question_answer

Assume that life has existed on the surface of the Moon, without changing its present acceleration due to gravity i.e., one sixth that on the surface of the Moon. If 5 kg weight of sugar is purchased on the Earth and the Moon, how many cups of tea can be made out of it on the Earth and the Moon respectively?                  Note: From 100 g of sugar 10 cups of tea can be made on the earth \[(\text{g}=\text{1}0\text{m}{{\text{s}}^{-\text{2}}})\]       

Answer:

Case (i) : Given, weight of the sugar on the Earth, \[{{w}_{Earth}}=50N\] We know, \[{{w}_{Earth}}={{m}_{Earth}}\times {{g}_{Earth}}\] \[\Rightarrow 50N={{m}_{Earth}}\times 10\,m{{s}^{-2}}\] \[\Rightarrow {{m}_{Earth}}=\frac{50}{10}=5kg\] And also given on Earth for \[100g=\frac{100}{1000}=0.1kg\] of sugar - 10 cups of tea can be made For 5 kg, how many cups can be made? \[\Rightarrow \frac{0.1kg}{5kg}=\frac{10}{x}\] \[\Rightarrow x\times 0.1=10\times 5\] \[\Rightarrow x=\frac{50}{0.1}=500\,cups\] Therefore, for 5 kg mass of sugar on the Earth we can make 500 cups Case (ii): Given, weight of the sugar on the Moon, \[{{w}_{Moon}}=50N\] We know, \[{{w}_{Moon}}={{m}_{Moon}}\times {{g}_{Moon}}\] \[\Rightarrow 50N={{m}_{Moon}}\times \frac{10}{6}m{{s}^{-2}}\,\,\left( \because {{g}_{Earth}}\frac{{{g}_{Moon}}}{6} \right)\] \[\Rightarrow {{m}_{Moon}}\frac{300}{10}=30kg\] And also given on the Moon for \[100g=\frac{100}{1000a}=0.1\] kg of sugar - 10 cups of tea can be made For 30 kg how many cups can be made? \[\Rightarrow \frac{0.1kg}{30kg}=\frac{10}{x}\] \[\Rightarrow 0.1\times x=30\times 10\Rightarrow \frac{300}{0.1}=3000\,cups\] From the cases, we can observe that on the Moon we can make more number of cups.