A thermometer graduated according to linear scale reads a value `x_(0)` when in contact with boiling water, and `X_(0)s//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`?

Options

A. (a)35
B. (b)40
C. (c)60
D. (d)25

Explanation

To solve the problem, we need to determine the temperature of an object based on the readings of a thermometer that has been calibrated with specific values for boiling water and ice. Here’s a step-by-step solution:

### Step 1: Understand the given values
- The thermometer reads \( x_0 \) when in contact with boiling water (100°C).
- The thermometer reads \( \frac{x_0}{3} \) when in contact with ice (0°C).
- The thermometer reads \( \frac{x_0}{2} \) when in contact with the object whose temperature we want to find.

### Step 2: Set up the linear relationship
From the information given, we can establish a linear relationship between the thermometer readings and the corresponding temperatures.

Let:
- \( T_1 = 100°C \) (temperature for boiling water)
- \( T_2 = 0°C \) (temperature for ice)
- \( R_1 = x_0 \) (reading for boiling water)
- \( R_2 = \frac{x_0}{3} \) (reading for ice)

### Step 3: Calculate the temperature difference
The temperature difference corresponding to the readings is:
\[ T_1 - T_2 = R_1 - R_2 \]
Substituting the values:
\[ 100 - 0 = x_0 - \frac{x_0}{3} \]
This simplifies to:
\[ 100 = x_0 - \frac{x_0}{3} \]
\[ 100 = \frac{3x_0 - x_0}{3} \]
\[ 100 = \frac{2x_0}{3} \]

### Step 4: Solve for \( x_0 \)
To find \( x_0 \), we multiply both sides by 3:
\[ 300 = 2x_0 \]
Now, divide by 2:
\[ x_0 = 150 \]

### Step 5: Find the temperature corresponding to \( \frac{x_0}{2} \)
Now we need to find the temperature \( T \) when the thermometer reads \( \frac{x_0}{2} \):
\[ T - 0 = \frac{x_0}{2} - \frac{x_0}{3} \]
This simplifies to:
\[ T = \frac{x_0}{2} - \frac{x_0}{3} \]
Finding a common denominator (which is 6):
\[ T = \frac{3x_0}{6} - \frac{2x_0}{6} \]
\[ T = \frac{x_0}{6} \]

### Step 6: Substitute \( x_0 \) to find \( T \)
Now substitute \( x_0 = 150 \):
\[ T = \frac{150}{6} = 25 \]

### Final Answer
The temperature of the object is \( 25°C \).

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