2 kg of ice at `-20^(@)C` is mixed with 5kg of water at `20^(@)C`. The water content of the final mixture is (Latent heat of ice `="80 kcal kg"^(-1)`, the specific heat of water `="1 kcal kg"^(-1).^(@)C^(-1)` and specific heat of ice `="0.5 kcal kg"^(-1).^(@)C^(-1)`)

Options

A. `x_(1):x_(2):x_(3) = sqrt(5):sqrt(6):sqrt(3)`
B. `x_(1):x_(2):x_(3)= sqrt(6):sqrt(5):sqrt(3)`
C. If `A_(1), A_(2), A_(3)` are the area of cross-section of streams when it hits the ground then `12A_(1)^(2):5A_(2)^(2):4A_(3)^(2)= 3:3:2`
D. If `A_(1), A_(2), A_(3)` are the area of cross-section of streams when it hits the ground then `12A_(1)^(2):5A_(2)^(2):4A_(3)^(2)= 2:2:3`

Answer

A, D

Explanation

In the figure …………
For case -1 :
`rho gh = (1)/(2)rho v_(1)^(2)`
`v_(1)= sqrt(2gh)`
`t_(1) = sqrt((10h)/(g))`
For case -2 :
`rho g2h+2rho gh = (1)/(2)2rho v_(2)^(2)`
`v_(2) = sqrt(4gh)`
`t_(2) = sqrt((6h)/(g))`
For case -3 :
`rho g2h+2rho g2h+3rho gh= (1)/(2) 3rho v_(3)^(2)`
`v_(3) = sqrt(6gh)`
`t_(3) = sqrt((2h)/(g))`
`x_(1)=v_(1)t_(1)= sqrt(20)` , `x_(2) = sqrt(24)`
`x_(3) = sqrt(12)`
`x_(1) : x_(2) : x_(3) = sqrt(20) : sqrt(24) : sqrt(12)`
`= sqrt(10) : sqrt(12) : sqrt(6)`
`= sqrt(5) : sqrt(6) : sqrt(3)` .
`a` = area of cross-section of hole.

`av_(1) = a'v_(1)^(1)`
`v'_(1)^(2)=v_(1)^(2)+2gh_(1)`
`a'=(av_(1))/(v_(1)^(1)) = (av_(1))/(sqrt(v_(1)^(2)+2gh_(1))`
`a^(2) = (a^(2)v_(1)^(2))/(v_(1)^(2)+2gh_(1))`
`A_(1)^(2) = (a^(2)xx2gh)/(2gh+2gxx3h) = (a^(2))/(6)`
`A_(2)^(2) = (a^(2)xx4gh)/(4gh+2gxx3h) = (2)/(5)a^(2)`
`A_(3)^(2) = (a^(2)xx6gh)/(6gh+2gxxh)=(3)/(4)a^(2)`
`12A_(1)^(2) : 5A_(2)^(2) : 4A_(3)^(2) = 12xx(a^(2))/(6) : 5xx(2)/(5)a^(2) : 4xx(3)/(4)a^(2)`
`= 2:2:3`