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Question

A metallic sphere floats in an immiscible mixture of water (ρW=103kg/m3) and a liquid (ρL=13.5×103) such that its 45thvolume is in water and 15thvolume is in the liquid. Find the density of the metal.


A

4.5×103kg/m3

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B

4.0×103kg/m3

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C

3.5×103kg/m3

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D

1.9×103kg/m3

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Solution

The correct option is C

3.5×103kg/m3


Step 1: Given parameters

The density of water, ρW=103kg/m3

The density of the liquid, ρL=13.5×103

Volume in water 45th

Volume in liquid 15th

Step 2: To find

We have to find the density of the metal.

Step 3: Calculation

We know that,

Totalupthrust=Weightofthemetalsphere

Hence,

45VρWg+15VρLg=(m)g

Here,

ρW is the density of water.

ρL is the density of the liquid.

V is the volume.

ρm is the density of the metal.

By substituting the value,

ρm=45×103+15×13.5×103=3.5×103kg/m3

Therefore, option B is the correct choice.


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