We know, m = V . $$\rho$$

where, V = volume and $$\rho$$ = density.

So, we have

$${{dQ} \over {dt}} = {h \over {ms}}(T - {T_0}) = {{h(T - {T_0})} \over {V\,.\,\rho s}}$$

Since, h, (T $$-$$ T₀) and V are constant for both beaker.

$$\therefore$$ $${{dQ} \over {dt}} \propto {1 \over {\rho s}}$$

We have given that $$\rho$$_A = 8 $$\times$$ 10² kgm^$$-$$3, $$\rho$$_B = 10³ kgm^$$-$$3, s_A = 2000 J kg^$$-$$1 K^$$-$$1 and s_B = 4000 J kg^$$-$$1 K^$$-$$1, $$\rho$$_As_A = 16 $$\times$$ 10⁵ and $$\rho$$_Bs_B = 4 $$\times$$ 10⁶

$$\rho$$_A < $$\rho$$_B, s_A < s_B and $$\rho$$_As_A < $$\rho$$_Bs_B

$$ \Rightarrow {1 \over {{\rho _A}{s_A}}} > {1 \over {{\rho _B}{s_B}}} \Rightarrow {{d{Q_A}} \over {dt}} > {{d{Q_B}} \over {dt}}$$

So, for container B, rate of cooling is smaller than the container A. Hence, graph of B lies above the graph of A and it is not a straight line (slope of A is greater than B).