Question

31. Three blocks are in equilibrium as shown in figure.
the tension in the string present between the blocks 1
kg & 5 kg is
(a) 100 N
(b) 90 N
(c) 50 N
(d) 10 N​

Answer 1

Answer:

the ans. is 90N

Explanation:

As, ma=mg-T

here ma=0

therefore, 0=5kg+4kg*10 -T

0=9kg*10-T

0=90-T

if we shift Tto left hand side

T=90N

therefore the tension between 1kg block and 5kg block is 90N

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Answer 2

Answer:

90N

Explanation:

FBD is used when in equilibrium-----

F(in direction of motion)-f(opposite direction of mation)=m(mass of the obj)*a(its accelaration)