A thermometer graduated according to a linear scale reads a value $$x_0$$ when in contact with boiling water, and $$x_0/3$$ when in contact with ice. What is the temperature of an object in °C, if this thermometer in the contact with the object reads $$x_0/2$$?

We are told that the thermometer follows a linear (uniform) scale, so its reading $$x$$ and the actual Celsius temperature $$T$$ are related by the straight-line equation

$$x = kT + c$$

where $$k$$ is the scale factor (slope) and $$c$$ is the intercept (the reading when the temperature is 0 °C). We will determine $$k$$ and $$c$$ using the two fixed points, and then find the unknown temperature.

First fixed point (boiling water): the temperature is $$100\,^{\circ}\text{C}$$ and the thermometer reads $$x_0$$. Substituting in the linear equation gives

$$x_0 = k(100) + c \quad\Longrightarrow\quad 100k + c = x_0 \quad -(1)$$

Second fixed point (ice): the temperature is $$0\,^{\circ}\text{C}$$ and the thermometer reads $$\dfrac{x_0}{3}$$. Substituting gives

$$\frac{x_0}{3} = k(0) + c \quad\Longrightarrow\quad c = \frac{x_0}{3} \quad -(2)$$

Now we substitute the value of $$c$$ from (2) into equation (1) to find $$k$$:

$$100k + \frac{x_0}{3} = x_0$$

Subtract $$\dfrac{x_0}{3}$$ from both sides:

$$100k = x_0 - \frac{x_0}{3}$$

Write the right side with a common denominator:

$$x_0 - \frac{x_0}{3} = \frac{3x_0}{3} - \frac{x_0}{3} = \frac{2x_0}{3}$$

So

$$100k = \frac{2x_0}{3}$$

Divide by 100:

$$k = \frac{\frac{2x_0}{3}}{100} = \frac{2x_0}{300} = \frac{x_0}{150}$$

Now the thermometer is placed in contact with the unknown object and reads $$\dfrac{x_0}{2}$$. Let the corresponding true temperature be $$T$$. Substituting $$x = \dfrac{x_0}{2}$$, $$k = \dfrac{x_0}{150}$$ and $$c = \dfrac{x_0}{3}$$ into the linear relation $$x = kT + c$$ gives

$$\frac{x_0}{2} = \left(\frac{x_0}{150}\right)T + \frac{x_0}{3}$$

Divide every term by $$x_0$$ to simplify:

$$\frac{1}{2} = \frac{T}{150} + \frac{1}{3}$$

Subtract $$\dfrac{1}{3}$$ from $$\dfrac{1}{2}$$:

$$\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$$

Hence

$$\frac{1}{6} = \frac{T}{150}$$

Multiply both sides by 150:

$$T = 150 \times \frac{1}{6} = 25$$

So the object’s temperature is $$25\,^{\circ}\text{C}$$.

Hence, the correct answer is Option A.