Question

A measuring cylinder contains 60cm3 of oil at 0 celcius. When a piece of ice was roped into the cylinder it sank completely in oil and the oil level rose to 90cm3 mark. When the ice is melted the oil level came down to 87 cm3 mark, The relative density of ice is,

Answer 1

`S_i=(9)/(10) =0.9`

Step-by-step explanation:

Given:

• volume of oil in the cylinder,
  `V_o=60\ cm^2`

• volume of the oil level when the ice is immersed,
  `V=90\ cm^3`

• the volume level of oil when the ice melted,
  `V'=87\ cm^3`

Now, therefore the volume of ice:

`V_i=V-V_o`

`V_i=90-60`

`V_i=30\ cm^3`

Now the volume of water:

`V_w=V'-V_o`

`V_w=87-60`

`V_w=27\ cm^3`

As we know that the relative density is the ratio of density of the substance to the density of water.

So, the relative density of ice:

`S_i=(\rho_i)/(\rho_w)` .....................(1)

as we know that density is given as:

`\rm \rho=(mass)/(volume)`

now eq. (1)

`S_i=(m)/(V_(i))/ (m)/(V_w)`

where, m = mass of the water or the ice which remains constant in any phase

`S_i=(V_w)/(V_i)`

`S_i=(27)/(30)`

`S_i=(9)/(10) =0.9`