$ - \frac{{dT}}{{dt}} = \frac{{e\sigma A}}{{ms}}\left( {{T^4} - T_0^4} \right)\,\,;\,\, - \frac{{dT}}{{dt}} = \frac{{4e\sigma AT_0^3}}{{ms}}\left( {\Delta T} \right)$
$T = {T_0} + \left( {{T_i} - {T_0}} \right){e^{ - kt}}$
$where\,k = \frac{{4e\sigma AT_0^3}}{{ms}}$
$k = \frac{{4e\sigma AT_0^3}}{{\rho vs}}\,\,;\,\,\left| {\frac{{dT}}{{dt}}} \right| \propto k$
$\therefore \left| {\frac{{dT}}{{dt}}} \right| \propto \frac{1}{{\rho s}}$
$\rho A{S_A} = 2000 \times 8 \times {10^2} = 16 \times {10^5}$
${\rho _B}{S_B} = 4000 \times {10^3} = 4 \times {10^6}$
${\rho _A}{S_A} < {\rho _B}{S_B}$
${\left| {\frac{{dT}}{{dt}}} \right|_A} > {\left| {\frac{{dT}}{{dt}}} \right|_B}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(a)$ The moment of inertia of cube about $z-$ axis is $I_z$ = $I_x + I_y$
$(b)$ The moment of inertia of cube about $A-$ axis is $I_A$=${I_z} + \frac{{m{a^2}}}{2}$
$(c)$ The moment of inertia of cube about $B-$ axis is $I_B$=${I_z} + \frac{{m{a^2}}}{2}$
$(d)$ $I_x$ = $I_z$
