type: worked_problem | topic: numerical-experiments

Problem 1: Vernier Caliper Reading with Zero Error

Given: 50 VSD = 49 MSD, 1 MSD = 1 mm, MSR = 35 mm, 24th VSD coincides, positive zero error = 3 divisions.

Solution: $\text{LC} = 1\text{ MSD} - 1\text{ VSD} = 1 - \frac{49}{50} = 1 - 0.98 = 0.02\text{ mm}$ $\text{Raw reading} = \text{MSR} + (n \times \text{LC}) = 35 + (24 \times 0.02) = 35 + 0.48 = 35.48\text{ mm}$ $\text{Zero error} = 3 \times 0.02 = 0.06\text{ mm (positive)} \Rightarrow \text{Corrected} = 35.48 - 0.06 = \boxed{35.42\text{ mm}}$

Problem 2: Resonance Tube — Speed and End Correction

Given: l_{1} = 17.0 cm = 0.170 m; l_{2} = 51.8 cm = 0.518 m; f = 480 Hz.

Solution: $v = 2f(l_2 - l_1) = 2 \times 480 \times (0.518 - 0.170) = 960 \times 0.348 = \boxed{334.1\text{ m/s}}$ $e = \frac{l_2 - 3l_1}{2} = \frac{0.518 - 3(0.170)}{2} = \frac{0.518 - 0.510}{2} = \frac{0.008}{2} = \boxed{0.004\text{ m} = 0.4\text{ cm}}$ Dimensional check: v: [Hz][m] = [$T^{-1}$][L] = [L$T^{-1}$] ✓

Problem 3: Metre Bridge — Finding S and Resistivity

Given: Balance at l = 36.0 cm, R = 6.0 $\Omega$ (left gap), wire length L = 1.2 m, diameter d = 0.4 mm.

Solution: $\frac{R}{S} = \frac{l}{100-l} = \frac{36}{64} \Rightarrow S = 6.0 \times \frac{64}{36} = 6.0 \times \frac{16}{9} = \boxed{10.67\text{ $\Omega$}}$ $\text{Interchange: } l' = 100 - 36 = \boxed{64.0\text{ cm}}$ $A = \pi\left(\frac{d}{2}\right)^2 = \pi\left(\frac{0.4 \times 10^{-3}}{2}\right)^2 = \pi \times (0.2 \times 10^{-3})^2 = 1.257 \times 10^{-7}\text{ m}^2$ $\rho = \frac{SA}{L} = \frac{10.67 \times 1.257 \times 10^{-7}}{1.2} = \frac{1.341 \times 10^{-6}}{1.2} \approx \boxed{1.12 \times 10^{-6}\text{ $\Omega$·m}}$

Problem 4: Simple Pendulum — Finding g

Given: Pendulum: string = 99 cm, bob radius = 1.0 cm. Time for 40 oscillations = 80.0 s.

Solution: $l = 99 + 1 = 100\text{ cm} = 1.00\text{ m}$ $T = \frac{80.0}{40} = 2.00\text{ s}$ $g = \frac{4\pi^2 l}{T^2} = \frac{4 \times 9.87 \times 1.00}{(2.00)^2} = \frac{39.48}{4.00} = \boxed{9.87\text{ m/s}^2}$ Note: If bob radius were ignored (l = 0.99 m): g = 4π^{2}×0.99/4 = 9.77 m/$s^{2}$ — underestimate of 0.10 m/$s^{2}$ (1% error).