The temperatures at the two ends A and B of a rod of length 25 cm and circular cross section are 100 -C and 0 -C respectively- In the steady state the temperature at a point 10 cm from the end B is- Ignore loss of heat from curved surface of the rod

The correct option is B 40 ^∘C Since the temperature gradient ( dTdx ) is maintained same at all points in the rod,the difference in temperature acros ...

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Standard XII

Physics

Capacitance of Parallel Plate Capacitor

The temperatu...

Question

The temperatures at the two ends A and B of a rod of length $25 c m$ and circular cross section are $100^{\circ} C$ and $0^{\circ} C$ respectively. In the steady state the temperature at a point $10 c m$ from the end B is:
(Ignore loss of heat from curved surface of the rod)

80^{\circ} C

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90^{\circ} C

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40^{\circ} C

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60^{\circ} C

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Solution

The correct option is B $40^{\circ} C$
Since the temperature gradient ( $\frac{d T}{d x}$ ) is maintained same at all points in the rod,
the difference in temperature across a point $10 c m$ from one end = $\frac{10}{25} \times 100^{\circ} C = 40^{\circ} C$

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The temperatures at the two ends A and B of a rod of length 25 cm and circular cross section are 100∘C and 0∘C respectively. In the steady state the temperature at a point 10 cm from the end B is: (Ignore loss of heat from curved surface of the rod)

The correct option is B 40∘CSince the temperature gradient (dTdx) is maintained same at all points in the rod,the difference in temperature across a point 10cm from one end =1025×100∘C=40∘C

The temperatures at the two ends A and B of a rod of length $25 c m$ and circular cross section are $100^{\circ} C$ and $0^{\circ} C$ respectively. In the steady state the temperature at a point $10 c m$ from the end B is:
(Ignore loss of heat from curved surface of the rod)

The correct option is B $40^{\circ} C$
Since the temperature gradient ( $\frac{d T}{d x}$ ) is maintained same at all points in the rod,
the difference in temperature across a point $10 c m$ from one end = $\frac{10}{25} \times 100^{\circ} C = 40^{\circ} C$