WBJEE · Physics · Motion In One Dimension

Ruma reached the metro station and found that the escalator was not working. She walked up the stationary escalator with velocity \(\mathrm{v}_1\) in time \(\mathrm{t}_1\). On other day if she remains stationary on the escalator moving with velocity \(\mathrm{v}_2\), then escalator takes her up in time \(t_2\). The time taken by her to walk up with velocity \(\mathrm{v}_1\) on the moving escalator will be

A \(\frac{t_1 t_2}{t_2-t_1}\)
B \(\frac{t_1 t_2}{t_2+t_1}\)
C \(\frac{t_1-t_2}{t_1+t_2}\)
D \(\frac{t_1+t_2}{2\left(t_1-t_2\right)}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{t_1 t_2}{t_2+t_1}\)

Step-by-step Solution

Detailed explanation

D = Distance to be covered for stationary escalator, \(\mathrm{V}_1 \mathrm{t}_1=\mathrm{D}\)....(1) for moving escalator, \(\mathrm{V}_2 \mathrm{t}_2=\mathrm{D}\)...(2) \(\qquad\) for ruma moving on moving escalator,…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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