A) \[\frac{2l}{\alpha \sqrt{{{T}_{1}}{{T}_{2}}}}\]
B) \[\alpha l\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]
C) \[\sqrt{{{T}_{1}}+{{T}_{2}}}.\alpha l\]
D) \[\frac{2l}{\alpha (\sqrt{{{T}_{2}}+\sqrt{{{T}_{1}}}})}\]
Correct Answer: D
Solution :
[d] \[<v>=\frac{{{v}_{1}}+{{v}_{2}}}{2}=\frac{\alpha \sqrt{{{T}_{1}}}+\alpha \sqrt{{{T}_{2}}}}{2}\] \[\Rightarrow \]Time taken\[=\frac{2l}{\alpha (\sqrt{{{T}_{1}}}+\sqrt{{{t}_{2}}})}\] Alternate Solution: \[\frac{dx}{dt}=V=\alpha \sqrt{{{T}_{1}}+\left( \frac{{{T}_{2}}-{{T}_{1}}}{l} \right)x}\] \[\int_{x=0}^{x=1}{\frac{dx}{\sqrt{{{T}_{1}}+\left( \frac{{{T}_{2}}-{{T}_{1}}}{l} \right)x}}}=\int_{0}^{t}{\alpha dt}\] On solving we get \[t=\frac{2l}{\alpha (\sqrt{{{T}_{1}}}+\sqrt{{{T}_{2}}})}\]
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