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$$ U_{dA}=\sqrt{\frac{\sum((d_i-\bar{d}))^2}{n(n-1)}}=0.008mm ,\ U_{dB}=\frac{0.004}{\sqrt{3}}mm $$

$$ U_d=\sqrt{U_{dA}^2+U_{dB}^2} = 0.008mm ,\ U_{rd}=\frac{U_d}{d} *100\%= 0.06\% $$

d = 13.707 ± 0.008mm

m = 11.18g

仪器误差Δ = 0.10g

$$ U_m=\frac{\Delta}{\sqrt{3}} = 0.06g,\, U_{rm}=\frac{U_m}{m}*100\%=0.5\% $$

m = 11.18 ± 0.06g

V = 2.25L = 2.25 * 10⁻³m³

$$ U_v = 0.0020*10^{-3}m^3,\, U_{rV}=\frac{U_V}{V}*100\%=0.09\% $$

V = (2.25 ± 0.0020) * 10⁻³m³

实验开始时：p_L1 = 101.6kPa,实验结束时：p_L2 = 101.6kPa

平均值$ p_L=101.6kPa $

仪器误差Δ = 0.01kPa

$$ U_{P_L} =\frac{\Delta}{\sqrt{3}} = 5*10^{-3}kPa, U_{rp_L}=\frac{U_{p_L}}{p_L}*100\%=5*10^{-3}\% $$

p_L = 101.6 ± 0.005kPa

$$ p=p_L+\frac{4mg}{\pi d^2} = 102.3kPa$$

$$U_p\approx U_{p_L}=5*10^{-3}kPa,\, U_{rp}=\frac{U_p}{p}*100\%=5*10^{-3}\% $$

p = 102.3 ± 0.005kPa

$$ U_{TA} = \sqrt{\frac{\sum (T_i-\bar{T})^2}{n(n-1)}} = 3.9*10^{-5}s,\, U_{TB}=\frac{0.1}{100}s=0.0010s $$

$$ U_T=\sqrt{U_{TA}^2+U_{TB}^2}=1.0*10^{-3}s,\, U_{rT}=\frac{U_T}{\bar{T}}*100\%=0.18\% $$

T = 0.5428 ± 0.001s

$$ \gamma = \frac{4mV}{T^2pr^4}=\frac{64mV}{T^2pd^4}=1.524 $$

$$ U_{r\gamma} = [U_{rm}^2+U_{rv}^2+(2U_{rT})^2+U_{rp}^2+(4U_{rd})^2]^{\frac{1}{2}} = 0.8\% $$

U_γ = γU_rγ = 0.012

实验结果表达式：$ = 1.524 $

γ₀ = 1.402

$$ E_0=\frac{|\gamma-\gamma_0|}{\gamma_0}*100\%=8.7\% $$