Question

find the ratio :
A cylinder is filled with water upto level ‘h’. The base radius of cylinder is 3/2 times the level of the water. A solid metal cuboid of length, breadth and height 2/5 times, 1/5 times and 2/3 times of the water level respectively, is immersed in water in the cylinder. This cuboid is then taken out and another metal cuboid of length, breadth and height 1/7 times, 3/7 times and 4/5 times of the water level respectively, is
immersed in the same cylinder. Find the ratio of the rise in water level in the two
cases.

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

A cylinder is filled with water upto level ‘h’. The base radius of cylinder is 3/2 times the level of the water.

So, Let assume that

Water level in the cylinder be h units.

So,

Radius of cylinder, r = 3/2 h units

Further, given that

A solid metal cuboid of length, breadth and height 2/5 times, 1/5 times and 2/3 times of the water level respectively, is immersed in water in the cylinder.

So,

$\rm \: Length \: of \: cuboid, \: l = \dfrac{2h}{5}$

$\rm \: Breadth \: of \: cuboid, \: b = \dfrac{h}{5}$

$\rm \: Height \: of \: cuboid, \: h = \dfrac{2h}{3}$

Now, we know that, Amount of water displaced = Volume of object submerged.

Let assume that, water level rises by H units.

Thus,

$\rm \: Volume_{(water displaced)} \: = \: Volume_{(cuboid)}$

$\rm \: \pi {(r)}^{2}H \: = \: l \times b \times h$

$\rm \: \pi {\bigg( \dfrac{3h}{2} \bigg)}^{2}H \: = \: \dfrac{2h}{5} \times \dfrac{h}{5} \times \dfrac{2h}{3}$

$\rm \: \pi {\bigg( \dfrac{9 {h}^{2} }{4} \bigg)}H \: = \: \dfrac{2h}{5} \times \dfrac{h}{5} \times \dfrac{2h}{3}$

$\rm\implies \:H = \dfrac{16h}{675\pi} \: units$

Now, Further given that

This cuboid is then taken out and another metal cuboid of length, breadth and height 1/7 times, 3/7 times and 4/5 times of the water level respectively, is immersed in the same cylinder.

$\rm \: Length \: of \: cuboid, \: l = \dfrac{h}{7}$

$\rm \: Breadth \: of \: cuboid, \: b = \dfrac{3h}{7}$

$\rm \: Height \: of \: cuboid, \: h = \dfrac{4h}{5}$

Now, we know that, Amount of water displaced = Volume of object submerged.

Let assume that, water level rises by H' units.

Thus,

$\rm \: Volume_{(water displaced)} \: = \: Volume_{(cuboid)}$

$\rm \: \pi {(r)}^{2}H' \: = \: l \times b \times h$

$\rm \: \pi {\bigg( \dfrac{3h}{2} \bigg)}^{2}H' \: = \: \dfrac{h}{7} \times \dfrac{3h}{7} \times \dfrac{4h}{5}$

$\rm \: \pi {\bigg( \dfrac{9 {h}^{2} }{4} \bigg)}H' \: = \: \dfrac{h}{7} \times \dfrac{3h}{7} \times \dfrac{4h}{5}$

$\rm\implies \:H' = \dfrac{16h}{735\pi} \: units$

Now, Consider

$\rm \: H : H'$

$\rm \: = \: \dfrac{16h}{675\pi} \: : \: \dfrac{16h}{735\pi}$

$\rm \: = \: 735 \: : \: 675$

$\rm \: = \: 147 \: : \: 135$

$\rm \: = \: 49 \: : \: 45$

Hence,

$\rm\implies \:H : H' \: = \: 49 : 45$

Short Cut Trick

Ratio of height of water displaced is Ratio of volume of two objects

That means,

$\rm \: H : H'$

$\rm \: = \: \dfrac{2h}{5} \times \dfrac{h}{5} \times \dfrac{2h}{3} \: : \dfrac{h}{7} \times \dfrac{3h}{7} \times \dfrac{4h}{5}$

$\rm \: = \: 49 \: : \: 45$

Hence,

$\rm\implies \:H : H' \: = \: 49 : 45$