
Which of the following cylindrical rods will conduct the most heat, when their ends are maintained at the same steady temperature.
A. \[l = 1m,r = 0.2m\]
B. \[l = 1m,r = 0.1m\]
C. \[l = 10m,r = 0.1m\]
D. \[l = 0.1m,r = 0.3m\]
Answer
162.3k+ views
Hint: In order to solve this problem we need to understand the amount of heat flow in the conductor. It is defined as the transfer of heat down a temperature gradient between two bodies in close physical contact.
Formula Used:
To find the amount of heat the formula is,
\[Q = - KA\dfrac{{\Delta T}}{L}\]
Where, A is cross sectional area, \[\Delta T\] is temperature difference between two ends and l is length of the cylinder.
Complete step by step solution:
Here, l is the length of the cylinder and r is radius. We need to find the amount of heat it would conduct and \[\Delta T\] is the temperature difference between the two ends. The amount of heat flow is,
\[Q = - KA\dfrac{{\Delta T}}{L}\]
Here, the area of cross section, \[A = \pi {r^2}\]
\[Q = - K\pi {r^2}\dfrac{{\Delta T}}{L}\]
Here, K, \[\pi \] and \[\Delta T\] are constants, therefore the above equation can be written as,
\[Q = C\dfrac{{{r^2}}}{L}\]……….. (1)
Now, we calculate for all the options,
For first option, equation (1) becomes,
\[{Q_1} = C\dfrac{{{{\left( {0.2} \right)}^2}}}{1}\]
\[\Rightarrow {Q_1} = 0.04C\]
For second option, equation (1) becomes,
\[{Q_2} = C\dfrac{{{{\left( {0.1} \right)}^2}}}{1}\]
\[\Rightarrow {Q_2} = 0.01C\]
For third option, equation (1) becomes,
\[{Q_3} = C\dfrac{{{{\left( {0.1} \right)}^2}}}{{10}}\]
\[\Rightarrow {Q_3} = 0.001C\]
For fourth option, equation (1) becomes,
\[{Q_4} = C\dfrac{{{{\left( {0.3} \right)}^2}}}{{0.1}}\]
\[\therefore {Q_4} = 0.9C\]
Since, the \[{Q_4}\] has more value compared to others, therefore the \[{Q_4}\] will conduct more heat.
Hence, Option D is the correct answer.
Note: The rate of conductive heat transfer depends on temperature gradient between the two bodies, the area of contact and the length of the conductor. So, here in this problem it is important to know which has more value of heat flow among the four options, then which cylindrical rod conducts more heat.
Formula Used:
To find the amount of heat the formula is,
\[Q = - KA\dfrac{{\Delta T}}{L}\]
Where, A is cross sectional area, \[\Delta T\] is temperature difference between two ends and l is length of the cylinder.
Complete step by step solution:
Here, l is the length of the cylinder and r is radius. We need to find the amount of heat it would conduct and \[\Delta T\] is the temperature difference between the two ends. The amount of heat flow is,
\[Q = - KA\dfrac{{\Delta T}}{L}\]
Here, the area of cross section, \[A = \pi {r^2}\]
\[Q = - K\pi {r^2}\dfrac{{\Delta T}}{L}\]
Here, K, \[\pi \] and \[\Delta T\] are constants, therefore the above equation can be written as,
\[Q = C\dfrac{{{r^2}}}{L}\]……….. (1)
Now, we calculate for all the options,
For first option, equation (1) becomes,
\[{Q_1} = C\dfrac{{{{\left( {0.2} \right)}^2}}}{1}\]
\[\Rightarrow {Q_1} = 0.04C\]
For second option, equation (1) becomes,
\[{Q_2} = C\dfrac{{{{\left( {0.1} \right)}^2}}}{1}\]
\[\Rightarrow {Q_2} = 0.01C\]
For third option, equation (1) becomes,
\[{Q_3} = C\dfrac{{{{\left( {0.1} \right)}^2}}}{{10}}\]
\[\Rightarrow {Q_3} = 0.001C\]
For fourth option, equation (1) becomes,
\[{Q_4} = C\dfrac{{{{\left( {0.3} \right)}^2}}}{{0.1}}\]
\[\therefore {Q_4} = 0.9C\]
Since, the \[{Q_4}\] has more value compared to others, therefore the \[{Q_4}\] will conduct more heat.
Hence, Option D is the correct answer.
Note: The rate of conductive heat transfer depends on temperature gradient between the two bodies, the area of contact and the length of the conductor. So, here in this problem it is important to know which has more value of heat flow among the four options, then which cylindrical rod conducts more heat.
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