Question

As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm² for each plate. The equivalent thermal conductivity of the compound plate is `(1+5/alpha)`K, then the value of a will be ______.

Options

• 22

• 21

• 23

• 24

Answer 1

As per the given figure, two plates A and B of thermal conductivity K and 2 K are joined together to form a compound plate. The thickness of plates are 4.0 cm and 2.5 cm respectively and the area of cross-section is 120 cm² for each plate. The equivalent thermal conductivity of the compound plate is `(1+5/alpha)`K, then the value of a will be 21.

Explanation:

Since, `"Q"/"t" = ("KA"Delta"T")/"d"`

Here, all alphabets are is their usual meanings

where `"Q"/"t"` = Rate of heat transfer

 ⇒ `"Q"_1/"t" = "Q"_2/"t"`

 ⇒ `(100-"T"_i)/(4//"KA") = ("T"_i -0)/(2.5//2"KA")`

 ⇒ `(100-"T"_i)/4 = (2"T"_i)/2.5`

 ⇒ 250 - 2.5 T_i = 8 T_i 

 T_i = `250/10.5`

As the two plates are in series, same thermal current will flow in composite rod as well DT for composite plate = 100°C

⇒ `(100-0)/(6.5/("K"_(eq)" A")) = (250/10.5-0)/(2.5/(2  "KA"))`

⇒ `("K"_(eq)xx100)/6.5 = (250xx2"K")/(10.5xx2.5)`

`"K"_(eq) = 130/105"K"  26/21"K" = (1+5/21)"K"`

on comparing with `"K"_(eq) = (1+5/21)"K"`, we get

α = 21.