Two plates of the same area and the same thickness having thermal conductivities $k_{1}$ and $k_{2}$ are placed one on top of the other. The top and bottom faces of the composite plate are maintained at different constant temperatures. The thermal conductivity of the composite plate will be

$(k_{1} + k_{2})$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{1}{2} (k_{1} + k_{2})$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{k_{1} k_{2}}{(k_{1} + k_{2})}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{2 k_{1} k_{2}}{(k_{1} + k_{2})}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$\frac{2 k_{1} k_{2}}{(k_{1} + k_{2})}$

Let d be the thickness and A the area of each plate. Let $T_{1}$ be the temperature of the top face of the upper plate I and $T_{2}$ that of the bottom face of the lower plate II. Let T be the temperature of the surface in contact.

In the steady state, the rate of flow of heat through plate I is the same as that through plate II, i.e.,
$\frac{Q}{t} = \frac{k_{1} A (T_{1} - T)}{d} = \frac{k_{2} A (T - T_{2})}{d}$
$\Rightarrow \frac{k_{1}}{k_{2}} = \frac{T - T_{2}}{T_{1} - T}$
$\Rightarrow \frac{k_{1}}{k_{2}} + 1 = \frac{T - T_{2}}{T_{1} - T} + 1$
$\Rightarrow \frac{k_{1} + k_{2}}{k_{2}} = \frac{T_{1} - T_{2}}{T_{1} - T}$ .....(1)
Now, the thickness of the composite plate is 2d. If k is the thermal conductivity of the composite plate, then the rate of flow of heat through it is
$\frac{Q}{t} = \frac{k A (T_{1} - T_{2})}{2 d}$
In the steady state, this must be the same as the rate of flow of heat through either plate I or plate II.
$∴ \frac{k A (T_{1} - T_{2})}{2 d} = \frac{k_{1} A (T_{1} - T)}{d}$
$\Rightarrow \frac{T_{1} - T_{2}}{T_{1} - T} = \frac{2 k_{1}}{k}$ .....(2)
Using (2) in (1), we get $k = \frac{2 k_{1} k_{2}}{(k_{1} + k_{2})}$ , which is choice (d).

Suggest Corrections

Similar questions

Two plates of the same area and the same thickness having thermal conductivities $k_{1}$ and $k_{2}$ are placed one on top of the other. The top and bottom faces of the composite plate are maintained at different constant temperatures. The thermal conductivity of the composite plate will be

Q. A composite slab have two layers of different materials, having thermal conductivity

k_{1}

and

k_{2}

. If each layer has same thickness, then what is the equivalent thermal conductivity of the slab?

Two rods A and B of different materials are welded together as shown in Figure. If their thermal conductivities are $k_{1}$ and $k_{2}$ , the thermal conductivity of the composite rod will be

Two slabs A and B of different materials but of the same thickness are joined as shown in figure. The thermal conductivities of A and B are $k_{1} a n d k_{2}$ respectively. The thermal conductivity of the composite slab will be

Two rods A and B of different materials are welded together as shown in the figure. If their thermal conductivities are $k_{1}$ and $k_{2}$ , the thermal conductivity of the composite rod will be

Join BYJU'S Learning Program

Two plates of the same area and the same thickness having thermal conductivities k1 and k2 are placed one on top of the other. The top and bottom faces of the composite plate are maintained at different constant temperatures. The thermal conductivity of the composite plate will be

Two plates of the same area and the same thickness having thermal conductivities $k_{1}$ and $k_{2}$ are placed one on top of the other. The top and bottom faces of the composite plate are maintained at different constant temperatures. The thermal conductivity of the composite plate will be