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NCERT Exemplar · Q23

Q.We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron (βvbrass=6×10−5\beta_{vbrass} = 6 \times 10^{-5}/K and βviron=3.55×10−5\beta_{viron} = 3.55 \times 10^{-5}/K) to create a volume of 100 cc. How do you think you can achieve this.

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Just as in the bimetallic-strip scale, we make the vessel's usable volume the difference between an iron container's volume and a brass insert's volume: Viron−Vbrass=100V_{\text{iron}} - V_{\text{brass}} = 100 cc. For that difference to stay fixed with temperature, the two volumes must expand by the same absolute amount, VβV\beta. Solving gives iron container ≈244.9\approx244.9 cc, brass insert ≈144.9\approx144.9 cc.

The hint points straight at the previous bimetallic-strip problem: there, a fixed length was created from the difference of two expanding lengths; here, a fixed volume is created the same way, from the difference of two expanding volumes.

Setting up the construction

Build an iron shell (container) of volume VironV_{\text{iron}} and place inside it a solid brass insert of volume VbrassV_{\text{brass}}. The empty (usable) space - the vessel's actual capacity - is

Vvessel=Viron−Vbrass=100 cc (fixed).V_{\text{vessel}} = V_{\text{iron}} - V_{\text{brass}} = 100\ \text{cc (fixed)}.

When temperature rises by ΔT\Delta T, each part expands by ΔV=Vβ ΔT\Delta V = V\beta\,\Delta T. For the difference - the usable volume - to stay unchanged, both parts must expand by the same absolute amount:

Viron βiron=Vbrass βbrass.V_{\text{iron}}\,\beta_{\text{iron}} = V_{\text{brass}}\,\beta_{\text{brass}}.

Deciding which volume is bigger

Since βiron=3.55×10−5\beta_{\text{iron}} = 3.55\times10^{-5}/K is smaller than βbrass=6×10−5\beta_{\text{brass}} = 6\times10^{-5}/K, iron expands less per unit volume - so it must be the larger volume to match brass's total expansion (exactly mirroring the earlier strip problem, where the metal with the smaller coefficient had to be the bigger piece).

Solving

Let Vbrass=xV_{\text{brass}} = x, so Viron=100+xV_{\text{iron}} = 100 + x. Substitute into the equal-expansion condition:

(100+x)(3.55×10−5)=x(6×10−5)(100+x)(3.55\times10^{-5}) = x(6\times10^{-5})

355+3.55x=6x355 + 3.55x = 6x

355=2.45x⇒x≈144.9.355 = 2.45x \quad\Rightarrow\quad x \approx 144.9. …

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