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JEE Main & Advanced Sample Paper JEE Main Sample Paper-13

  • question_answer
    The displacement \[x\] of a particle of mass m kg moving in one dimension, under the action of a force is related to the time t by the equation \[t=\sqrt{x}+3,\] where x is in metre and t is in second. The work done by the force in the first six seconds in joule is

    A)  zero                                     

    B)  3 m              

    C)  6 m                                       

    D)  9 m

    Correct Answer: A

    Solution :

     \[=2\times 2=\,In\,2\] \[=2\times 2\times 0.693\]          \[[\because \,I{{n}^{2}}\,or\,{{\log }_{e}}2=0.693]\] Velocity at \[FeC{{l}_{3}}+3NaOH\xrightarrow{\,}\,Fe{{(OH)}_{3}}+3NaCl\] is \[NaOH\xrightarrow{\,}N{{a}^{\oplus }}+O{{H}^{\odot -}}\] Velocity at \[O{{H}^{\odot -}}\] is \[Fe{{(OH)}_{3}}\] According to work-energy theorem. Work done by all the forces = Change in KE \[Fe{{(OH)}_{3}}\]


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