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A composite block is made of slabs A,B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length. L) as shown in the figure. All slabs are of same with. Heat Q flows only from left to right through the blocks. Then, in steady state

A
heat flown trough A and E slabs are same
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B
heat flow through slab e is maximum
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C
temperature difference across slab E is smallest
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D
heat flow through C = heat flow through B + heat flow through D
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Solution

The correct option is D heat flow through C = heat flow through B + heat flow through D
Thermal resistance R=R=1KA
RA=L(2K)(4LW) (Here w = width)
=18Kw
RB=4L3k(LW)=43Kw
RC=4L(4K)(Lw)=12Kw
RD=4L(5K)(Lw)=45Kw
RE=L(6K)(Lw)=16Kw
RA:RB:RC:RD:RE
=15:160:60:96:12
So, let us write, RA=15 R, RB=160 R etc and draq a simple electrical circuit as shown in figure

H = Heat current = Rate of heat flow.
HA=HE=H (let)
In parallel current distributes in inverse ratio of resistance.
HB:HC:HD=1RB:1RC:1RD
=1160:160:196
=9:24:15
HB=(99+24+15)H=316HHC=(249+24+15)H=12Hand HD=(159+24+15)H=516H
Temperature difference (let us call it T)
= (Heat current) × ( Thermal resistance)
TA=HARA=(H)(15R)=15HR
TB=HBRB=(316H)(160 R)=30 HR
TC=HCRC=(12H)(60 R)=30 HR
TD=HDRD=(516H)(96 R)=30 HR
TE=HERE=(H)(12 R)=12 HR
Here, TE is minimum. Therefore option (c) is also correct.
Correct options are (a), (c), and (d).

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