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JAMB Physics 2021 Paper

Question : 31

Total: 40

Solution:

Concept:

Temperature gradient is the rate at which temperature changes per unit length along the rod.

Explanation:

Temperature difference:

\Delta \theta = 80^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C} = 60^{\circ}\mathrm{C}

A change of

1^{\circ}\mathrm{C}

is equal to a change of

1\ \mathrm{K}

, so

\Delta \theta = 60\ \mathrm{K}

Thickness of the rod:

\Delta x = 0.02\ \mathrm{m}

Temperature gradient is given by

\dfrac{\Delta \theta}{\Delta x}

Substitute values:

\dfrac{60\ \mathrm{K}}{0.02\ \mathrm{m}} = 3000\ \mathrm{K\,m^{-1}}

So the temperature gradient is

3.0 \times 10^{3}\ \mathrm{K\,m^{-1}}

, which corresponds to option B.

Answer:

3.0 \times 10^{3}\ \mathrm{K\,m^{-1}}

(Option B).

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