Home
Class 12
PHYSICS
A stone of mass 1 kg is tied to a string...

A stone of mass 1 kg is tied to a string 2m long and it's rotated at constant speed of 40 `ms^(-1)` in a vertical circle. The ratio of the tension at the top and the bottom is
[Take g=10m`s^(2)`]

A

`81/79`

B

`79/81`

C

`19/12`

D

`12/19`

Text Solution

Verified by Experts

The correct Answer is:
B
Free body diagram (FBD) of a stone moving in a vertical circular path, which has tension force at point A and B as represented by TA and Ta respectively as given below in the figure,

Given, mass of stone (m) = 1 kg, length of the string ( R) = 2 m and
rotating· linear speed (v) = 40m`s^(-1)`
As, we know that the tension at position A,
`T_(A)=(mv^(2))/R+mg (thereforeF_(c)=(mv^(2))/R)`
`rArrT_(A)=(1xx(40)^(2))/2+1xx10=810N`
Similarly, tension at position B,
`rArrT_(B)=(mv^(2))/R-mg=(1xx(40)^(2))/R-mg=(1xx(40)^(2))/2-1xx10=790` N
So, the ratio of `T_(B)` and `T_(A)` i.e.,
`(T_(B))/(T_(A))=790/810=79/81`
Hence: the ratio of tension at position B and tension at position A is 79 : 81 .
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MHTCET 2018

    MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS|Exercise Physics|50 Videos

  • MHTCET 2019 PAPER 2

    MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS|Exercise PHYSICS|50 Videos

Similar Questions

Explore conceptually related problems

A stone of mass 50 g is tied to the end of a string 2 m long and is set into rotation in a horizontal circle with a uniforn speed of 2m//s .Then tension in the string is

A paricle of mass 2 kg is rotating by means of a string in a verticle circle. The difference in the tensions at the bottom and the top would be

Knowledge Check

  • A stone of mass 1 kg tied to a string 4 m long and is rotated at constant speed of 40 m/s a vertical circle. The ratio of the tension at the top and the bottom, is (g=10" m"//"s"^(2))

    A
    `11 : 12`
    B
    `39 : 41`
    C
    `41 : 39`
    D
    `12 : 41`

  • A stone of mass m is tied to a string of length l and rotated in a circle with a constant speed v . If the string is released, the stone flies

    A
    Radially outward
    B
    Radialy inward
    C
    Tangentially outward
    D
    With acceleration `(mv^(2))/l`

  • A particle of mass m is rotating by means of a string in a vertical circle . The difference in tension at the top and the bottom revolution is

    A
    6 mg
    B
    4 mg
    C
    2 mg
    D
    3 mg

    • Similar Questions

    Explore conceptually related problems

    A stone of mass 1 kg tied at the end of a string of length 1 m and is whirled in a verticle circle at a constant speed of 3 ms^(-1) . The tension in the string will be 19 N when the stone is (g=10 ms^(-1))

    A body of mass 1 kg is suspended from a string 1 m long it is rotated in a vertical circle , what is the tension in the string when it is horizontal and the speed of the body is 2m/s ?

    A particle of mass m is rotating by means of a string in a vertical circle. The difference in the tension at the bottom and top would be-

    A mass of 5 kg is tied to a string of length 1.0 m is rotated in vertical circle with a uniform speed of 4 m//s . The tension in the string will be 170 N when the mass is at (g=10 m//s^(2))

    A mass of 5 kg is tied to a string of length 1.0 m and is rotated in vertical circle with a uniform speed of 4m//s . The tension in the string will be 130 N when the mass is at (g= 10 m//s^2 )

    Home
    Profile