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 6

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Qstn #8
A steel rod of cross-sectional area 4 cm² and 2 m shrinks by 0.1 cm as the temperature decreases in night. If the rod is clamped at both ends during the day hours, find the tension developed in it during night hours. Young modulus of steel = 1.9 × 10¹¹ N m^-2.
Ans : Given:
Cross-sectional area of steel rod A = 4 cm² = 4 × 10^-4 m²
Length of steel rod L = 2 m
Compression during night hours ΔL = 0.1 cm = 10^-3 m
Young modulus of steel Y = 1.9 × 10¹¹ N m^-2
Let the tension developed at night be F.
`` Y=\frac{F}{A}\times \frac{L}{∆L}``
`` \Rightarrow F=\frac{YA∆L}{L}``
`` =\frac{1.9\times {10}^{11}\times 4\times {10}^{-4}\times {10}^{-3}}{2}``
`` =3.8\times {10}^{4}\,\mathrm{\,N\,}``
∴ Required tension developed in steel rod during night hours = 3.8 × 10⁴ N.
Page No 300:

Qstn #9
Consider the situation shown in figure. The force F is equal to the m₂g/2. If the area of cross section of the string is A and its Young modulus Y, find the strain developed in it. The string is light and there is no friction anywhere.

Figure
Ans : Given:
Force (F) = m₂g/2
Area of cross-section of the string = A
Young's modulus = Y
Let a be the acceleration produced in block m₂ in the downward direction and T be the tension in the string.
From the free body diagram:
`` {m}_{2}g-T={m}_{2}a...\left(\,\mathrm{\,i\,}\right)``
`` T-F={m}_{1}a...\left(\,\mathrm{\,ii\,}\right)``

From equations (i) and (ii), we get:
`` a=\frac{{m}_{2}g-F}{{m}_{1}+{m}_{2}}``
`` \,\mathrm{\,Applying\,}F=\frac{{m}_{2}g}{2}``
`` \Rightarrow a=\frac{{m}_{2}g}{2\left({m}_{1}+{m}_{2}\right)}``
Again, T = F + m₁a
On applying the values of F and a, we get:
`` \Rightarrow T=\frac{{m}_{2}g}{2}+{m}_{1}\frac{{m}_{2}g}{2\left({m}_{1}+{m}_{2}\right)}``
We know that:
`` Y=\frac{FL}{A∆L}``
`` \Rightarrow \,\mathrm{\,Strain\,}=\frac{∆L}{L}=\frac{F}{AY}``
`` \Rightarrow \,\mathrm{\,Strain\,}=\frac{\left({m}_{2}^{2}+2{m}_{1}{m}_{2}\right)g}{2\left({m}_{1}+{m}_{2}\right)AY}``
`` =\frac{{m}_{2}g\left(2{m}_{1}+{m}_{2}\right)}{2AY\left({m}_{1}+{m}_{2}\right)}``
∴ Required strain developed in the string `` =\frac{{m}_{2}g\left(2{m}_{1}+{m}_{2}\right)}{2AY\left({m}_{1}+{m}_{2}\right)}``.
Page No 300:

Qstn #10
A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young modulus of the metal of the wire is 2.0 × 10¹¹ N m^-2. Make appropriate approximations.
Ans : Given:
Mass of sphere (m) = 20 kg
Length of metal wire (L) = 4 m
Diameter of wire (d = 2r) = 1 mm
⇒ r = 5 × 10^-4 m
Young's modulus of the metal wire = 2.0 × 10¹¹ N m^-2
Tension in the wire in equilibrium = T
T= mg
When it is moved at an angle θ and released, let the tension at the lowest point be T'.
$$ \Rightarrow T\text{'}=mg+\frac{m{v}^{2}}{r}$$
The change in tension is due to the centrifugal force.
∴ `` ∆T=T\text{'}-T``
$$ ∆T=\frac{m{v}^{2}}{r}...\left(\,\mathrm{\,i\,}\right)$$
Now, using work energy principle:
$$ \frac{1}{2}m{v}^{2}-0=mgr\left(1-\,\mathrm{\,cos\,}\theta \right)$$
`` \Rightarrow {v}^{2}=2gr\left(1-\,\mathrm{\,cos\,}\theta \right)...\left(2\right)``
`` ``
Applying the value of v² in (i):
$$ ∆T=\frac{m\left[2gr\left(1-cos\theta \right)\right]}{r}$$
`` =2mg\left(1-cos\theta \right)``
`` ``
`` \,\mathrm{\,Now\,},F=∆T``
`` \,\mathrm{\,Also\,},F=\frac{YA∆L}{L}``
`` \Rightarrow \frac{YA∆L}{L}=2mg\left(1-\,\mathrm{\,cos\,}\theta \right)``
`` \Rightarrow \,\mathrm{\,cos\,}\theta =1-\frac{YA∆L}{L\left(2mg\right)}``
$$ \Rightarrow cos\theta =1-\left[\frac{2\times {10}^{11}\times 4\times 3.14\times {\left(5\right)}^{2}\times {10}^{-8}\times 2\times {10}^{-3}}{4\times 2\times 20\times 10}\right]$$
`` \Rightarrow \,\mathrm{\,cos\,}\theta =0.80``
`` \,\mathrm{\,Or\,},\theta =36.4°``
Hence, the required maximum value of θ is 35.4˚.
Page No 301:

