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If a 5 Kg mass is suspended by a spring balance in a lift which is accelerated downward at 10 m/s .The reading of the balance is 

Option: 1

more than 5Kg weight 


Option: 2

is less than 5Kg weight  and greater than zero


Option: 3

is equal to 5 Kg weight 


Option: 4

zero 


Answers (1)

best_answer

 

 

Apparent weight of body in a lift  when Lift is moving down with a = g

As a=g

mg-R=mg

R=0

Apparent weight = 0 (weightlessness)

 Draw FBD of block 

mg- T = mg 

 T= 0 

 so reading of balance is zero 

 

Posted by

shivangi.shekhar

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