Here, h = 1 m, volume = 15.4 litres

= 15.4/ 1000 m^{3 }= 0.0154 m^{3}

Also, volume of the cylinderical vessel = π r^{2}h

⇒ 0.0154 = 22/ 7 × r^{2} × 1

⇒ r^{2} = 0.0154 × 7/ 22 = 0.0049

⇒ r = 0.07 m

∴ Total surface area of the cylinder = 2πr (h + r)

= 2 × 22/ 7 × 0.07 (1 + 0.07) m^{2}

= 44 × 0.01 × 1.07 m^{2 }= 0.4708 m^{2 }

Hence, 0.4708 m^{2} of metal sheet would be needed **Ans.**