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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!!!!

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!!!!

Grade:9

3 Answers

shubham pansare
15 Points
10 years ago

initially,, kx=mg   -(1)

later double the reading...

2kx=m(g+a)          

2mg=m(g+a)....use (1)

therefore a=g.

vikas askiitian expert
509 Points
10 years ago

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

Fawz Naim
37 Points
10 years ago

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 

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