A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.

#### Solution

Internal diameter of the bowl = 36 cm

Internal radius of the bowl, r = 18 cm

Volume of the liquid, V =(2/3)𝜋r3 =(2/3)× 𝜋 × 183

Let the height of the small bottle be ‘h’.

Diameter of a small cylindrical bottle = 6 cm

Radius of a small bottle, R = 3 cm

Volume of a single bottle = 𝜋R^{2}h = 𝜋 × 32 × h

No. of small bottles, n = 72

Volume wasted in the transfer =(10/100)×(2/3)× 𝜋 × 18^{3}

Volume of liquid to be transferred in the bottles

`=2/3xxpixx18^3-10/100xx2/3xxpixx18^3`

`=2/3xxpixx18^3(1-10/100)`

`=2/3xxpixx18^3xx90/100`

we know that volume of cylinder =`pir^2h` so we get

`72(pir^2h)=(2/3xxpixx18^3xx90/100)`

`72=(2/3xxpixx18^3xx90/100)/(pixx3^2xxh)`

`72=(2/3xx18^3xx9/10)/(3^2xxh)`

`h=(2/3xxpixx18xx18xx18xx9/10)/(pixx3^2xx72) `

h=5.4 cm

Height of the small cylindrical bottle = 5.4 cm