A thermometer graduated according to a linear scale reads a value $x_{0}$ when in contact with boiling water…
- 25
- 60
- 40
- 35
Solution

$\Rightarrow \mathrm{T}^{\circ} \mathrm{C}=\frac{\mathrm{x}_{0}}{2}-\frac{\mathrm{x}_{0}}{3}=\frac{\mathrm{x}_{0}}{6}$ $\&\left(\mathrm{x}_{0}-\frac{\mathrm{x}_{0}}{3}\right)=\left(100-0^{\circ} \mathrm{C}\right)$ $\Rightarrow \frac{2 \mathrm{x}_{0}}{3}=100 \Rightarrow \mathrm{x}_{0}=\frac{300}{2}$ $\Rightarrow \mathrm{T}^{\circ} \mathrm{C}=\frac{\mathrm{x}_{0}}{6}=\frac{150}{6}=25^{\circ} \mathrm{C}$
Asked in: JEE Main 2019 (11 Jan Shift 2)
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