JEE MAIN - Physics (2020 - 6th September Evening Slot - No. 2)
Three rods of identical cross-section and lengths are made of three different materials of thermal
conductivity K1
, K2
and K3
, respecrtively. They are joined together at their ends to make a long rod
(see figure). One end of the long rod is maintained at 100oC and the other at 0oC (see figure). If
the joints of the rod are at 70oC and 20oC in steady state and there is no loss of energy from the
surface of the rod, the correct relationship between K1
, K2
and K3
is :
_6th_September_Evening_Slot_en_2_2.png)
K1 : K3 = 2 : 3,
K2 : K3 = 2 : 5
K2 : K3 = 2 : 5
K1 < K2 < K3
K1 : K2 = 5 : 2,
K1 : K3 = 3 : 5
K1 : K3 = 3 : 5
K1 > K2 > K3
Explanation
All three rods have same length($$l$$) and area of cross-section(A).
They are in series combination so heat current is same for all rods.
$$ \therefore $$ $${\left( {{{\Delta Q} \over {\Delta t}}} \right)_1} = {\left( {{{\Delta Q} \over {\Delta t}}} \right)_2} = {\left( {{{\Delta Q} \over {\Delta t}}} \right)_3}$$
$$ \Rightarrow $$ $${{\left( {100 - 70} \right){K_1}A} \over l} = {{\left( {70 - 20} \right){K_2}A} \over l} = {{\left( {20 - 0} \right){K_3}A} \over l}$$
$$ \Rightarrow $$ 30K1 = 50K2 = 20K3
$$ \therefore $$ K1 : K3 = 2 : 3,
K2 : K3 = 2 : 5
They are in series combination so heat current is same for all rods.
$$ \therefore $$ $${\left( {{{\Delta Q} \over {\Delta t}}} \right)_1} = {\left( {{{\Delta Q} \over {\Delta t}}} \right)_2} = {\left( {{{\Delta Q} \over {\Delta t}}} \right)_3}$$
$$ \Rightarrow $$ $${{\left( {100 - 70} \right){K_1}A} \over l} = {{\left( {70 - 20} \right){K_2}A} \over l} = {{\left( {20 - 0} \right){K_3}A} \over l}$$
$$ \Rightarrow $$ 30K1 = 50K2 = 20K3
$$ \therefore $$ K1 : K3 = 2 : 3,
K2 : K3 = 2 : 5
Comments (0)
