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*Simulator*

**Previous Years AIEEE/JEE Mains Questions**

**[AIEEE – 2009]**

**Sol.**(4)

The heat flow rate is given by

$\frac{d Q}{d t}=\frac{k A\left(\theta_{1}-\theta\right)}{x}$

$\Rightarrow \theta=\theta_{1}-\frac{x}{k A} \frac{d Q}{d t}$

where $\theta_{1}$ is the temp of hot end and $\theta$ is temp at a distance x from hot end.

**[AIEEE 2012]**

**Sol.**(2)

Newtons law of cooling

$\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right)$

$\Rightarrow \frac{d \theta}{\left(\theta-\theta_{0}\right)}=-k d t$

Integrating

$\ln \left(\theta-\theta_{0}\right)=-k t+C$

**[JEE-Main- 2013]**

**Sol.**(3)

According to Newtons law of cooling, The temp goes on decreasing with time non-linearly.

(1) 4.8 cal/s (2) 6.0 cal/s (3) 1.2 cal/s (4) 2.4 cal/s

**[JEE-Main-2014]**

**Sol.**(1)

Heat flow per unit time through copper rod

$=\frac{(100-40)}{\ell_{\mathrm{c}}}\left(\mathrm{K}_{\mathrm{c}} \mathrm{A}_{\mathrm{c}}\right)$

$=\frac{60}{46} \times 0.92 \times 4$

$=4.8 \mathrm{cal} / \mathrm{s}$