Home
Class 12
PHYSICS
The mass of the liquid flowing per secon...

The mass of the liquid flowing per second per unit area of cross section of the tupe 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:

A

`x=y`

B

`x=-y`

C

`y^(2)=x`

D

`y=-x^(2).`

Text Solution

Verified by Experts

The correct Answer is:
B
`(M)/(At)=ML^(-2)T^(-1)=[ML^(-1)T^(-2)]^(x)[L^(1)T^(-1)]^(y)`
`=M^(x)L^(-x+y)T^(-2x-y)`
`x=1`
`-x+y=-2`
`y=-1`
Thus `x=-y.`
Hence correct choice is `(b).`
Promotional Banner

Topper's Solved these Questions

  • UNITS, MEASURMENTS, DIMENSIONS, ERRORS OF MEASUREMENTS

    MODERN PUBLICATION|Exercise Multiple Choice Questions (Level-II)|47 Videos

  • UNITS, MEASURMENTS, DIMENSIONS, ERRORS OF MEASUREMENTS

    MODERN PUBLICATION|Exercise Multiple Choice Questions (Level-III)|6 Videos

  • UNIT TEST PAPER NO.4

    MODERN PUBLICATION|Exercise MCQ|31 Videos

  • WAVE MOTION

    MODERN PUBLICATION|Exercise RECENT COMPETITIVE QUESTIONS|18 Videos

Similar Questions

Explore conceptually related problems

Water is in streamline flow along a horizontal pipe where the area of cross-section is 10" cm"^(2) , the velocity of water is 1" m s"^(-1) and the pressure is 2000 Pa. The pressure at another point where the cross-sectional area is 5" cm"^(2) is

P( a, b) is the mid-point of a line segment between axes. Show that equation of the line is x/a + y/b = 2 .

The area enclosed between the curves, y = ax^(2) and x = ay^(2) ( a gt 0) is 1 square units, then the value of a is