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NEET-XII-Physics

14: Some Mechanical Properties of Matter

with Solutions - page 9
Qstn# iv-26 Prvs-QstnNext-Qstn
  • #26
    The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of mercury column in the capillary. (b) If the length dipped inside is half the answer of part
    (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury = 0.465 N m-1 and the contact angle of mercury with glass -135 °. (a) Find the depression of mercury column in the capillary. (b) If the length dipped inside is half the answer of part
    (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury = 0.465 N m-1 and the contact angle of mercury with glass -135 °.
    Ans : Given:
    Radius of tube r = 1 mm = 10-3 m
    Contact angle of mercury with glass θ = 135°
    Surface tension of mercury T = 0.465 N/m
    Let ρ be the density of mercury. (a) Depression (h) of mercury level is expressed as follows:
    `` h=\frac{2T\,\mathrm{\,cos\,}\theta }{r\rho g}`` ...(i)
    `` \Rightarrow h=\frac{2\times 0.465\times \,\mathrm{\,cos\,}135°}{{10}^{-3}\times 13600\times \left(9.8\right)}``
    `` =0.0053\,\mathrm{\,m\,}=5.3\,\mathrm{\,mm\,}`` (b) If the length dipped inside is half the result obtained above:
    New depression h'= `` \frac{h}{2}``
    Let the new contact angle of mercury with glass be θ'.
    ∴ `` h\text{'}=\frac{2T\,\mathrm{\,cos\,}\theta \text{'}}{r\,\mathrm{\,\rho \,}g}`` ...(ii)
    Dividing equation (ii) by (i), we get:
    `` \frac{h\text{'}}{h}=\frac{\,\mathrm{\,cos\,}\theta \text{'}}{\,\mathrm{\,cos\,}\theta }``
    `` \Rightarrow \,\mathrm{\,cos\,}\theta \text{'}=\frac{\,\mathrm{\,cos\,}\theta }{2}``
    `` \Rightarrow \theta =112°``
    Page No 301: (a) Depression (h) of mercury level is expressed as follows:
    `` h=\frac{2T\,\mathrm{\,cos\,}\theta }{r\rho g}`` ...(i)
    `` \Rightarrow h=\frac{2\times 0.465\times \,\mathrm{\,cos\,}135°}{{10}^{-3}\times 13600\times \left(9.8\right)}``
    `` =0.0053\,\mathrm{\,m\,}=5.3\,\mathrm{\,mm\,}`` (b) If the length dipped inside is half the result obtained above:
    New depression h'= `` \frac{h}{2}``
    Let the new contact angle of mercury with glass be θ'.
    ∴ `` h\text{'}=\frac{2T\,\mathrm{\,cos\,}\theta \text{'}}{r\,\mathrm{\,\rho \,}g}`` ...(ii)
    Dividing equation (ii) by (i), we get:
    `` \frac{h\text{'}}{h}=\frac{\,\mathrm{\,cos\,}\theta \text{'}}{\,\mathrm{\,cos\,}\theta }``
    `` \Rightarrow \,\mathrm{\,cos\,}\theta \text{'}=\frac{\,\mathrm{\,cos\,}\theta }{2}``
    `` \Rightarrow \theta =112°``
    Page No 301:
  • #26-a
    Find the depression of mercury column in the capillary.
    Ans : Depression (h) of mercury level is expressed as follows:
    `` h=\frac{2T\,\mathrm{\,cos\,}\theta }{r\rho g}`` ...(i)
    `` \Rightarrow h=\frac{2\times 0.465\times \,\mathrm{\,cos\,}135°}{{10}^{-3}\times 13600\times \left(9.8\right)}``
    `` =0.0053\,\mathrm{\,m\,}=5.3\,\mathrm{\,mm\,}``
  • #26-b
    If the length dipped inside is half the answer of part
    (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury = 0.465 N m-1 and the contact angle of mercury with glass -135 °.
    digAnsr:   I
    Ans : If the length dipped inside is half the result obtained above:
    New depression h'= `` \frac{h}{2}``
    Let the new contact angle of mercury with glass be θ'.
    ∴ `` h\text{'}=\frac{2T\,\mathrm{\,cos\,}\theta \text{'}}{r\,\mathrm{\,\rho \,}g}`` ...(ii)
    Dividing equation (ii) by (i), we get:
    `` \frac{h\text{'}}{h}=\frac{\,\mathrm{\,cos\,}\theta \text{'}}{\,\mathrm{\,cos\,}\theta }``
    `` \Rightarrow \,\mathrm{\,cos\,}\theta \text{'}=\frac{\,\mathrm{\,cos\,}\theta }{2}``
    `` \Rightarrow \theta =112°``
    Page No 301: