JEE Main & Advanced JEE Main Paper Phase-I (Held on 08-1-2020 Evening)

  • question_answer
    A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stopwatch with 1 s resolution measures the time taken for 40 oscillation to be 50 s. The accuracy in g is                                                        [JEE MAIN Held on 08-01-2020 Evening]

    A) 3.40%  

    B) 2.40%

    C) 5.40%  

    D) 4.40%

    Correct Answer: D

    Solution :

    l = 25.0 cm Time of 40 oscillation is 50 sec \[\therefore \,\,\,\,\,\,\,\,g=\frac{4{{\pi }^{2}}l}{{{T}^{2}}}\]     \[\Rightarrow \,\,\,\,\,\,\,\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2\Delta T}{T}\] \[\Rightarrow \,\,\,\,\,\,\,\frac{\Delta g}{g}=\left( \frac{0.01}{25.0} \right)+2\left( \frac{1}{50} \right)\] \[\Rightarrow \,\,\,\,\,\,\,\left( \frac{\Delta g}{g}\times 100 \right)=4.4%\]


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