Two identical square rods of metal are welded end to end as shown in figure (i), 20 calories of heat flows through it in 4 minutes. If the rods are welded as shown in figure (ii), the same amount of heat will flow through the rods in

1 minute

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2 minutes

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4 minutes

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16 minutes

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Solution

The correct option is A
1 minute

$\frac{Q}{t} = \frac{K A △ θ}{l} = \frac{△ θ}{(l / K A)} = \frac{△ θ}{R}$ ( R = Thermal resistance )

$\Rightarrow$ t ∝ R ( $∴ a n d △ θ$ are same )

$\Rightarrow \frac{t_{p}}{t_{s}} = \frac{R_{P}}{R_{S}} = \frac{R / 2}{2 R} = \frac{1}{4} \Rightarrow t_{p} = \frac{t_{s}}{4} = \frac{4}{4} = 1 m i n .$

( Series resistance $R_{S} = R_{1} + R_{2}$ and parallel

resistance R_P = $\frac{R_{1} R_{2}}{R_{1} + R_{2}}$ )

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Two identical square rods of metal are welded end to end as shown in figure (i), 20 calories of heat flows through it in 4 minutes. If the rods are welded as shown in figure (ii), the same amount of heat will flow through the rods in

The correct option is A 1 minute Qt=KA△θl=△θ(l/KA)=△θR ( R = Thermal resistance ) ⇒ t ∝ R ( ∴and△θ are same ) ⇒tpts=RPRS=R/22R=14⇒tp=ts4=44=1min. ( Series resistance RS=R1+R2 and parallel resistance R_P = R1R2R1+R2 )