MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
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!!!!

6 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.

6 years ago
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

6 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 

6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
Get extra Rs. 551 off
USE CODE: CART20
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details
Get extra Rs. 594 off
USE CODE: CART20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details