JEE Main chapters · Class 12 · Physics

Experimental Skills: JEE Main Physics formulas and notes

Experimental Skills gives 2 to 3 questions in JEE Mains: a vernier or screw gauge reading with zero error, the least count, a percentage error in a formula, and one question on a standard experiment (metre bridge, potentiometer, galvanometer conversion, diode, pendulum graph). The formulas are short and the marks are easy to get once the rules are clear. About 5 hours spread over 4 days is enough.

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Quick revision

Vernier: LC = 1 MSD/N; reading = MSR + VSR × LC; true = observed − zero error (negative error is added)
Screw gauge: LC = pitch/divisions; reading = MSR + CSR × LC; closed reading above/below the line gives ∓e
Errors: sum and difference add ΔA + ΔB; product and quotient add relative errors; power Aⁿ gives n × relative error
Sig figs: sum by decimal places, product by significant figures; many oscillations reduce the error in T
Pendulum: g = 4π²L/T², L = l + r; small amplitude; T²–L line has slope 4π²/g
Searle: Y = MgL/(πr²l); capillary: T = rhρg/2; Stokes: η = 2r²(ρ − σ)g/9v
Resonance tube: v = 2f(L₂ − L₁); e = (L₂ − 3L₁)/2; mixtures: m_s c_s(T₁ − T) = (m_w c_w + w)(T − T₀)
Mirror 1/v + 1/u = 1/f; lens 1/v − 1/u = 1/f; displacement f = (D² − d²)/4D; slab n = sin i/sin r; prism n = sin((A + δm)/2)/sin(A/2)
Metre bridge R/S = l/(100 − l); ρ = πd²R/4L; potentiometer E₁/E₂ = l₁/l₂, r = R(l₁ − l₂)/l₂
Galvanometer: G = RS/(R − S), k = E/[(R + G)θ]; shunt S = I_gG/(I − I_g); series R = V/I_g − G; diode knee 0.7 V (Si)

Formulas with solved examples

Vernier callipers and least count

Least count of a vernier

LC = 1 MSD − 1 VSD = (1 MSD)/(N) = (value of one main scale division)/(number of vernier divisions)
for N VSD = (N − 1) MSD, LC = (1 MSD)/(N)
with 1 MSD = 1 mm and N = 10, LC = 0.1 mm = 0.01 cm
if N VSD = (N + 1) MSD (reverse vernier), LC = 1 VSD − 1 MSD, again (1 MSD)/(N)
the least count is the smallest length the instrument can read

How to apply. Write what 1 MSD is, then the number of vernier divisions N. The least count is 1 MSD divided by N. Do not use the length of the vernier scale.

Example. 1 MSD = 0.5 mm and N = 50 gives LC = 0.5 ÷ 50 = 0.01 mm. A vernier with 1 MSD = 1 mm and 20 divisions has LC = 0.05 mm.

Reading a vernier

reading = MSR + (VSR × LC)
MSR = the main scale reading just before the vernier zero
VSR = the number of the vernier line that coincides with a main scale line
reading in cm: MSR (cm) + VSR × 0.01 cm
the final answer is written with the same number of decimal places as the least count

How to apply. Read the main scale to the left of the vernier zero. Then find the one vernier line that is in line with a main scale line. Multiply that line number by LC and add.

Example. Zero of the vernier just after 23 mm and the 6th line coinciding: reading = 23 + 6 × 0.1 = 23.6 mm = 2.36 cm.

Zero error and zero correction

zero error e: the reading when the jaws are closed
zero correction = −e
true reading = observed reading − e
positive error (vernier zero to the right of the main zero, nth line coincides): e = + n × LC
negative error (vernier zero to the left, nth line coincides): e = −(N − n) × LC
the same rules hold for a screw gauge, with the circular scale in place of the vernier

How to apply. Close the jaws first and note e with its sign. Always subtract e from every observed reading: a positive e is subtracted, a negative e is added.

Example. Closed jaws: 4th line coincides with the vernier to the left of zero, N = 10: e = −(10 − 4) × 0.1 = −0.6 mm. An observed 23.6 mm gives 23.6 − (−0.6) = 24.2 mm. For a positive e = +0.2 mm the true value is 23.6 − 0.2 = 23.4 mm.

What the vernier measures and how

outside jaws: external diameter or length of a block
inside jaws: internal diameter of a tube or a hollow
the strip at the end: depth of a cavity or a beaker
volume of a cylinder V = πr²l = (πd²l)/(4)
hold the object gently between the jaws; do not press, and take readings at several places

How to apply. For a cylinder, measure d at two perpendicular directions and l at three places, then average. The percentage error in V adds 2 × (Δd/d) and Δl/l.

Example. d = 2.36 cm and l = 5.40 cm give V = π × 2.36² × 5.40 ÷ 4 = 23.6 cm³ (three significant figures).

Screw gauge and metre scale

Pitch and least count of a screw gauge

pitch p = (distance moved on the main scale)/(number of full turns)
LC = (pitch)/(number of divisions on the circular scale)
pitch 0.5 mm and 50 divisions: LC = 0.01 mm = 0.001 cm
pitch 1 mm and 100 divisions: LC = 0.01 mm
one full turn of the thimble moves the screw by one pitch

How to apply. Find the pitch from the data (distance moved in a few turns divided by the number of turns). Then divide by the circular scale divisions.

Example. The screw advances 2.5 mm in 5 turns: pitch = 0.5 mm. With 50 divisions, LC = 0.5 ÷ 50 = 0.01 mm.

Reading of a screw gauge

reading = MSR + (CSR × LC)
MSR = the last main scale (linear scale) mark uncovered by the thimble, including a half-millimetre mark if it is visible
CSR = the circular scale division that is on the reference line
corrected reading = reading − zero error
the screw gauge is read in mm and changed to cm or m at the end

How to apply. Count the whole mm and the half-mm mark visible on the sleeve, then add CSR × LC. Take care when the thimble edge is almost on a mark: use the circular reading to decide.

