Experimental Skills — Easy Practice (Set 1)

Quick Recap — Vernier, Screw Gauge & Pendulum

• Vernier calliper: least count $= 1\ \text{MSD} - 1\ \text{VSD}$; reading $=$ MSR $+$ (vernier coincidence $\times$ LC).
• Screw gauge (micrometer): LC $= \dfrac{\text{pitch}}{\text{number of circular-scale divisions}}$; reading $=$ MSR $+$ (circular reading $\times$ LC).
• Zero error: always corrected — subtract a positive zero error and add a negative one.
• Simple pendulum: $g = \dfrac{4\pi^2 L}{T^2}$; time many oscillations to cut the timing error, and a graph of $T^2$ vs $L$ is a straight line.
• A smaller least count means a more precise instrument.

Beyond-NCERT JEE Formulae

This chapter is about measurement and error inside named experiments. Below are the working formulae, the exact percentage-error rule for each, and when to reach for them.

1. Least count and a corrected reading (every instrument)

• Vernier: $\text{LC} = 1\,\text{MSD} - 1\,\text{VSD}$. If $n$ vernier divisions cover $(n-1)$ main divisions, then $\text{LC} = \dfrac{1\,\text{MSD}}{n}$.
• Screw gauge and spherometer: $\text{LC} = \dfrac{\text{pitch}}{N}$, where $N$ is the number of circular-scale divisions.
• Every reading: value $= \text{MSR} + \text{LC} \times (\text{coincidence})$, then correct the zero error — subtract a positive zero error, add a negative one.
• Spherometer (beyond NCERT) — radius of curvature of a spherical surface from the sagitta $h$ and the leg spacing $l$:

$$R = \dfrac{l^{2}}{6h} + \dfrac{h}{2}.$$

[JEE Tip] A negative zero error shows up as a large coincidence near $N$; its magnitude is $(N - \text{coincidence}) \times \text{LC}$, and you ADD it back to the reading.

2. Master rule for propagating errors

For any product-power law $Q = k\,\dfrac{A^{a} B^{b}}{C^{c}}$ the maximum relative error adds each term weighted by its power:

$$\dfrac{\Delta Q}{Q} = |a|\dfrac{\Delta A}{A} + |b|\dfrac{\Delta B}{B} + |c|\dfrac{\Delta C}{C}.$$

For a sum or a difference the ABSOLUTE errors add: $\Delta(A \pm B) = \Delta A + \Delta B$.

[JEE Tip] Subtracting two nearly equal readings is dangerous — the absolute error stays fixed while the result shrinks, so the percentage error explodes. Never build a measurement around a small difference of two large numbers.

3. Young's modulus — Searle's method

$$Y = \dfrac{FL}{A\,\Delta L} = \dfrac{MgL}{\pi r^{2}\ell},$$

with load $F = Mg$, original length $L$, cross-section $A = \pi r^{2}$ and extension $\ell = \Delta L$. Its error:

$$\dfrac{\Delta Y}{Y} = \dfrac{\Delta M}{M} + \dfrac{\Delta L}{L} + 2\dfrac{\Delta r}{r} + \dfrac{\Delta \ell}{\ell}.$$

When to use: a wire stretched by a hanging load; the twin reference wire cancels temperature drift and support sag.

[JEE Tip] The radius enters squared, so its 1 percent error becomes 2 percent in $Y$ — almost always the dominant term. Measure $r$ with a screw gauge, never a metre scale.

4. Resistivity by a metre bridge

The balance gives the unknown $X = R\,\dfrac{l}{100 - l}$; the resistivity of the wire then follows from

$$\rho = \dfrac{X A}{L} = \dfrac{\pi r^{2} X}{L},$$

with wire radius $r$ and length $L$.

When to use: finding $\rho$ of a resistance wire; keep the balance point near mid-wire.

[JEE Tip] Because $l$ appears in both the top and the bottom of $X$, its error enters twice: $\dfrac{\Delta X}{X} = \dfrac{\Delta l}{l} + \dfrac{\Delta l}{100 - l}$, which is smallest at $l = 50$ cm. That is exactly why you tune $R$ to balance in the middle.

