A) 0.4
B) 40
C) 400
D) 0.04
Correct Answer: B
Solution :
| Mass, \[M=dV\] |
| \[d=\frac{M}{{{L}^{3}}}\] |
| \[d=4\,g/c{{m}^{3}}\] |
| If unit of mass is 100 g and unit of distance is |
| 10 cm. |
| So, density \[=\frac{4\left( \frac{100g}{100} \right)}{{{\left( \frac{10}{10}cm \right)}^{3}}}\] |
| \[=\frac{\left( \frac{4}{100} \right)}{{{\left( \frac{1}{10} \right)}^{3}}}.\frac{(100\,g)}{{{(10\,cm)}^{3}}}\] |
| \[=40\,g\,c{{m}^{-3}}\] |
| Aliter \[{{u}_{2}}={{u}_{1}}\left( \frac{{{M}_{1}}}{{{M}_{2}}} \right){{\left( \frac{{{L}_{1}}}{{{L}_{2}}} \right)}^{-3}}\] |
| \[=4\left( \frac{1}{100} \right){{\left( \frac{1}{10} \right)}^{-3}}\] |
| \[=40\] |
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