Example. Pitch 0.5 mm, 50 divisions, MSR = 4.5 mm and CSR = 28: reading = 4.5 + 28 × 0.01 = 4.78 mm. With zero error +0.03 mm the true diameter is 4.75 mm.

Zero error of a screw gauge

close the gauge and read the circular scale against the reference line
reading n between 0 and about 25: positive error e = +n × LC
reading n near the end of the scale (for example 97 on 100 divisions): negative error e = −(N − n) × LC, N = number of circular divisions
true diameter = observed reading − e
so a negative error is added to the observed reading

How to apply. Read the closed gauge first and write e with its sign. Small reading: positive. Reading close to N: negative, equal to (n − N) × LC.

Example. Closed gauge reads 97 on a 100-division scale (LC 0.01 mm): e = (97 − 100) × 0.01 = −0.03 mm. An observed 4.78 mm becomes 4.78 + 0.03 = 4.81 mm. A closed reading of 3 divisions gives e = +0.03 mm.

Backlash and the ratchet

backlash error: the screw turns a little without moving when the direction of turning is reversed
always turn the screw in the same direction while taking readings
use the ratchet (friction head) so that the pressure on the wire is the same each time
do not press the wire: a soft wire gets flattened and the reading is too small

How to apply. A screw gauge reading is a measurement of one diameter at one place. Rotate the wire and take readings in two perpendicular directions, at three places, then find the mean.

Example. Diameters of 0.42, 0.41, 0.43, 0.42 mm give a mean of 0.42 mm. Using the mean in the area A = πd²/4 keeps the random error small.

Metre scale, spherometer and parallax

metre scale: LC = 1 mm, so a length of 50.0 cm has an uncertainty of 0.1 cm at each end (read both ends)
parallax: keep the eye directly above the mark
a worn end of the scale gives an error: start from the 1.00 cm mark and subtract
spherometer: R = (l²)/(6h) + (h)/(2), with l the distance between two legs and h the height of the central screw

How to apply. Read both ends of a length and subtract so that the zero error of the scale cancels. For the spherometer, always take h from the difference of the readings on a flat plate and on the curved surface.

Example. Legs 3.0 cm apart (l = 3.0 cm) and h = 0.1 cm: R = (l²)/(6h) + (h)/(2) = 9 ÷ 0.6 + 0.05 = 15.05 cm.

Errors and significant figures

Absolute, relative and percentage error

mean a = (a₁ + a₂ + … + aₙ)/(n)
absolute error in one reading Δaᵢ = |a − aᵢ|
mean absolute error Δa = (Σ|Δaᵢ|)/(n)
relative (fractional) error = (Δa)/(a)
percentage error = (Δa)/(a) × 100 %
the result is written as a ± Δa
the least count error of an instrument is its LC

How to apply. Find the mean first, then the deviation of each reading from the mean, then the mean of the deviations. Report the error with 1 or 2 significant figures.

Example. Readings 2.63, 2.56, 2.42, 2.71, 2.80: mean = 2.62; deviations 0.01, 0.06, 0.20, 0.09, 0.18; mean deviation = 0.11. The result is 2.62 ± 0.11, a percentage error of 4.1 %.

Error in a sum or a difference

Z = A + B or Z = A − B: ΔZ = ΔA + ΔB
the absolute errors add in both cases
a difference of two nearly equal numbers has a large relative error because Z becomes small
the maximum error is the sum, never the difference

How to apply. Add the absolute errors (not the percentages) for a sum or a difference. Convert to a percentage only at the end.

Example. Temperatures 20.0 ± 0.5 °C and 50.0 ± 0.5 °C: the rise is 30.0 ± 1.0 °C, a relative error of 3.3 %.

Error in a product or a quotient

Z = A × B or Z = (A)/(B): (ΔZ)/(Z) = (ΔA)/(A) + (ΔB)/(B)
relative errors add (for both multiplication and division)
for a constant factor (like 4π) no extra error comes in
for several quantities, add all the relative errors

How to apply. Write the relative errors as fractions or percentages and add them. Do not subtract for a quotient.

Example. R = V/I with V = 5.0 ± 0.1 V and I = 2.0 ± 0.05 A: (ΔR)/(R) = 0.02 + 0.025 = 0.045, so R = 2.5 Ω with ΔR = 0.11 Ω (4.5 %).

Error in a power

Z = Aⁿ: (ΔZ)/(Z) = |n| (ΔA)/(A)
Z = (Aᵖ Bᵠ)/(Cʳ): (ΔZ)/(Z) = p(ΔA)/(A) + q(ΔB)/(B) + r(ΔC)/(C)
the sign of the power does not matter: the factor is the size of the power
a square root has a factor (1)/(2)

How to apply. Multiply each relative error by the size of its power and add everything. A quantity with a high power has the largest effect on the error.

Example. Z = x²y³/z with errors 1 %, 2 %, 3 % in x, y, z: error in Z = 2 × 1 + 3 × 2 + 1 × 3 = 11 %.

Error in g, density and similar results

g from the pendulum: (Δg)/(g) = (ΔL)/(L) + 2(ΔT)/(T)
density of a cylinder: (Δρ)/(ρ) = (Δm)/(m) + 2(Δd)/(d) + (Δl)/(l)
resistivity ρ = (πd²R)/(4L): (Δρ)/(ρ) = 2(Δd)/(d) + (ΔR)/(R) + (ΔL)/(L)
the quantity with the highest power or the largest relative error dominates

How to apply. Write the formula, list the power of each measured quantity, and add the relative errors with those powers. Then decide which measurement should be improved first.

Example. L = 100.0 ± 0.1 cm and T = 2.00 ± 0.005 s (20 oscillations in 40.0 ± 0.1 s): (Δg)/(g) = 0.001 + 2 × 0.0025 = 0.006, that is 0.6 %.

