Home
Class 12
PHYSICS
Two point like objects each with mass m ...

Two point like objects each with mass m are connected by a massless rope of length `l` the object are suspended vertically near the surface of earth, so that one object hanging below the other then the objects are relased. Show that the tension in the rope is `T=(GMml)/(R^(3))` where M is the mass of the earth and R is its radius `|l lt lt R|`

Text Solution

Verified by Experts

The correct Answer is:
n/a

`mg-T=ma` ..(1)
`T+mg'=ma` …(2)
by adding (1) and (2)
`a=(g+g')/(2),T=(m(g+g))/(2)-mg'`
`T=m((g-g')/(2))=(m)/(2)[g-g[1-(2h)/(R)]]`
`T=(mg)/(2)(2l)/(R)=(mgl)/(R)=(GMml)/(R^(3))`
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    ALLEN |Exercise Exercise 4 B (Brain Storming Subjective Exercise)|20 Videos

  • GRAVITATION

    ALLEN |Exercise Exercise 5 A (Previous Year Questions)|26 Videos

  • GRAVITATION

    ALLEN |Exercise Exercise 3 (Miscellaneous Type Questions)|20 Videos

  • GEOMETRICAL OPTICS

    ALLEN |Exercise subjective|14 Videos

  • KINEMATICS-2D

    ALLEN |Exercise Exercise (O-2)|47 Videos

Similar Questions

Explore conceptually related problems

Show that if a body is projected vertically upward from the surface of the earth so as to reach a height nR above the surface (i) the increase in its potential energy is ((n)/(n + 1)) mgR , (ii) the velocity with which it must be projected is sqrt((2ngR)/(n + 1)) , where r is the radius of the earth and m the mass of body.

A larger spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M. The point masses are connected by rigid massless rod of length l and this assembly is free to move along the line connecting them. All three masses interact only throght their mutual gravitational interaction. When the point mass nearer to M is at a distance r =3l form M, the tensin in the rod is zero for m =k((M)/(288)). The value of k is

A ball of mass m is fired vertically upwards from the surface of the earth with velocity nv_(e) , where v_(e) is the escape velocity and nlt1 . Neglecting air resistance, to what height will the ball rise? (Take radius of the earth=R):-

An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to 2R from the centre of the earth. What is the gain in its potential energy ?

Two metallic balls of mass m are suspended by two strings of length L . The distance between upper ends is l . The angle at which the string will be inclined with vertical due to attraction is (m lt lt M where M is the mass of Earth)

If g is the acceleration due to gravity on earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is

Two identical rods each of mass m and length L , are tigidly joined and then suspended in a vertical plane so as to oscillate freely about an axis normal to the plane of paper passing through 'S' (point of supension). Find the time period of such small oscillations

Two point masses m_1 and m_2 are connected by a spring of natural length l_0 . The spring is compressed such that the two point masses touch each other and then they are fastened by a string. Then the system is moved with a velocity v_0 along positive x-axis. When the system reached the origin, the string breaks (t=0) . The position of the point mass m_1 is given by x_1=v_0t-A(1-cos omegat) where A and omega are constants. Find the position of the second block as a function of time. Also, find the relation between A and l_0 .

A body of mass m is situated at a distance 4R_(e) above the earth's surface, where R_(e) is the radius of earth. How much minimum energy be given to the body so that it may escape

Figure shows a wire of length 2L and crosssectional area A, stretched horizontally between two clamps. When an object of mass M is suspended from the mid point of the wire, the downward displacement of the young's modulus of the material of the wire is x. Shown that M = (YA x ^(3))/(gL ^(3)) (where theta is small)