct.
Answer
Statement I is incorrect, but statement II is correct
Why this option and not the others
Option
Verdict
Reason
Both correct
rule out
Statement I fails for unlike charges, where U rises towards zero as the separation grows.
I correct, II incorrect
rule out
The reverse of the truth. II is simply the formula.
I incorrect, II correct
keep
The word always breaks statement I; statement II is exactly U ∝ 1/r.
Both incorrect
rule out
Statement II is correct as written.
Shortcut
Hunt for absolute words. “Always”, “never”, “all” and “only” make a statement fail if a single counterexample exists — and in electrostatics the counterexample is nearly always the other sign. Whenever a statement about charges says “always”, test it on the opposite sign combination before accepting it. That one habit is worth several marks a paper.
Where it went wrong
Both statements were accepted, which means the magnitude of U was being tracked rather than its signed value. For unlike charges U is negative and moving apart makes it less negative — a genuine increase, even though the force weakens. This is the absolute-word trap in its standard form, and it is the same pattern flagged in earlier papers: a statement containing always needs one counterexample, not confirmation.
Q 107Left blankBuilding a capacitor bank
A number of capacitors, each of capacitance 1 μF and each one of which gets punctured if a potential difference just exceeding 500 volt is applied, are provided. Then an arrangement suitable for giving a capacitor of capacitance 2 μF across which 3000 volt may be applied requires at least
18 component capacitors
36 component capacitors
Correct72 component capacitors
144 component capacitors
Given
Component capacitors: 1 μF each, breakdown voltage 500 V.
Required: an equivalent capacitance of 2 μF.
Required: it must withstand 3000 V.
Asked
The minimum number of component capacitors.
Concept to use
Two requirements have to be met by two different arrangements, and they are handled in a fixed order. Series shares out the voltage, so a series chain solves the breakdown problem. Parallel adds capacitance, so parallel rows solve the capacitance problem. Build one row long enough to survive the voltage, then add as many identical rows as it takes to reach the capacitance.
DiagramAnimatedSix in series for the voltage, twelve rows in parallel for the capacitance.
Formula to usen = V_total / V_each ; row capacitance = C/n ; m = C_required / (C/n) ; total = n × m
Baby steps
Voltage first. Each capacitor takes 500 V, and 3000 V must be shared, so n = 3000/500 = 6 in series per row.
A row of 6 identical 1 μF capacitors in series has capacitance 1/6 μF.
Capacitance next. To reach 2 μF from rows of 1/6 μF: m = 2 ÷ (1/6) = 12 rows in parallel.
Total = 6 × 12 = 72 capacitors.
Check: 12 rows × 1/6 μF = 2 μF ✓, and each capacitor sees 3000/6 = 500 V ✓.
Answer
72 component capacitors
Why this option and not the others
Option
Verdict
Reason
18
rule out
6 × 3 — only 3 rows, giving 3 × 1/6 = 0.5 μF. Four times short of the required capacitance.
36
rule out
6 × 6 = 36 gives 6 × 1/6 = 1 μF. Half of what is needed.
72
keep
6 in series × 12 in parallel: 2 μF, and 500 V per capacitor.
144
rule out
Twice as many as necessary. It would give 4 μF — the requirement is met but the question asks for the minimum.
Shortcut
The order is fixed and never varies: voltage decides the length of a row; capacitance decides the number of rows. n = V_total/V_each, then m = C_required × n / C_each. Here that is 6 and then 2 × 6 / 1 = 12, giving 72. Two divisions, about thirty seconds — well worth attempting rather than leaving blank.
Q 118Left blankSeries and parallel · finding the pair
Two capacitors when connected in series have a capacitance of 3 μF, and when connected in parallel have a capacitance of 16 μF. Their individual capacities are
1 μF, 2 μF
6 μF, 2 μF
Correct12 μF, 4 μF
3 μF, 16 μF
Given
Series combination: Cₛ = 3 μF.
Parallel combination: Cₚ = 16 μF.
Asked
The two individual capacitances.
Concept to use
Parallel gives you the sum directly: C₁ + C₂ = 16. Series gives you the product over the sum, and since you already know the sum, that hands you the product: C₁C₂ = 3 × 16 = 48. Sum and product together define a quadratic, and its two roots are the answer.
DiagramAnimatedSum from the parallel value, product from the series value.
Sum 16 ✓, and series value 12×4/16 = 3 ✓. Both conditions met.
3, 16
rule out
These are the two combination values quoted back as the individual ones. Sum 19, and the series value would be 2.53 — neither matches.
Shortcut
Do not solve anything. Add each option and look for 16. Only one pair sums to the parallel value, and that is the answer — four additions, about ten seconds. This is the cheapest question on the paper, and it was left blank. Whenever the options are pairs of numbers and the stem gives you a sum, testing beats solving every time.
Q 131Left blankBreakdown voltage in series
A capacitor of capacitance C₁ = 1 μF can withstand a maximum voltage V₁ = 6 kV, while a capacitor of capacitance C₂ = 3.0 μF can withstand the maximum voltage V₂ = 4 kV. The maximum voltage which the capacitors can withstand when connected in series, is
Correct8 kV
9 kV
10 kV
5 kV
Given
C₁ = 1 μF, rated 6 kV.
C₂ = 3 μF, rated 4 kV.
The two are connected in series.
Asked
The maximum voltage the combination can withstand.
