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“1991年全国初中应用物理知识竞赛试题”填空题的第2题,题目是“某工厂生产的酒精要含水量不超过10%.用抽测密度的方法检查产品的质量,密度在 ——kg/m~3至——kg/m~3范围内为合格产品”其答案是“密度在 0.8 X 10~3kg/m~2至 0.82 X10~3kg/m~3范围内为合格产品.”有些竞赛指导书(注①)对以上题目的解答是(以下简称“解法一”):“解:产品的含水量最小为0时,产品的密度即为酒精的密度 0.8X10~3kg/m~3.产品的含水量最多为 10%时,为研究问题,我们取产品的体积为1m~3.由此可得水的质量为m_水=ρ_水V_水=1.0 X10~3kg/m~3X0.1m~3= 0.1X10~3kg,酒精的质量为 m_酒精=ρ_酒精×V_酒精=0.8 X 10~3kg/m~2X 0.9m~3=0.72 x10~3kg,水和酒精的总质量为0.82X10~3kg,这时产品的密度为 ρ=m/v=0.82X10~3kg/m~2.所以密度在 0.8 X10~3kg/m~3至0.82X10~3kg/m~2范围内为合格产品.”
“The first question of the 1991 National Junior Physics Applied Knowledge Contest” fills in the blank title, “The alcohol content of a certain factory produces no more than 10% of water content. The quality of the product is checked by the method of density measurement, and the density is - kg/ The answer is ”The density is in the range of 0.8 X 10~3kg/m~2 to 0.82 X10~3kg/m~3.“ Some competitions are qualified products in the range of m~3 to - kg/m~3. The answer to the above question in the guide book (Note 1) is (hereinafter referred to as ”solution one“): ”Solution: When the minimum moisture content of the product is 0, the density of the product is the density of alcohol 0.8X10~3kg/m~3. When the moisture content of the product is at most 10%, for the sake of research, we take the volume of the product as 1m~3. The available water quality is m_water=ρ_water V_water=1.0×10~3kg/m~ 3X0.1m~3= 0.1X10~3kg, the mass of alcohol is m_alcohol=ρ_alcohol×V_alcohol=0.8 X 10~3kg/m~2X 0.9m~3=0.72 x10~3kg, water and alcohol The total mass is 0.82X10~3kg. At this time, the density of the product is ρ=m/v=0.82X10~3kg/m~2. So the density is in the range of 0.8 X10~3kg/m~3 to 0.82X10~3kg/m~2. Is a qualified product."