Least count error and the measurement of time

least count error = LC of the instrument
for n oscillations timed together, the error in T is (LC of the watch)/(n)
timing 20 oscillations reduces the error in T by a factor of 20
more readings reduce the random error but not the systematic error

How to apply. To measure a small quantity, repeat or multiply it (time many oscillations, count many turns of a wire, use a thick stack of sheets) so that the least count error becomes a small fraction.

Example. A stopwatch with LC 0.1 s: for one oscillation of 2.0 s the error is 5 %; for 20 oscillations (40.0 s) the error in T is 0.1 ÷ 20 = 0.005 s, 0.25 %.

Systematic and random errors

systematic errors have one sign and one cause: instrument (zero error, wrong calibration), personal (parallax) or imperfection of method; they can be removed by correction
random errors change in sign and size from reading to reading; they are reduced by taking many readings and averaging
gross errors are mistakes (a wrong reading); they are discarded

How to apply. Name the type of error from its behaviour: always the same direction means systematic; scattered both ways means random.

Example. A screw gauge with zero error +0.03 mm: every reading is high by 0.03 mm (systematic, removed by correction). Differences of ±0.01 mm between readings are random.

Significant figures and rounding

all non-zero digits are significant
zeros between digits are significant
leading zeros are not (0.0042 has 2)
trailing zeros after the decimal point are significant (4.20 has 3)
sum or difference: keep as many decimal places as the number with the fewest
product or quotient: keep as many significant figures as the number with the fewest
exact numbers (2, π, counts) do not limit the result

How to apply. Do the arithmetic in full and round only at the end. In an addition count decimal places; in a multiplication count significant figures. Round up when the next digit is 5 or more.

Example. 12.5 + 3.27 = 15.77, written 15.8 (one decimal place). 2.5 × 3.41 = 8.525, written 8.5 (two significant figures).

Simple pendulum and graphs

Time period and the value of g

T = 2π√((L)/(g))
g = (4π²L)/(T²)
L = effective length = length of the thread l + radius r of the bob (to the centre of mass)
T is found from n oscillations: T = (t)/(n) (n = 20 to 30), counted from the mean position
T does not depend on the mass of the bob

How to apply. Measure the length from the point of suspension to the centre of the bob. Put the values in g = 4π²L/T². Time more than 20 oscillations.

Example. L = 1.00 m and 20 oscillations take 40.2 s: T = 2.01 s, g = 4 × 9.87 × 1.00 ÷ 2.01² = 9.77 m/s².

Effect of the bob, the thread and the amplitude

use a small, heavy, spherical bob and a thin, light, inextensible thread
the bob radius must be added to the thread length
the amplitude should be small (θ below about 10° to 15°), since T ≈ T₀(1 + (θ²)/(16)) rises for large θ
the thread should be fixed between two split cork pieces so that the point of suspension is fixed
if r is ignored, g comes out too small
a thread that stretches gives an error in L

How to apply. Check each condition of the formula: small angle, point mass, fixed support, no air resistance. A violation means a systematic error.

Example. For θ = 0.2 rad (about 11.5°): T is larger than T₀ by (θ²)/(16) = 0.04 ÷ 16 = 0.25 %, which makes g lower by about 0.5 %.

The T²–L graph

T² = (4π²)/(g)L
the graph of T² against L is a straight line through the origin
slope = (T²)/(L) = (4π²)/(g), so g = (4π²)/(slope)
if T² is plotted against the thread length l, the line cuts the l axis at −r (negative) because L = l + r
the graph gives a better g than a single reading because the random errors average out
the graph of T against √L is also a straight line

How to apply. Take two far-apart points on the best-fit line (not data points) and find the slope. If the line does not pass through the origin, the intercept shows a constant error in L (such as the bob radius).

Example. Slope = 4.0 s²/m gives g = 4 × 9.87 ÷ 4.0 = 9.87 m/s². The length of a seconds pendulum (T = 2 s) is L = g/π² = 0.993 m.

Seconds pendulum and effect of changes

a seconds pendulum has T = 2 s and L = 0.993 m (about 99.3 cm) at g = 9.8 m/s²
T ∝ √L, T ∝ 1/√g
if L is made four times, T doubles
a bob filled with a liquid that drains out first raises and then lowers T (the centre of mass moves)
the pendulum in a lift accelerating up: g′ = g + a

How to apply. Use ratios for a change of L or g: T′/T = √(L′/L) or √(g/g′).

Example. L is changed from 25 cm to 100 cm: T doubles. In a lift with a = 2.0 m/s² upward, T becomes T × √(9.8 ÷ 11.8) = 0.91 T.

Young’s modulus, surface tension and viscosity

Young’s modulus by Searle’s method

Y = (F/A)/(ΔL/L) = (MgL)/(πr²l) = (4MgL)/(πD²l)
M = load added, L = original length of the experimental wire, r or D = radius or diameter (screw gauge), l = extension (spirit level and micrometer screw)
a second, identical wire (reference wire) carries a constant load and keeps the spirit level horizontal against temperature and support changes
plot load against extension: the slope M/l gives Y = (gL)/(πr²) × slope
take readings for loading and unloading and find the mean to remove elastic fatigue

How to apply. Find the slope of the load–extension graph (M against l), then put it in Y = (gL/πr²) × (M/l). The diameter appears squared, so it must be measured carefully.

Example. L = 2.0 m, r = 0.4 mm, a 1 kg load gives l = 0.20 mm: Y = 9.8 × 1 × 2.0 ÷ (π × (0.4 × 10⁻³)² × 0.20 × 10⁻³) = 1.95 × 10¹¹ N/m² (the value for steel).

