A) \[25\text{ }m/{{s}^{2}}\]
B) \[2.5\text{ }m/{{s}^{2}}\]
C) \[5\text{ }m/{{s}^{2}}\]
D) \[10\text{ }m/{{s}^{2}}\]
Correct Answer: B
Solution :
Maximum bearable tension in the rope |
\[T=25\times 10=250\text{ }N\] |
From the figure, |
\[T-mg=ma\] |
or \[a=\frac{T-mg}{m}\] |
Given, |
\[m=20\text{ }kg,\] |
\[g=10\text{ }m/{{s}^{2}},\] |
\[T=250\text{ }N\] |
Hence, |
\[a=\frac{250-20\times 10}{20}\] |
\[=\frac{50}{20}=2.5\,m/{{s}^{2}}\] |
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