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.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]
A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_0\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _______. 