Concept to use
In series every capacitor carries the same charge. So work out how much charge each one can hold before it breaks down, take the smaller of the two — that is the one that fails first and therefore sets the limit — and then find the voltage each actually carries at that charge. Adding the two ratings straight away is wrong, because the weaker capacitor would break long before the stronger reached its rating.
DiagramAnimatedThe smaller CV product decides when the chain gives way.
Formula to useQ = CV ; Q_max = min(C₁V₁, C₂V₂) ; V_total = Q_max/C₁ + Q_max/C₂
Baby steps
Maximum charge on C₁: Q = 1 × 6 = 6 μC.
Maximum charge on C₂: Q = 3 × 4 = 12 μC.
In series the charge is common, so the limit is the smaller: Q = 6 μC. C₁ is the weak link.
At that charge: V₁ = 6/1 = 6 kV (at its limit) and V₂ = 6/3 = 2 kV (well below its 4 kV limit).
Total = 6 + 2 = 8 kV.
Answer
8 kV
Why this option and not the others
Option
Verdict
Reason
8 kV
keep
The limiting charge is 6 μC, giving 6 kV across C₁ and 2 kV across C₂.
9 kV
rule out
No combination of the numbers produces this.
10 kV
rule out
This is 6 + 4, the two ratings simply added. It assumes both capacitors reach their limit at the same time, which series operation makes impossible unless C₁V₁ = C₂V₂.
5 kV
rule out
Below even the rating of C₁ alone, so the combination is being underestimated.
Shortcut
One line does it: find the smaller CV product, then divide it by each capacitance and add. Q_min = min(1×6, 3×4) = 6, and 6/1 + 6/3 = 8 kV. And note the sanity check: the answer must lie above the larger single rating (6 kV) but below the sum of the two (10 kV), which already leaves only 8 and 9.
Q 134Left blankEnergy stored · a percentage change
If the charge on a capacitor is increased by 3 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C):
10 C
Correct15 C
20 C
30 C
Given
Charge increased by 3 C.
Energy stored increases by 44%, so U₂/U₁ = 1.44.
The capacitance C is unchanged.
Asked
The original charge Q.
Concept to use
Energy stored at constant capacitance goes as the square of the charge: U = Q²/2C. So a ratio of energies is the square of the ratio of charges — and the way in is to take the square root of the energy ratio, not to use it directly. The number 44% is chosen deliberately, because 1.44 is a perfect square.
DiagramAnimatedEnergy goes as Q², so ratios need a square root.
Formula to useU = Q²/2C → U₂/U₁ = (Q₂/Q₁)²
Baby steps
U₂/U₁ = 1.44, since the energy rose by 44%.
Take the square root: Q₂/Q₁ = √1.44 = 1.2.
The new charge is Q + 3, so (Q + 3)/Q = 1.2.
Q + 3 = 1.2Q, giving 0.2Q = 3.
Q = 15 C.
Answer
15 C
Why this option and not the others
Option
Verdict
Reason
10 C
rule out
13²/10² = 1.69 — a 69% rise, not 44%.
15 C
keep
18²/15² = 324/225 = 1.44 exactly — a 44% rise. ✓
20 C
rule out
23²/20² = 1.3225 — about a 32% rise.
30 C
rule out
33²/30² = 1.21 — a 21% rise. The larger the starting charge, the smaller the percentage effect of adding 3 C.
Shortcut
√1.44 = 1.2 is worth recognising on sight, and it turns the whole question into one line: a 20% rise in charge for a 44% rise in energy. Then Q + 3 = 1.2Q gives Q = 15 immediately. Alternatively, since the options are only four numbers, just test them: (Q+3)²/Q² and look for 1.44. Either route takes well under a minute, which makes this an expensive question to leave blank.
What the eight have in common
Reading the paper as a whole
The four blanks were all quick
Q
What it needed
Realistic time
118
Add each option; look for the pair summing to 16.
about 10 seconds
134
Recognise √1.44 = 1.2, then Q + 3 = 1.2Q.
under a minute
131
Smaller CV product, then divide by each capacitance and add.
under a minute
107
Two divisions: 3000/500 = 6, then 2×6/1 = 12.
about 30 seconds
Sixteen marks in roughly three minutes of work. None of these needed a concept that was
missing — they needed the decision to start.
Three of the four wrong answers contradict a formula already known
Q100. Q constant and C decreasing, with V also decreasing — but V = Q/C makes
that impossible. The chosen option is self-contradictory.
Q105. (Q₁+Q₂)/2C does not reduce to Q/C when the plates carry equal and
opposite charge. Testing a formula on a case you already know is the fastest check there is.
Q91. V contains −2x, so V falls as x grows, so the field must point towards +x.
The chosen answer has it pointing the other way.
In each case the check takes seconds and uses only what was already on the page. Test the
option against the defining equation before selecting it.
Two rules that cover most of this chapter
Decide what is nailed down first. ISOLATED means Q is fixed — work with
V = Q/C. CONNECTED TO A BATTERY means V is fixed — work with Q = CV. Getting this backwards
is the commonest error in capacitance, and it decided Q100.
Series shares voltage; parallel adds capacitance. Every capacitor-bank question
(Q107, Q118, Q131) is built from that one sentence. In series the charge is common, so the
smaller CV product always sets the limit.
And one habit for the statement questions: hunt for absolute words. Q106 hinges
entirely on “always”, and the counterexample in electrostatics is nearly always the
opposite sign.