Q176 / 180 · KCET 2020 · Physics
At a metro station, a girl walks up a stationary escalator in $20 \mathrm{~s}$. If she remains stationary on the escalator, then the escalator take her up in $30 \mathrm{~s}$. The time taken by her to walk up on the moving escalator will be
- A$25 \mathrm{~s}$
- B$60 \mathrm{~s}$
- Correct answerC$12 \mathrm{~s}$
- D$10 \mathrm{~s}$
Solution
Let $h$ be the height of platform, then speed of walking $\left(v_{g}\right)=\frac{h}{20}$
Speed of escalator $\left(v_{e}\right)=\frac{h}{30}$
Time taken by girl when she walk up on the moving escalator is
$t=\frac{h}{v_{g}+v_{e}}=\frac{h}{\frac{h}{20}+\frac{h}{30}}=\frac{20 \times 30}{(30+20)}=12 \mathrm{~s}$
Hence, the correct answer is option: 'C'.