NEET-XII-Physics
14: Some Mechanical Properties of Matter
- #26-aFind 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 : 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°``
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- #26-bIf 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: IAns : 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°``
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