a uniform circular coin required too much metal relative to its value.. it is decided that the diameter of the coin can be reduced by 10% by how much will weight of the coin reduce if everything else about the coin remains unchanged.

• It will be in same ratio as change in volume of coin.
  considering thickness remains unchanged,
  change in volume = change in area of one side.
  Earlier area = pi*r^2
  new area = pi*(0.9*r)^2 =pi*0.81*r^2 as radius is also reduced by 10%
  
  % reduction in area = 100*0.19/1 = 19%
• 11 years agoHelpfull: Yes(23) No(12)
• how could u tell area=weight of the material?
• 11 years agoHelpfull: Yes(4) No(0)
• % of weight reduced=19%
• 11 years agoHelpfull: Yes(1) No(3)
• %reduce in weight = %reduce in volume
  This is because density remains constant
  i.e mass/volume = constant
  => mass ∝ volume ∝ weight.
  
  Also volume = area x thickness,
  thickness is constant
  => volume ∝ area
  
  Therefore
  Weight ∝ area
• 11 years agoHelpfull: Yes(1) No(0)
• m=d/v
  here d means denisity remain constant
  v means volume
  m1m2=(v2)^3/(v1)3
  m1=.729m2
  so the reduced weight is 1-.729=.271
  
• 11 years agoHelpfull: Yes(0) No(0)
• 19% weight will be reduced diameter is reduced by 10% means diameter 2r will be 1.8 r. Therefore new radius = 0.9 r Volume formula Pir r^2h/Pie (0.9r^2) h.It implies it is 0.81 means 81% Weight reduced is 100-81 19%
• 11 years agoHelpfull: Yes(0) No(0)
• There is short cut method for this types i.e net-equivalent
  for reduce -
  increase +
  10% reduce so
  -10-10+(-10*-10/100)
  -20+100/100
  -20+1=-19
  ie19%
• 8 years agoHelpfull: Yes(0) No(0)