NCHMCT JEE 2012 Question Paper with solutions for online practice

Let the height of the cylindrical tank = x cm
∴ Radius of tank = $\frac{3}{4}$ x cm
We know that
1 ml = 1 $\mathit{c}{\mathit{m}}^3$
38808 ml = 38808 $\mathit{c}{\mathit{m}}^3$
∴ Volume of the cylindrical tank = $\mathit{\pi }{\mathit{r}}^2\mathit{h}$
⇒ 38808 = $\frac{22}{7}\times {\left(\frac{3}{4}\mathit{x}\right)}^2$ × x
⇒ 38808 = $\frac{22}{7}\times \frac{9}{16}\times {\mathit{x}}^3$
⇒ ${\mathit{x}}^3$ = $\frac{38808\times 7\times 16}{22\times 9}$ = 21952
∴ x = 28 xm
∴ Radius = $\frac{3}{4}$ × 28 = 28 = 21 cm
∴ Diameter = 2 × 21 = 42 cm.