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This is one of our many quizzes on Conduction and induction. Click on this to access more such competitive quizzes on Thermal Physics.

Conduction and induction Quiz 2

physics

 A wall whose dimensions are $1m\times 2m$ has thickness $3m$, thermal conductivity of $1.4$ W/mK and temperature of the two surfaces are $1200K$ and $1000K$, respectively. What will be the rate of heat transfer through this wall? 

 Heat transfer is given by $Q=\dfrac{KA\Delta T}{l}$.By substituting these values in the above equation we can find the unknown. 

  Consider a wall of dimensions,$1m\times 2m\times 3m$, as shown in the figure below. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4880b97c-3ddd-49fc-97b5-0c0b42533e923762827139057871208.png"/> $ \\ $ Heat transfer is given by, $ \\ Q=\dfrac{KA\Delta T}{l}   ….(i) \\ $ Putting the values given in the question, we get, $ \\ Q=\dfrac{1.4\times 1\times 2\times (1200-1000)}{3} \\ \Rightarrow Q =186.67 \\ \therefore Q =187$ W

 How do you permanently charge a body by induction? 

 Due to electrostatic induction the charges are separated like charges repel like charges. 

  Electrostatic induction takes place, the front end of the uncharged sphere becomes negatively charged as the positive charges repel each other. However, when the charged sphere A is removed the bound charge mix with the free charges and sphere becomes neutral, so to permanently charge it we ground the sphere B, as shown in the figure below, $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/705aaa3b-5df1-49df-a6cf-2242246da9102665468972186967869.png"/> $ \\ $ Electrons from ground neutralize the positive charges and thus the net charge of sphere B becomes negative and it becomes permanently charged even after sphere A is removed.

 The maximum value of charge induced in a body by a charged body with net charge ${{Q}_{0}}$ is given by, 

 The induced charge can never be greater than the inducing charge. 

 A body is charged by induction in two steps, first the charge is induced by a charged body then to retain the charge the body is grounded, so as to it remains charged even when the inducing charge is removed. The induced charge can be less than the inducing charge or equal to the inducing charge but never will it be greater than the net charge of the inducing body $($conservation of charge$)$.

 Two cylindrical rods A and B, the length of A is twice the length of B and the cross section area of A is thrice of the cross section area of B. Then, the rate of heat conduction for a given difference in temperature is same for both rods, then the thermal conductivity of rod A and rod B are related as 

 Thermal conductivity is given by $K=\dfrac{Ql}{A\Delta T}$.By substituting these values in the above equation we can find the unknown.	 

 Thermal conductivity of a substance is its capacity to conduct heat and is given by, $ \\ K=\dfrac{Ql}{A\Delta T} ….(i) \\ $ We see from equation $(i)$ that thermal conductivity $K$ is directly proportional to length $l$ and inversely proportional to area, so we can write, $ \\ \dfrac{{{K}_{A}}}{{{K}_{B}}}=\dfrac{{{l}_{A}}}{{{l}_{B}}} \times \dfrac{{{A}_{B}}}{{{A}_{A}}} \\ \Rightarrow\dfrac{{{K}_{A}}}{{{K}_{B}}}=\dfrac{r_{B}^{2}}{r_{A}^{2}}  ….(ii) \\$ And, $\dfrac{{{l}_{A}}}{{{l}_{B}}}=\dfrac{2}{1}   ….(iii)$    $[$Since, length of A is twice of the length of B$]$ $ \\ \Rightarrow\dfrac{{{A}_{B}}}{{{A}_{A}}}=\dfrac{1}{{3}}\\ \Rightarrow\dfrac{{{A}_{B}}}{{{A}_{A}}}=\dfrac{1}{3}  ….(iv)$     $[$Since, cross section area of A is thrice of the length of B$]$ $ \\ $ Now, comparing equation $(iii)$ and $(iv)$, we get, $ \\ \dfrac{{{K}_{A}}}{{{K}_{B}}}=\dfrac{2\times 1}{3} \\ \Rightarrow {{K}_{A}}=\dfrac{2}{3}{{K}_{B}} $

 Two wires of same material and equal dimensions of length, having the radius equal to ${{r}_{1}}=3$ m and ${{r}_{2}}=4$ m. If the temperature difference is the same, then state which wire has a greater rate of conduction. 

 The rate of conduction is $\dfrac{Q}{t}=\dfrac{KA\Delta \theta }{l}$.By substituting these values in the above equation of we can find the rate of conduction of the two wires.

Thermal conductivity is given by,  $ \\ K=\dfrac{Ql}{A\Delta T} \\ \Rightarrow K=\dfrac{Ql}{At\Delta \theta } \\ $  Therefore, rate of conduction is given by, $ \\ \dfrac{Q}{t}=\dfrac{KA\Delta \theta }{l}  ….(i) \\ $  For first wire rate of conduction is, $ \\ \dfrac{{{Q}_{1}}}{t}=\dfrac{K{{A}_{1}}\Delta \theta }{l} \\ \Rightarrow\dfrac{{{Q}_{1}}}{t}=\dfrac{K\pi {{r}_{1}}^{2}\Delta \theta }{l} \\ \Rightarrow\dfrac{{{Q}_{1}}}{t}=\dfrac{9K\Delta \theta }{l}  ….(ii) \\ $ And, for second wire rate of conduction is, $ \\ \dfrac{{{Q}_{2}}}{t}=\dfrac{K{{A}_{2}}\Delta \theta }{l} \\ \Rightarrow\dfrac{{{Q}_{2}}}{t}=\dfrac{K\pi {{r}^{2}}_{2}\Delta \theta }{l} \\ \therefore\dfrac{{{Q}_{2}}}{t}=\dfrac{16K\Delta \theta }{l}  ….(iii) \\ $ Comparing equations $(ii)$ and $(iii)$, the rate of conduction of wire with greater radius is greater.  