A) \[[M{{L}^{3}}{{l}^{1}}{{T}^{-3}}]\]
B) \[[M{{L}^{-3}}{{l}^{-1}}{{T}^{-3}}]\]
C) \[\sqrt{\frac{{{M}_{p}}}{{{M}_{e}}}}\]
D) \[\sqrt{\frac{{{M}_{e}}}{{{M}_{p}}}}\]
Correct Answer: B
Solution :
The situation can be figured as Taking horizontal direction as x-axis and vertical direction as y-axis resolving the forces along the axes, we get \[{{\tan }^{-1}}\left[ \frac{{{p}^{2}}+{{Q}^{2}}}{2PQ} \right]\] \[{{\sin }^{-1}}\left[ \frac{{{p}^{2}}+{{Q}^{2}}+PQ}{2PQ} \right]\] \[N=m{{\omega }^{2}}r\] ...(i) and \[\text{A}=\text{1}0\text{m},\text{L}=\text{2m}\,\,\text{and}\,\,\text{d}=\text{3}0{}^\circ .\] ...(ii) Dividing Eq. (i) by Eq. (ii), we get \[{{\mu }_{s}}=5,{{\mu }_{x}}=0.4,g=10m/{{s}^{2}}\] Angular speed of the ball \[\text{1 m}/{{\text{s}}^{\text{2}}}\]You need to login to perform this action.
You will be redirected in
3 sec