For maximum elongation charges on the blocks must be equal to $$\frac{Q}{2}$$ on each block.
$$\therefore \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\frac{Q}{2}\frac{Q}{2}}}{{{{\left( {{\ell _0} + x} \right)}^2}}} = kx,\,\,Q = 2\left( {{\ell _0} + x} \right)\sqrt {4\pi {\varepsilon _0}kx} .$$

Since the sheet is metallic, the charges must be redistributed over its surface so that the field in the bulk of the sheet is zero. In the first approximation, we can assume that this distribution is uniform and has density $$ - \sigma $$ and $$\sigma $$ over the upper and the lower surface respectively of the sheet. According to the superposition principle, we obtain the condition for the absence of the field in the bulk of the sheet :
$$\frac{q}{{4\pi {\varepsilon _0}{\ell ^2}}} - \frac{\sigma }{{{\varepsilon _0}}} = 0$$
Let us now take into consideration the non uniformity of the field produced by the point charge since it is the single cause of the force $$F$$ of interaction. The upper surface of the sheet must be attracted with a force $$\frac{{\sigma Sq}}{{4\pi {\varepsilon _0}{\ell ^2}}},$$  while the lower surface must be repelled with a force $$\frac{{\sigma Sq}}{{4\pi {\varepsilon _0}{{\left( {\ell + d} \right)}^2}}}.$$
Consequently, the force of attraction of the sheet to the charge is
$$F = \frac{{\sigma Sq}}{{4\pi {\varepsilon _0}{\ell ^2}}}\left[ {1 - \frac{1}{{{{\left( {1 + \frac{d}{\ell }} \right)}^2}}}} \right] \approx \frac{{{q^2}Sd}}{{8{\pi ^2}{\varepsilon _0}{\ell ^5}}}$$

$$\eqalign{ & \Rightarrow {F_{net{\text{ }}}} = 2F\cos \theta \cr & {F_{net}} = \frac{{2kq\left( {\frac{q}{2}} \right)}}{{{{\left( {\sqrt {{y^2} + {a^2}} } \right)}^2}}} \cdot \frac{y}{{\sqrt {{y^2} + {a^2}} }} \cr & {F_{net}} = \frac{{2kq\left( {\frac{q}{2}} \right)y}}{{{{\left( {{y^2} + {a^2}} \right)}^{\frac{3}{2}}}}} \Rightarrow \frac{{k{q^2}y}}{{{a^3}}} \cr & {\text{So, }}F \propto y \cr} $$

The frequency of SHM performed by wooden block is

when electric field is switched on, the value of $$k$$ and $$m$$ is not affected and therefore the frequency of SHM remains the same. But as an external force $$QE$$  starts acting on the block towards right, the mean position of SHM shifts towards right by $$\frac{{QE}}{k}$$  correct option is (A).
Note : In SHM if a constant additional force is applied then it only shift the quilibrium position and does not change the frequency of SHM.

The charges on the surfaces of the metallic spheres are shown in the diagram. It is given that the surface charge densities on the outer surfaces of the shells are equal. Therefore

$$\eqalign{ & \frac{{{Q_1}}}{{4\pi {R^2}}} = \frac{{{Q_1} + {Q_2}}}{{4\pi {{\left( {2R} \right)}^2}}} = \frac{{{Q_1} + {Q_2} + {Q_3}}}{{4\pi {{\left( {3R} \right)}^2}}} = x\left( {{\text{say}}} \right) \cr & \therefore {Q_1} = 4\pi {R^2}x \cr & {Q_1} + {Q_2} = 4\pi {\left( {2R} \right)^2}k = 4\left[ {4\pi {R^2}x} \right] \cr & \Rightarrow {Q_2} = 4\left[ {4\pi {R^2}x} \right] - {Q_1} \cr & = 4\left[ {4\pi {R^2}x} \right] - 4\pi {R^2}x = 3\left[ {4\pi {R^2}x} \right] \cr & {\text{Also}}\,{Q_1} + {Q_2} + {Q_3} = 4\pi {\left( {3R} \right)^2}x = 9\left[ {4\pi {R^2}x} \right] \cr & \therefore {Q_3} = 9\left[ {4\pi {R^2}x} \right] - {Q_1} - {Q_2} = 9\left[ {4\pi {R^2}x} \right] - \left[ {4\pi {R^2}x} \right] - 3\left[ {4\pi {R^2}x} \right] = 5\left[ {4\pi {R^2}x} \right] \cr & \Rightarrow {Q_1}:{Q_2}:{Q_3} = 1:3:5 \cr} $$

Initial energy

$$\eqalign{ & {U_f} = \frac{{Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{{a + x}} + \frac{1}{{a - x}}} \right] = \frac{{2Qqa}}{{4\pi {\varepsilon _0}\left( {{a^2} - {x^2}} \right)}} \cr & {U_i} - {U_f} = \frac{{2Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{a} - \frac{a}{{\left( {{a^2} - {x^2}} \right)}}} \right] \cr & = \frac{{2Qq}}{{4\pi {\varepsilon _0}}}\left[ {\frac{{{a^2} - {x^2} - {a^2}}}{{a\left( {{a^2} - {x^2}} \right)}}} \right] = \frac{{ - 2Qq{x^2}}}{{4\pi {\varepsilon _0}{a^3}}} \cr} $$
when $$x < < a$$   then $${x^2}$$ is neglected in denominator.
$${U_i} - {U_f} = \left( {\frac{{ - Qq}}{{2\pi {\varepsilon _0}{a^3}}}} \right){x^2}$$

Statement 1 is true.
Statement 2 is true and is the correct explanation of (1)

The charge on disc $$A$$ is $${10^{ - 6}}\mu C.$$   The charge on disc $$B$$ is $$10 \times {10^{ - 6}}\mu C.$$    The total charge on both $$ = 11\mu C.$$  When touched, this charge will be distributed equally i.e. $$5.5\mu C$$  on each disc.

For equilibrium of charge $$Q$$
$$K\frac{{Q \times Q}}{{{{\left( {2x} \right)}^2}}} + K\frac{{Qq}}{{{x^2}}} = 0 \Rightarrow q = - \frac{Q}{4}$$

When a charge density is given to the inner cylinder, the potential developed at its surface is different from that on the outer cylinder. This is because the potential decreases with distance for a charged conducting cylinder when the point of consideration is outside the cylinder.
But when a charge density is given to the outer cylinder, it will change its potential by the same amount as that of the inner cylinder. Therefore no potential difference will be produced between the cylinders in this case.