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NEET-XII-Physics

11: Gravitation

with Solutions - page 6
Qstn# iv-28-a Prvs-QstnNext-Qstn
  • #28-a
    Find the speed of the ship due to rotation of the earth about its axis. (b) Find the difference between T0 and the earth’s attraction on the bob. (c) If the ship sails at speed v, what is the tension in the string? Angular speed of earth’s rotation is ω and radius of the earth is R.
    Ans : Speed of the ship due to rotation of the Earth is v = ωR, where R is the radius of the Earth and ω is its angular speed. (b) The tension in the string is given by
    T0 = mg - mω2R
    ∴ T0 - mg = mω2R (c) Let the ship move with a speed v.
    Then the tension in the string is given by
    `` T=mg-m{\,\mathrm{\,\omega \,}}_{1}^{2}R``
    `` ={T}_{0}-\frac{{\left(v-\omega R\right)}^{2}}{{R}^{2}}R``
    `` ={T}_{0}-\frac{\left({v}^{2}+{\,\mathrm{\,\omega \,}}^{2}{R}^{2}-2\omega Rv\right)}{R}R``
    `` \therefore T={T}_{0}+2\,\mathrm{\,\omega \,}vm``
    Page No 227:
  • #28-b
    Find the difference between T0 and the earth’s attraction on the bob.
    Ans : The tension in the string is given by
    T0 = mg - mω2R
    ∴ T0 - mg = mω2R
  • #28-c
    If the ship sails at speed v, what is the tension in the string? Angular speed of earth’s rotation is ω and radius of the earth is R.
    Ans : Let the ship move with a speed v.
    Then the tension in the string is given by
    `` T=mg-m{\,\mathrm{\,\omega \,}}_{1}^{2}R``
    `` ={T}_{0}-\frac{{\left(v-\omega R\right)}^{2}}{{R}^{2}}R``
    `` ={T}_{0}-\frac{\left({v}^{2}+{\,\mathrm{\,\omega \,}}^{2}{R}^{2}-2\omega Rv\right)}{R}R``
    `` \therefore T={T}_{0}+2\,\mathrm{\,\omega \,}vm``
    Page No 227: