Question

A lift is going up with an acceleration of $0.5g$. If a man $M$ is standing on a spring balance placed in the lift then the reading of the balance (in Newton) will be:

• A. $Mg$
• B. $0.5Mg$
• C. $1.5Mg$
• D. $zero$

Answer 1

The correct option is C $1.5Mg$

From Newton's 2nd Law, $R-Mg=Ma$
$S_{red}=R=Mg+M\left(0.5g\right)=1.5MgN$
where $R$ is the normal reaction when the lift is accelerating. The reading of the spring balance is same as the normal reaction $R$ when the lift is accelerating. Here, $R$ is different from $Mg$. $Mg$ is the normal reaction when the lift is stationary.