The weight of a body is G = 3275 g. Find its specific

The weight of a body is $G = 3275\ \text{g}$. Find its specific gravity if the volume is $V = (430\pm1)\ \text{cm}^{3}$.

Step 1: The nominal value

\[ \gamma = \frac{G}{V}, \qquad \gamma_{0} = \frac{3275}{430} = \frac{655}{86} \approx 7.6163\ \frac{\text{g}}{\text{cm}^{3}}. \]

The weight is given exactly, so the volume is the only source of uncertainty.

Step 2: Differentiate with respect to $V$

\[ \frac{d\gamma}{dV} = -\frac{G}{V^{2}}, \qquad \bigl|\Delta\gamma\bigr| \approx \frac{G}{V^{2}}\,\bigl|\Delta V\bigr|. \]

The minus sign only says that a larger volume means a smaller specific gravity; for an error bound the magnitude is what matters.

Step 3: Evaluate

\[ 430^{2} = 184\,900, \qquad \bigl|\Delta\gamma\bigr| = \frac{3275}{184\,900}\cdot 1 = \frac{131}{7396} \approx 0.01771\ \frac{\text{g}}{\text{cm}^{3}}. \]

Step 4: The relative error

Dividing the error by $\gamma_{0} = G/V$, the weight cancels entirely: \[\frac{\bigl|\Delta\gamma\bigr|}{\gamma_{0}} = \frac{G\bigl|\Delta V\bigr|/V^{2}}{G/V} = \frac{\bigl|\Delta V\bigr|}{V} = \frac{1}{430} \approx 0.233\%.\]

For a reciprocal the relative error passes through unchanged — consistent with the rule $\dfrac{dy}{y} = n\dfrac{dx}{x}$ of Problem 250 with $n = -1$.

Step 5: Verification

\[\frac{3275}{429} \approx 7.63403, \qquad \frac{3275}{431} \approx 7.59861\]

\[7.63403-7.61628 = 0.01775; \qquad 7.61628-7.59861 = 0.01767 \;\;\checkmark\]

\[\text{Both bracket the estimate } 0.01771, \text{ the small asymmetry being the second-order term} \;\;\checkmark\]