Chapter 1 - Problems - Page 32: 1.38

Given data Weight of oxygen in tank at sea level $W=1 \mathrm{lb}$ The acceleration due to gravity at sea level $\mathrm{g}_{\mathrm{sea}}=32.174 \frac{\mathrm{ft}}{\mathrm{s}^{2}}$ The acceleration due to gravity at the top of the Everest $g_{\text {top }}=32.082 \frac{\mathrm{ft}}{\mathrm{s}^{2}}$ We know that Weight $=$ mass $\times$ acceleration due to gravity $W=m g$ $W_{\text {sealevel }}=m_{\text {sealevel }} \times g_{\text {sea level }}$ And $W_{\text {top }}=m_{\text {top }} \times g_{\text {top }}$ But mass of any substance remains same anywhere i.e, $m_{\text {sea level }}=m_{\text {top }}$ $\frac{W_{\text {sea level }}}{g_{\text {sea level }}}=\frac{W_{\text {top }}}{g_{\text {top }}}$ $W_{\text {top }}=\frac{1 \times 32.082}{32.174}$ $W_{\text {top }}=0.997 \mathrm{lb}$ $\therefore$ Weight at the top of Everest $0.997 \mathrm{lb}$

Given data Weight of oxygen in tank at sea level $W=1 \mathrm{lb}$ The acceleration due to gravity at sea level $\mathrm{g}_{\mathrm{sea}}=32.174 \frac{\mathrm{ft}}{\mathrm{s}^{2}}$ The acceleration due to gravity at the top of the Everest $g_{\text {top }}=32.082 \frac{\mathrm{ft}}{\mathrm{s}^{2}}$ We know that Weight $=$ mass $\times$ acceleration due to gravity $W=m g$ $W_{\text {sealevel }}=m_{\text {sealevel }} \times g_{\text {sea level }}$ And $W_{\text {top }}=m_{\text {top }} \times g_{\text {top }}$ But mass of any substance remains same anywhere i.e, $m_{\text {sea level }}=m_{\text {top }}$ $\frac{W_{\text {sea level }}}{g_{\text {sea level }}}=\frac{W_{\text {top }}}{g_{\text {top }}}$ $W_{\text {top }}=\frac{1 \times 32.082}{32.174}$ $W_{\text {top }}=0.997 \mathrm{lb}$ $\therefore$ Weight at the top of Everest $0.997 \mathrm{lb}$