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) &radic;5g (b) 2g (c) g (d) &radic;3g pls explain!!!!

Shivam Bhagat Grade: 9

        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!!!!

7 years ago

Answers : (3)

shubham pansare

15 Points

										initially,, kx=mg   -(1)
later double the reading...
2kx=m(g+a)          
2mg=m(g+a)....use (1)
therefore a=g.

7 years ago

Approved

vikas askiitian expert

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

7 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

7 years ago

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