Water flows @ 10m per minute …

Radius of the pipe is {tex}\begin{array}{l}\frac52\mathrm{mm}\;=\;\frac52\times\frac1{10}\;\mathrm{cm}\\\end{array}{/tex}

= {tex}\begin{array}{l}\;\frac14\;\mathrm{cm}\\\end{array}{/tex}

Speed of water =  {tex}\begin{array}{l}\;\frac{10\;\mathrm m}\min\;=\;\frac{1000\;\mathrm{cm}}\min\\\end{array}{/tex}

⇒ v = {tex} \frac { 10 } { 60 }{/tex} m/s = {tex} \frac { 1 } { 6 }{/tex} m/s

∴ Volume of flowing water = Volume of cone

⇒ Area of base × height (dist.) = {tex} \frac { 1 } { 3 } \pi R ^ { 2 } H{/tex}

⇒ A × vt = {tex} \frac { 1 } { 3 } \pi R ^ { 2 } H{/tex} [as distance=speed ×time]

⇒ πr².vt= {tex} \frac { 1 } { 3 }{/tex}πR²H

⇒ r².vt= {tex} \frac { 1 } { 3 }{/tex}R²H

⇒ {tex} \frac { 1 } { 400 } \times \frac { 1 } { 400 } \times \frac { 1 } { 6 } t = \frac { 1 } { 3 } \times 0.2 \times 0.2 \times 0.24{/tex}

⇒ {tex}t = \frac { 2 \times 2 \times 24 \times 400 \times 400 } { 3 \times 10000 }{/tex}

⇒ t = 4 × 24 × 4 × 4 × 2 sec

{tex} = \frac { 4 \times 24 \times 4 \times 4 \times 2 } { 60 } \mathrm { min } = \frac { 512 } { 10 }{/tex} = 51.2 min

= 51 min + 0.2 min = 51 min + 0.2 × 60sec

⇒ t = 51 min  and 12 sec.

Hence, conical tank will fill in 51 min and 12 sec.