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AMU Medical AMU Solved Paper-2005

  • question_answer
    Three blocks of masses \[{{m}_{1}},\]\[{{m}_{2}}\] and \[{{m}_{3}}\]are connected by massless strings as shown on africtionless table. They are pulled with a  force of 40 then tension T: will be

    A)  10N                                      

    B)  20 N

    C)   32 N                                    

    D)  40 N

    Correct Answer: B

    Solution :

                     From Newtons law, the force (F) is given by                                 F = Ma where a is acceleration and M the mass. Given, \[M={{m}_{1}}+{{m}_{2}}+{{m}_{3}}\]                 \[=10+6+4=20\,kg\],                 F= 40 N \[\therefore \]  \[a=\frac{F}{M}=\frac{40}{20}=2\,m/{{s}^{2}}\] Tension between 10 kg and 6 kg masses is given by                 \[{{T}_{2}}=({{m}_{2}}+{{m}_{3}})a\]                 \[=(6+4)2\]                 \[=10\times 2=20\,N\]


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