HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

14: Some Mechanical Properties of Matter

with Solutions - page 9
Qstn# iv-30-a Prvs-QstnNext-Qstn
  • #30-a
    the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm s-1, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine = 1260 kg m-3 and its coefficient of viscosity at room temperature = 8.0 poise.
    Ans : Viscous force exerted by glycerine on the sphere F = 6πηrv
    ⇒ F= 6 × (3.14) × (0.8) × 10-3 × (10-2)
    = 1.50 × 10-4 N (b) Let V be the volume of the sphere.
    Hydrostatic force exerted by glycerin on the sphere `` F\text{'}=V\sigma g``
    `` \Rightarrow F\text{'}=\frac{4}{3}\pi {r}^{2}\sigma g``
    `` =\left(\frac{4}{3}\right)\times \left(3.14\right)\times \left({10}^{-6}\right)\times 1260\times 10``
    `` =5.275\times {10}^{-5}\,\mathrm{\,N\,}`` (c) Let the terminal velocity of the sphere be v'.
    The forces acting on the drops are
    (i) The weight mg acting downwards
    (ii) The force of buoyance, i.e., `` \frac{4}{3}\pi {r}^{3}\sigma g`` acting upwards
    (iii) The force of viscosity, i.e., 6πηrv' acting upwards
    From the free body diagram:

    `` 6\pi \eta rv\text{'}+\frac{4}{3}\pi {r}^{3}\sigma g=mg``
    `` \Rightarrow v=\frac{mg-{\displaystyle \frac{4}{3}}\pi {r}^{2}\sigma g}{6\pi \eta r}``
    `` =\frac{50\times {10}^{-3}-{\displaystyle \frac{4}{3}}\times 3.14\times {10}^{-6}\times 1260\times 10}{6\times 3.14\times 0.8\times {10}^{-3}}``
    `` =\frac{500-{\displaystyle \frac{4}{3}}\times 3.14\times {10}^{-3}\times 1260\times 10}{6\times 3.14\times 0.8}``
    `` =2.3\,\mathrm{\,cm\,}/s``
    Page No 301:
  • #30-b
    the hydrostatic force exerted by the glycerine on the sphere and
    Ans : Let V be the volume of the sphere.
    Hydrostatic force exerted by glycerin on the sphere `` F\text{'}=V\sigma g``
    `` \Rightarrow F\text{'}=\frac{4}{3}\pi {r}^{2}\sigma g``
    `` =\left(\frac{4}{3}\right)\times \left(3.14\right)\times \left({10}^{-6}\right)\times 1260\times 10``
    `` =5.275\times {10}^{-5}\,\mathrm{\,N\,}``
  • #30-c
    the terminal velocity with which the sphere will move down without acceleration. Density of glycerine = 1260 kg m-3 and its coefficient of viscosity at room temperature = 8.0 poise.
    Ans : Let the terminal velocity of the sphere be v'.
    The forces acting on the drops are
    (i) The weight mg acting downwards
    (ii) The force of buoyance, i.e., `` \frac{4}{3}\pi {r}^{3}\sigma g`` acting upwards
    (iii) The force of viscosity, i.e., 6πηrv' acting upwards
    From the free body diagram:

    `` 6\pi \eta rv\text{'}+\frac{4}{3}\pi {r}^{3}\sigma g=mg``
    `` \Rightarrow v=\frac{mg-{\displaystyle \frac{4}{3}}\pi {r}^{2}\sigma g}{6\pi \eta r}``
    `` =\frac{50\times {10}^{-3}-{\displaystyle \frac{4}{3}}\times 3.14\times {10}^{-6}\times 1260\times 10}{6\times 3.14\times 0.8\times {10}^{-3}}``
    `` =\frac{500-{\displaystyle \frac{4}{3}}\times 3.14\times {10}^{-3}\times 1260\times 10}{6\times 3.14\times 0.8}``
    `` =2.3\,\mathrm{\,cm\,}/s``
    Page No 301: