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a) The density of hollow sphere of inner and outer radii of R1 and R2 is 8g/cc.If it is just immersed in water the find the ratio of the radii. b) Two liquids A and B in different test tubes of same length and negligible mass just immerse in water. The length of liquid column of A and B are L/4 and L/3 respectively. Find their density ratio. (L = length of the tube.)

a) The density of hollow sphere of inner and outer radii of R1 and R2 is 8g/cc.If it is just immersed in water the find the ratio of the radii.

b) Two liquids A and B in different test tubes of same length and negligible mass just immerse in water. The length of liquid column of A and B are L/4 and L/3 respectively. Find their density ratio. (L = length of the tube.)

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3 Answers

Chetan Mandayam Nayakar
312 Points
13 years ago

let R2=kR1, 8*(4pi/3)(R1)3(k3-1)=(4pi/3)(R1)3k3, 8k3-8=k3,k=7k3=8,k=2/(7)1/3

(DA)gL/4=(DB)gL/3, DA/DB=4/3

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AJIT AskiitiansExpert-IITD
68 Points
13 years ago

Dear Ajit ,

a) just immersed means that the weight of liquid displaced by it is just equal to its weight

       mass of sphere  =  volume * density

                                   =  4/3*∏* (R23-  R13) * 8  =  mass of water displaced   =  =  4/3*∏* (R23) * 1

             so , 7R23  =8  R13    so  , R1/R2 =  (7/8)1/3

b) if the test tube are of equal lengths , then they must displace same amount of water and they both just immerse so their mass must also be equal

                     AL/4 * D1  =  AL/3*D2

                   D1/D2 =  4 /3

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Chetan Mandayam Nayakar
312 Points
13 years ago

Dear Ajit, 

8*(4pi/3)(R23-R13)=(4pi/3)R23, 7R23=8R13,R2/R1=2/(71/3)

let dA and dB be the densities of the two liquids, and let a and b be the cross-sectional areas of A and B respectively.w=density of water

 dA(L/4)ag=Lawg, dA/4=w. similarly, we get dB/3=w, therefore dA/dB=4/3

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Chetan Mandayam Nayakar – IIT Madras  

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