Physics, asked by MUHAMMED10LEO, 10 months ago

A CHAIN OF MASS 10 KG AND LENGTH 10 METER IS HELD VERTICAL BY FIXING ITS UPPER END TO A RIGID SUPPORT. THE TENSION IN THE CHAIN AT A DISTANCE 3 METRE FROM THE RIGID SUPPORT IS
[A]. 100N.
[B]. 70N.
[C]. 143N.
[D]. 76N.

Answers

Answered by arenarohith
1

Answer:

(B) 70 N

Explanation:

-------------

TOTAL WEIGHT OF THE CHAIN =10KG

MASS PER UNIT LENGTH=10/10KG PER METER

                                          =1 KG PER METER

NOW AT A DISTANCE OF 3 METRE FROM THE SUPPORT THE FORCE EXPERIENCED IS DUE TO THE REMAINING CHAIN HANGING DOWN i.e 7 METRE

WEIGHT OF THE CHAIN =1 KG/M *7=7 KG

THE FORCE EXPERIENCED = F=mg=7*10 = 70 N

Answered by AadilPradhan
0

The tension in the chain at a distance 3 mts from the rigid support is 70N (c).

1) The mass of the chain is given as 10 kg and the length is given as 10 mts.

2) The mass per unit length of the chain is given as 10/10 kg/m = 1 kg/m.

3) The tension in the chain at a distance of 3 mts from the support is due to the lower half of the chain. That is, due to the lower 7 mts of the chain.

4) The tension is thus given as 7 * 1 * 10 N = 70 N. (Taking acceleration due to gravity = 10 m/s^2).

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