A cylindrical rod having temperature T1 and T2 at its ends. The rate of flow of heat is Q1 cal / sec. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat Q2 will be

Question

A cylindrical rod having temperature T1 and T2 at its ends. The rate of flow of heat is Q1 cal / sec. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat Q2 will be (A) 4 Q1 (B) 2 Q1 (C) (Q1/4) (D) (Q1/2)

Tardigrade Admin · Accepted Answer

Rate of heat flow ((Q/t))=(k π r2(θ1-θ2)/L) propto (r2/L) ∴ (Q1/Q2)=((r1/r2))2((l2/l1)) =((1/2))2 ×((2/1))=(1/2) ⇒ Q2=2 Q1

Q. A cylindrical rod having temperature $T_{1}$ and $T_{2}$ at its ends. The rate of flow of heat is $Q_{1} c a l / sec$ . If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat $Q_{2}$ will be

$4 Q_{1}$

$2 Q_{1}$

$\frac{Q _{1}}{4}$

$\frac{Q _{1}}{2}$

Rate of heat flow $(\frac{Q}{t}) = \frac{kπ r ^{2} ( θ _{1} - θ _{2} )}{L} \propto \frac{r ^{2}}{L}$
$∴ \frac{Q _{1}}{Q _{2}} = (\frac{r _{1}}{r _{2}})^{2} (\frac{l _{2}}{l _{1}})$
$= (\frac{1}{2})^{2} \times (\frac{2}{1}) = \frac{1}{2}$
$\Rightarrow Q_{2} = 2 Q_{1}$

Q. A cylindrical rod having temperature T1​ and T2​ at its ends. The rate of flow of heat is Q1​cal/sec. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat Q2​ will be

4Q1​

2Q1​

4Q1​​

2Q1​​