Surface tension by capillary rise

T = (rhρg)/(2 cosθ) ≈ (rhρg)/(2) for a wetting liquid (θ ≈ 0°)
r = radius of the tube (travelling microscope or screw gauge), h = rise measured from the free surface
more exactly h is corrected for the liquid above the lowest point of the meniscus: h + (r)/(3)
the tube must be clean and vertical; a narrower tube gives a larger h
for mercury, θ ≈ 135° and the level is depressed

How to apply. Measure h from the level of the liquid outside the tube to the bottom of the meniscus, add r/3 for accuracy, and use the formula. The tube radius must be measured at the level of the meniscus.

Example. Water in a tube of r = 0.50 mm rises to h = 3.0 cm: T = 0.5 × 10⁻³ × 0.030 × 1000 × 9.8 ÷ 2 = 0.0735 N/m = 7.35 × 10⁻² N/m.

Viscosity by Stokes’ method

viscous force F = 6πηrv
terminal velocity v = (2r²(ρ − σ)g)/(9η)
η = (2r²(ρ − σ)g)/(9v)
ρ = density of the ball, σ = density of the liquid
v is the speed measured between two marks after the ball has reached terminal speed (time over a fixed distance)
the radius of the ball is found with a screw gauge
the tube must be wide and the ball small; the viscosity falls as temperature rises

How to apply. Check terminal speed first (equal time over equal distances). Then use η = 2r²(ρ − σ)g/9v with the radius in metres and the densities in kg/m³.

Example. Steel ball (ρ = 7800 kg/m³, r = 1.0 mm) in glycerine (σ = 1260 kg/m³) with v = 0.016 m/s: η = 2 × 10⁻⁶ × 6540 × 9.8 ÷ (9 × 0.016) = 0.89 Pa s.

Stokes’ law checks and the effect of size

when v is terminal: weight = buoyancy + viscous force, (4)/(3)πr³ρg = (4)/(3)πr³σg + 6πηrv
v ∝ r² (a double radius gives four times v)
the unit of η is Pa s = kg m⁻¹ s⁻¹
the ratio of terminal speeds of two balls of the same material in the same liquid is (v₁)/(v₂) = (r₁²)/(r₂²)

How to apply. If a question gives two radii in one liquid, write the ratio of the speeds as the square of the ratio of the radii.

Example. A ball of twice the radius in the same liquid reaches a terminal speed 4 times as large.

Speed of sound and specific heat

Resonance tube: first and second resonance

closed pipe: the first resonance at L₁ + e = (λ)/(4) and the second at L₂ + e = (3λ)/(4)
λ = 2(L₂ − L₁)
v = fλ = 2f(L₂ − L₁)
end correction e = (L₂ − 3L₁)/(2) ≈ 0.6 r (r = radius of the tube)
the end correction is removed when the difference of the two lengths is used
the fork must be vibrating freely; the water level is adjusted until the loudest sound is heard

How to apply. Use two resonances, so that e does not enter. Find λ from L₂ − L₁ and v from f × λ. The temperature of the air matters: v increases about 0.6 m/s per °C.

Example. f = 512 Hz, L₁ = 16.0 cm and L₂ = 50.0 cm: λ = 68 cm, v = 512 × 0.68 = 348 m/s, and e = (50.0 − 48.0) ÷ 2 = 1.0 cm.

Corrections and sources of error in the resonance tube

v at 0 °C: v₀ = v − 0.61t (t in °C) with v = 331 + 0.61t
v₀ ≈ 331 m/s at 0 °C
v ∝ √T (absolute temperature)
neglecting e gives a wrong v if only L₁ is used: λ = 4L₁ is too small by 4e
humidity increases v
the position of the resonance is found by the loudest note, not by the first sound

How to apply. Always use the two-length formula in exact work. If an answer uses only L₁, correct it with L₁ + e = λ/4.

Example. At 27 °C, v = 331 + 0.61 × 27 = 347 m/s. The measured 348 m/s agrees within the error.

Specific heat by the method of mixtures

heat lost by the hot solid = heat gained by the water and the calorimeter
m_s c_s(T₁ − T) = (m_w c_w + m_c c_c)(T − T₀)
T₁ = temperature of the hot solid, T₀ = initial temperature of the water, T = the highest steady temperature of the mixture
water equivalent of the calorimeter w = m_c c_c
c_s = ((m_w + w)(T − T₀))/(m_s(T₁ − T))

How to apply. Write the energy balance in one line, include the calorimeter, and solve for c_s. Transfer the solid quickly and stir.

Example. 100 g solid at 100 °C in a calorimeter (50 g, c = 0.09 cal/g °C) with 150 g water at 25.0 °C; the final temperature is 30.0 °C: c_s = (150 + 4.5) × 5.0 ÷ (100 × 70) = 0.110 cal/g °C.

Specific heat of a liquid and the losses

liquid: m_s c_s(T₁ − T) = (m_l c_l + w)(T − T₀) with a solid of known c_s
c_l = (m_s c_s(T₁ − T) − w(T − T₀))/(m_l(T − T₀))
heat loss by radiation is reduced by a polished calorimeter placed in a cotton-lined outer jacket, a lid, and a final temperature close to the room temperature

How to apply. Give the liquid the position of the water in the balance equation. Sources of error: heat lost during the transfer, an unstirred liquid and a thermometer that is too slow.

Example. Copper block (c = 0.093 cal/g °C, 200 g) at 100 °C into 150 g of liquid at 20 °C with w = 10 g, final temperature 30 °C: c_l = (200 × 0.093 × 70 ÷ 10 − 10) ÷ 150 = 0.80 cal/g °C.

Mirror, lens, slab and prism

Focal length of a concave mirror: u–v method

1/v + 1/u = 1/f with the Cartesian sign convention (object distance u negative)
real image on the same side: v negative
f = (uv)/(u + v)
the average of several (u, v) pairs gives f
the 1/u–1/v graph is a straight line of slope −1 meeting both axes at 1/f (magnitude)
the u–v graph is a hyperbola; the line v = u meets it at u = v = 2f = R (the centre of curvature)

How to apply. Use the Cartesian sign convention throughout and find f from many readings. Place the object at the centre of curvature to get an image of the same size, then R = u.

