A cube of edge 9cm is immersed completely in rectangular vessel containing water. If the dimensions of base are 15 cm ×12m , find the rise in water level in vessel.
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rise in water =volumeof cube/base of vessel=
9*9*9/15*12=
81/20=
4.05
9*9*9/15*12=
81/20=
4.05
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Question:A cube of edge 9cm is immersed completely in rectangular vessel containing water. If the dimensions of base are 15 cm ×12m , find the rise in water level in vessel.
Answer:-
Given, edge of the cube = 9 cm
Dimension of the base of the rectangular vessel = 15 cm × 12 cm
Let the rise in water level in the rectangular vessel be h cm.
Since, the cube is completely immersed in the rectangular vessel,
∴ Volume of water displaced in the rectangular vessel = Volume of the cube
15 cm × 12 cm × h = (9 cm)3
15 cm × 12 cm × h = 729 cm3
h = (729/180) cm
⇒ h ≈ 4.05 cm
Thus, the rise in water level in the rectangular vessel is 4.05 cm approximately.
I hope this helps you. If yes then please mark as brainliest answer. Thanks for giving me a chance to help you. Have a marvelous day ahead.
Question:A cube of edge 9cm is immersed completely in rectangular vessel containing water. If the dimensions of base are 15 cm ×12m , find the rise in water level in vessel.
Answer:-
Given, edge of the cube = 9 cm
Dimension of the base of the rectangular vessel = 15 cm × 12 cm
Let the rise in water level in the rectangular vessel be h cm.
Since, the cube is completely immersed in the rectangular vessel,
∴ Volume of water displaced in the rectangular vessel = Volume of the cube
15 cm × 12 cm × h = (9 cm)3
15 cm × 12 cm × h = 729 cm3
h = (729/180) cm
⇒ h ≈ 4.05 cm
Thus, the rise in water level in the rectangular vessel is 4.05 cm approximately.
I hope this helps you. If yes then please mark as brainliest answer. Thanks for giving me a chance to help you. Have a marvelous day ahead.
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