To convert the voltmeter so that it can measure up to 250 V, we use a series resistor.
Let the series resistance be \( R_s \), and the total resistance should be such that the voltmeter can measure 250 V when the original meter reads 25 V.
Using the formula for voltage division:
\[
V_{\text{new}} = V_{\text{max}} \times \frac{R_{\text{meter}}}{R_{\text{meter}} + R_s}
\]
Where:
- \( V_{\text{new}} \) is the new voltage range (250 V),
- \( V_{\text{max}} \) is the original maximum voltage (25 V),
- \( R_{\text{meter}} \) is the resistance of the voltmeter (1000 \( \Omega \)).
We can rearrange the formula to solve for \( R_s \):
\[
\frac{250}{25} = \frac{1000}{1000 + R_s}
\]
Simplifying:
\[
10 = \frac{1000}{1000 + R_s}
\]
Now, solve for \( R_s \):
\[
10 (1000 + R_s) = 1000
\]
\[
10000 + 10 R_s = 1000
\]
\[
10 R_s = 1000
- 10000 = 9000
\]
\[
R_s = 9000 \, \Omega
\]
Thus, the required series resistance is \( 9000 \, \Omega \).