[JEE Main Online Paper ( Held On 09 Apirl 2014 )
A) \[\sqrt{\frac{8}{7}}\]
B) \[\sqrt{\frac{15}{14}}\]
C) \[\frac{8}{7}\]
D) \[\frac{15}{14}\]
Correct Answer: D
Solution :
As we know, Acceleration,\[a=\frac{mg\sin \theta }{m+\frac{I}{{{r}^{2}}}}\] For cylinder, \[{{a}_{c}}=\frac{{{M}_{c}}.g.\sin {{\theta }_{c}}}{{{M}_{c}}+\frac{1}{2}\frac{{{M}_{c}}{{R}^{2}}}{{{R}^{2}}}}\] \[=\frac{{{M}_{c}}.g.\sin {{\theta }_{c}}}{{{M}_{c}}+\frac{{{M}_{c}}{{R}^{2}}}{2{{R}^{2}}}}\]or\[{{a}_{c}}=\frac{2}{3}g\sin {{\theta }_{c}}\] For sphere,\[{{a}_{s}}=\frac{{{M}_{s}}g\sin {{\theta }_{s}}}{{{M}_{s}}+\frac{{{I}_{s}}}{{{r}^{2}}}}\] \[=\frac{{{M}_{s}}g\sin {{\theta }_{s}}}{{{M}_{s}}+\frac{2}{5}\frac{M{{R}^{2}}}{{{R}^{2}}}}\]or\[{{a}_{s}}=\frac{5}{7}g\sin {{\theta }_{s}}\]given,\[{{a}_{c}}={{a}_{s}}\] i.e.,\[\frac{2}{3}g\sin {{\theta }_{c}}=\frac{5}{7}g\sin {{\theta }_{s}}\] \[\therefore \]\[\frac{\sin {{\theta }_{c}}}{\sin {{\theta }_{s}}}=\frac{\frac{5}{7}g}{\frac{2}{3}g}=\frac{15}{14}\]
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