In an experment ot verify Newton's law of cooling, a graph is plotted between, the temperature difference $(\Delta T )$ of the water and surroundings and time as shown in figure. The initial temperature of water is taken as $80^{\circ} \,C$. The value of $t _{2}$ as mentioned in the graph will be...........
JEE MAIN 2022, Diffcult
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$T - T _{0}=\left( T _{ i }- T _{0}\right) e ^{-\frac{ Bt }{ ms }}$
$6 \lambda=\ln 1.5$
$40=60 \,e ^{-\lambda(6)} \Rightarrow 6 \lambda=\ln 1.5$
$20=60 \,e ^{-\lambda t _{2}} \Rightarrow t _{2} \lambda=\ln 3$
$\frac{ t _{2}}{6}=\frac{\ln 3}{\ln 1.5}$
$\therefore t _{2}=16.25 min$
So $\approx 16$
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