A) 1 J
B) 1.25 J
C) 1.5 J
D) 1.75 J
Correct Answer: B
Solution :
[b] \[{{\operatorname{v}}^{2}}={{u}^{2}}+2gh ={{\left( 10 \right)}^{2}}+2\times 10\times 19.5 =490\] K.E. at the ground \[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times \frac{5}{1000}\times 490=\frac{49}{40}\operatorname{J}\] \[\operatorname{P}.E=mgh=\frac{5}{1000}\times 10\times \left( \frac{-50}{100} \right)=-\frac{1}{40}\operatorname{J}\] \[\therefore \] Change in energy \[=\frac{49}{40}-\left( -\frac{1}{40} \right)\] \[=\frac{50}{40}=1.25\operatorname{J}\]
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