IIT-JEE 2011 · Paper 1 · Physics Q33

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PhysicsMultiple correct+4Question 33 of 69

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 width. Heat 'Q' flows only from left to right through the blocks. Then in steady state

A composite block 6L long and 4L tall, heat flowing left to right. Slab A (2K) spans x = 0 to L over the full height 4L; slab E (6K) spans x = 5L to 6L over the full height. Between them (x = L to 5L) three layers in parallel: B (3K) of height L at the top, C (4K) of height 2L in the middle, D (5K) of height L at the bottom.

heat flow through A and E slabs are same.

heat flow through slab E is maximum.

temperature difference across slab E is smallest.

heat flow through C = heat flow through B + heat flow through D.

Answer and worked solution

Official answer: A, C, D

Thermal resistances (width $w$ ): A: $\frac1{8Kw}$ ; E: $\frac1{24Kw}$ ; middle layers in parallel with conductances $\frac{3Kw}4$ (B), $2Kw$ (C), $\frac{5Kw}4$ (D), total $4Kw$ .

(A) A and E are in series with everything: same heat current. ✓
(B) E carries the same current as A, not more. ✗
(C) $\Delta T=QR$ and E has the smallest resistance. ✓
(D) B, C, D share the current in proportion 0.75 : 2 : 1.25, so C carries half = B + D. ✓

(A), (C), (D)

Topics:Conduction

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IIT-JEE 2011 · Paper 1, Physics question 33