2. Solved Papers
  3. NEET

NEET NEET SOLVED PAPER 2019

  • question_answer
    A force \[F=20+10y\]acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from \[y=0\text{ }to\text{ }y=1\]m is:      [NEET 5-5-2019]

    A) 25 J     

    B) 20 J

    C) 30 J                             

    D) 5 J

    Correct Answer: A

    Solution :

    \[F=20+10y\] \[W=\int\limits_{0}^{1}{(20\text{ }10y)dy}\] \[=20[y]_{0}^{1}+\left[ \frac{10{{y}^{2}}}{2} \right]_{0}^{-1}\] \[=20+5\] \[=25\]             


You need to login to perform this action.
You will be redirected in 3 sec spinner