5. Focal length

Mirror or lens by the $u$-$v$ method: $\dfrac{1}{f} = \dfrac{1}{v} \pm \dfrac{1}{u}$. Because $f$ is built from reciprocals, its error is

$$\Delta f = f^{2}\left(\dfrac{\Delta u}{u^{2}} + \dfrac{\Delta v}{v^{2}}\right).$$

Displacement (Bessel) method, beyond NCERT — object and screen fixed a distance $D > 4f$ apart; the lens forms a sharp image at two positions separated by $d$:

$$f = \dfrac{D^{2} - d^{2}}{4D}.$$

[JEE Tip] Do NOT write $\dfrac{\Delta f}{f} = \dfrac{\Delta u}{u} + \dfrac{\Delta v}{v}$; that product rule fails for a reciprocal relation. Use the $f^{2}$ form above.

6. Simple pendulum

$$g = \dfrac{4\pi^{2} L}{T^{2}}, \qquad \dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T}.$$

Time $n$ oscillations so that $T = \dfrac{t}{n}$; since the count $n$ is exact, $\dfrac{\Delta T}{T} = \dfrac{\Delta t}{t}$.

[JEE Tip] Timing dominates because $T$ is squared — timing 50 swings instead of 1 divides the timing error by 50 at no cost.

7. Sonometer (vibrating string)

$$n = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}},$$

with tension $T$ and mass per unit length $\mu$. At fixed tension $n \propto \dfrac{1}{L}$, so resonance against a fork of known frequency fixes $L$.

[JEE Tip] Tension sits under a square root, so a 2 percent error in $T$ is only 1 percent in $n$; but $\mu$ hides there too — find $\mu$ by weighing a measured length, not from the label.

8. Reading a graph — slope and intercept

Straight-line experiments hand you a constant through the slope:

• Ohm's law: plot $V$ against $I$; the slope is the resistance $R$.
• Pendulum: plot $T^{2}$ against $L$; the slope is $\dfrac{4\pi^{2}}{g}$, so $g = \dfrac{4\pi^{2}}{\text{slope}}$.
• Searle's: plot extension $\ell$ against load $M$; the slope is $\dfrac{gL}{\pi r^{2} Y}$, giving $Y$.
• Resonance tube: plot length $l$ against $\dfrac{1}{f}$; the slope is $\dfrac{v}{4}$ and the intercept is $-e$, the end correction.

[JEE Tip] Read the quantity from the SLOPE of a best-fit line, not from a single point — the line averages out random error, and a nonzero intercept exposes a systematic (zero) error at a glance.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Screw gauge reading with a zero error

Given: pitch $= 0.5$ mm, $N = 50$ circular divisions; a positive zero error with the 4th division coinciding when the studs touch; measuring a wire the main scale reads 4.0 mm and the 27th circular division coincides.

Formula: $\text{LC} = \dfrac{\text{pitch}}{N}$; reading $= \text{MSR} + \text{LC} \times (\text{coincidence})$; a positive zero error is subtracted.

Working: $\text{LC} = \dfrac{0.5}{50} = 0.01$ mm. Zero error $= 4 \times 0.01 = +0.04$ mm. Observed $= 4.0 + 27 \times 0.01 = 4.27$ mm. Corrected $= 4.27 - 0.04 = 4.23$ mm.

Answer: diameter $= 4.23$ mm.

Example 2 — Radius of curvature with a spherometer (beyond NCERT)

Given: a spherometer of least count 0.01 mm; the mean spacing between its three legs is $l = 3.0$ cm; the central screw records a sagitta $h = 0.50$ mm.

Formula: $R = \dfrac{l^{2}}{6h} + \dfrac{h}{2}$.

Working: in millimetres $l = 30$ mm and $h = 0.50$ mm, so $R = \dfrac{30^{2}}{6 \times 0.50} + \dfrac{0.50}{2} = \dfrac{900}{3.0} + 0.25 = 300.25$ mm.

Answer: $R \approx 30.0$ cm — the $\dfrac{h}{2}$ term adds only 0.25 mm.

