SSC CHSL Model Paper 14

$4\left(\mathit{l}+\mathit{b}+\mathit{h}\right)=112$
⇒ $\left(\mathit{l}+\mathit{b}+\mathit{h}\right)=28$
and
$2\left(\mathit{l}\mathit{b}+\mathit{b}\mathit{h}+\mathit{h}\mathit{l}\right)=384$
⇒ $\mathit{l}\mathit{b}+\mathit{b}\mathit{h}+\mathit{h}\mathit{l}=192$
we know, ${\left(\mathit{l}+\mathit{b}+\mathit{h}\right)}^2={\mathit{l}}^2+{\mathit{b}}^2+{\mathit{h}}^2$ $+2\mathit{l}\mathit{b}+2\mathit{b}\mathit{h}+2\mathit{h}\mathit{l}$
⇒ ${\mathit{l}}^2+{\mathit{b}}^2+{\mathit{h}}^2={\left(\mathit{l}+\mathit{b}+\mathit{h}\right)}^2$ $-2\left(\mathit{l}\mathit{b}+\mathit{b}\mathit{h}+\mathit{h}\mathit{l}\right)$
⇒ ${\mathit{l}}^2+{\mathit{b}}^2+{\mathit{h}}^2={\left(28\right)}^2-384$
$=784-384$
$=400$
The relation between the radius of the sphere and the dimensions of the rectangular box is that ‘diagonal of rectangular box will be diameter of sphere’.
⇒ $2\mathit{r}=\left({\mathit{l}}^2+{\mathit{b}}^2+{\mathit{h}}^2\right)\frac{1}{2}$
⇒ $2\mathit{r}=\left(400\right)\frac{1}{2}$
⇒ $\mathit{r}=\frac{20}{2}=10\mathrm{cm}$