Heat flows radially outward through a spherical shell of outside radius R1 and inner radius R2.The temperature of inner surface of shell is q1 and that of outer is q2.The radial distance from center of shell where the temperature is just half way between q1 and q2 is:

Qt=K 4πR1R2R2−R1(T1−T2)→(1) Qt=K.4πR1.RR−R1T1−T1+T22→(2) (1)=(2) R1R2R2−R1(T1−T2)=R1.RR−R1T1−T1+T22 R2R2−R1(T1−T2)=RR−R1T1−T22 2R2(R−R1)=R(R2−R1) R=2R1R2R1+R2

Heat flows radially outward through a spherical shell of outside radius R1 and inner radius R2.The temperature of inner surface of shell is q1 and that of outer is q2.The radial distance from center of shell where the temperature is just half way between q1 and q2 is:

Qt=K 4πR1R2R2−R1(T1−T2)→(1) Qt=K.4πR1.RR−R1T1−T1+T22→(2) (1)=(2) R1R2R2−R1(T1−T2)=R1.RR−R1T1−T1+T22 R2R2−R1(T1−T2)=RR−R1T1−T22 2R2(R−R1)=R(R2−R1) R=2R1R2R1+R2

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Heat flows radially outward through a spherical shell of outside radius R₁ and inner radius R₂.The temperature of inner surface of shell is q₁ and that of outer is q₂.The radial distance from center of shell where the temperature is just half way between q₁ and q₂ is:

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$\frac{R_{1} + R_{2}}{2}$

$\frac{R_{1} R_{2}}{R_{1} + R_{2}}$

$\frac{2 R_{1} R_{2}}{R_{1} + R_{2}}$

$R_{1} + \frac{R_{2}}{2}$

answer is C.

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Detailed Solution

$\frac{Q}{t} = K 4 π \frac{R_{1} R_{2}}{R_{2} - R_{1}} (T_{1} - T_{2}) \to (1) \frac{Q}{t} = K . 4 π \frac{R_{1} . R}{R - R_{1}} (T_{1} - \frac{T_{1} + T_{2}}{2}) \to (2) (1) = (2) \frac{R_{1} R_{2}}{R_{2} - R_{1}} (T_{1} - T_{2}) = \frac{R_{1} . R}{R - R_{1}} (T_{1} - \frac{T_{1} + T_{2}}{2}) \frac{R_{2}}{R_{2} - R_{1}} (T_{1} - T_{2}) = \frac{R}{R - R_{1}} (\frac{T_{1} - T_{2}}{2}) 2 R_{2} (R - R_{1}) = R (R_{2} - R_{1}) R = \frac{2 R_{1} R_{2}}{R_{1} + R_{2}}$