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Previous Years AIEEE/JEE Mains Questions
Q. A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $\theta$ along the length x of the bar from its hot end is best described by which of the following figures ?
[AIEEE - 2009]
[AIEEE - 2009]
Ans. (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.
Q. A liquid in a beaker has temperature $\theta(\mathrm{t})$ at time t and $\theta_{0}$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between $\log _{\mathrm{e}}\left(\theta-\theta_{0}\right)$ and t is :-
[AIEEE 2012]
[AIEEE 2012]
Ans. (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
$\ln \left(\theta-\theta_{0}\right)=-k t+C$
Q. If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature $\theta_{0}$ the graph between the temperature T of the metal and time t will be closed to:
[JEE-Main- 2013]
[JEE-Main- 2013]
Ans. (3)
According to Newtons law of cooling, The temp goes on decreasing with time non-linearly.
Q. Three rods of Copper, Brass and Steel are welded together to form a Y-shaped structure. Area of cross-section of each rod = $4 \mathrm{cm}^{2}$. End of copper rod is maintained at $100^{\circ} \mathrm{C}$ where as ends of brass and steel are kept at $0^{\circ} \mathrm{C}$. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is :
(1) 4.8 cal/s (2) 6.0 cal/s (3) 1.2 cal/s (4) 2.4 cal/s
[JEE-Main-2014]
Ans. (1)
Heat flow per unit time through copper rod
$=\frac{(100-40)}{\ell_{\mathrm{c}}}\left(\mathrm{K}_{\mathrm{c}} \mathrm{A}_{\mathrm{c}}\right)$
$=\frac{60}{46} \times 0.92 \times 4$
$=4.8 \mathrm{cal} / \mathrm{s}$
Heat flow per unit time through copper rod
$=\frac{(100-40)}{\ell_{\mathrm{c}}}\left(\mathrm{K}_{\mathrm{c}} \mathrm{A}_{\mathrm{c}}\right)$
$=\frac{60}{46} \times 0.92 \times 4$
$=4.8 \mathrm{cal} / \mathrm{s}$
Comments
SA
Aug. 8, 2020, 6:59 p.m.
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