Question

The acceleration of a train travelling with speed of 400 m/s as it goes round a curve of radius 160 m, is :

• A. $ \overrightarrow{S}=\left( \text{3\hat{j}}+\text{5\hat{k}} \right)\text{m} $
• B. $ {{\upsilon }_{A}} $
• C. $ {{\upsilon }_{B}} $
• D. $ {{\upsilon }_{A}}:{{\upsilon }_{B}} $

Answer 1

Answer: (A)

Here : $ {{\upsilon }_{A}}=\tan {{30}^{\circ }} $ Using the relation $ {{\upsilon }_{B}}=\tan {{60}^{\circ }} $ $ \frac{{{\upsilon }_{A}}}{{{\upsilon }_{B}}}=\frac{\tan {{30}^{\circ }}}{\tan {{60}^{\circ }}} $ Or $ =\frac{1/\sqrt{3}}{\sqrt{3}}=\frac{1}{3} $ So $ {{E}_{K}}\left( =\frac{1}{2}m{{\upsilon }^{2}} \right)=\frac{3}{2}KT $ $ {{\text{E}}_{i}}=\text{ 1}0.\text{2 eV }=\text{1}0.\text{2}\times \text{1}.\text{6}\times \text{1}{{0}^{-\text{19}}}\text{ J} $