Imagine a person standing on a weighing machine placed inside an elevator. The elevator first accelerates, then moves with a constant velocity and finally decelerates to stop. The maximum and minimum weight recorded are $80 \mathrm{~kg}$ and $64 \mathrm{~kg}$, respectively. Find out the true weight of that person considering $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$.

• A

  70 kg
• B

  85 kg
• Correct answer
  C

  72 kg
• D

  65 kg

Solution

When the lift accelerates with acceleration $a$, then the normal reaction force $N$ by the elevator on the person is given as
$\therefore N-m g=m a$ (where, $m$ is the actual mass of that person)
When the lift deaccelerates with acceleration $a$ then, $m g-N=m a$

$\Rightarrow$ Dividing Eq. (ii) by Eq. (i), we get $$ \begin{aligned} & \frac{g-a}{g+a}=\frac{64 g}{80 g} \\ & \Rightarrow \quad \frac{g-a}{g+a}=\frac{4}{5} \\ & \Rightarrow \quad 5 g-5 a=4 g+4 a \\ & \Rightarrow \quad g=9 a \\ & \Rightarrow \quad a=\frac{g}{9} \\ & \end{aligned} $$ $\therefore$ From Eq. (i), we get, $80 g=m(g+a)=m\left(g+\frac{g}{9}\right)$ $$ \begin{array}{ll} \Rightarrow & 80 g=\frac{10 g}{9} m \\ \Rightarrow & m=\frac{80 \times 9}{10}=72 \mathrm{~kg} \end{array} $$ So, the true weight of that person is $72 \mathrm{~kg}$.

Hence, the correct answer is option: 'C'.