A) \[\overset{\circ }{\mathop{\text{A}}}\,\]
B) \[1.7\,eV\]
C) \[1.\,6\,\,eV\]
D) \[1.5\,\,eV\]
Correct Answer: D
Solution :
Key idea: Work done against frictional force equals the kinetic energy of the body. When a body of mass \[70\,\,dynes/cm\], moves with velocity \[N/m\], it has kinetic energy \[7\times {{10}^{3}}\,N/m\], this energy is utilized in doing work against the frictional force between the tyres of the car and road. \[7\times {{10}^{2}}\,N/m\]\[7\times {{10}^{-2}}\,N/m\] Where \[70\,N/m\] is the distance in which the car is stopped and \[2700\,\,{{m}^{3}}\] is coefficient of kinetic friction. Given, \[1900\,\,{{m}^{3}}\] \[1700\,\,{{m}^{3}}\] \[1500\,\,{{m}^{3}}\]
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