A) 29.4 N
B) 39.2 N
C) 18.6 N
D) 42.6 N
Correct Answer: A
Solution :
\[{{F}_{l}}\,\propto \,\,R\,\,\,\,\therefore \,\,{{F}_{1}}\,\propto m\], i.e., limiting friction depends upon the mass of body. So \[\frac{({{F}_{1}})'}{({{F}_{1}})}\,=\,\frac{m'}{m}\,\,=\,\,\frac{10+5}{10}\] \[\Rightarrow \,\,\,\,\,\,({{F}_{1}}{)}'\,\,=\,\,\frac{3}{2}\times {{F}_{1}}\times \frac{3}{2}\,\times \,\,19.6\,\,=\,\,29.4\,\,N\]
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