Home
Class 12
PHYSICS
A stone of mass 0.1 kg tied to one end o...

A stone of mass 0.1 kg tied to one end of a string lm long is revolved in a horizontal circle at the rate of `(10)/(pi)" rev/s"`. Calculate the tension in the string.

Text Solution

AI Generated Solution

To solve the problem of calculating the tension in the string when a stone is revolved in a horizontal circle, we can follow these steps: ### Step 1: Identify the given values - Mass of the stone, \( m = 0.1 \, \text{kg} \) - Length of the string (radius of the circle), \( r = 1 \, \text{m} \) - Rate of revolution, \( f = \frac{10}{\pi} \, \text{rev/s} \) ### Step 2: Convert the rate of revolution to angular velocity ...
Promotional Banner

Topper's Solved these Questions

  • Motion in Straight Line

    VMC MODULES ENGLISH|Exercise SOLVED EXAMPLES|20 Videos

  • Motion in Straight Line

    VMC MODULES ENGLISH|Exercise PRACTICE EXERCISE|80 Videos

  • MOTION IN A STRAIGHT LINE & PLANE

    VMC MODULES ENGLISH|Exercise IMPECCABLE|52 Videos

  • Motion in Two Dimensions

    VMC MODULES ENGLISH|Exercise MCQ|2 Videos

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))

Stone of mass 1 kg tied to the end of a string of length 1m , is whirled in horizontal circle with a uniform angular velocity 2 rad s^(-1) . The tension of the string is (in newton)

A mass is tied to a, string 1 m long and rotated in a horizontal circle at the rate of 600 rpm. If the tension inthe string is 3943.8 N, calculate the mass.

A stone of mass 2.0 kg is tied to the end of a string of 2m length. It is whirled in a horizontal circle. If the breaking tension of the string is 400 N, calculate the maximum velocity of the stone.

A stone of mass 100 g tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed.If the stone makes 14 revolutions in 25 s, calculate the centripetal force acting on the stone.

A mass of 100 gm is tied to one end of a string 2 m long. The body is revolving in a horizontal circle making a maximum of 200 revolutions per min. The other end of the string is fixed at the centre of the circle of revolution. The maximum tension that the string can bear is (approximately)

A string breaks under a load of 4.8kg A mass of 0.5 kg is attached to one end of a string 2 m long and is rotated in a horizontal circle . Calculate the greatest number of revolutions that the mass can make without breaking the string .

A stone of mass 0.3 kg tied to the end of a string in a horizontal plane in whirled round in a circle of radius 1 m withp. speed of 40 rev/min. What is the tension in the string ? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N ?

A stone of mass of 1 kg is tied to the end of a string 1 m long. It is whirled in a vertical circle. The velocity of the stone at the bottom of the circle is just sufficient to take it to the top of circle without slackening of the string. What is the tension in the string at the top of the circle? (Take, g =10 ms^(-2) )

A body of mass 200g tied to one end of string is revolved in a horizontal circle of radius 50cm with angular speed 60 revolution per minute ("rpm") on a smooth horizontal surface. Find (a) linear speed, (b) the acceleration and ( c) tension in the sting. what will happen if string is broken? (Take pi^(2)=10)