The ends of copper rod of length 1m and area of cross-section 1 cm2 are maintained at 00C and 1000C and steady state achieved, at the centre of rod, there is a source of heat of 25 watt. The thermal conductivity of copper is 400 wattmK. Then,

400 × 10 − 4 ( 100 − 0 ) l 2 + 25 = 400 × 10 − 4 ( θ ) l 2 θ = 206 0 C in left half = 206 − 100 l 2 = 212 0 C , in right half = 206 − 0 l 2 = 412 0 Cm − 1

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The ends of copper rod of length 1m and area of cross-section 1 cm² are maintained at 0⁰C and 100⁰C and steady state achieved, at the centre of rod, there is a source of heat of 25 watt. The thermal conductivity of copper is 400 $\frac{watt}{mK}$ . Then,

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The temperature of rod is at centre is nearly 50°C

The temperature gradient in left half is nearly 212°C/m

The temperature gradients in right half is nearly 412°C/m

The temperature of rod is at centre is nearly 206°C

answer is A, B, C.

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Detailed Solution

$\frac{400 \times 10^{- 4} (100 - 0)}{\frac{l}{2}} + 25 = \frac{400 \times 10^{- 4} (θ)}{\frac{l}{2}} θ = 206^{0} C in left half = \frac{206 - 100}{\frac{l}{2}} = 212^{0} C, in right half = \frac{206 - 0}{\frac{l}{2}} = 412^{0} {Cm}^{- 1}$

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