question_answer

The weight of a rider, driving scooter is assumed to be evenly distributed on both the tyres. The area of contact of each type with the ground is \[10\text{ }c{{m}^{2}}\]. If the pressure inside the tyres is 3 bar, find the mass of the rider (\[g\text{ }=\text{ }10\text{ }m{{s}^{-2}}\])           

Answer:

     Area of contact of each tyre\[=10\text{ }c{{m}^{2}}\] Area of contact of two tyres \[(A)=2\times 10\text{ }c{{m}^{2}}\] \[=20\text{ }c{{m}^{2}}\] \[=2\times 10\text{ }c{{m}^{4}}{{m}^{2}}\]          \[(\because 1c{{m}^{2}}={{10}^{4}}{{m}^{2}})\] Pressure (P) = 3 bar \[=3\times {{10}^{5}}\]Pa                                                 \[(\because 1\text{ }bar={{10}^{5}}Pa)\] Weight of the person = pressure × area. \[W=P\times A=3\times {{10}^{5}}\times 20\times {{10}^{4}}=600\text{ }N\] Weight of the person is 600 N. Mass of the person (m) \[=\frac{W}{g}=\frac{600}{10}=60\,kg\] Therefore, the mass of the person is 60 kg.