1. Initial capacitance with air

For a parallel plate capacitor, $C_0=\frac{\varepsilon_0 A}{d}$

Given:

• Plate area: $A=12\,\text{cm}^2$
• Initial separation: $d=0.6\,\text{cm}$

So initially, $C_0=\frac{\varepsilon_0 A}{0.6}$

(We do not need to convert units since only ratios are involved.)

2. After inserting dielectric slab

The dielectric slab has:

• area equal to plate area,
• thickness $0.6\,\text{cm}$, so it completely fills the space between the plates if the plate separation remains $0.6\,\text{cm}$.

But to keep the capacitance same as before, one plate is moved by $0.2\,\text{cm}$.

Hence new plate separation becomes $d'=0.6+0.2=0.8\,\text{cm}$

Now the arrangement consists of:

• dielectric slab of thickness $0.6\,\text{cm}$ and dielectric constant $K$,
• air gap of thickness $0.2\,\text{cm}$.

These act like capacitors in series.

The effective capacitance is $C=\frac{\varepsilon_0 A}{\frac{0.6}{K}+0.2}$

3. Condition: capacitance remains same

Given that new capacitance equals old capacitance: $\frac{\varepsilon_0 A}{\frac{0.6}{K}+0.2}=\frac{\varepsilon_0 A}{0.6}$

Cancel $\varepsilon_0 A$ from both sides: $\frac{1}{\frac{0.6}{K}+0.2}=\frac{1}{0.6}$

So, $\frac{0.6}{K}+0.2=0.6$

$\frac{0.6}{K}=0.4$

$K=\frac{0.6}{0.4}=1.5$

4. Checking options

• A: $1.50$ ✅
• B: $0.66$ ❌
• C: $1.33$ ❌
• D: $1$ ❌

Thus the correct answer is: $\boxed{1.50}$