NEET-XII-Physics
14: Some Mechanical Properties of Matter
- #25-aFind 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 : 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
- `` \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\,}``
`` ``
`` 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: - `` \Rightarrow h=\frac{2\times \left(0.076\right)}{{10}^{-3}\times 10\times 100}``
- #25-bIf 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: bAns : 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: