The mass of the liquid flowing per second per unit area of cross section of the tube is proportional to `P^(x) and v^(y)` , where `P` is the pressure difference and `v` is the velocity , then the relation between x and Y is

Options

A. ` x = y `
B. ` x = -y`
C. ` y^(2) = x`
D. ` y = -x^(2)`

Answer

B

Explanation

`(M)/(At) prop P^(x) v^(y)`
rArr `ML^(-2) T^(-1) = [ML^(-1)T^(-2)]^(x) [L^(1) T^(-1)]^(y) = M^(x) L^(-x + y )T^(-2 x - y)`
` x = 1 , -x + y = -2 , and -2x -y = -1`
From here, we get `y = -1` . Thus , `x = -y`.