If $T, T_0, \sigma $ and t are same for both bodies then $\frac{{{Q_{sphere}}}}{{{Q_{cube}}}} = \frac{{{A_{sphere}}}}{{{A_{cube}}}} = \frac{{4\pi {r^2}}}{{6{a^2}}}$ …..$(i)$
But according to problem, volume of sphere = Volume of cube
==> $\frac{4}{3}\pi {r^3} = {a^3}$
==> $a = {\left( {\frac{4}{3}\pi } \right)^{1/3}}r$
Substituting the value of a in equation $(i)$ we get
$\frac{{{Q_{sphere}}}}{{{Q_{cube}}}} = \frac{{4\pi {r^2}}}{{6{a^2}}} = \frac{{4\pi {r^2}}}{{6{{\left\{ {{{\left( {\frac{4}{3}\pi } \right)}^{1/3}}r} \right\}}^2}}}$
$ = \frac{{4\pi {r^2}}}{{6\,{{\left( {\frac{4}{3}\pi } \right)}^{2/3}}{r^2}}} = {\left( {\frac{\pi }{6}} \right)^{1/3}}:1$
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Assertion A :Two identical balls $A$ and $B$ thrown with same velocity '$u$ ' at two different angles with horizontal attained the same range $R$. If $A$ and $B$ reached the maximum height $h_{1}$ and $h_{2}$ respectively, then $R =4 \sqrt{ h _{1} h _{2}}$
Reason R: Product of said heights.
$h _{1} h _{2}=\left(\frac{u^{2} \sin ^{2} \theta}{2 g }\right) \cdot\left(\frac{u^{2} \cos ^{2} \theta}{2 g }\right)$
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