Two unequal bars in length and cross-section with thermal conductivities K1 and K2 are connected in series so that one end is maintained at - 1 -C and other end at - 2 -C- If the ratio of length of two bars is n1 and ratio of cross-sectional areas of two bars is n2- then find the value of n1-n2 for which temperature of the junction point is - 1- - 22 -

Formula used: I=Δθ /R Given: nbsp; θ _J= θ _1+ θ _22, l_1l_2=n_1, A_1A_2=n_2 2 θ _J= θ _1+ θ _2 θ _1 θ _J= θ _J θ _2 …(i) nbsp; If bars are in serie ...

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Standard X

Physics

Electric Current

Two unequal b...

Question

Two unequal bars in length and cross-section with thermal conductivities $K_{1}$ and $K_{2}$ are connected in series so that one end is maintained at $θ_{1}^{\circ} C$ and other end at $θ_{2}^{\circ} C$ . If the ratio of length of two bars is $n_{1}$ and ratio of cross-sectional areas of two bars is $n_{2}$ , then find the value of $n_{1} / n_{2}$ for which temperature of the junction point is $\frac{θ_{1} + θ_{2}}{2}$ :

\frac{K_{1}}{K_{2}}

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\frac{K_{1} + K_{2}}{K_{2}}

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\frac{K_{1} + K_{2}}{K_{1}}

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\frac{K_{2}}{K_{1} + K_{2}}

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Solution

The correct option is A $\frac{K_{1}}{K_{2}}$
Formula used: $I = Δ θ / R$

Given:
$θ_{J} = \frac{θ_{1} + θ_{2}}{2}$ ,

$\frac{l_{1}}{l_{2}} = n_{1}$ ,

$\frac{A_{1}}{A_{2}} = n_{2}$

$2 θ_{J} = θ_{1} + θ_{2}$

$θ_{1} - θ_{J} = θ_{J} - θ_{2} \dots (i)$

If bars are in series,
Thermal current

$I = \frac{θ_{1} - θ}{R_{1}} = \frac{θ - θ_{2}}{R_{2}} \dots (i i)$

Formula used: $R = l / K A$

From Eqs. $(i)$ and $(i i)$ ,

$R_{1} = R_{2}$

$\frac{l_{1}}{K_{1} A_{1}} = \frac{l_{2}}{K_{2} A_{2}}$

$\frac{l_{1}}{l_{2}} = \frac{K_{1}}{K_{2}} . \frac{A_{1}}{A_{2}}$

$n_{1} = \frac{K_{1}}{K_{2}} . n_{2}$

$\frac{n_{1}}{n_{2}} = \frac{K_{1}}{K_{2}}$

Final answer: (d)

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