The solubility product constant, \(K_{sp}\), is a measure of the solubility of a compound; the smaller the \(K_{sp}\), the less soluble the compound is. To determine the increasing order of solubility product for the given compounds: \(Ca(OH)_2\), \(AgBr\), \(PbS\), and \(HgS\), we compare their \(K_{sp}\) values.
1. \({HgS}\): It has a very low \(K_{sp}\) with a value approximately in the order of \(10^{-54}\), indicating extremely low solubility.
2. \({PbS}\): This compound also has a low \(K_{sp}\) but is slightly more soluble than \(HgS\), with \(K_{sp}\) around \(10^{-28}\).
3. \({AgBr}\): It is more soluble than both \(HgS\) and \(PbS\), with a \(K_{sp}\) around \(10^{-13}\).
4. \({Ca(OH)}_2\): This compound has the highest \(K_{sp}\) among the given compounds, approximately \(10^{-6}\), making it the most soluble.
Based on these \(K_{sp}\) values, the increasing order of solubility product is:
\(HgS<PbS<AgBr<Ca(OH)_2\)
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
