Home
Class 11
PHYSICS
If dimensions of critical velocity of a ...

If dimensions of critical velocity of a liquid `v_c` flowing through a tube are expressed as `[eta^xrho^yr^z]`, where `eta`, `rho` and `r` are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of `x`, `y` and `z` are given by :

A

-1,-1,1

B

-1,-1,-1

C

1, 1, 1

D

1,-1,-1

Text Solution

Verified by Experts

The correct Answer is:
D
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • NATURE OF PHYSICAL WORLD AND MEASUREMENT

    PREMIERS PUBLISHERS|Exercise OTHER IMPORTANT QUESTIONS & ANSWERS (V. Numerical Problems.)|102 Videos

  • NATURE OF PHYSICAL WORLD AND MEASUREMENT

    PREMIERS PUBLISHERS|Exercise OTHER IMPORTANT QUESTIONS & ANSWERS (VI. Conceptual Questions.)|20 Videos

  • NATURE OF PHYSICAL WORLD AND MEASUREMENT

    PREMIERS PUBLISHERS|Exercise TEXTBOOK QUESTIONS & ANSWERS (V. Conceptual Questions.)|10 Videos

  • MOTION OF SYSTEM OF PARTICLES AND BODIES

    PREMIERS PUBLISHERS|Exercise OTHER IMPORTANT QUESTIONS & ANSWER (NUMERICAL PROBLEMS)|60 Videos

  • OSCILLATIONS

    PREMIERS PUBLISHERS|Exercise OTHER IMPORTANT QUESTIONS & ANSWERS ( CONCEPTUAL QUESTIONS.)|24 Videos

Similar Questions

Explore conceptually related problems

When a solid moves through a liquid, the liquid opposes the motion with a force F. The magnitude of F depends on the coefficient of viscosity eta of the liquid, the radius r of the sphere and the speed v of the sphere. Assuming that F is proportional to different powers of these quantities, guess a formula for F using the method of dimension.

A liquid flows throug ha pipe of 1.0 mm radius and 10 cm length under a pressure 10^(4) dyne cm^(-2) . Calculate the rate of flow and the speed of the liquid coming out of the tube. The coefficient of viscosity of the liquid is 1.25 centipoise.

Knowledge Check

  • The rate of low of liquids in a tube of radius V length l, whose ends are maintained at a pressure difference p is V = (pi Q p r^(4))/(eta l) , where eta is coefficient of viscosity & Q is

    A
    8
    B
    `(1)/(8)`
    C
    16
    D
    `(1)/(16)`

  • Two capillaries of radii r_(1)andr_(2) and lengths l_(1)andl_(2) respectively are in series. A liquid of viscosity eta is flowing through the combination under a pressure difference P. What is the rate of volume flow of liquid?

    A
    `(8piP)/(eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))`
    B
    `(piP)/(8eta)((r_(1)^(4))/(l_1)+(r_(2)^(4))/(l_2))^(-1)`
    C
    `(8piP)/(eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))^(-1)`
    D
    `(piP)/(8eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))^(-1)`

  • Two capillaries of radii r_(1)andr_(2) and lengths l_(1)andl_(2) respectively are in series. A liquid of viscosity eta is flowing through the combination under a pressure difference P. What is the rate of volume flow of liquid?

    A
    `(8piP)/(eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))`
    B
    `(piP)/(8eta)((r_(1)^(4))/(l_1)+(r_(2)^(4))/(l_2))^(-1)`
    C
    `(8piP)/(eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))^(-1)`
    D
    `(piP)/(8eta)((l_1)/(r_(1)^(4))+(l_2)/(r_(2)^(4)))^(-1)`

    • Similar Questions

    Explore conceptually related problems

    A liquid flows throug ha pipe of 1.0 mm radius and 10 cm length under a pressure 10^(4) dyne cm^(-2) . Calculate the rate of flow and the speed of the liquid coming out of the tube. The coefficient of viscosity of the liquid is 1.25 centipoise.

    The cylindrical tube of a spray pump has radius R, one end of which has 'n' fine holes, each of radius r. If the speed of the liquid in the tube is V, what is the speed of the ejection of the liquid through the holes.

    Find the dimensional formulae of the following quantities: a. the universal constant of gravitation G, b. the surface tension S, c. the thermal conductivity k and d. the coeficient of viscosity eta . Some equation involving these quantities are F=(Gm_1m_2)/r^2, S= (rho g r h)/2, Q=k(A(theta_2-theta_1)t)/d and F=- eta A (v_2-v_1)/(x_2-x_1) where the symbols have their usual meanings.

    A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n_(2)= n_(1) "exp"[-mg(h_(2)-h_(1))//k_(B)T] where n_(2), n_(1) refer to number density at heights h_(2) and h_(1) respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n_(2)= n_(1)"exp"[ -mgN_(A)(rho-rho')(h_(2)-h_(1))//rhoRT)] where rho is the density of the suspended particle, and rho' , that of surrounding medium. [ N_(A) is Avogadro's number, and R the universal gas constant.] [Hint : Use Archimedes principle to find the apparent weight of the suspended particle.]

    Three capillary tubes of the same radius r but of lengths l_(1),l_(2)andl_(3) are fitted horizontally to the bottom of a tall vessel containing a liquid at constant head and flowing through these tubes. Calculate the length of a single outflow tube of the same radius r which can replace the three capillaries.

    Home
    Profile