Example. u = −30 cm and v = −60 cm: f = (−30)(−60) ÷ (−90) = −20 cm (concave, f = 20 cm in magnitude). For u = v = −40 cm, f = −20 cm.

Focal length of a convex lens: u–v method

1/v − 1/u = 1/f
u negative, v positive for a real image
f = (uv)/(u − v)
plot 1/v against 1/u: a straight line with intercepts 1/f on both axes (v intercept = +1/f, u intercept = −1/f)
plot v against u: a hyperbola with u = v = ±2f at the point where the image is as large as the object
the image is real and inverted for u greater than f

How to apply. Use the sign convention and many readings. Avoid the readings very close to f. For a graph, find the intercept on the 1/v axis: it gives 1/f at once.

Example. u = −30 cm and v = +60 cm: 1/f = 1/60 + 1/30 = 3/60, so f = 20 cm. A 1/v–1/u graph with the 1/v intercept 0.050 cm⁻¹ gives f = 20 cm.

Displacement method (conjugate foci)

object and screen fixed at distance D > 4f
the lens has two positions, a distance d apart, which give a sharp image
f = (D² − d²)/(4D)
magnification at the two positions: m₁ m₂ = 1
the size of the object is √(I₁I₂)
for D = 4f only one position exists (d = 0)

How to apply. Measure D (object to screen) and d (the shift of the lens). The formula removes the need to find the exact centre of the lens, which makes this method more accurate.

Example. D = 100 cm and d = 20 cm: f = (10000 − 400) ÷ 400 = 24 cm.

Refractive index of a glass slab

n = (sin i)/(sin r)
the sin i–sin r graph is a straight line through the origin with slope n
lateral shift = (t sin(i − r))/(cos r) (t = thickness)
with a travelling microscope: n = (real thickness)/(apparent thickness) (the apparent depth is smaller)
the emergent ray is parallel to the incident ray

How to apply. Draw the normal at the pins with a fine pencil. Plot sin i against sin r and take the slope; for the microscope method note three readings (mark on the paper, on the slab, on the powder).

Example. i = 45° and r = 28.1°: n = 0.7071 ÷ 0.4714 = 1.50. A slab of real thickness 3.0 cm with an apparent thickness of 2.0 cm: n = 1.5.

Prism: the i–δ curve and the refractive index

δ = i + e − A
the i–δ graph is U-shaped with the lowest point at the minimum deviation δm, where i = e = (A + δm)/(2) and r = (A)/(2)
n = (sin((A + δm)/2))/(sin(A/2))
for a thin prism, δ = (n − 1)A
every deviation larger than δm is obtained at two values of i (one is i, the other is e)

How to apply. Draw the graph and read δm at the lowest point. Then use the formula for n. Pins must be 6 to 10 cm apart for the lines to be accurate.

Example. A = 60° and δm = 37°: n = sin 48.5° ÷ sin 30° = 0.749 ÷ 0.5 = 1.50.

Metre bridge and potentiometer

Metre bridge: resistance of a wire

Wheatstone balance: (R)/(S) = (l)/(100 − l)
R = (S l)/(100 − l)
the unknown R is in the left gap, the known resistance box S in the right gap
the balance point should be near the middle (30 to 70 cm) for the best sensitivity
swap R and S and average to remove the end errors; do not press the jockey hard or heat the wire by a long current

How to apply. Write R/S = l/(100 − l), put l in cm and solve for R. Adjust S so that l lies between 40 cm and 60 cm.

Example. S = 3.0 Ω and l = 40.0 cm: R = 3.0 × 40 ÷ 60 = 2.0 Ω.

Resistivity of the wire by the metre bridge

R = ρ(L)/(A) = (4ρL)/(πd²) so ρ = (πd²R)/(4L)
d is measured with a screw gauge, L with a metre scale, R with the bridge
the percentage error in ρ is 2(Δd)/(d) + (ΔR)/(R) + (ΔL)/(L) (d has the largest weight)
the specific resistance rises with temperature

How to apply. Measure d at several places with the screw gauge and use the mean. Then ρ = πd²R/4L with all lengths in metres.

Example. R = 2.0 Ω, d = 0.40 mm, L = 1.0 m: ρ = 2.0 × π × (0.40 × 10⁻³)² ÷ (4 × 1.0) = 2.5 × 10⁻⁷ Ω m.

Potentiometer: comparison of emfs

potential gradient k = (V)/(L) = (I ρ)/(A) (volt per metre) along the wire
at the null point E = k l
(E₁)/(E₂) = (l₁)/(l₂)
the driver cell has emf greater than both E₁ and E₂ and the positive poles of all cells are joined to the same end of the wire
no current flows from the test cell at balance, so the potentiometer measures the emf itself, not the terminal voltage

How to apply. Use the same wire and the same current: the ratio of the balance lengths equals the ratio of the emfs. The two-way key swaps E₁ and E₂ without changing anything else.

Example. l₁ = 75.0 cm for one cell and l₂ = 50.0 cm for the other: E₁ ÷ E₂ = 1.5. If the first emf is 1.5 V, the second is 1.0 V.

Potentiometer: internal resistance of a cell

balance length l₁ with the cell on open circuit (emf E = k l₁)
with a resistance R across the cell, balance length l₂ (terminal voltage V = k l₂)
r = R((l₁ − l₂)/(l₂)) = R((E)/(V) − 1)
the internal resistance is found without drawing a current from the cell at balance
choose R so that l₂ is not very small

How to apply. Write E = k l₁ and V = k l₂, and divide to get E/V = l₁/l₂. Then r = (E/V − 1)R.

