question_answer

Three blocks of masses \[{{m}_{1}},{{m}_{2}}\] and \[{{m}_{3}}\] are connected by massless strings as shown on a frictionless table. They are pulled with a force of 40N. If\[{{m}_{1}}=10kg\], \[{{m}_{2}}=6kg\] and \[{{m}_{3}}=4kg\], then tension \[{{T}_{2}}\] will be:

A) 10N                             

B) 20N   

C) 32N                             

D) 40N

Correct Answer: B

Solution :

[b]   \[a=\frac{F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}=\frac{40}{10+6+4}=2m/{{s}^{2}}\] \[40-{{T}_{2}}={{m}_{1}}a\] \[40-{{T}_{2}}=10\times 2\] \[{{T}_{2}}=20N\]