91Ó°ÊÓ

The blocks have a volume of \(8.00 \,\mathrm{cm}^{3}\) which we must convert to \(\mathrm{m}^{3}\). To do this, we use the conversion factor \(1 \mathrm{m} = 100 \mathrm{cm}\). Volume of blocks in \(\mathrm{m}^{3}\): $$V = 8.00 \,\mathrm{cm}^{3} \cdot \left(\frac{1 \,\mathrm{m}}{100 \,\mathrm{cm}}\right)^3 = 8.00 \times 10^{-6} \,\mathrm{m}^{3}$$

Next, we compute the mass of the pine and steel blocks using their respective densities and the volume we calculated in the previous step. Let \(m_p\) be the mass of the pine block and \(m_s\) be the mass of the steel block. Density of pine, \(\rho_p = 420 \,\mathrm{kg} / \mathrm{m}^{3}\) Density of steel, \(\rho_s = 7850 \,\mathrm{kg} / \mathrm{m}^{3}\) Mass of pine block, \(m_p = V \cdot \rho_p\) $$m_p = 8.00 \times 10^{-6} \,\mathrm{m}^{3} \cdot 420 \,\mathrm{kg} / \mathrm{m}^{3} = 0.00336 \,\mathrm{kg}$$ Mass of steel block, \(m_s = V \cdot \rho_s\) $$m_s = 8.00 \times 10^{-6} \,\mathrm{m}^{3} \cdot 7850 \,\mathrm{kg} / \mathrm{m}^{3} = 0.0628 \,\mathrm{kg}$$

When the blocks are placed in the beakers, they displace an amount of water equal to their volume. We need to compare the weight of the displaced water with the weight of the blocks being added. Density of water, \(\rho_w = 1000 \,\mathrm{kg} / \mathrm{m}^{3}\) Mass of water displaced by both blocks, \(m_w = V \cdot \rho_w\) $$m_w = 8.00 \times 10^{-6} \,\mathrm{m}^{3} \cdot 1000 \,\mathrm{kg} / \mathrm{m}^{3} = 0.008 \,\mathrm{kg}$$ Now, we calculate the differences in weight between the displaced water and the blocks: Weight difference for pine block: \(\Delta W_p = m_w - m_p\): $$\Delta W_p = 0.008 \,\mathrm{kg} - 0.00336 \,\mathrm{kg} = 0.00464 \,\mathrm{kg}$$ Weight difference for steel block: \(\Delta W_s = m_w - m_s\): $$\Delta W_s = 0.008 \,\mathrm{kg} - 0.0628 \,\mathrm{kg} = -0.0548\,\mathrm{kg}$$

Now we have the weight differences for each beaker, considering the weight of the blocks and the weight of the displaced water. For the beaker with the pine block, the scale reading will decrease by \(0.00464\,\mathrm{kg}\) because the weight of the pine block is less than the weight of the displaced water. For the beaker with the steel block, the scale reading will increase by \(0.0548\,\mathrm{kg}\). This is because the weight of the steel block is greater than the weight of the displaced water. In summary, the scale reading will decrease by 0.00464 kg for the pine block and increase by 0.0548 kg for the steel block.