A) \[\frac{1}{2}\]
B) \[2\]
C) \[\frac{2}{5}\]
D) \[\frac{5}{6}\]
E) \[\frac{1}{5}\]
Correct Answer: B
Solution :
\[{{a}_{1}}=\frac{F}{{{m}_{1}}}=\frac{4\times 10}{20}=2m{{s}^{-2}}\] \[{{a}_{2}}=\frac{F}{{{m}_{2}}}=\frac{4\times 10}{5}=8m{{s}^{-2}}\] Given that, \[{{K}_{A}}={{K}_{B}}\] ie, \[\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{1}{2}{{m}_{2}}v_{2}^{2}\] or \[{{m}_{1}}{{(u+{{a}_{1}}{{t}_{1}})}^{2}}={{m}_{2}}{{(u+{{a}_{2}}{{t}_{2}})}^{2}}\] \[(\because v=u+at)\] or \[{{m}_{1}}a_{1}^{2}t_{1}^{2}={{m}_{2}}a_{2}^{2}t_{2}^{2}\] \[(\because u=0)\] or \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{{{m}_{2}}}{{{m}_{1}}}\times \frac{a_{2}^{2}}{a_{1}^{2}}}\] \[=\sqrt{\frac{5}{20}\times \frac{{{(8)}^{2}}}{{{(2)}^{2}}}}=\sqrt{\frac{5\times 64}{20\times 4}}=2\]
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