The work done against electric in increasing the potential difference of a condenser from 20 V to 40 V is W. The work done in increasing its potential difference from 40 V to 50 V will be

Options

A. 4W
B. `(3W)/(4)`
C. 2W
D. `(W)/(2)`

Answer

To solve the problem, we need to find the work done in increasing the potential difference of a condenser from 40 V to 50 V, given that the work done in increasing the potential from 20 V to 40 V is W.

### Step-by-Step Solution:

1. **Understanding Work Done in a Capacitor:**
The work done (W) in charging a capacitor (or condenser) when the potential difference changes from \( V_1 \) to \( V_2 \) can be expressed using the formula:
\[ W = \frac{1}{2} C (V_2^2 - V_1^2) \]
where \( C \) is the capacitance of the condenser.

2. **Calculating Work Done from 20 V to 40 V:**
Given that the work done in increasing the potential from 20 V to 40 V is \( W \):
\[ W = \frac{1}{2} C (40^2 - 20^2) \]
Simplifying this:
\[ W = \frac{1}{2} C (1600 - 400) = \frac{1}{2} C \cdot 1200 = 600C \]
Thus, we have:
\[ W = 600C \]

3. **Finding Capacitance (C):**
From the equation \( W = 600C \), we can express capacitance \( C \) as:
\[ C = \frac{W}{600} \]

4. **Calculating Work Done from 40 V to 50 V:**
Now, we need to find the work done \( W' \) in increasing the potential from 40 V to 50 V:
\[ W' = \frac{1}{2} C (50^2 - 40^2) \]
Simplifying this:
\[ W' = \frac{1}{2} C (2500 - 1600) = \frac{1}{2} C \cdot 900 = 450C \]

5. **Substituting the Value of C:**
Now, we substitute the value of \( C \) from step 3 into the equation for \( W' \):
\[ W' = 450 \left(\frac{W}{600}\right) = \frac{450W}{600} \]
Simplifying this gives:
\[ W' = \frac{3W}{4} \]

6. **Final Result:**
Therefore, the work done in increasing the potential difference from 40 V to 50 V is:
\[ W' = \frac{3W}{4} \]

### Answer:
The work done in increasing the potential difference from 40 V to 50 V is \( \frac{3W}{4} \).