A) \[\frac{{{t}_{2}}}{{{\operatorname{t}}_{1}}}=1\]
B) \[\frac{{{t}_{2}}}{{{\operatorname{t}}_{1}}}={{\left[ \frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}} \right]}^{{}^{1}/{}_{2}}}\]
C) \[\frac{{{t}_{2}}}{{{\operatorname{t}}_{1}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}\]
D) \[\frac{{{t}_{2}}}{{{\operatorname{t}}_{1}}}=\frac{{{\sin }^{2}}{{\theta }_{1}}}{{{\sin }^{2}}{{\theta }_{2}}}\]
Correct Answer: B
Solution :
In both cases black starts from rest u = 0 \[\operatorname{S}=ut+\frac{1}{2}{{a}_{1}}{{t}^{2}}s=ut+\frac{1}{2}{{a}_{2}}{{t}^{2}}\] \[{{\operatorname{S}}_{1}}=0+\frac{1}{2}gsin\theta {{t}^{2}}{{s}_{2}}=0+\frac{1}{2}gsin{{\theta }_{2}}{{t}^{2}}\]Two inclined plane have same height \[{{\operatorname{S}}_{1}}={{\operatorname{S}}_{2}}\] \[\frac{1}{2}gsin{{\theta }_{1}}{{t}_{1}}^{2}=\frac{1}{2}gsin{{\theta }_{2}}{{t}^{2}}_{2}\] \[\frac{{{t}^{2}}_{1}}{{{t}^{2}}_{2}}=\frac{sin{{\theta }_{2}}}{sin{{\theta }_{1}}}\] \[\frac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\frac{sin{{\theta }_{2}}}{sin{{\theta }_{1}}}}\] \[\frac{{{t}_{2}}}{{{t}_{1}}}={{\left[ \frac{sin{{\theta }_{2}}}{sin{{\theta }_{1}}} \right]}^{{}^{1}/{}_{2}}}\]You need to login to perform this action.
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