1. Initial capacitance and energy

For a parallel plate capacitor of plate area $A = lw$ and separation $d$:

$C_0 = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 lw}{d}$

Since it is connected to a battery of emf $V$, the potential difference remains constant.

Initial energy stored is:

$U_0 = \frac{1}{2} C_0 V^2$

2. Capacitance after partial insertion of dielectric

Let the dielectric slab be inserted up to length $x$ inside the capacitor.

Then the capacitor behaves like two capacitors in parallel:

• One part with dielectric, area $xw$
• One part without dielectric, area $(l-x)w$

So,

$C = \frac{k\varepsilon_0 xw}{d} + \frac{\varepsilon_0 (l-x)w}{d}$

Given $k=4$,

$C = \frac{\varepsilon_0 w}{d}(4x + l - x)$

$C = \frac{\varepsilon_0 w}{d}(l + 3x)$

3. Condition for energy to become twice

Since battery remains connected, $V$ is constant, hence

$U = \frac{1}{2}CV^2$

Given final energy is twice the initial energy:

$U = 2U_0$

Thus,

$\frac{1}{2}CV^2 = 2\left(\frac{1}{2}C_0V^2\right)$

$C = 2C_0$

Now,

$\frac{\varepsilon_0 w}{d}(l+3x) = 2\left(\frac{\varepsilon_0 lw}{d}\right)$

Cancel common factors $\frac{\varepsilon_0 w}{d}$:

$l+3x = 2l$

$3x = l$

$x = \frac{l}{3}$

4. Option check

• A: $\frac{l}{4}$ ✗
• B: $\frac{l}{2}$ ✗
• C: $\frac{2l}{3}$ ✗
• D: $\frac{l}{3}$ ✓

Therefore, the required length of slab inserted is:

$\boxed{\frac{l}{3}}$