A boy purchases 2 litre of milk from a shop. To find out the extent of adulteration he constructed a device by using a capillary tube and a cylindrical container. In order to make the device float upto a mark M , in pure milk he adds 10 lead shots each of mass 5 g into it. To make the device float to the same mark in a sample of milk purchased, he removes one lead shot from it. The total weight of device is 200 gwt. Determine the volume of water that the shopkeeper adds to the milk. Density of pure milk is 1.045 g/cm^3 and density of water is 1 g/cm^3.

Asked by KSHITIJ AGRAWAL | 30th Jun, 2013, 09:27: AM

Expert Answer:

Weight of the device + Weight of the lead shots = Volume of the milk displaced*density of the milk. 
Let W be the weight of the device, 
w be the weight of 1 lead shot, 
Vm be the volume of the milk displaced (constant in 2 cases as the device is floating to the same level M)
dim be the density of the impure milk
dp be the density of the pure milk
So, in the case of pure milk
W + 10*w = v*d
200+ 10*5 = v*1.045
250/1.045 = v 
v = 239.23 cm3
For impure milk, W+9w = v*dim
200+ 9*5 = 239.23*dim
245/239.23 = dim
dim  = 1.0241 g/cm3

Answered by  | 30th Jun, 2013, 05:12: PM

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