The resistance of the four arms of a Wheatstone bridge are P=`10Omega` , Q = `100Omega`, R=`40Omega` and S `=10Omega`. What resistance in series or parallel with the last one will be required to obtain no deflection in the galvanometer?

Answer

To solve the problem, we need to find the resistance that should be added in series with the resistance S (which is 10Ω) in order to achieve a balance in the Wheatstone bridge. The balance condition for a Wheatstone bridge is given by the equation:

\[ \frac{P}{Q} = \frac{R}{S'} \]

where \( S' \) is the new resistance in series with S.

### Step-by-Step Solution:

1. **Identify the given resistances:**
- \( P = 10 \, \Omega \)
- \( Q = 100 \, \Omega \)
- \( R = 40 \, \Omega \)
- \( S = 10 \, \Omega \)

2. **Set up the balance condition:**
Using the balance condition of the Wheatstone bridge:
\[ \frac{P}{Q} = \frac{R}{S'} \]

Substitute the known values:
\[ \frac{10}{100} = \frac{40}{S'} \]

3. **Cross-multiply to solve for \( S' \):**
\[ 10 \cdot S' = 100 \cdot 40 \]
\[ 10 \cdot S' = 4000 \]
\[ S' = \frac{4000}{10} = 400 \, \Omega \]

4. **Determine the required resistance in series:**
Since we have \( S = 10 \, \Omega \), we need to find the additional resistance \( X \) that we need to add in series with \( S \):
\[ S' = S + X \]
\[ 400 = 10 + X \]
\[ X = 400 - 10 = 390 \, \Omega \]

### Final Answer:
The resistance required in series with the last one (S) to obtain no deflection in the galvanometer is **390Ω**.