    Question

# 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 DThermal 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).  Suggest Corrections  0      Explore more