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Question Number 36855 by Tinkutara last updated on 06/Jun/18

Commented by Tinkutara last updated on 06/Jun/18

Answered by ajfour last updated on 11/Jun/18

$$\frac{{\tau }_A}{{\tau }_B}=\frac{6}{\sqrt{11}}.$$

Commented by Tinkutara last updated on 11/Jun/18

How?

Commented by ajfour last updated on 11/Jun/18

Commented by ajfour last updated on 11/Jun/18

$$\frac{{\tau }_A}{{\tau }_B}=\frac{\frac{D}{v+u}+\frac{D}{v-u}}{\frac{2D}{\sqrt{v^2-u^2}}}$$$$withv=1.2u=\frac{6u}{5}$$$$v+u=\frac{11u}{5};v-u=\frac{u}{5};$$$$\sqrt{v^2-u^2}=\frac{\sqrt{11}}{5}$$$$\frac{{\tau }_A}{{\tau }_B}=\frac{\frac{1}{11}+1}{\frac{2}{\sqrt{11}}}=\frac{6}{\sqrt{11}}.$$

Commented by Tinkutara last updated on 14/Jun/18

Thanks Sir!

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