14: Some Mechanical Properties Of Matter : Neet-xii-physics

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NEET-XII-Physics

14: Some Mechanical Properties of Matter

with Solutions - page 7

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#18-b
by the mercury below it and
Ans : Force exerted by mercury below the surface area:
`` \,\mathrm{\,Pressure\,}P\text{'}={P}_{0}+\frac{2T}{r}``
`` F=P\text{'}A=\left({P}_{0}+\frac{2T}{r}\right)A``
`` =\left(0.1+\frac{2\times 0.465}{4\times {10}^{-3}}\right)\times {10}^{-6}``
`` =0.1+0.00023=0.10023\,\mathrm{\,N\,}``

#18-c
by the mercury surface in contact with it. Atmospheric pressure = 1.0 × 10⁵ Pa and surface tension of mercury = 0.465 N m^-1. Neglect the effect of gravity. Assume all numbers to be exact.
Ans : Force exerted by mercury surface in contact with it:
`` P=\frac{2T}{r}``
`` F=PA=\frac{2T}{r}A``
`` =\frac{2\times 0.465}{4\times {10}^{-3}}\times {10}^{-6}=0.00023\,\mathrm{\,N\,}``
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Qstn #19
The capillaries shown in figure have inner radii 0.5 mm, 1.0 mm and 1.5 mm respectively. The liquid in the beaker is water. Find the heights of water level in the capillaries. The surface tension of water is 7.5 × 10^-2 N m^-1.

Figure
Ans : Given:
Surface tension of water T = 7.5 × 10^-2 N/m
Taking cos θ = 1:
Radius of capillary A (r_A) = 0.5 mm = 0.5 × 10^-3 m
Height of water level in capillary A:
`` {h}_{\,\mathrm{\,A\,}}=\frac{2T\,\mathrm{\,cos\,}\theta }{{r}_{\,\mathrm{\,A\,}}\rho g}``
`` =\frac{2\times 7.5\times {10}^{-2}}{0.5\times {10}^{-3}\times 1000\times 10}``
`` =3\times {10}^{-2}\,\mathrm{\,m\,}=3\,\mathrm{\,cm\,}``
Radius of capillary B (r_B) = 1 mm = 1 × 10^-3 m
Height of water level in capillary B:
`` {h}_{\,\mathrm{\,B\,}}=\frac{2T\,\mathrm{\,cos\,}\theta }{{r}_{\,\mathrm{\,B\,}}\rho g}``
`` =\frac{2\times 7.5\times {10}^{-2}}{1\times {10}^{-3}\times {10}^{3}\times 10}``
`` =15\times {10}^{-3}\,\mathrm{\,m\,}=1.5\,\mathrm{\,cm\,}``
Radius of capillary C (r_C) = 1.5 mm = 1.5 × 10^-3 m
Height of water level in capillary C:
`` {h}_{\,\mathrm{\,C\,}}=\frac{2T\,\mathrm{\,cos\,}\,\mathrm{\,\theta \,}}{{r}_{\,\mathrm{\,C\,}}\rho g}``
`` =\frac{2\times 7.5\times {10}^{-2}}{1.5\times {10}^{-3}\times {10}^{3}\times 10}``
`` =\frac{15}{1.5}\times {10}^{-3}\,\mathrm{\,m\,}=1\,\mathrm{\,cm\,}``
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Qstn #20
The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, up to what height will the water rise in the capillary?
Ans : Let T be the surface tension, r be the inner radius of the capillary tube and ρ be the density of the liquid.
For cos θ = 1, height (h) of the liquid level is given as:
`` h=\frac{2T\,\mathrm{\,cos\,}\theta }{r\,\mathrm{\,\rho \,}g}``
Now, for mercury:
`` {h}_{\,\mathrm{\,Hg\,}}=\frac{2{T}_{\,\mathrm{\,Hg\,}}}{r{\,\mathrm{\,\rho \,}}_{\,\mathrm{\,Hg\,}}g}`` ...(i)
For water:
`` {h}_{w}=\frac{2{T}_{w}}{r{\,\mathrm{\,\rho \,}}_{w}g}`` ...(ii)
Dividing (ii) by (i), we get:
`` \frac{{h}_{w}}{{h}_{\,\mathrm{\,Hg\,}}}=\frac{{T}_{w}}{{T}_{\,\mathrm{\,Hg\,}}}\times \frac{{\rho }_{\,\mathrm{\,Hg\,}}}{{\rho }_{w}}``
`` =\left(\frac{0.075}{0.465}\right)\times \left(13.6\right)``
`` =2.19``
Height of the water level:
h_w = 2 × 2.19 = 4.38 cm
Hence, the required rise in the water level in the capillary tube is 4.38 cm.
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Qstn #21
A barometer is constructed with its tube having radius 1.0 mm. Assume that the surface of mercury in the tube is spherical in shape. If the atmospheric pressure is equal to 76 cm of mercury, what will be the height raised in the barometer tube? The contact angle of mercury with glass = 135° and surface tension of mercury = 0.465 N m^-1. Density of mercury = 13600 kg m^-3.
Ans : Given:
Radius of tube r = 1.0 mm
Atmospheric pressure = 76 cm of Hg
Contact angle of mercury with glass θ = 135°
Surface tension of mercury T = 0.465 N/m
Density of mercury = 13600 kg m^-3
Let h be the rise in level in the barometer.
`` h=\frac{2T\,\mathrm{\,cos\,}\theta }{r\rho g}``
`` =\frac{2\times 465\times \left(1/\sqrt{2}\right)}{{10}^{-3}\times 13600\times 10}=0.0048\,\mathrm{\,m\,}``
`` =0.48\,\mathrm{\,cm\,}``
∴ Net rise in level in the barometer tube = H - h
= 76 - 0.48
= 75.52 cm
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Qstn #22
A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. Find the difference between the pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure. Surface tension of water = 0.075 N m^-1.
Ans : Given:
Radius of capillary tube r = 0.5 mm = 5 × 10^-4 m
Depth (where pressure is to be found) h = 5.0 cm = 5 × 10^-2 m
Surface tension of water T = 0.075 N/m
Excess pressure at 5 cm before the surface:
P = ρhg = 1000 × (5 × 10^-2) × 9.8 = 490 N/m²
Excess pressure at the surface is given by:
`` {P}_{0}=\frac{2T}{r}=\frac{2\times \left(0.75\right)}{\left(5\times {10}^{-4}\right)}``
`` =300\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
Difference in pressure: P₀- P `` =490-300=190\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
Hence, the required difference in pressure is 190 N/m².
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Qstn #23
Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water = 0.075 J m^-2.
Ans : Given:
Radius of cylindrical vessel, r = 6.0 cm = 0.06 m
Surface tension of water, T = 0.075 J/m²
Area, A = πr² = π × (0.06)²
Surface energy = T × A
= (0.075) × (3.14) × (0.06)²
= 8.5 × 10^-4 J
Therefore, the surface energy of water kept in a cylindrical vessel is 8.5 × 10^-4 J.
Page No 301:

