JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to     AIEEE  Solved  Paper-2003

    A) \[{{t}^{3/4}}\]                                   

    B)       \[{{t}^{3/2}}\]   

    C)       \[{{t}^{1/4}}\]                   

    D)       \[{{t}^{1/2}}\]

    Correct Answer: B

    Solution :

                                    In case of velocity changing with distance, we apply integration method to determine the relationship between distance and time, if force is given. As, power, P = constant \[\Rightarrow \]   \[Fv=P\]                   (\[\because \] P = force \[\times \] velocity) \[\Rightarrow \]   \[Ma\times v=P\]            \[(\because F=Ma)\] \[\Rightarrow \]   \[va=\frac{P}{M}\] \[\Rightarrow \]   \[v\times \left[ \frac{v\,dv}{ds} \right]=\frac{P}{M}\]                     \[\left( \because a=\frac{v\,dv}{ds} \right)\] \[\Rightarrow \]   \[\int_{0}^{v}{{{v}^{2}}}dv=\int_{0}^{s}{\frac{P}{M}}\,ds\] (assuming at t = 0 it starts from rest, i.e., from s = 0) \[\Rightarrow \]   \[\frac{{{v}^{3}}}{s}=\frac{P}{M}s\] \[\Rightarrow \]   \[v={{\left( \frac{3P}{M} \right)}^{1/3}}{{s}^{1/3}}\] \[\Rightarrow \]   \[\frac{ds}{dt}=K\,{{s}^{1/3}}\] \[\left[ \because K={{\left( \frac{3P}{M} \right)}^{1/3}} \right]\] \[\Rightarrow \]   \[\int_{0}^{s}{\frac{ds}{{{s}^{1/3}}}=\int_{0}^{t}{K\,dt}}\] \[\Rightarrow \]   \[\frac{{{s}^{2/3}}}{2/3}=Kt\] \[\Rightarrow \]   \[{{s}^{2/3}}=\frac{2}{3}Kt\] \[\Rightarrow \]   \[s={{\left( \frac{2}{3}K \right)}^{3/2}}\times {{t}^{3/2}}\] \[\Rightarrow \]   \[s\propto {{t}^{3/2}}\]


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