12-25 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 2R&=4.8\ut{cm}\\ L&=5.3\ut{cm}\\ m&=1500\ut{kg}\\ G&=3.0\times10^{10}\ut{N/m^2}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} {F\over A}&={mg\over \pi R^2}\\ &=\frac{7812500 g}{3 \pi }\\ &\approx 8.129045951417372\times10^6\ut{N/m^2}\\ &\approx 8.1\ut{MN/m^2}\\ \end{aligned} $$ $$\ab{b}$$ $${F\over A}=G{\Delta x\over L},$$ $$ \begin{aligned} \Delta x&={FL\over AG}\\ &={Lmg\over \pi R^2 G}\\ &=\frac{53 g}{11520000 \pi }\ut{m}\\ &\approx 1.436131451417069\times10^{-5}\ut{m}\\ &\approx 1.4\times10^{-5}\ut{m}\\ &\approx 14\ut{\mu m}\\ \end{aligned} $$