Question

When a lift is moving upward with acceleration of 5 $m/s^2$ then percentage change in the weight of the person in the lift is x%.If the lift is moving down with acceleration 5 $m/s^2$ then percentage change in the weight of same person is y%.Then ratio.$\frac{x}{y}is\left(g=10m/s^2\right)$

• A. 1:1
• B. 3:1
• C. 1:3
• D. 3:4

Answer 1

The correct option is C 1:3

Lift moving upward $=5m/s^2$

Percentage of change of its weight of the person $=x$%

lift moving down $=5m/s^2$

Percentage of change of its weight of the person $=y$%

Since the lift is moving upwards, the equation will be

$ma=T-mg$

$T=ma+mg$ $\underset{m}{100}$ $100+x$

$T=m(a+g)$ $\frac{m(100+x)}{100}$

$T=\frac{m(1000-x)}{100}(65+10)$

$=\frac{m(100-x)}{100}\left(25\right)=\frac{m(100-x)}{4}$

For the lift is moving downwards,

$ma=T+mg$

or, $T=ma-mg$

$=m(a-g)$

$=\frac{m(100+y)}{100}(15-10)$

$=\frac{m(100+y)}{100}\times 5$

$=\frac{m(100+y)}{20}$

$\frac{m(100-x)}{4}=\frac{m(100+y)}{20}$

or, $5(100-x)=100+y$

or, $500-5x=100+y$

or, $y+5x=400$

or, $y(1+\frac{5x}{y})=400$

or, $5y(\frac{1}{5}+\frac{x}{y})=400$

or, $y(\frac{1}{5}+\frac{x}{y})=80$

or, $\frac{y}{5}+x=80$

or, $\frac{y+5x}{5}=80$

or, $\frac{x}{y}=1:3$