2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    A body of mass 5 kg under the action of constant force \[\overset{\to }{\mathop{F}}\,={{F}_{x}}\hat{i}+{{F}_{y}}\hat{j}\]has velocity at\[t=0s\]as\[\overset{\to }{\mathop{v}}\,=\left( 6\hat{i}-2\hat{j} \right)m/s\]and at t = 10s as \[\overset{\to }{\mathop{v}}\,=+6\hat{j}\,m/s.\] The force \[\overset{\to }{\mathop{F}}\,\]is:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) \[\left( -3\hat{j}+4\hat{j} \right)N\]

    B) \[\left( -\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j} \right)N\]

    C) \[\left( 3\hat{i}-4\hat{j} \right)N\]                          

    D) \[\left( \frac{3}{5}\hat{i}-\frac{4}{5}\hat{j} \right)N\]

    Correct Answer: A

    Solution :

                    From question, Mass of body, m = 5 kg Velocity at t = 0, \[u=(6\hat{i}-2\hat{j})m/s\] Velocity at t = 10s, \[v=+6\hat{j}\,m/s\] Force, F = ? Acceleration, \[a=\frac{v-u}{t}\] \[=\frac{6\hat{j}-(6\hat{i}-2\hat{j})}{10}=\frac{-3\hat{i}+4\hat{j}}{5}m/{{s}^{2}}\] Force, F = ma \[=5\times \frac{(-3\hat{i}+4\hat{j})}{5}=(-3\hat{i}+4\hat{j})N\]                


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