Verified answer

Answer:

[tex]\qquad\qquad\boxed { \sf \: Height\:of\:water\:level = 1.5 \: m \: } \\ \\ [/tex]

Step-by-step explanation:

Given that, swimming pool is 260 m long and 140 m wide and 54600 cubic metres of water is pumped into it.

So, we have

[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:Length \: of \: swimming \:pool=260 \: m \qquad \: \\ \\& \qquad \:\sf \:Breadth \: of \: swimming \:pool=140 \: m\\ \\& \qquad \:\sf \:Volume \: of \: water \:in\:pool=54600 \: m^3\end{aligned}} \qquad \: \\ \\ [/tex]

Now, we have to find the height of water level in pool.

[tex]\sf \: Height\:of\:water\:level \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{Volume}{length \times breadth} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{54600}{260 \times 140} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{546}{26 \times 14} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{39}{26} \\ \\ [/tex]

[tex]\sf \: = \: \dfrac{3}{2} \\ \\ [/tex]

[tex]\sf \: = \: 1.5 \: m \\ \\ [/tex]

Hence,

[tex]\sf\implies \sf \: Height\:of\:water\:level = 1.5 \: m \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

GIVEN :

• Length = 260 m

• Breadth = 140 m

• Volumne = 54600 m³

TO FIND :

To find the Height of the water pumped in it

FORMULA USED :

V = L × B × H

SOLUTION :

: [tex] \implies[/tex] L = 260 m

: [tex] \implies[/tex] B = 140 m

: [tex] \implies[/tex] V = 54600 m³

We get ,

: [tex] \implies[/tex] 54600 m³ = 260m × 140 m × H

:[tex] \implies[/tex] H = 54600 m³ / 260 m × 140 m

:[tex] \implies[/tex] H = 54600 m³ / 36400 m³

: [tex] \implies[/tex] H = 1.5 m

FINAL ANSWER :

Hence , The Hight of swimming pool is H = 1.5 m

More to know :

• The surface area of sphere = 2πrh

• TheVolume of a cuboid = lbh

• The total surface area of cuboid = 2 (lb + bh + hl)

• The lateral surface area of cuboid = 2h (l+b)