The heat flows through a rod of length 50 cm and area of cross-section 5 ~cm^2 . Its ends are respectively at 25^° C and 125^° C . The coefficient of thermal conductivity of the material rod is 0.092 kcal / m^° C . The temperature gradient of the rod is

Correct answer: (A) 2^° C / cm. Temperature gradient of the rod (Δ θ)/(Δ x)=(θ_2-θ_1)/l ∴ (Δ θ)/(Δ x)=(125-25)/50 =100/50=2^° C / cm

Q165 / 200 · Manipal (MET) 2011 · Physics

The heat flows through a rod of length 50 cm and area of cross-section $5 \mathrm{~cm}^2$. Its ends are respectively at $25^{\circ} \mathrm{C}$ and $125^{\circ} \mathrm{C}$. The coefficient of thermal conductivity of the material rod is $0.092 \mathrm{kcal} / \mathrm{m}^{\circ} \mathrm{C}$. The temperature gradient of the rod is

Correct answer
A
$2^{\circ} \mathrm{C} / \mathrm{cm}$
B
$2^{\circ} \mathrm{C} / \mathrm{m}$
C
$20^{\circ} \mathrm{C} / \mathrm{cm}$
D
$20^{\circ} \mathrm{C} / \mathrm{m}$

Solution

Temperature gradient of the rod $\frac{\Delta \theta}{\Delta x}=\frac{\theta_2-\theta_1}{l}$ $\therefore \quad \frac{\Delta \theta}{\Delta x}=\frac{125-25}{50}$ $=\frac{100}{50}=2^{\circ} \mathrm{C} / \mathrm{cm}$

Hence, the correct answer is option: 'A'.

Solution

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