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```
A boy stands on a weighing machine inside a lift.When the lift is going down with an acceleration of g/4 the machine shows a reading of 30kg.when the lift goesupward at acceleration g/4the reading will be............pls solve

```
9 years ago

```							When the lift is going in downward direction with an acc. of g/4, net acc. of the lift
g`=g-g/4=3g/4. Weight of the body=mg`=m3g/4
Now the weight of the body is due to the normal reaction of the floor if the lift which is
N=mg`=3mg/4
3mg/4=30
m=12
Now when the lift is going up with an acc. of g/4 then net acc. is g+g/4=5g/4
N=m5g/4
N=12*5*10/4=150kg
```
9 years ago
```							when lift goes downwards then seudo force acts on boy in upward direction...
if a is the accleration of lift then
mg - ma = net weight=30g         ...........1         (given)        g is accleration due to gravity
m(g-a) = 30g
m=40kg
when lift goes upward then seudo force acts in downward direction so now net weight
mg + ma = W
40(a+g) = W
50kg = W                 ans

```
9 years ago
```							when lift goes downwards then seudo force acts on boy in upward direction...
if a is the accleration of lift then
mg - ma = net weight=30g         ...........1         (given)        g is accleration due to gravity
m(g-a) = 30g
m=40kg
when lift goes upward then seudo force acts in downward direction so now net weight
mg + ma = W
40(a+g) = W
50kg = W                 ans

```
9 years ago
```							While goin down:
mg-N=mg/4
N=3mg/4=30g
Therefore, m=40
While going up:
N-mg=mg/4
N=5/4mg
=50Kg(m being 40)
```
9 years ago
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