- Home
- Standard 11
- Physics
Two identical square rods of metal are welded end to end as shown in figure $(a),\, 20$ $calories$ of heat flows through it in $4\,minutes$ . If the rods are welded as shown in figure $(b)$ , the same amount of heat will flow through the rods in ...... $\min$
$1$
$2$
$4$
$16$
Solution
$Q=\frac{K_{e q} A\left(\theta_{1}-\theta_{2}\right) t}{L}$
For fig. $(a)$ $:$
$\mathrm{Keq}=\mathrm{HM}$ of $\mathrm{K}_{1} \& \mathrm{K}_{2}$
$\mathrm{Keq}=\mathrm{K}$
for fig. $(b) ; K_{e q}=\frac{K A+K A}{A+A}=K$
$\left(\mathrm{AM} \text { of } \mathrm{K}_{1} \& \mathrm{K}_{2}\right)$
$Q_{1}=Q_{2}$
$\Rightarrow \frac{\mathrm{K}(\mathrm{A})}{2 \mathrm{L}}\left(\theta_{1}-\theta_{2}\right) \mathrm{t}_{1}=\frac{\mathrm{K}(2 \mathrm{A})}{\mathrm{L}}\left(\theta_{1}-\theta_{2}\right) \mathrm{t}_{2}$
Given $t_{1}=4 \mathrm{min}$
$\mathrm{q}_{1}-\mathrm{q}_{2}=100 \Rightarrow \mathrm{t}_{2}=1 \mathrm{min}$