Qstn #24
A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J m^-2.
Ans : Given:
Initial radius of mercury drop R = 2 mm = 2 × 10^-3 m
Surface tension of mercury T = 0.465 J/m²
Let the radius of a small drop of mercury be r.
As one big drop is split into 8 identical droplets:
volume of initial drop = 8 × (volume of a small drop)
`` \left(\frac{4}{3}\right)\pi {R}^{3}=\left(\frac{4}{3}\right)\,\mathrm{\,\pi \,}{r}^{3}\times 8``
`` ``
Taking cube root on both sides of the above equation:
`` r=\frac{R}{2}=10``^-3 m
Surface energy = T × surface area
∴ Increase in surface energy = TA' - TA
= (8 × 4πr² - 4πR²) T
`` =4\pi T\left[8\times \left(\frac{{R}^{2}}{4}\right)-{R}^{2}\right]``
`` =4\pi T{R}^{2}``
= 4 × (3.14) × (0.465) × (4 × 10^-6)
= 23.36 × 10^-6
= 23.4 μJ
Hence, the required increase in the surface energy of the mercury droplets is 23.4 μJ.
Page No 301:

Qstn #25
A capillary tube of radius 1 mm is kept vertical with the lower end in water.
Ans : Given:
Radius of capillary tube r = 1 mm = 10^-3 m

#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

`` \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:

Qstn #26
The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury.
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.

#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°``
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Qstn #27
Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = 0.075 Nm^-1.
Ans : Given:
Surface tension of water T = 0.075 N/m
Separation between the glass plates d = 1 mm = 10^-3 m
Density of water ρ = 10³ kg/m³
Applying law of conservation of energy:
T (2L) = [1 × (10^-3) × h] ρg
`` \Rightarrow h=\frac{2\times \left(0.075\right)}{{10}^{-3}\times {10}^{3}\times 10}``
`` =0.015\,\mathrm{\,m\,}=1.5\,\mathrm{\,cm\,}``
Therefore, the rise of water in the space between the plates is 1.5 cm.
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