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Read the following statements :
A. When small temperature difference between a liquid and its surrounding is doubled, the rate of loss of heat of the liquid becomes twice.

B. Two bodies \mathrm{P} and \mathrm{\mathrm{Q}} having equal surface areas are maintained at temperature \mathrm{10^{\circ} \mathrm{C}} and \mathrm{20^{\circ} \mathrm{C}. }The thermal radiation emitted in a given time by \mathrm{P } and \mathrm{ Q } are in the ratio \mathrm{ 1: 1.15. }

C. A Carnot Engine working between \mathrm{ 100 \! \mathrm{~K}} and \mathrm{ 400\! \mathrm{~K}} has an efficiency of \mathrm{ 75 \%.}
D. When small temperature difference between a liquid and its surrounding is quadrupled, the rate of loss of heat of the liquid becomes twice.
Choose the correct answer from the options given below :

Option: 1

\mathrm{\mathrm{A}, \mathrm{B}, \mathrm{C} \text { only }}


Option: 2

\mathrm{\text { A, B only }}


Option: 3

\mathrm{\text { A. C only }}


Option: 4

\mathrm{\text { B, C, D only }}


Answers (1)

best_answer

\mathrm{Rate \: of\: loss \: of\: heat(H) =\frac{K A \Delta T}{\ell}}
(i) If temperature difference is doubled, the rate of loss of heat of the liquid becomes twice
(ii)
\mathrm{\Delta T_1 =T_P-T_0 }

\mathrm{\Delta T_2 =T_Q=T_0 }
\mathrm{T_0 \rightarrow }Temperature of surrounding

\mathrm{\frac{H_1}{H_2} =\frac{1}{1.15}=\frac{\left(10-T_0\right)}{\left(20-T_0\right)} }

\mathrm{20-T_0 =11.5-1.15 T_0 }

\mathrm{0.15 T_0 =-8.5 }

\mathrm{T_0 =-\frac{8.5}{0.15}=\frac{-1.7}{0.03} }

\mathrm{T_0 =-\frac{170\degree}{3} \mathrm{C}}

(iii)

\mathrm{\eta =1-\frac{T_C}{T_H} }

\mathrm{=1-\frac{100}{400} }

\mathrm{\eta =\frac{3}{4} }

\mathrm{\eta \% =75 \%}

(iv)

\mathrm{\Delta T_f=4 \Delta T_i }

\mathrm{H_f=4 H_i}
If the temperature difference is quadrupled, the rate of loss of heat of the liquid becomes four times.The fourth statement is not correct.
The A, B and C are correct statements.

Hence 1 is correct option.




 

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Rishabh

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