$A$ sample of paramagnetic salt contains $2.0 \times 10^{24}$ atomic dipoles each of dipole moment $1.5 \times 10^{-23} \; J \, T^{-1}$. The sample is placed under a homogeneous magnetic field of $0.64 \; T$ and cooled to a temperature of $4.2 \; K$. The degree of magnetic saturation achieved is $15 \%$. What is the total dipole moment of the sample for a magnetic field of $0.98 \; T$ and a temperature of $2.8 \; K$? (Assume Curie's law)

(D) Number of atomic dipoles,$n = 2.0 \times 10^{24}$.
Dipole moment of each atomic dipole,$M = 1.5 \times 10^{-23} \; J \, T^{-1}$.
Total dipole moment of the sample if fully saturated,$M_{\text{total}} = n \times M = 2.0 \times 10^{24} \times 1.5 \times 10^{-23} = 30 \; J \, T^{-1}$.
Given saturation at $B_1 = 0.64 \; T$ and $T_1 = 4.2 \; K$ is $15 \%$.
Effective dipole moment $M_1 = 0.15 \times 30 = 4.5 \; J \, T^{-1}$.
According to Curie's law,the magnetization $M \propto \frac{B}{T}$,so $\frac{M_2}{M_1} = \frac{B_2}{B_1} \times \frac{T_1}{T_2}$.
Substituting the values for $B_2 = 0.98 \; T$ and $T_2 = 2.8 \; K$:
$M_2 = M_1 \times \frac{B_2}{B_1} \times \frac{T_1}{T_2} = 4.5 \times \frac{0.98}{0.64} \times \frac{4.2}{2.8}$.
$M_2 = 4.5 \times 1.53125 \times 1.5 = 10.336 \; J \, T^{-1}$.