In the given arrangement, friction is absent and
strings, pulleys are ideal. The difference of
tension (T1- T2) is
(1) mg
(2) mg/2
(3) 3mg
(4) 2mg
Answers
Answer:
the answer is mg/2
Explanation:
I can't explain but the answer is right
Answer: mg/2
Given: Friction is absent and strings, and pulleys are ideal
To Find: Difference in Tension (T1- T2)
Step-by-step explanation:
As per the given arrangement let the tension in the first string be 2mg and let the tension in the second string be mg.
Also, mass at the different places is m, 2m, and 3m.
As per the given formula
a= 2mg- mg/ m+2m+3m
a= mg/6m
a=g/6 ……….(1)
2mg- T1= 2ma
T1= 2mg- 2ma
Placing value of a from (1)
T1= 2mg- 2m(g/6)
T1= 5mg/3 ………… (2)
T2- mg= ma
T2= ma+mg
Placing value of a from (1)
T2= m(g/6)+mg
T2= 7mg/6 ……….. (3)
Now difference of (2) and (3)
T1- T2 = 5mg/3- 7mg/6
= mg/2
So, the difference of T1- T2 is mg/2
Project code: #SPJ2