A rod of thermal conductivity K and area A and length L is there- The initial temperature distribution along the rod is given by T - To sin - xL where To is in oC- Suppose both the end of the rod kept at 0oC-What is the initial temperature gradient at the end of rods-

The correct option is A T_o πL T= T _ o sin( π x L #160; ) #160; Temperature gradient, dT dx = T _ o π #160; L cos( π x L #160; ) ...

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Byju's Answer

Standard XII

Physics

Law of Conservation of Mechanical Energy

A rod of ther...

Question

A rod of thermal conductivity $K$ and area $A$ and length $L$ is there. The initial temperature distribution along the rod is given by $T = T_{o} sin (\frac{π x}{L})$ where $T_{o}$ is in $^{o} C$ .
Suppose both the end of the rod kept at $0^{o} C$ .

What is the initial temperature gradient at the end of rods?

\frac{T_{o} π}{L}

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T_{o} π

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\frac{T_{o}}{L}

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\frac{T_{o}}{π L}

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Solution

The correct option is A $\frac{T_{o} π}{L}$
$T = T_{o} sin (\frac{π x}{L})$

Temperature gradient, $\frac{d T}{d x} = T_{o} \frac{π}{L} cos (\frac{π x}{L})$

Because initially length of rod is $0$
${(\frac{d T}{d x})}_{x = 0} = T_{o} \frac{π}{L} cos (\frac{π \cdot 0}{L}) = T_{o} \frac{π}{L}$

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A rod of thermal conductivity K and area A and length L is there. The initial temperature distribution along the rod is given by T=Tosin(πxL) where To is in oC. Suppose both the end of the rod kept at 0oC.What is the initial temperature gradient at the end of rods?

The correct option is A ToπLT=Tosin(πxL)Temperature gradient, dTdx=ToπLcos(πxL)Because initially length of rod is 0(dTdx)x=0=ToπLcos(π⋅0L)=ToπL

The correct option is A $\frac{T_{o} π}{L}$
$T = T_{o} sin (\frac{π x}{L})$

Temperature gradient, $\frac{d T}{d x} = T_{o} \frac{π}{L} cos (\frac{π x}{L})$

Because initially length of rod is $0$
${(\frac{d T}{d x})}_{x = 0} = T_{o} \frac{π}{L} cos (\frac{π \cdot 0}{L}) = T_{o} \frac{π}{L}$