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```        A spring balance hanging in steady state inside a lift accelerating upwards shows a reading of double the actual mass of an object. Then the acceleration of the lift is-
(a)    √5g
(b)    2g
(c)     g
(d)    √3g              pls explain!!!!```
6 years ago

shubham pansare
15 Points
```										initially,, kx=mg   -(1)
2kx=m(g+a)
2mg=m(g+a)....use (1)
therefore a=g.
```
6 years ago
510 Points
```										when lift acclerates it upward direction then ,then particle experinces a seudo force in downward direction
magnitude of this force = ma                     (a is accleration of lift)
net force in downward direction = net weight
weight = m(g+a)
weight is given 2mg so
2mg = m(g+a)
a = g
```
6 years ago
Fawz Naim
37 Points
```										as the lift is accelerating upwards a pseudo force is acting on the mass hanging from the spring balance.
let the acc of the lift be a    and the mass of the body be m
so the total downward force acting on the object =weightof the object+pseudo force(ma)
F=mg+ma
now the spring balance is showing the mass of the object which is double its actual mass
so  F=2mg
mg+ma=2mg
ma=mg
a=g
```
6 years ago
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