Home
Class 11
PHYSICS
An object is moving through the liquid. ...

An object is moving through the liquid. The viscous damping force acting on it is proportional to the velocity. Then dimensions of constant of proportionality are

A

`ML^(-1)T^(-1)`

B

`MLT^(-1)`

C

`M^(0)LT^(-1)`

D

`ML^(0)T^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • TRANSMISSION OF HEAT

    ERRORLESS |Exercise ET Self Evaluation Test|23 Videos
  • VECTORS

    ERRORLESS |Exercise Exercise|223 Videos

Similar Questions

Explore conceptually related problems

The damping force on an oscillator is directly proportional to the velocity. The units of the constant to proportionality are

The damping force on an oscillator is directly proportional to the velocity .The units of the constant of propotionality are

The viscous force acting on a body moving through a liquid is proportional to its velocity. What is the dimensional formula for the constant of proportional to its velocity. What is the dimensional formula for the constant of proportionality?

The damping force acting on an oscillating body is proportional to its velocity i.e. F = Kv. What is the dimensional formula for K?

The viscous force acting on adjacent layers of liquid is

An oil drop of mass m fall through a viscous medium. The viscous drag force. F is proportional to the velocity of the drop . At the instant it begins to fall the force that acts on the oil drop is (neglect buoyancy)

An object falling from rest in air is subject not only to the gravitational force but also to air resistance.Assume that the air resistance is proportional to the velocity with constant of proportionality as k>0, and acts in a direction opposite to motion (g=9.8(m)/(s^(2))) Then velocity cannot exceed.

If an object is moving with a constant acceleration, the net force acting on that body is

Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} If an object of mass 2 kg and constant b = 4 (N-s)/(m) has terminal speed v_(T) in a liquid then time required to reach 0.63 v_(T) from start of the motion is :

The viscous force acting on a solid ball of surface area. A moving with terminal velocity v is proportional to