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NCERT Exemplar · Q22

Q.We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their lengths remain constant. If αiron=1.2×10−5\alpha_{iron} = 1.2 \times 10^{-5}/K and αbrass=1.8×10−5\alpha_{brass} = 1.8 \times 10^{-5}/K, what should we take as length of each strip?

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To keep the difference between two strip lengths fixed as temperature changes, the two strips must expand by the same absolute amount - that requires L αL\,\alpha to be equal for both. Since brass expands faster per unit length (αbrass>αiron\alpha_{\text{brass}} > \alpha_{\text{iron}}), the iron strip must be made longer to compensate. Solving gives iron = 30 cm, brass = 20 cm.

The scale's "length" is defined as the difference in length between the iron and brass strips, fixed at 1010 cm. For that difference to stay 1010 cm at every temperature, both strips must lengthen by exactly the same number of centimetres when heated - if one grew more than the other, the gap between them would change, and the "10 cm" mark would drift.

Setting up the condition

When temperature rises by ΔT\Delta T, a strip of length LL and expansion coefficient α\alpha grows by

ΔL=L α ΔT.\Delta L = L\,\alpha\,\Delta T.

For the two strips' changes to be equal (so their difference is unchanged):

Liron αiron=Lbrass αbrass.L_{\text{iron}}\,\alpha_{\text{iron}} = L_{\text{brass}}\,\alpha_{\text{brass}}.

Deciding which strip is longer

Since αbrass=1.8×10−5\alpha_{\text{brass}} = 1.8\times10^{-5}/K is larger than αiron=1.2×10−5\alpha_{\text{iron}} = 1.2\times10^{-5}/K, brass expands more per centimetre of its own length. To make its total expansion match iron's, brass must be the shorter strip, and iron the longer one.

Solving the two equations

Let Lbrass=xL_{\text{brass}} = x. Since iron is the longer strip by 1010 cm: Liron=x+10L_{\text{iron}} = x + 10.

Substitute into the equal-expansion condition:

(x+10)(1.2×10−5)=x(1.8×10−5)(x+10)(1.2\times10^{-5}) = x(1.8\times10^{-5})

1.2x+12=1.8x1.2x + 12 = 1.8x

12=0.6x⇒x=20.12 = 0.6x \quad\Rightarrow\quad x = 20. …

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