A) 13 N
B) 32N
C) 25 N
D) 35 N
Correct Answer: B
Solution :
Let a be the acceleration of each blocks. Then, \[{{\operatorname{T}}_{3}}=\left( {{m}_{1}}+{{m}_{2}}+{{m}_{3}} \right)a\] ? (i) and \[{{\operatorname{T}}_{2}}=\left( {{m}_{1}}+{{m}_{2}} \right)a\] ... (ii) From Eqs. (i) and (ii), we get \[{{\operatorname{T}}_{2}}=\left( \frac{{{m}_{1}}+{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}} \right)\,\times \,{{T}_{3}}\] \[=\,\,\,\,\left( \frac{10+6}{10+6+4} \right)\,\,\times \,\,40\] \[=\,\,\,32\,\,N\]You need to login to perform this action.
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