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Standard XII

Physics

Latent Heat

Three identic...

Question

Three identical rods of same material are joined to form an equilateral triangle. The temperature of end A and b is maintained constant as $\sqrt{3} T$ and $T$ . The ratio of T_C/T_B will be (Assuming no loss of heat from surfaces)

\frac{1 + \sqrt{3}}{2}

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\frac{1 - \sqrt{3}}{2}

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\frac{1 + \sqrt{2}}{2}

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\frac{1 - \sqrt{2}}{2}

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Solution

The correct option is A $\frac{1 + \sqrt{3}}{2}$

$H =$ rate of heat flow or heat current

In the figure: $H_{1} = H_{2}$

$∴ \frac{\sqrt{3 T} - T_{C}}{R} = \frac{T_{C} - T}{R}$

$= 2 T_{C} = (\sqrt{3} + 1) T$

$∴ \frac{T_{C}}{T} = \frac{T_{C}}{T_{B}} = \frac{\sqrt{3} + 1}{2}$

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Three identical rods of same material are joined to form an equilateral triangle. The temperature of end A and b is maintained constant as √3T and T. The ratio of TC/TB will be (Assuming no loss of heat from surfaces)

The correct option is A 1+√32 H= rate of heat flow or heat current In the figure: H1=H2 ∴√3T−TCR=TC−TR =2TC=(√3+1)T ∴TCT=TCTB=√3+12

Three identical rods of same material are joined to form an equilateral triangle. The temperature of end A and b is maintained constant as $\sqrt{3} T$ and $T$ . The ratio of T_C/T_B will be (Assuming no loss of heat from surfaces)

The correct option is A $\frac{1 + \sqrt{3}}{2}$

$H =$ rate of heat flow or heat current

In the figure: $H_{1} = H_{2}$

$∴ \frac{\sqrt{3 T} - T_{C}}{R} = \frac{T_{C} - T}{R}$

$= 2 T_{C} = (\sqrt{3} + 1) T$

$∴ \frac{T_{C}}{T} = \frac{T_{C}}{T_{B}} = \frac{\sqrt{3} + 1}{2}$