Example. l₁ = 60 cm, l₂ = 50 cm, R = 10 Ω: r = 10 × 10 ÷ 50 = 2.0 Ω.

Sensitivity of the potentiometer

sensitivity rises with a longer wire (a smaller potential gradient k)
k = (V)/(L) falls when the series resistance in the driver circuit is raised
a potentiometer is better than a voltmeter for emf because it draws no current at balance (infinite resistance)
the null (balance) point is found without a deflection, which needs a sensitive galvanometer with a protective resistance in series at first

How to apply. Use a long wire and a low k for small emfs; remove the protective resistance when the balance point is close.

Example. A 10 m wire with 2.0 V across it has k = 0.2 V/m. A 1 mm shift in the balance length therefore means 0.2 mV.

Galvanometer, ammeter and voltmeter

Galvanometer resistance by half deflection

close K₁ only and adjust R to bring the deflection to θ (full scale): I_g = (E)/(R + G)
then close K₂ and adjust the shunt S until the deflection is (θ)/(2)
G = (R S)/(R − S)
if R ≫ G the shunt is nearly equal to G (G ≈ S)
the cell and resistance R are chosen so that R is a few thousand ohm

How to apply. Use the two steps: R sets the full deflection; S halves it. Put R and S in G = RS/(R − S). Do not use a very small R, because G would not be negligible.

Example. R = 3000 Ω and S = 100 Ω: G = 3000 × 100 ÷ 2900 = 103 Ω (about S).

Figure of merit of a galvanometer

figure of merit k = (I)/(θ) = (E)/((R + G)θ) (ampere per division)
current sensitivity = (θ)/(I) = (1)/(k)
the full-scale current I_g = kN for N divisions
k is the current needed for one division of deflection

How to apply. Find θ at the first step (without the shunt), then k = E/[(R + G)θ] with R + G in ohm and E in volt.

Example. E = 2 V, R = 4000 Ω, G = 100 Ω, θ = 20 divisions: k = 2 ÷ (4100 × 20) = 2.4 × 10⁻⁵ A per division.

Conversion to an ammeter

shunt S in parallel with the galvanometer
I_g G = (I − I_g)S
S = (I_g G)/(I − I_g)
the resistance of the ammeter is (GS)/(G + S) (very small)
the ammeter is joined in series
a smaller shunt gives a larger range

How to apply. The shunt takes the extra current I − I_g. Write I_g G = (I − I_g)S and solve for S.

Example. G = 100 Ω, I_g = 1 mA and range 1 A: S = 0.001 × 100 ÷ 0.999 = 0.100 Ω.

Conversion to a voltmeter

a high resistance R in series with the galvanometer
V = I_g(G + R)
R = (V)/(I_g) − G
the resistance of the voltmeter is G + R (very large)
the voltmeter is joined in parallel with the part of the circuit
its reading is exact only if its resistance is much more than the resistance across which it is connected

How to apply. Find the total resistance V/I_g and subtract G.

Example. G = 100 Ω, I_g = 1 mA and range 10 V: R = 10 ÷ 0.001 − 100 = 9900 Ω.

Checking a converted meter

compare the converted ammeter with a standard ammeter in series (or a converted voltmeter with a standard voltmeter in parallel) at several readings
the error in the reading = reading of the converted meter − standard reading
the percentage error = (error)/(full-scale value) × 100 %
a sensitive galvanometer needs a rheostat to vary the current smoothly

How to apply. Choose several points along the scale, note both readings, and plot one against the other. The line should have a slope of 1.

Example. Converted 0–1 A ammeter reads 0.48 A while the standard reads 0.50 A: the error is −0.02 A, which is 2 % of full scale.

Diodes, multimeter and graphs

I–V characteristic of a p-n junction diode

forward bias: the current is almost zero up to the knee voltage (about 0.7 V for silicon, 0.3 V for germanium) and then rises quickly
reverse bias: only a very small leakage current (μA) up to the breakdown
the characteristic is non-linear (not Ohm’s law)
the dynamic resistance r = (ΔV)/(ΔI) at a point of the forward curve
a series resistance in the forward circuit protects the diode

How to apply. Use a milliammeter in the forward circuit and a microammeter in the reverse circuit. Take many points near the knee. Draw the graph with the forward side on the right.

Example. Between 0.70 V and 0.75 V the current rises by 10 mA: r = 0.05 ÷ 0.010 = 5 Ω.

Zener diode characteristic

in the forward bias it behaves as an ordinary diode
in the reverse bias the current is tiny until the Zener (breakdown) voltage V_z, then it increases sharply at nearly constant voltage
this property is used for voltage regulation
always use a series resistance to limit the current
a Zener diode regulator works with V_z across the load; the series resistance takes the extra voltage

How to apply. Plot the reverse voltage against the reverse current. The very steep part of the curve is the Zener region, and V_z is read at the start of that region.

Example. For a 5.6 V Zener, the reverse current is below 10 μA up to 5 V and then rises to several mA in a very small voltage range around 5.6 V.

Using a multimeter

as a voltmeter: in parallel, as an ammeter: in series (move the leads to the current socket)
set the range first to the largest value, then reduce
as an ohmmeter: only with the component out of the circuit and no source connected
the DC and AC settings are separate
diode test: forward-biased reading is low (about 0.5 V to 0.7 V shown), the reverse reading is OL (open)

How to apply. Select the quantity, then the largest range, then connect. Never connect an ohmmeter across a live supply.

Example. To test a resistor marked 100 Ω ± 5 %, the reading should be between 95 Ω and 105 Ω; a reading of OL means the resistor is open.

