QJEEM19Ph64b8: JEE Main 2019 Physics - Class 11 JEEM19Ph64b8 Solution: Two identical beakers A and B contain equal volumes of two…

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}}} &gt; {1 \over {{\rho _B}{s_B}}} \Rightarrow {{d{Q_A}} \over {dt}} &gt; {{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).