A) \[\frac{{{t}_{2}}}{{{t}_{1}}}={{\left( \frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}} \right)}^{1/2}}\]
B) \[\frac{{{t}_{2}}}{{{t}_{1}}}=1\]
C) \[\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\]
D) \[\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}\]
Correct Answer: C
Solution :
From first inclined plane \[\sin {{\theta }_{1}}=\frac{h}{{{l}_{1}}}\] Or \[{{l}_{1}}=\frac{h}{\sin {{\theta }_{1}}}\] From second inclined plane \[\sin {{\theta }_{2}}=\frac{h}{{{l}_{2}}}\]or \[{{l}_{2}}=\frac{h}{\sin {{\theta }_{2}}}\] Acceleration of the block down the two planes are \[{{a}_{1}}=g\sin {{\theta }_{1}}\] and \[{{a}_{2}}=g\sin {{\theta }_{2}}\] For the first inclined plane, distance \[{{l}_{1}}=\frac{1}{2}{{a}_{1}}t_{1}^{2}\] \[(\because u=0)\] For the second inclined plane, distance \[{{l}_{2}}=\frac{1}{2}{{a}_{2}}t_{2}^{2}\] \[\therefore \] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{a}_{1}}t_{1}^{2}}{{{a}_{2}}t_{2}^{2}}\] \[\therefore \] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{g\sin {{\theta }_{1}}t_{1}^{2}}{g\sin {{\theta }_{2}}t_{2}^{2}}\] Or \[\frac{t_{1}^{2}}{t_{2}^{2}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\times \frac{{{l}_{2}}}{{{l}_{1}}}\] \[=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\times \frac{h}{\sin {{\theta }_{2}}}\times \frac{\sin {{\theta }_{1}}}{h}\] \[\therefore \] \[\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\]You need to login to perform this action.
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