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