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A hot placed in air cools down to a lower temperature. The rate of decrease of temperature isproportionalto the temperaturedifference from the surrounding. The body loses 60% and 80% of maximum heat it can loose in time t_(1) and t_(2) respectively. The ratio t_(2)//t_(1) will be |
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Answer» `(ln(10))/(ln(2))` `-(dT)/(dt) alpha DeltaT ` or `(dT)/(dt)=-K DeltaT` Solved the above Eq., we get `T(t)=T_(0)+(T_(1)-T_(0))E^(kt) rArr e^(-kt)=(T(t)-T_(0))/(T_(1)-T_(0))` or `t=l/k ln (T_(1)-T_(0))/(T(t)-T_(0))` Where, `T_(1)=` temperature of BODY at `t=0` `T_(0)=` temperature of SURROUNDING. Let `T_(1)=0^(@)C` So, `t=1/k ln (-T_(0))/(T(t)-T_(0))= 1/k ln (-Q_(0))/(Q(t)-Q_(0))` [ `:.Q(t)=mcT(t)`] According to the question, the body loses 60% and 80% of maximum heat in time `t_(1)` and `t_(2)` then `t_(1)=1/k ln ((-Q_(0))/((60)/(100)Q_(0)-Q_(0)))` ...(i) `t_(2)=1/k ln ((-Q_(0))/((80)/(100)Q_(0)-Q_(0)))` ...(ii) DIVIDE Eqs. (ii) by (i), we get `t_(2)/t_(1)=(1/k ln ((-Q_(0))/((80)/(100)Q_(0)-Q_(0))))/(1/k ln ((-Q_(0))/(100-Q_(0))))` `rArr (t_(2))/(t_(1))=(ln((-Q_(0))/((80Q_(0)-100Q_(0))/(100))))/(ln((-Q_(0))/((60Q_(0)-100Q_(0))/100)))=(ln((-Q_(0)xx100)/(-20Q_(0))))/(ln((Q_(0)xx5)/(2Q_(0))))` `=(ln(5))/(ln(5/2))` Hence, the correct OPTION is (d). |
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