A) 200mA
B) 150 m/s
C) 400 m/s
D) 300 m7s
Correct Answer: A
Solution :
[a] Initial, \[\operatorname{K}.E.=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times \frac{20}{1000}\times 600\times 600\] Change in \[K.E.=P.E.\] \[\frac{1}{2}m\left( {{v}^{2}}-v{{'}^{2}} \right)=mgh\] \[\Rightarrow 3600-\frac{1}{2}\times \frac{20}{1000}\times {{v}_{1}}^{2}=4\times 10\times 80\] \[\Rightarrow {{\operatorname{v}}_{1}}=200m/s\]
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