Question

Look at this cylinder:

Answer 1

• Height=h=8cm
• Radius=r=4cm

We know

$\boxed{\sf \star TSA_{(Cylinder)}=2\pi r(h+r)}$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=2\times \dfrac{22}{7}\times 4(8+4)$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{176}{7}(12)$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=\dfrac{2112}{7}$

$\\ \sf\longmapsto TSA_{(Old\:Cylinder)}=301.7cm^2$

Now

• New Radius=2(4)=8cm
• New Height=2(8)=16cm

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=2\times \dfrac{22}{7}\times 8(16+8)$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{352}{7}(24)$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=\dfrac{8448}{7}$

$\\ \sf\longmapsto TSA_{(New\:Cylinder)}=1204.7cm^2$

So

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{1204.7}{301.7}$

$\\ \sf\longmapsto \dfrac{TSA_{(New\:Cylinder)}}{TSA_{(Old\:Cylinder)}}=\dfrac{4}{1}$

$\\ \sf\longmapsto\underline{\boxed{\bf{ {TSA_{(New\:Cylinder)}}:{TSA_{(Old\:Cylinder)}}=4:1}}}$

Hence our correct option is Option C