A duster is of dimension 5*5*7 is sinked into a cylindrical vessel of radius 5 cm and height 10 cm .After insertion the level of the water is inccresed by 2.14 cm ..Then by what percentage the part of the duster is above/below(something like that) the layer?

• i guess answer is 96% . by relation volume of water displaced we can get the volume of immersed part and we know the toatal volume therefore we can get the answer
  vol of water displaced = 3.14*5*5*2.14= 167.99 ==168
  this is nothing but the volume of immersed part
  now toatal volume of duster=l*b*h= 25*7=175
  %below= 16800/175= 96%
  
• 10 years agoHelpfull: Yes(11) No(0)
• volume of rise water=pi*5*5*2.14=167.99 ~168
  volume of duster=5*5*7=175
  %part of duster below the level is=(175-168)*100/175
• 10 years agoHelpfull: Yes(5) No(7)
• volume of duster=175cm^3
  v of cylnder=3.14*25*10
  volume increase by duster=3.14*25*2.14
  now suppose x volume of duster is either above or below layer
  that much x volume of duster is responsible for gain in volume of cylinder.
  so, x=3.14*25*2.14(volume of duster above or below)
  tptal volume of duster-x=175-3.14*25*2.14=6.9cm^3
  ans-6.9%
• 10 years agoHelpfull: Yes(2) No(1)
• which option was correct..
• 10 years agoHelpfull: Yes(0) No(0)
• i think in this question all available answer were wrong
• 10 years agoHelpfull: Yes(0) No(1)
• Since the level of water level increased is 2.14cm.
  volume occupied by duster in water is =pi*r^2*2.14=pi*5^2*2.14=53.50
  percentage of part of the duster inside the layer is =53.50/2(lb+bh+hl)
  =53.50/190=.28= 28%
  
• 10 years agoHelpfull: Yes(0) No(5)
• Dimension of the duster is assumed as 5*5*
  Volume of the duster below the water layer
  = volume of water increased
  = pir2h = 3.14×52×2.14 =168
  Total volume of the duster = 5×5×7 = 175
  
  Percentage of the duster part below the layer = 168×100175 = 96%
  
  Percentage of the duster part above the layer = 4%
• 9 years agoHelpfull: Yes(0) No(0)
• Dimension of the duster is assumed as 5*5*7 cm3 (unit is not given in the question)
  
  Volume of the duster below the water layer
  = volume of water increased
  = πr2h = 3.14×52×2.14 ≈ 168 cm3
  
  Total volume of the duster = 5×5×7 = 175 cm3
  
  Percentage of the duster part below the layer =
  168
  ×
  100
  175
  168×100175 = 96%
  
  Percentage of the duster part above the layer = 4%
• 7 years agoHelpfull: Yes(0) No(0)