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Standard XII

Mathematics

Derivative from First Principle

ABCDE is a re...

Question

ABCDE is a regular pentagon of uniform wire. The rate of heat entering at A and leaving at C is equal.
$T_{B}$ and $T_{D}$ are temperature of B and D. Find the temperature $T_{C}$

\frac{3 T_{B} + 2 T_{D}}{5}

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3 T_{D} - 2 T_{B}

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3 T_{D} - 2 T_{B}

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can have any value

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Solution

The correct option is B $3 T_{D} - 2 T_{B}$
$T_{B} = \frac{T_{E} + T_{D}}{2} 2 T_{B} = T_{E} + T_{D} . . . (i)$
$T_{D} = \frac{T_{E} + T_{C}}{2} T_{E} = 2 T_{D} - T_{C} . . . . (i i)$
$2 T_{B} = 2 T_{D} - T_{C} + T_{D} T_{C} = 3 T_{D} - 2 T_{B}$

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ABCDE is a regular pentagon of uniform wire. The rate of heat entering at A and leaving at C is equal. TB and TD are temperature of B and D. Find the temperature TC

The correct option is B 3TD−2TBTB=TE+TD22TB=TE+TD...(i) TD=TE+TC2TE=2TD−TC....(ii) 2TB=2TD−TC+TDTC=3TD−2TB

ABCDE is a regular pentagon of uniform wire. The rate of heat entering at A and leaving at C is equal.
$T_{B}$ and $T_{D}$ are temperature of B and D. Find the temperature $T_{C}$

The correct option is B $3 T_{D} - 2 T_{B}$
$T_{B} = \frac{T_{E} + T_{D}}{2} 2 T_{B} = T_{E} + T_{D} . . . (i)$
$T_{D} = \frac{T_{E} + T_{C}}{2} T_{E} = 2 T_{D} - T_{C} . . . . (i i)$
$2 T_{B} = 2 T_{D} - T_{C} + T_{D} T_{C} = 3 T_{D} - 2 T_{B}$