Determine the ratio of the weights of a 1 kg block of iron and a 1...

The spring balance measures the apparent weight of an object, which is its true weight minus the buoyant force exerted by the displaced air.

The apparent weight $W$ of a block of mass $m$ and density $\rho$ is given by:

W = mg - F_{b} = mg - V \rho_{\text{air}} g = mg \left( 1 -\dfrac{\rho_{\text{air}}}{\rho} \right)

For the iron block, the apparent weight is:

$W_{\text{iron}} = mg \left( 1 - \dfrac{\rho_{\text{air}}}{\rho_{\text{iron}}} \right)$

For the wood block, the apparent weight is:

W_{\text{wood}} = mg \left( 1 -\dfrac{\rho_{\text{air}}}{\rho_{\text{wood}}} \right)

The ratio of their weights as measured by the spring balance is:

\dfrac{W_{\text{iron}}}{W_{\text{wood}}} = \dfrac{1 - \dfrac{\rho_{\text{air}}}{\rho_{\text{iron}}}}{1 - \dfrac{\rho_{\text{air}}}{\rho_{\text{wood}}}}

Since $\rho_{\text{air}}$ is much smaller than both $\rho_{\text{iron}}$ and $\rho_{\text{wood}}$, we can use the binomial approximation $(1 - x)^{-1} \approx 1 + x$:

\dfrac{W_{\text{iron}}}{W_{\text{wood}}} \approx \left($1 -$\dfrac{\rho_{\text{air}}}{\rho_{\text{iron}}} \right) \left($1 +$\dfrac{\rho_{\text{air}}}{\rho_{\text{wood}}} \right) \approx$1 + \rho_{\text{air}} \left( \dfrac{1}{\rho_{\text{wood}}} - \dfrac{1}{\rho_{\text{iron}}} \right)$

Substituting the given values $\rho_{\text{air}} = 1.293 \text{ kg m}^{-3}$, $\rho_{\text{wood}} = 800 \text{ kg m}^{-3}$, and $\rho_{\text{iron}} = 7800 \text{ kg m}^{-3}$:

\dfrac{W_{\text{iron}}}{W_{\text{wood}}} \approx 1 + 1.293 \left(\dfrac{1}{800} - \dfrac{1}{7800} \right)

\dfrac{W_{\text{iron}}}{W_{\text{wood}}} \approx$1 + 1.293 \left( \dfrac{7800 - 800}{800 \times 7800} \right)$

\dfrac{W_{\text{iron}}}{W_{\text{wood}}} \approx 1 + 1.293 \left(\dfrac{7000}{6240000} \right)

\dfrac{W_{\text{iron}}}{W_{\text{wood}}} \approx$1 +$\dfrac{9051}{6240000} \approx$1 + 0.00145$

\dfrac{W_{\text{iron}}}{W_{\text{wood}}} \approx$1.0015$