Physics, asked by vijayaghatol30, 9 months ago

When an object is made to float in two different
liquids of densities d1, and d2, the length of the object
seen above the liquid surface are 4 cm and 5 cm
respectively. Which of the following is the correct
alternative?
A. d1>d2
B. d1<d2
C. d1= d2
D. None of these

pls answer only if you are sure​

Answers

Answered by IamIronMan0
2

A

Explanation:

If the density of liquid less then object's it will float or more part will remain outside water . So

d2 < d1

You can do exact calculation by balancing buoyancy and gravition forces .

Answered by sonuvuce
1

When an object is made to float in two different  liquids of densities d1 and d2, the length of the object  seen above the liquid surface are 4 cm and 5 cm  respectively.

d1 < d2

Therefore, Option B is correct

Explanation:

We know that the object will float in the liquid if the weight of the liquid displaced by the object is is sufficient enough to balance the weight of the object.

The liquid displaced by the object in the case of liquid with density d1 will be more than that of the liquid displaced in case of liquid of density d2

Therefore, the volume of liquid in case of d1 density is more than that in case of d2 density

Therefore, d1 < d2 as they both are balancing the same weight of the object.

Hope this helps.

Know More:

Q: An object is put in three liquids having different densities one by one. The object floats with 1/9, 2/11 and 3/7 parts of its volume outside the surface of liquids of densities d1, d2 and d3 respectively. Which of the following is the correct order of the densities of the three liquids?

Click Here: https://brainly.in/question/2362336

Q: Same Question as above.

Click Here: https://brainly.in/question/13647742

Similar questions