JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    Speeds of two identical cars are u and 4 u at a specific instant. The ratio of the respective distances at which the two cars are stopped from that instant is                                 [AIEEE 2002]

    A) \[1:1\]

    B) \[1:4\]

    C) \[1:8\]

    D) \[1:16\]

    Correct Answer: D

    Solution :

    [d] In this question, the cars are identical means coefficient of friction between the tyre and the ground is same for both the cars, as a result retardation is same for both the cars equal to\[\mu g\].          \[(\because mg=\mu mg\Rightarrow a=\mu g)\]
    Let first car travel distance \[{{s}_{1}}\], before stopping while second car travel distance \[{{s}_{2}}\], then from
    \[{{v}^{2}}={{u}^{2}}-2\] as, we get
    \[O={{u}^{2}}-2\mu g\times {{s}_{1}}\]
    \[\Rightarrow \]   \[{{s}_{1}}=\frac{{{u}^{2}}}{2\mu g}\]
    and      \[0={{(4u)}^{2}}-2\mu g\times {{s}_{2}}\]
    \[\Rightarrow \]   \[{{s}_{2}}=\frac{16{{u}^{2}}}{2\mu g}=16{{s}_{1}}\]
    \[\Rightarrow \]   \[\frac{{{s}_{1}}}{{{s}_{2}}}=\frac{1}{16}\]

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