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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25rad/s, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take \[g=10m{{s}^{-2}}\])   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 20 N                      

    B) 10 N

    C) 60 N                      

    D) 40 N

    Correct Answer: C

    Solution :

                    Mass of bigger body M = 4 kg Mass of smaller body m = 1 kg Smaller mass (m = 1 kg) executes S.H.M of angular frequency \[\omega =25\]rad/s Amplitude \[x=1.6cm=1.6\times {{10}^{-2}}\] As we know,\[T=2\pi \sqrt{\frac{m}{K}}\]or\[\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{K}}\] or\[\frac{1}{25}=\sqrt{\frac{1}{K}}[\because m=1kg;\omega =25rad/s]\] or\[K=625N{{m}^{-1}}.\] The maximum force exerted by the system on the floor \[=Mg+Kx+mg\] \[=4\times 10+625\times 1.6\times {{10}^{-2}}+1\times 10\] \[=40+10+10\]


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