1. Metre bridge balance condition

For a metre bridge at balance point, $\frac{R_1}{R_2}=\frac{l}{100-l}$ where $l$ is the balance length from end $A$.

Initially, null point is at $40\,\text{cm}$ from $A$. So, $\frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3}$ Hence, $R_1=\frac{2}{3}R_2 \quad ...(1)$

2. After adding $10\,\Omega$ in series with $R_1$

New resistance in left gap becomes $(R_1+10)\,\Omega$ The null point shifts by $10\,\text{cm}$. Since resistance on the left has increased, the balance length from $A$ must increase from $40$ cm to $50$ cm.

Thus, $\frac{R_1+10}{R_2}=\frac{50}{50}=1$ So, $R_1+10=R_2 \quad ...(2)$

Using (1): $R_1=\frac{2}{3}R_2$ Substitute into (2): $\frac{2}{3}R_2+10=R_2$ $10=\frac{1}{3}R_2$ $R_2=30\,\Omega$ Therefore, $R_1=20\,\Omega$

So after adding $10\,\Omega$ in series, $R_1+10=30\,\Omega$

3. Now connect a resistance $X$ in parallel with $(R_1+10)=30\,\Omega$

Equivalent resistance should bring balance point back to initial $40\,\text{cm}$. That means the effective left-gap resistance must again be equal to original $R_1=20\,\Omega$.

So, $\frac{30X}{30+X}=20$

Solve: $30X=20(30+X)$ $30X=600+20X$ $10X=600$ $X=60\,\Omega$

4. Check options

• A: $40\,\Omega$ ❌
• B: $30\,\Omega$ ❌
• C: $20\,\Omega$ ❌
• D: $60\,\Omega$ ✅

Therefore, the required resistance is $\boxed{60\,\Omega}$