A) \[1:9\]
B) \[3:1\]
C) \[3:2\]
D) \[4:1\]
Correct Answer: A
Solution :
The maximum height reached by a body \[h=\frac{{{u}^{2}}}{2g}\Rightarrow {{h}_{1}}=\frac{{{u}^{2}}}{2g}\] \[\Rightarrow \] \[{{h}_{1}}=\frac{u_{1}^{2}}{2g}\Rightarrow {{h}_{1}}=\frac{{{u}^{2}}}{2g}\] \[\Rightarrow \] \[{{h}_{2}}=\frac{u_{2}^{2}}{2g}\Rightarrow {{h}_{2}}=\frac{{{(3u)}^{2}}}{2g}=\frac{9{{u}^{2}}}{2g}\] \[\therefore \]\[P.E.={{m}_{1}}{{g}_{1}}h\]and\[P.E{{.}_{2}}=\frac{9{{u}^{2}}}{2g}\] \[\Rightarrow \] \[m\times g\times \frac{{{u}^{2}}}{2g}=\frac{m{{u}^{2}}}{2}\] ? (1) \[m\times g\times \frac{9{{u}^{2}}}{2g}=\frac{9m{{u}^{2}}}{2}\] ? (2) Dividing (1) by (2) we get, \[\frac{P.{{E}_{1}}}{P.{{E}_{2}}}=\frac{\frac{m{{u}^{2}}}{2}}{\frac{9m{{u}^{2}}}{2}}=\frac{1}{9}\]You need to login to perform this action.
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