Thermometer A and B have ice points marked at `15^@` and `25^@` and steam points at `75^@` and `125^@` respectively. When thermometer A measures the temperature of a bath as `60^@`, the reading of B for the same bath is

Options

A. (a)`60^@`
B. (b)`75^@`
C. (c)`100^@`
D. (d)`90^@`

Answer

To solve the problem, we need to find the reading of thermometer B when thermometer A reads 60°C. We will use the concept of the linear relationship between the readings of two thermometers based on their ice and steam points.

### Step-by-Step Solution:

1. **Identify the Ice and Steam Points**:
- For thermometer A:
- Ice point (A_ice) = 15°C
- Steam point (A_steam) = 75°C
- For thermometer B:
- Ice point (B_ice) = 25°C
- Steam point (B_steam) = 125°C

2. **Calculate the Temperature Range for Each Thermometer**:
- For thermometer A:
- Temperature range (A_range) = A_steam - A_ice = 75°C - 15°C = 60°C
- For thermometer B:
- Temperature range (B_range) = B_steam - B_ice = 125°C - 25°C = 100°C

3. **Set Up the Proportion**:
- We know that the readings of both thermometers for the same temperature can be set up in a ratio:
\[ \frac{T_A - A_{ice}}{A_{range}} = \frac{T_B - B_{ice}}{B_{range}} \]
- Here, \(T_A\) is the reading of thermometer A (60°C), and \(T_B\) is the reading of thermometer B that we need to find.

4. **Substitute the Known Values**:
- Substitute \(T_A = 60°C\), \(A_{ice} = 15°C\), \(A_{range} = 60°C\), \(B_{ice} = 25°C\), and \(B_{range} = 100°C\):
\[ \frac{60 - 15}{60} = \frac{T_B - 25}{100} \]

5. **Simplify the Left Side**:
- Calculate the left side:
\[ \frac{45}{60} = \frac{3}{4} \]

6. **Set Up the Equation**:
- Now we have:
\[ \frac{3}{4} = \frac{T_B - 25}{100} \]

7. **Cross Multiply**:
- Cross multiplying gives:
\[ 3 \times 100 = 4 \times (T_B - 25) \]
- This simplifies to:
\[ 300 = 4T_B - 100 \]

8. **Solve for \(T_B\)**:
- Rearranging gives:
\[ 4T_B = 300 + 100 = 400 \]
- Dividing by 4:
\[ T_B = 100°C \]

### Final Answer:
The reading of thermometer B for the same bath is **100°C**.