Q 1 :    

A sphere of relative density σ and diameter D has concentric cavity of diameter d. The ratio of Dd, if it just floats on water in a tank is:            [2024]

  • (σσ-1)13

     

  • (σ+1σ-1)13

     

  • (σ-1σ)13

     

  • (σ-2σ+2)13

     

(1)   

        Weight of sphere, W=43π(D3-d38)σg

         Buoyant force Fb=43π(D38)ρwg

         For Just Float, W=Fb

        43π(D3-d38)σg=43π(D38)ρwg

        1-d3D3=ρwσ(ρw=1)

          1-1σ=(dD)3Dd=(σσ-1)1/3

 



Q 2 :    

A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of kerosene oil = 0.8, specific gravity of ice = 0.9)      [2024]

  • 5 : 4

     

  • 9 : 10

     

  • 1 : 1

     

  • 8 : 9

     

(3)

V1=volume immersed in water

V2=volume immersed in oil

Buoyancy force balance weight at equilibrium

FB1+FB2=mg

V1ρwg+V2ρog=(V1+V2)ρcg

V1+V2ρoρw=(V1+V2)ρcρw

V1+0.8V2=0.9V1+0.9V2

0.1V1=0.1V2V1:V2=1:1V1:V2=1:1



Q 3 :    

A 400 g solid cube having an edge of length 10 cm floats in water. How much volume of the cube is outside the water? (Given: density of water = 1000 kg m3)          [2025]

  • 1400 cm3

     

  • 4000 cm3

     

  • 400 cm3

     

  • 600 cm3

     

(4)

Volume of cube inside water

Vcube, in=(Density of cubeDensity of water)×Volume of cube

                     =Mass of cubeDensity of water=400 gm1 gm/cm3=400 cm3

Volume of cube outside water

Vcube, out = Volume of cube – Volume of cube inside water

                       = 1000 cm3 – 400 cm3 = 600 cm3



Q 4 :    

A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 cm. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water.

(Take the density of unknown mass more than that of the water, the mass did not absorb water and water density is 1 gm/cm3.)The unknown mass is ______ kg.          [2025]



(3)

Given, volume of block =(10×102)3=103 m3

Let density of block ρ kg/m3

mass of block =ρ×103 kg

Buoyant Force (FB)=1000×1032×10=5 N

F.B.D. of blocks

Balancing torque about point O, we get

mg(2×102)FB(2×102)=0.2 g(25×102)

ρ×103×10×210=50  ρ=3000 kg/m3

Hence, mass of block, m=3000×103=3 kg