聯考物理T24

Solution — T24

題目描述

（本PDF為 JJL 所制）

24. 如圖所示，一厚壁玻璃容器放在水平面桌面上,容器底內底面積為 $50\ cm^2 $,外底面積為 $100 \ cm^2 $。將一定質量的水倒入容器中,水的深度為 \(10 \ cm\) 。求:\((p_水=1.0 \times 10^3 kg/m^3 , g \text{取} 10 N/kg)\)

(1) 水對容器底的壓強

\[P = ρ_水gh_水 = 1.0 \times 10^3 kg/m^3 \times 10N/kg \times 10 \times 10^{-2} m(1 cm = 1 \times 10^{-2}m) = 1 \times 10^3 Pa \]

(2) 水的重力

1. 解法一：

   \(V_水 = Sh = 50cm^2 \times 10cm = 5 \times 10^2 cm^3 = 5 \times 10^{-4} m^3(1cm^3 = 1 \times 10^{-6}m^3)\)

   \(m_水 = V_水ρ_水 = 5 \times 10^{-4} m^3 \times 1.0 \times 10^3 kg/m^3 = 5 \times 10^{-1} kg\)

   \(G_水 = m_水g = 5 \times 10^{-1} kg \times 10N/kg = 5N\)

2. 解法二：

   \(G = F = PS = 1 \times 10^3 Pa \times 50 \times 10^{-4} m^2 (1cm^2 = 10^{-4} m^2) = 5N\)

(3) 將體積為 \(500 \ cm\) 的金屬塊放入水中後,金屬塊浸沒並下沉到水底(此過程水未溢位容器),待水面靜止後,水對容器底的壓強的變化量與容器對桌面的壓強的變化量相差 \(350\ Pa\),求金屬塊的密度

\(△h_水 = \frac{V_金}{S} = 500cm^3 \div 50cm^2 = 10cm\)

\(△P_水 = ρg{△h} = 1.0 \times 10^3 kg/m^3 \times 10N/kg \times 10cm = 1 \times 10^3 Pa\)

• 注意分析題目意思，來進行分類討論：

1. \(△P_容 = △P_水 - 350 Pa = 650Pa = \frac{G_金}{S} = \frac{G_金}{100 \times 10^-4 m^2}\)

   解得 \(G_金 = 6.5 N\)

   \(ρ_金 = \frac{m_金}{V_金} = \frac{G_金 \div g}{V_金} = 6.5N \div 10N/kg \div(500 \times 10^{-6} m^3) = 1.3 \times 10^3 kg/m^3\)

2. \(△P_容 = △P_水 + 350 Pa = 1350Pa = \frac{G_金}{S} = \frac{G_金}{100 \times 10^-4 m^2}\)

   解得 \(G_金 = 13.5 N\)

   \(ρ_金 = \frac{m_金}{V_金} = \frac{G_金 \div g}{V_金} = 13.5N \div 10N/kg \div(500 \times 10^{-6} m^3) = 2.7 \times 10^3 kg/m^3\)