A) 68%
B) 41%
C) 200%
D) 100%
Correct Answer: B
Solution :
Velocity \[v=\sqrt{2gh}\] ...(i) and momentum p = mv ... (ii) From Eqs. (i) and (ii), we have \[p\propto \sqrt{h}\] Here \[\frac{{{p}_{2}}}{{{p}_{1}}}=\sqrt{\frac{{{h}_{2}}}{{{h}_{1}}}}\] \[\therefore \] \[\frac{{{p}_{2}}}{{{p}_{1}}}=\sqrt{\frac{2h}{h}}=\sqrt{2}\] \[{{p}_{2}}=1.414{{p}_{1}}\] % change \[=\frac{{{p}_{2}}-{{p}_{1}}}{{{p}_{1}}}\times 100=41%\]You need to login to perform this action.
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