Let the length of the escalator be \( L \), the speed of the girl be \( v_g \), and the speed of the escalator be \( v_e \).
When the girl walks up the stationary escalator, she takes 20 sec. So, the speed of the girl is: \[ v_g = \frac{L}{20} \]
When the girl remains stationary on the escalator, the escalator takes her up in 30 sec. So, the speed of the escalator is: \[ v_e = \frac{L}{30} \] Now, when the girl walks on the moving escalator, the relative speed is \( v_g + v_e \). The time taken for her to reach the top is: \[ \text{Time} = \frac{L}{v_g + v_e} = \frac{L}{\frac{L}{20} + \frac{L}{30}} = \frac{L}{\frac{5L}{60}} = 12 \, \text{sec} \]
Thus, the time taken by her to walk up on the moving escalator is 12 sec.
The correct answer is (C) : 12 sec.
A body of mass \( (5 \pm 0.5) \, \text{kg} \) is moving with a velocity of \( (20 \pm 0.4) \, \text{m/s} \). Its kinetic energy will be:
A solid sphere of mass $1 \,kg$ rolls without slipping on a plane surface Its kinetic energy is $7 \times 10^{-3} J$. The speed of the centre of mass of the sphere is ___$cm s ^{-1}$.