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

14: Some Mechanical Properties of Matter

with Solutions - page 7
Qstn# iv-25 Prvs-QstnNext-Qstn
  • #25
    A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of water raised in the capillary. (b) If the length of the capillary tube is half the answer of part
    (a), find the angle θ made by the water surface in the capillary with the wall.
    Ans : Given:
    Radius of capillary tube r = 1 mm = 10-3 m (a) Let T be the surface tension and ρ be the density of the liquid.
    Then, for cos θ = 1, height (h) of liquid level:
    `` h=\frac{2T}{r\rho g}`` ...(i),
    where g is the acceleration due to gravity

    1. `` \Rightarrow h=\frac{2\times \left(0.076\right)}{{10}^{-3}\times 10\times 100}``
      `` =1.52\,\mathrm{\,cm\,}``
      `` =1.52\times {10}^{-2}\,\mathrm{\,m\,}``
      `` =1.52\,\mathrm{\,cm\,}``
      `` ``

    (b) Let the new length of the tube be h'.
    `` h\text{'}=\frac{2T\,\mathrm{\,cos\,}\theta }{r\rho g}``
    `` \,\mathrm{\,cos\,}\theta =\frac{h\text{'}r\rho g}{2T}``
    `` \,\mathrm{\,Using\,}\,\mathrm{\,equation\,}\left(\,\mathrm{\,i\,}\right),\,\mathrm{\,we\,}\,\mathrm{\,get\,}:``
    `` \,\mathrm{\,cos\,}\theta =\frac{h\text{'}}{h}=\frac{1}{2}\left(\,\mathrm{\,Because\,}h\text{'}=\frac{h}{2}\right)``
    `` \Rightarrow \theta ={\,\mathrm{\,cos\,}}^{-1}\left(\frac{1}{2}\right)=60°``
    `` ``
    The water surface in the capillary makes an angle of 60∘with the wall.
    Page No 301:
  • #25-a
    Find the height of water raised in the capillary.
    Ans : Let T be the surface tension and ρ be the density of the liquid.
    Then, for cos θ = 1, height (h) of liquid level:
    `` h=\frac{2T}{r\rho g}`` ...(i),
    where g is the acceleration due to gravity

    1. `` \Rightarrow h=\frac{2\times \left(0.076\right)}{{10}^{-3}\times 10\times 100}``
      `` =1.52\,\mathrm{\,cm\,}``
      `` =1.52\times {10}^{-2}\,\mathrm{\,m\,}``
      `` =1.52\,\mathrm{\,cm\,}``
      `` ``

  • #25-b
    If the length of the capillary tube is half the answer of part
    (a), find the angle θ made by the water surface in the capillary with the wall.
    digAnsr:   b
    Ans : Let the new length of the tube be h'.
    `` h\text{'}=\frac{2T\,\mathrm{\,cos\,}\theta }{r\rho g}``
    `` \,\mathrm{\,cos\,}\theta =\frac{h\text{'}r\rho g}{2T}``
    `` \,\mathrm{\,Using\,}\,\mathrm{\,equation\,}\left(\,\mathrm{\,i\,}\right),\,\mathrm{\,we\,}\,\mathrm{\,get\,}:``
    `` \,\mathrm{\,cos\,}\theta =\frac{h\text{'}}{h}=\frac{1}{2}\left(\,\mathrm{\,Because\,}h\text{'}=\frac{h}{2}\right)``
    `` \Rightarrow \theta ={\,\mathrm{\,cos\,}}^{-1}\left(\frac{1}{2}\right)=60°``
    `` ``
    The water surface in the capillary makes an angle of 60∘with the wall.
    Page No 301: