Verified Solution • 2025–26 Session moderate QJEEM25Ph4528: JEE Main 2025 Physics - Class 11 JEEM25Ph4528 Solution: The temperature of a body in air falls from to in 4 minutes.… The temperature of a body in air falls from 40∘C40^{\circ} \mathrm{C}40∘C to 24∘C24^{\circ} \mathrm{C}24∘C in 4 minutes. The temperature of the air is 16∘C16^{\circ} \mathrm{C}16∘C . The temperature of the body in the next 4 minutes will be : Multiple Choice Options A 283∘C\frac{28}{3}{ }^{\circ} \mathrm{C}328∘C B 563∘C\frac{56}{3}{ }^{\circ} \mathrm{C}356∘C C 423∘C\frac{42}{3}{ }^{\circ} \mathrm{C}342∘C D 143∘C\frac{14}{3}{ }^{\circ} \mathrm{C}314∘C Step-by-Step Verified Solution amp;T2−T1t=K[ Tavg −Ts]amp;T1=24∘C;T2=40∘C,t=4, Ts=16∘Camp;40−244=K[32−16]amp;K=416=14amp; Now 24−T4=K[ T+242−16]amp;24−T=T−162+16amp;3 T2=28amp; T=563C\begin{aligned}& \frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{t}}=\mathrm{K}\left[\mathrm{~T}_{\text {avg }}-\mathrm{T}_{\mathrm{s}}\right] \\& \mathrm{T}_1=24^{\circ} \mathrm{C} ; \mathrm{T}_2=40^{\circ} \mathrm{C}, \mathrm{t}=4, \mathrm{~T}_{\mathrm{s}}=16^{\circ} \mathrm{C} \\& \frac{40-24}{4}=\mathrm{K}[32-16] \\& \mathrm{K}=\frac{4}{16}=\frac{1}{4} \\& \text { Now } \frac{24-\mathrm{T}}{4}=\mathrm{K}\left[\frac{\mathrm{~T}+24}{2}-16\right] \\& 24-\mathrm{T}=\frac{\mathrm{T}-16}{2}+16 \\& \frac{3 \mathrm{~T}}{2}=28 \\& \mathrm{~T}=\frac{56}{3} \mathrm{C}\end{aligned}amp;tT2−T1=K[ Tavg −Ts]amp;T1=24∘C;T2=40∘C,t=4, Ts=16∘Camp;440−24=K[32−16]amp;K=164=41amp; Now 424−T=K[2 T+24−16]amp;24−T=2T−16+16amp;23 T=28amp; T=356C Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring