Paramagnetic sample shows a net magnetisation of  4A/m ,  when placed in an external magnetic field of 0.3T  at a temperature of 8K. When the same sample is placed in an external magnetic field of 0.4T at a temperature of  4K, the magnetisation will be $\frac{p}{3}(A/\mathrm{m})$. Find $p$

As Curie’s law explains, we can deduce a formula for the relation between magnetic field induction temperature and magnetisation

$\begin{array}{c}I \propto \frac{B}{T}\\ \Rightarrow \frac{I_2}{I_1}=\frac{B_2}{B_1}\times \left(\frac{T_1}{T_2}\right)\end{array}$

                              $$\begin{gathered} \begin{array}{c}Put I_1=4A/m{ , \mathrm{B}}_1=0.3T , {\mathrm{T}}_1=8K\end{array} \\ \begin{array}{l}{I}_2=?{ , \mathrm{B}}_2=0.4T , {\mathrm{T}}_2=4K\end{array} \end{gathered}$$                                

$\begin{array}{l}\therefore \frac{I_2}{4}=\frac{0.4}{0.3}\times \frac{8}{4}\\ \Rightarrow I_2=\frac{4}{3}\times 2\times 4=\frac{32}{3}(A/\mathrm{m})\end{array}$