由ρ ＝
| m |
| V |
| m |
| ρ |
则VA＝
| mA |
| ρA |
VB＝
| mB |
| ρB |
根据题意：
| mA |
| mB |
| 1 |
| 2 |
| ρA |
| ρB |
| 5 |
| 4 |
所以
| VA |
| VB |
| mAρB |
| mBρA |
| 1 |
| 2 |
| 4 |
| 5 |
| 2 |
| 5 |
故选B
方法二：取特殊值法
设mA=1kg； mB=2kg；ρA=5kg/m3； ρB=4kg/m3．
选用变形式V＝
| m |
| ρ |
根据题意得出
| VA |
| VB |
| 0.2 m3 |
| 0.5m3 |
| 2 |
| 5 |
故选B
| m |
| V |
| m |
| ρ |
| mA |
| ρA |
| mB |
| ρB |
| mA |
| mB |
| 1 |
| 2 |
| ρA |
| ρB |
| 5 |
| 4 |
| VA |
| VB |
| mAρB |
| mBρA |
| 1 |
| 2 |
| 4 |
| 5 |
| 2 |
| 5 |
| m |
| ρ |
| VA |
| VB |
| 0.2 m3 |
| 0.5m3 |
| 2 |
| 5 |