Question

150 spherical marbles, each of diameter 1.4 cm are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.​

Answer 1

Answer:

The  rise in level of water in the vessel is 5.6 cm.

Step-by-step explanation:

Let 'h’ be the rise in level of water in the cylindrical vessel.

Given :  

Diameter of the cylindrical vessel = 7 cm

Radius of the cylindrical vessel , R = 7/2 = 3.5 cm

Diameter of the spherical marble = 1.4 cm

Radius of the spherical marble , r = 1.4/2 = 0.7 cm

Volume of each marble = 4/ 3 × π r³

Volume of 150 marble = 150 ×  4/ 3 × π (0.7)³ ………(1)

Volume of water displaced in the cylindrical vessel  = πR²h  = π(3.5)²× h ……….(2)

Since, volume of water displaced in the cylindrical vessel is equal to the volume of 150 spherical marbles

Volume of water displaced in the cylindrical vessel = Volume of 150 spherical marbles

π(3.5)²× h = 150 ×  4/ 3 × π (0.7)³

(3.5)² × h = 150 ×  4/ 3 × (0.7)³

h = [150 ×  4 × (0.7)³] /[ (3.5)² × 3]

h = [150 ×  4 × 0.7 × 0.7 × 0.7] /[ 3.5 × 3.5 × 3]

h = (50 × 4 ×  0.7) / ( 5 × 5)

h = (200× 0.7)/ 25 = 8 × 0.7 = 5.6  

h = 5.6 cm  

Hence, the  rise in level of water in the vessel is 5.6 cm.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

5.6cm

Step-by-step explanation:

$sphere \\ d = 1.4cm \\ r = \frac{1.4}{2}cm \\ cylinder \ \\ d = 7cm \\ r = \frac{7}{2} cm \\ vol.o f \: 150 \: \: spherical \: marbles \: = vol.of \: cylinder \\ 150 \times \frac{4}{3} \pi {r}^{3} = \pi {r}^{2} h \\ 50 \times 4 {r}^{3} = {r}^{2} h \\ 50 \times 4 \times \frac{1.4}{2} \times \frac{1.4}{2} \times \frac{1.4}{2} = \frac{7}{2} \times \frac{7}{2} \times h \\ h = \frac{50 \times 4 \times 1.4 \times 1.4 \times 1.4 \times 2 \times 2}{2 \times 2 \times 2 \times 7 \times 7} \\ h = 5.6cm$