Identifying a diode, LED, resistor and capacitor

diode: low resistance one way and very high the other way (the cathode is marked by a band)
LED: same as a diode, but needs a higher forward voltage (about 1.8 V to 3 V) and glows in the forward direction in the test
resistor: a steady reading in both directions equal to its value
capacitor: the ohmmeter reading starts low and rises slowly towards OL as it charges (reverse the leads, and the effect repeats); a dead capacitor reads 0 (short) or stays at OL immediately (open)
a continuity test buzzes for a short circuit

How to apply. Look at what the reading does as the leads are reversed or as time passes. Steady and the same in both ways means a resistor; one-way means a diode or LED; a rising reading means a capacitor.

Example. A component reads 500 Ω in one direction and OL in the other: a diode (or LED if it glows). A component whose reading rises from 2 kΩ to OL over a few seconds: a capacitor.

Plotting graphs and finding the slope

choose a scale so that the points fill the sheet (equal divisions per cm, easy values like 1, 2 or 5 per cm)
mark the points with a small dot or cross and draw a smooth best-fit line so that points lie on both sides
slope = (Δy)/(Δx) using two points on the LINE far apart, not two data points
the intercept is read at x = 0 (state when the origin is not shown)
the units of the slope are the units of y divided by the units of x

How to apply. Label the axes with quantity and unit. Draw the triangle for the slope with a large base. A line that should pass through the origin but does not shows a constant (systematic) error.

Example. A T²–L line passes through (0.20 m, 0.80 s²) and (1.00 m, 4.00 s²): slope = 3.20 ÷ 0.80 = 4.0 s²/m, so g = 4π²/4.0 = 9.87 m/s².

Most asked question types

How do I solve “Least count of a vernier or a screw gauge” questions in JEE Main?

The main scale division, the number of vernier divisions or the pitch and the circular divisions are given. Find the least count or the length of a division. 1. Vernier: LC = 1 MSD ÷ N. 2. Screw gauge: LC = pitch ÷ circular divisions. 3. Pitch = distance ÷ number of turns. Example: 1 MSD = 0.5 mm, 50 VSD: LC = 0.01 mm.

How do I solve “Reading of a vernier” questions in JEE Main?

A figure or the numbers show the main scale reading and the coinciding vernier line. Find the length. 1. Take the main scale reading before the vernier zero. 2. Find the coinciding line number. 3. Reading = MSR + VSR × LC. Example: 23 mm and the 6th line: 23.6 mm.

How do I solve “Reading of a screw gauge with zero error” questions in JEE Main?

The linear scale, the circular scale and the closed-gauge reading are given. Find the true diameter. 1. Reading = MSR + CSR × LC. 2. Find e from the closed gauge, with its sign. 3. True reading = observed − e. Example: 4.78 mm with e = +0.03 mm: 4.75 mm.

How do I solve “Zero error and zero correction” questions in JEE Main?

The closed vernier or screw gauge shows a reading. Say whether the error is positive or negative and find the correction. 1. Zero to the right or the circular zero below the line: positive. 2. Negative error: e = −(N − n) × LC. 3. Correction = −e; true = observed − e. Example: 4th line, N = 10, vernier zero on the left: e = −0.6 mm, true = observed + 0.6 mm.

How do I solve “Percentage error in a formula” questions in JEE Main?

Quantities with their errors are combined in a formula with powers. Find the percentage error in the result. 1. Write the powers of each quantity. 2. Add power × relative error for each. 3. Convert to a percentage. Example: x²y³/z with 1 %, 2 %, 3 %: 11 %.

How do I solve “Error in a sum, a difference or a quotient” questions in JEE Main?

Two measured values with absolute errors are added, subtracted, multiplied or divided. Find the error in the result. 1. Sum or difference: add the absolute errors. 2. Product or quotient: add the relative errors. 3. State the answer as value ± error. Example: V = 5.0 ± 0.1 V, I = 2.0 ± 0.05 A: R = 2.5 Ω with a 4.5 % error.

How do I solve “Mean error and significant figures” questions in JEE Main?

A series of readings is given. Find the mean, the mean absolute error and write the result with the right significant figures. 1. Find the mean. 2. Mean of the |deviations| is Δa. 3. Round the result to the decimal place of the error. Example: 2.63, 2.56, 2.42, 2.71, 2.80: 2.62 ± 0.11.

How do I solve “Simple pendulum: g and the effect of changes” questions in JEE Main?

The length, the number of oscillations and the time are given. Find T and g, or say how a change of the bob, the amplitude or the length changes T. 1. T = t ÷ n. 2. g = 4π²L/T² with L = l + r. 3. T depends on L and g only. Example: 1.00 m, 20 oscillations in 40.2 s: g = 9.77 m/s².

How do I solve “T²–L graph of a pendulum” questions in JEE Main?

A straight line is drawn or points are given. Find g from the slope, or say what an intercept means. 1. Slope = ΔT²/ΔL. 2. g = 4π² ÷ slope. 3. An intercept on the L axis is a constant error in L (such as r). Example: Slope 4.0 s²/m: g = 9.87 m/s².

How do I solve “Young’s modulus by Searle’s method” questions in JEE Main?

The length, the radius of the wire, the load and the extension are given. Find Y, or its percentage error, or the slope of the graph. 1. Y = MgL/(πr²l). 2. Slope of M–l gives M/l. 3. Percentage error includes 2Δr/r. Example: 2.0 m, 0.4 mm, 1 kg, 0.20 mm: 1.95 × 10¹¹ N/m².

How do I solve “Surface tension by capillary rise” questions in JEE Main?

The radius of the tube and the height of the rise are given. Find T, or the effect of a smaller radius. 1. T = rhρg/2. 2. h can be corrected by adding r/3. 3. A narrower tube gives a larger h. Example: r = 0.5 mm, h = 3 cm: 0.0735 N/m.

How do I solve “Viscosity by Stokes’ method” questions in JEE Main?

