Home
Class 12
PHYSICS
In a continuous printing process paper i...

In a continuous printing process paper is drawn into the press at a constant speed v. Denoting by 'r' the radius of the paper roll at any given time and by 'b' the thickness of the paper. What is the angular acceleraion of the paper roll?

A

`(bv^(2))/(2pir^(3))`

B

`(2bv^(2))/(pir^(3))`

C

`(bv^(2))/(4pir^(3))`

D

`(4bv^(2))/(pir^(3))`

Text Solution

Verified by Experts

The correct Answer is:
A

Let one layer of paper be unrolled
`vDeltat =2pir and Deltar=-b`
`(Deltar)/(Deltat)=(-bv)/(2pir)=(dr)/(dt)`
`alpha=(domega)/(dt)=(d)/(dt)((v)/(r ))`
`=(1)/(r)(dv)/(dt)+v(d)/(dt)((1)/(r))`
`= 0 -(v)/(r^(2))(dr)/(dt)`
`=(-(v)/(r^(2)))(-(bv)/(2pir))`
`(bv^(2))/(2pir^(3))`
`alpha=(bv^(2))/(2pir^(3)`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • TEST PAPERS

    ALLEN |Exercise PHYSICS|16 Videos
  • TEST PAPERS

    ALLEN |Exercise MATHS|1 Videos
  • TEST PAPERS

    ALLEN |Exercise part-2 physics|73 Videos
  • TEST PAPER 4

    ALLEN |Exercise PHYSICS|45 Videos
  • UNIT & DIMENSIONS, BASIC MATHS AND VECTOR

    ALLEN |Exercise Exercise (J-A)|7 Videos

Similar Questions

Explore conceptually related problems

A disc of the radius R is confined to roll without slipping at A and B. If the plates have the velocities v and 2v as shown, the angular velocity of the disc is

Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed omega . The discs are in the same horizontal plane. At time t = 0 , the points P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is v_(r) . In one time period (T) of rotation of the discs , v_(r) as a function of time is best represented by

As shown in figure, a long wire kept vertically on the plane of paper carries electric current-I. A conducting ring moves towards the wire with velocity v with its plane conducting with the plane of paper. Find the induced emf produced in the ring when it is at a perpendicular distance r from the wire. Radius of the ring is a and a lt lt r.

A magnetic field B is confined to a region rle a and points out of the paper (the z-axis), r = 0 being the centre of the circular region. A charged ring (charge Q) of radius b, b gt a and mass m lies in the xy-plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time Deltat . Find the angular velocity omega of the ring after the field vanishes.

An accelration produces a narrow beam of protons, each having an initial speed of v_(0) . The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. One the potential of the sphere has reached its final, constant value, the minimum speed v of a proton along its trajectory path is given by

An accelration produces a narrow beam of protons, each having an initial speed of v_(0) . The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. After a long time, when the potential of the sphere reaches a constant value, the trajectory of proton is correctly sketched as

An accelration produces a narrow beam of protons, each having an initial speed of v_(0) . The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. If the initial kinetic energy of a proton is 2.56 ke V , then the final potential of the sphere is

An accelration produces a narrow beam of protons, each having an initial speed of v_(0) . The beam is directed towards an initially uncharges distant metal sphere of radius R and centered at point O. The initial path of the beam is parallel to the axis of the sphere at a distance of (R//2) from the axis, as indicated in the diagram. The protons in the beam that collide with the sphere will cause it to becomes charged. The subsequentpotential field at the accelerator due to the sphere can be neglected. The angular momentum of a particle is defined in a similar way to the moment of a force. It is defined as the moment of its linear momentum, linear replacing the force. We may assume the angular momentum of a proton about point O to be conserved. Assume the mass of the proton as m_(P) and the charge on it as e. Given that the potential of the sphere increases with time and eventually reaches a constant velue. The total energy (E) of a proton in the beam travelling with seed v at a distance of r (r ge R) from point O. Assuming that the sphere has acquired an electrostatic charge Q is

When an object moves through a fluid, as when a ball falls through air or a glass sphere falls through water te fluid exerts a viscous foce F on the object this force tends to slow the object for a small sphere of radius r moving is given by stoke's law, F_(w)=6pietarv . in this formula eta in the coefficient of viscosity of the fluid which is the proportionality constant that determines how much tangential force is required to move a fluid layer at a constant speed v, when the layer has an area A and is located a perpendicular distance z from and immobile surface. the magnitude of the force is given by F=etaAv//z . For a viscous fluid to move from location 2 to location 1 along 2 must exceed that at location 1, poiseuilles's law given the volumes flow rate Q that results from such a pressure difference P_(2)-P_(1) . The flow rate of expressed by the formula Q=(piR^(4)(P_(2)-P_(1)))/(8etaL) poiseuille's law remains valid as long as the fluid flow is laminar. For a sfficiently high speed however the flow becomes turbulent flow is laminar as long as the reynolds number is less than approximately 2000. This number is given by the formula R_(e)=(2overline(v)rhoR)/(eta) In which overline(v) is the average speed rho is the density eta is the coefficient of viscosity of the fluid and R is the radius of the pipe. Take the density of water to be rho=1000kg//m^(3) Q. What is the viscous force on a glass sphere of radius r=1mm falling through water (eta=1xx10^(-3)Pa-s) when the sphere has speed of 3m/s?

When an object moves through a fluid, as when a ball falls through air or a glass sphere falls through water te fluid exerts a viscous foce F on the object this force tends to slow the object for a small sphere of radius r moving is given by stoke's law, F_(w)=6pietarv . in this formula eta in the coefficient of viscosity of the fluid which is the proportionality constant that determines how much tangential force is required to move a fluid layer at a constant speed v, when the layer has an area A and is located a perpendicular distance z from and immobile surface. the magnitude of the force is given by F=etaAv//z . For a viscous fluid to move from location 2 to location 1 along 2 must exceed that at location 1, poiseuilles's law given the volumes flow rate Q that results from such a pressure difference P_(2)-P_(1) . The flow rate of expressed by the formula Q=(piR^(4)(P_(2)-P_(1)))/(8etaL) poiseuille's law remains valid as long as the fluid flow is laminar. For a sfficiently high speed however the flow becomes turbulent flow is laminar as long as the reynolds number is less than approximately 2000. This number is given by the formula R_(e)=(2overline(v)rhoR)/(eta) In which overline(v) is the average speed rho is the density eta is the coefficient of viscosity of the fluid and R is the radius of the pipe. Take the density of water to be rho=1000kg//m^(3) Q. Blood vessel is 0.10 m in length and has a radius of 1.5xx10^(-2) m blood flows at rate of 10^(-7)m^(3)//s through this vessel. The pressure difference that must be maintained in this flow between the two ends of the vessel is 20 Pa what is the viscosity sufficient of blood?