question_answer

A sphere B is in air sandwiched between blocks A and C which are compressed by two spring of spring constant \[{{k}_{1}}\] and \[{{k}_{2}}\] compressed by \[{{x}_{1}}\] and \[{{x}_{2}}\]respectively. The coefficient of friction between sphere and blocks is same and other surfaces are smooth. The sphere is at equlibrium then the ratio\[\frac{{{K}_{1}}}{{{K}_{2}}}\] will be:

A)  \[\frac{{{x}_{1}}+{{x}_{2}}}{{{x}_{1}}}\]               

B)  \[\frac{{{x}_{1}}+{{x}_{2}}}{{{x}_{2}}}\]

C)  \[\frac{{{x}_{1}}}{{{x}_{1}}+{{x}_{2}}}\]               

D)  \[\frac{{{x}_{1}}}{{{x}_{1}}+{{x}_{2}}}\]

Correct Answer: D

Solution :

Torque about contact at A = 0 \[mgR\,=\mu R_{B}^{'}\,2R\]                         Or \[mg=2\mu R_{B}^{'}\] Torque about contact at B = 0 Gives \[mg=2\mu R_{A}^{'}\] So        \[R_{A}^{'}\,=R_{B}^{'}\] But       \[R_{A}^{'}\,={{k}_{1}}{{x}_{1}}\] And      \[R_{B}^{'}\,={{k}_{2}}{{x}_{2}}\] So,       \[{{k}_{1}}{{x}_{1}}\,={{k}_{2}}{{x}_{2}}\]             \[\frac{{{k}_{1}}}{{{k}_{2}}}\,=\frac{{{x}_{2}}}{{{x}_{1}}}\]