• # question_answer Three blocks of masses${{m}_{1}},{{m}_{2}}$and${{m}_{3}}$are connected by massless string as shown in the figure on a friction- less table. They are pulled with a force${{T}_{3}}=40\,N.$If ${{m}_{1}}=10\,kg,\,{{m}_{2}}=6kg$and ${{m}_{3}}=4\,kg,$tension ${{T}_{2}}=$will be A)  20 N              B)  40 N              C)  10 N              D)  32 N

Common acceleration, $a=\frac{F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}=\frac{40}{10+6+4}=2\,m/{{s}^{2}}$ Equation of motion of ${{m}_{3}}$is ${{T}_{3}}-{{T}_{2}}={{m}_{3}}a$ $40-{{T}_{2}}=4\times 2,{{T}_{2}}=32\,N$ Hence, the correction option is [d].