Question

A steel vessel has a base of length 60 cm and breadth 30 cm. Water is poured in the vessel. A cubical steel box having edge of 30 cm is immersed completely in the vessel. How much will the water rise?

Answer 1

Answer:

15 cm

Step-by-step explanation:

as we know thatwhen we immerse object in liquid , the liquid level rises. This is because the object displaces liquid and therefore

the volume of liquid displaced= volume of object

let the rise be H cm.

volume of cubical object=vessel length*vessel breadth* vessel H(rise in water level)

and the volume of cube as per formula is a^3

therefore

60*30*H=30^3

by solving we get

H=15 cm

hope  this answer helps...

Answer 2

Given:

• Length, Breadth of a tank is 60cm and 30cm respectively .
• A cubical steel vessel having edge of 30cm is immersed completely in the tank.

$\begin{gathered}\mapsto\sf{Length=60\:cm}\\ \mapsto\sf{Breadth=30\:cm} \\\mapsto\sf{Side\:of\:Cube\:(a)=30\:cm} \end{gathered}$

To find:

• The water in the vessel will rise by/Height.

Required Formula:

$\orange \bigstar\boxed {\frak{Volume\:of\:Cuboid=Volume\:of\:Cube}}$

Solution:

• Now, put the values.

$\begin{gathered}: \implies\sf{l\times{b}\times{h}={a}^{3}} \\\\ \\ : \implies\sf{60\times{30}\times{h}=30\times{30}\times{30}} \\ \\\\ :\implies\sf{h=\dfrac{{\cancel{30}}\times{{\cancel{30}}}\times{30}}{{\cancel{60}}\times{{\cancel{30}}}}} \\\\ \\ : \implies\sf{h=\cancel\dfrac{30}{2}} \\ \\\\ : \implies   \underline{\boxed{ \frak{h = 15 \: cm}}} \pink \bigstar \end{gathered}$

$\underline {\sf Hence, the \: water \: will \: rise \: by \textsf{\textbf{ 15cm}}}.$

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