Example 3 — Percentage error in Young's modulus (Searle's)

Given: $M = 2.00 \pm 0.01$ kg, $L = 1.500 \pm 0.005$ m, wire diameter $d = 0.50 \pm 0.01$ mm, extension $\ell = 0.80 \pm 0.01$ mm.

Formula: $Y = \dfrac{4MgL}{\pi d^{2}\ell}$, so $\dfrac{\Delta Y}{Y} = \dfrac{\Delta M}{M} + \dfrac{\Delta L}{L} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta \ell}{\ell}$.

Working: the terms are $\dfrac{\Delta M}{M} = 0.50$ percent, $\dfrac{\Delta L}{L} = 0.33$ percent, $2\dfrac{\Delta d}{d} = 2(2.0) = 4.0$ percent and $\dfrac{\Delta \ell}{\ell} = 1.25$ percent. Adding, $\dfrac{\Delta Y}{Y} = 0.50 + 0.33 + 4.0 + 1.25 = 6.1$ percent.

Answer: $\dfrac{\Delta Y}{Y} \approx 6.1$ percent (here $Y \approx 1.9 \times 10^{11}$ Pa); the squared diameter dominates the error.

Example 4 — Percentage error in g from a pendulum

Given: $L = 1.000$ m with $\Delta L = 1$ mm; 20 oscillations take $t = 40.2$ s on a stopwatch of least count 0.1 s.

Formula: $g = \dfrac{4\pi^{2} L}{T^{2}}$ with $T = \dfrac{t}{n}$, so $\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta t}{t}$.

Working: $\dfrac{\Delta L}{L} = \dfrac{0.001}{1.000} = 0.10$ percent and $\dfrac{\Delta t}{t} = \dfrac{0.1}{40.2} = 0.25$ percent, so $\dfrac{\Delta g}{g} = 0.10 + 2(0.25) = 0.60$ percent. Here $T = 2.01$ s gives $g = 9.77$ m/s$^2$.

Answer: $\dfrac{\Delta g}{g} \approx 0.60$ percent, so $g = 9.77 \pm 0.06$ m/s$^2$.

Example 5 — Resistivity from a metre bridge

Given: a resistance wire of radius $r = 0.25$ mm and length $L = 1.0$ m lies in the left gap and balances a known $R = 3\ \Omega$ in the right gap at $l = 62.5$ cm.

Formula: $X = R\dfrac{l}{100 - l}$, then $\rho = \dfrac{\pi r^{2} X}{L}$.

Working: $X = 3 \times \dfrac{62.5}{37.5} = 5.0\ \Omega$. Then $\rho = \dfrac{\pi (0.25 \times 10^{-3})^{2}(5.0)}{1.0} = 9.8 \times 10^{-7}\ \Omega$ m.

Answer: $\rho \approx 9.8 \times 10^{-7}\ \Omega$ m.

Example 6 — Slope of a V-I graph gives the resistance

Given: for a metallic resistor a plot of $V$ (vertical axis) against $I$ (horizontal axis) is a straight line through the origin, passing through the points $(0.20, 1.0)$ and $(0.80, 4.0)$, with $I$ in ampere and $V$ in volt.

Formula: Ohm's law $V = IR$ has the form $y = (\text{slope})\,x$ with slope $= R$.

Working: slope $= \dfrac{4.0 - 1.0}{0.80 - 0.20} = \dfrac{3.0}{0.60} = 5.0$. Since the axes read volt over ampere, the slope is the resistance itself.

Answer: $R = 5.0\ \Omega$, read from the slope rather than any single point.

Example 7 — Focal length by the displacement (Bessel) method (beyond NCERT)

Given: an object and a screen are fixed $D = 100$ cm apart; a convex lens between them throws a sharp image at two positions separated by $d = 40$ cm.

Formula: $f = \dfrac{D^{2} - d^{2}}{4D}$.

Working: $f = \dfrac{100^{2} - 40^{2}}{4 \times 100} = \dfrac{10000 - 1600}{400} = \dfrac{8400}{400} = 21.0$ cm.

Answer: $f = 21.0$ cm, found without measuring $u$ and $v$ separately.