Question

A thermometer graduated according to linear scale reads a value $t$ when in contact with boiling water and $\frac{t}{4}$ when kept in contact with ice. When the thermometer is in contact with an object it reads $\frac{t}{3}$. What is the temperature of the object in ${}^{\circ }C?$

Answer 1

Answer: 11.11

(Boiling point of water) - (ice point of water) $=100^{\circ }C$
$\therefore \left(t-\frac{t}{4}\right)=100{}^{\circ }C$
$\therefore \frac{3t}{4}=100$
$\therefore t=\left(\frac{400}{3}\right){}^{\circ }C$ ... (i)
Let, $T$ be the temperature of the object in ${}^{\circ }C$.
$\therefore \left(\frac{t}{3}-\frac{t}{4}\right)=(T-0)$
$\therefore \frac{t}{12}=T$
$\therefore \frac{400}{3\times 12}=T$ .... [From (i)]
$\therefore T=\frac{100}{9}=11.11^{\circ }C$