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