Home
Class 11
PHYSICS
A chain consisting of 5 links each of ma...

A chain consisting of `5` links each of mass `0.1kg` is lifted vertically with a constant acceleration of `2m//s^(2)` as shown in The force of interaction (in newton) between the top link and the link immediately below it will be Take `g = 10m//s^(2)`
.

Text Solution

Verified by Experts

The correct Answer is:
`3`
If `F` is the upward force applied then
`F - 5 mg =5 ma`
`F = 5 m (g + a)`
If `T` is force of interaction between the top link and link immediately below it then
`ma = F - mg -T`
`T = F - mg - ma = 5 m (g + a) - m (g + a)`
`T = 4 m (g + a) = 4 xx 0.1 (10 + 2) = 3N` .
Promotional Banner

Topper's Solved these Questions

  • LAWS OF MOTION

    PRADEEP|Exercise Assertion- Reason Type Questions|17 Videos

  • LAWS OF MOTION

    PRADEEP|Exercise JEE (Main and Advanced)/ Medical Entrance Special|93 Videos

  • KINEMATICS

    PRADEEP|Exercise 1 NCERT Comprehension|4 Videos

  • MATHEMATICAL TOOLS

    PRADEEP|Exercise Fill in the blanks|5 Videos

Similar Questions

Explore conceptually related problems

A chain consisting of 5 links each of mass 0.1 kg is lifted vertically with a constant acceleration of 2.5 m//s^(2) as shown in the figure. The force of interaction between the top link and the link immediately below it, will be

A chain consisting of 5 links each of mass d 0.1 kg is lifted vertically up with a constant acceleration of 2.5m//s^(2) .The force of interaction between 1st and 2nd links as shown

A block of mass 1 kg is at rest relative to a smooth wedge moving leftwards with constant acceleration a = 5 m//s^(-2) .Let N be the normal reation between the block and the wedge . Then (g = 10 m//s^(-2))

A block of mass 1 kg is at rest relative to a smooth wedge leftwards with constant acceleration a =5 m//s^(2) .Let N be the normal reaction between the block and the wedge. Then ( g = 10 m//s^(2) )

The mass of a lift is 600 kg and it is moving upwards with a uniform acceleration of 2m//s^(2) . Then the tension in the cable of the lift is-

The mass of lift is 500 kg. When it ascends with an acceleration of 2m//s^(2) ,the tension in the cable will be [g=10m//s^(2)]