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!!!!

initially,, kx=mg   -(1)

later double the reading...

2kx=m(g+a)          

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

therefore a=g.

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

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