Question

Half of a tank of 21dcm. length,11dcm. breadth and 6dcm. depth is full of water. Now if 100 iron spheres of 21cm. diameter is immersed completely into water of this tank then calculate the rise of water level in decimeter( dcm.).

Answer 1

Hope it is helpful..................

Answer 2

$\bf\huge\underline{Explanation :-}$

Given :

• Length of tank = 21 dcm
• Breadth of tank = 11 dcm
• Depth of tank = 6 dcm
• Diameter of iron spheres => 21 cm = 2.1 dcm
• No. of iron spheres = 100

To find :

• Rise of water level when sphere is completely immersed into the water tank.

Solution :

We know that,

Volume of tank :

= (length × breadth × height) dcm³

= (21 × 11 × 3) dcm³

= 693 dcm³

[Note : Depth of the tank is halved since the water is only half filled.]

Now, Volume of sphere :

$\sf\longrightarrow \dfrac{4}{3} \times \pi \times {r}^{3}$

[Since, Diameter = 21 dcm, Radius is halved to 1.05 dcm]

$\sf\longrightarrow \dfrac{4}{3} \times \dfrac{22}{7} \times {1.05}^{3}$

$\sf\longrightarrow \dfrac {4 \times 22 \times 105 \times 105 \times 105}{7 \times 3 \times 100 \times 100 \times 100}$

$\sf\longrightarrow 4.851$

.°. Volume of 100 spheres :

= (4.851 × 100) dcm³

= 485.1 dcm³

Since,

Volume of tank = Volume of sphere

$\sf\longrightarrow 21 \times 11 \times x = 485.1$

$\sf\longrightarrow x = \dfrac{4851}{21 \times 11 \times 10}$

$\sf\huge\longrightarrow x = 2.1$

Hence, The rise of water level will be 2.1 dcm respectively.

$\bf\huge\boxed{2.1 dcm}$