Home
Class 11
PHYSICS
A stone of mass 0.3 kg attched to a 1.5 ...

A stone of mass `0.3 kg` attched to a `1.5 m` long stirng is whirled around in a horizontal cirlcle at a speed of 6 m/s The tension in the string is

A

`10N`

B

`20 N`

C

`7.2N`

D

`30 N`

Text Solution

Verified by Experts

The correct Answer is:
C

Here, mass of the stone, m=0.3 kg
Length of the string, R= 1.5 m
Speed, v= 6 m/s
Tension in the string = centripetal force
` therefore T=(mv^2)/(R ) = ((0.3 ) (6)^2)/(1.5 ) = 7.2 N`
Promotional Banner

Similar Questions

Explore conceptually related problems

A 1kg stone at the end of 1m long string is whirled in a vertical circle at a constant speed of 4m//s . The tension in the string is 6N , when the stone is at (g=10m//s^(2))

A stone of mass 250 gram , attached at the end of a string of length 1.25 m is whirled in a horizontal circle at a speed of 5 m/s . What is the tension in the string ?

A stone of mass 16 kg is attached to a string 144 m long and is whirled in a horizontal circle .The maximum tension the string can stand is 16 newton .The maximum velocity of revolution that can be given to the stone without breaking it will be

A stone of mass of 16 kg is attached to a string 144 m long and is whirled in a horizontal circle. The maximum tension the string can withstand is 16 Newton . The maximum velocity of revolution that can be given to the stone without breaking it, will be

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 stone of mass 0.5 kg tied with a string of length 1 metre is moving in a circular path with a speed of 4 m/sec. The tension acting on the string in Newton is-

A stone of mass m tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by T = Ar^2 where A is a constant, r is the instantaneous radius fo the circle and n=....

A stone of mass m, tied to the end of a string, is. whirled around in a horizontal circle (neglect gravity). The length of the string is reduced gradually such that mvr = constant. Then, the tension in the string is given · by T = Ar '', where A is a constant and r is the instantaneous radius of the circle. Then, n is equal to:

A point mass of 2 kg tied to a string of 1 m length is rotated in a vertical circle with uniform speed of 4m/s. The tension in the string is nearly 32 N when mass is at