The size and density of the ball, the density of the liquid and the terminal speed are given. Find η, or compare two balls. 1. η = 2r²(ρ − σ)g/9v. 2. Use SI units. 3. Terminal speed v ∝ r². Example: r = 1 mm, ρ = 7800, σ = 1260, v = 0.016 m/s: 0.89 Pa s.

How do I solve “Resonance tube: speed of sound and end correction” questions in JEE Main?

The frequency and two resonance lengths are given. Find the speed of sound, the wavelength or the end correction. 1. λ = 2(L₂ − L₁). 2. v = fλ. 3. e = (L₂ − 3L₁)/2. Example: 512 Hz, 16.0 cm and 50.0 cm: 348 m/s, e = 1.0 cm.

How do I solve “Specific heat by the method of mixtures” questions in JEE Main?

The masses, the temperatures and the water equivalent are given. Find the specific heat of a solid or of a liquid. 1. Heat lost = heat gained. 2. Include the calorimeter (water equivalent). 3. Solve for the unknown c. Example: 100 g, 100 °C, 150 g water at 25 °C, final 30 °C: 0.110 cal/g °C.

How do I solve “Focal length of a mirror or a lens from u and v” questions in JEE Main?

Pairs of object and image distances or a graph are given. Find f, or the type of image and the intercepts. 1. Use the Cartesian sign convention. 2. 1/f = 1/v + 1/u (mirror) or 1/v − 1/u (lens). 3. The 1/v–1/u graph has intercepts 1/f. Example: u = −30 cm, v = +60 cm, convex lens: f = 20 cm.

How do I solve “Displacement method for a convex lens” questions in JEE Main?

The distance between object and screen and the shift of the lens are given. Find f. 1. D must be more than 4f. 2. f = (D² − d²)/4D. 3. Magnifications at the two positions multiply to 1. Example: D = 100 cm, d = 20 cm: 24 cm.

How do I solve “Refractive index of a slab or a prism” questions in JEE Main?

The angles of incidence and refraction, or A and δm, or real and apparent depths, are given. Find n. 1. Slab: n = sin i/sin r. 2. Prism: n = sin((A + δm)/2)/sin(A/2). 3. Microscope: n = real depth/apparent depth. Example: A = 60°, δm = 37°: n = 1.50.

How do I solve “Metre bridge: resistance and resistivity” questions in JEE Main?

The known resistance and the balance length are given. Find the unknown resistance, or the resistivity from the wire dimensions. 1. R/S = l/(100 − l). 2. ρ = πd²R/4L. 3. Interchange R and S to remove end errors. Example: S = 3.0 Ω, l = 40 cm: R = 2.0 Ω.

How do I solve “Potentiometer: emfs and internal resistance” questions in JEE Main?

Balance lengths for two cells, or with and without a shunt resistance, are given. Find the ratio of emfs or the internal resistance. 1. E₁/E₂ = l₁/l₂. 2. r = R(l₁ − l₂)/l₂. 3. The driver emf must be more than the test emf. Example: l₁ = 60, l₂ = 50, R = 10 Ω: r = 2.0 Ω.

How do I solve “Galvanometer resistance, figure of merit and conversion” questions in JEE Main?

R, S, E and the deflection are given for the half deflection method, or a range is given for an ammeter or voltmeter. Find G, k, the shunt or the series resistance. 1. G = RS/(R − S) and k = E/[(R + G)θ]. 2. Shunt: S = I_gG/(I − I_g). 3. Series: R = V/I_g − G. Example: G = 100 Ω, I_g = 1 mA: 0–1 A needs 0.100 Ω shunt; 0–10 V needs 9900 Ω.

How do I solve “Diode, Zener, LED and multimeter” questions in JEE Main?

A characteristic curve or the readings of a multimeter are given. Identify the component, find the knee or breakdown voltage or the dynamic resistance. 1. A diode conducts one way; Zener breaks down in reverse. 2. Dynamic resistance = ΔV/ΔI. 3. Resistor: same both ways; capacitor: reading rises. Example: ΔV = 0.05 V, ΔI = 10 mA: 5 Ω.

Common myths and the truth

Myth: The least count of a vernier is the length of one vernier division. Truth: It is the difference between one main scale division and one vernier division, 1 MSD ÷ N.
Myth: A positive zero error is added to the reading. Truth: The true reading is the observed reading minus the zero error, so a positive error is subtracted.
Myth: A larger number of readings removes a zero error. Truth: Averaging reduces random errors only. A systematic error has to be corrected.
Myth: In a quotient the relative errors cancel. Truth: They add, because both quantities can go the wrong way together.
Myth: The time period of a pendulum depends on the mass of the bob. Truth: It depends on L and g only (for a small amplitude).
Myth: The T²–L graph must be drawn through the data points. Truth: Draw the best-fit line with points on both sides and read the slope from two points on the line.
Myth: The potentiometer reads the terminal voltage of the cell. Truth: At balance no current flows through the cell, so it gives the emf.
Myth: A voltmeter needs a low resistance in series. Truth: A voltmeter needs a high resistance in series with the galvanometer; an ammeter needs a low resistance in parallel.
Myth: The metre bridge works equally well at any balance length. Truth: It is most accurate near the middle of the wire, at 40 cm to 60 cm.
Myth: A diode obeys Ohm’s law. Truth: The I–V curve is non-linear; there is a knee voltage and the resistance changes with the voltage.

Mistakes to avoid

Do not mix up the sign of the zero error: a negative error is added to the observed reading.
The screw gauge reading includes the half-millimetre mark if it is visible.
The absolute errors add in a difference; they do not subtract.
The relative errors add in a quotient; they do not subtract.
The bob radius is added to the thread length in the pendulum.
Take the slope from two points on the line, not from two data points.
In the resonance tube e is not needed if λ = 2(L₂ − L₁) is used.
For a lens, 1/v − 1/u = 1/f; for a mirror, 1/v + 1/u = 1/f.

More: Class 12 Physics simulations · all JEE Main chapters.