A) 22.5m
B) 15m
C) 10m
D) 5m
Correct Answer: B
Solution :
In order to climb a height h the boy utilizes potential energy\[=mgh\] In order to climb he will use the efficient energy. Also \[1\,J={{10}^{3}}kJ\] Energy of one bread\[=21\,kJ=21\times {{10}^{3}}J\] Efficiency of boy = 28% Hence, energy consumed by boy is \[=\frac{28}{100}\times 21000\] \[=5880J\] ...(i) From law of conservation of energy, this energy is utilized in giving potential energy \[mgh\]where g is acceleration due to gravity. \[\therefore \] \[mgh=40\times 9.8\times h\] ...(ii) Equating Eqs. (i) and (ii), we have \[40\times 9.8\times h=5880\] \[\Rightarrow \] \[h=\frac{5880}{40\times 9.8}=15\,m\]You need to login to perform this action.
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