Qstn #11
A steel wire of original length 1 m and cross-sectional area 4.00 mm² is clamped at the two ends so that it lies horizontally and without tensions. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression?
Y of the steel = 2.0 × 10¹¹ N m^-2. Take g = 10 m s^-2.
Ans : Given:
Original length of steel wire (L) = 1 m
Area of cross-section (A) = 4.00 mm² = 4 × 10^-2 cm²
Load = 2.16 kg
Young's modulus of steel (Y) = 2 × 10¹¹ N/m²
Acceleration due to gravity (g) = 10 m s^-2
Let T be the tension in the string after the load is suspended and θ be the angle made by the string with the vertical, as shown in the figure:

`` \,\mathrm{\,cos\,}\theta =\frac{x}{\sqrt{{x}^{2}+{l}^{2}}}=\frac{x}{l}{\left\{1+\frac{{x}^{2}}{{l}^{2}}\right\}}^{-1/2}``
`` ``
Expanding the above equation using the binomial theorem:
`` \,\mathrm{\,cos\,}\theta =\frac{x}{l}\left\{1-\frac{1}{2}\frac{{x}^{2}}{{l}^{2}}\right\}\left(\,\mathrm{\,neglecting\,}\,\mathrm{\,the\,}\,\mathrm{\,higher\,}\,\mathrm{\,order\,}\,\mathrm{\,terms\,}\right)``
`` \,\mathrm{\,Since\,}x<<l,\frac{{x}^{2}}{{l}^{2}}\,\mathrm{\,can\,}\,\mathrm{\,be\,}\,\mathrm{\,neglected\,}.``
`` \Rightarrow \,\mathrm{\,cos\,}\theta =\frac{x}{l}``
`` ``
`` ``
`` ``
Increase in length:
ΔL = (AC + CB) - AB
AC = (l² + x²)^1/2
ΔL = 2 (l² + x²)^1/2 - 2l
We know that:
`` Y=\frac{F}{A}\frac{L}{∆L}``
`` \Rightarrow 2\times {10}^{12}=\frac{T\times 100}{\left(4\times {10}^{-2}\right)\times \left[2{\left({50}^{2}+{x}^{2}\right)}^{1/2}-100\right]}``
From the free body diagram:
`` 2T\,\mathrm{\,cos\,}\theta =mg``
`` 2T\left(\frac{x}{50}\right)=2.16\times {10}^{3}\times 980``
`` \Rightarrow \frac{2\times \left(2\times {10}^{12}\right)\times \left(4\times {10}^{-2}\right)\times \left[2\left({50}^{2}+{x}^{2}{\displaystyle \frac{1}{2}}\right)-100\right]x}{100\times 50}=\left(2.16\right)\times {10}^{3}\times 980``
On solving the above equation, we get x = 1.5 cm.
Hence, the required vertical depression is 1.5 cm.
Page No 301:

Qstn #12
A copper wire of cross-sectional area 0.01 cm² is under a tension of 20N. Find the decrease in the cross-sectional area. Young modulus of copper = 1.1 × 10¹¹ N m^-2 and Poisson ratio = 0.32.
Ans : Given:
Cross-sectional area of copper wire A = 0.01 cm² = 10^-6 m²
Applied tension T = 20 N
Young modulus of copper Y = 1.1 × 10¹¹ N/m²
Poisson ratio σ = 0.32
We know that:
`` Y=\frac{FL}{A∆L}``
`` \Rightarrow \frac{∆L}{L}=\frac{F}{AY}``
`` =\frac{20}{{10}^{-6}\times 1.1\times {10}^{11}}=18.18\times {10}^{-5}``
`` ``
`` \,\mathrm{\,Poisson\,}\text{'}\,\mathrm{\,s\,}\,\mathrm{\,ratio\,},\sigma =\frac{{\displaystyle \frac{∆d}{d}}}{{\displaystyle \frac{∆L}{L}}}=0.32``
`` \,\mathrm{\,Where\,}d\,\mathrm{\,is\,}\,\mathrm{\,the\,}\,\mathrm{\,transverse\,}\,\mathrm{\,length\,}``
`` \,\mathrm{\,So\,},\frac{∆d}{d}=\left(0.32\right)\times \frac{∆L}{L}``
`` =0.32\times \left(18.18\right)\times {10}^{-5}=5.81\times {10}^{-5}``
`` \,\mathrm{\,Again\,},\frac{∆A}{A}=\frac{2∆r}{r}=\frac{2∆d}{d}``
`` \Rightarrow ∆A=\frac{2∆d}{d}A``
`` \Rightarrow ∆A=2\times \left(5.8\times {10}^{-5}\right)\times \left(0.01\right)``
`` =1.164\times {10}^{-6}{\,\mathrm{\,cm\,}}^{2}``
Hence, the required decrease in the cross -sectional area is `` 1.164\times {10}^{-6}{\,\mathrm{\,cm\,}}^{2}``.
Page No 301:

Qstn #13
Find the increase in pressure required to decrease the volume of a water sample by 0.01%. Bulk modulus of water = 2.1 × 10⁹ N m^-2.
Ans : Given:
Bulk modulus of water (B) = `` 2.1\times {10}^{9}{\,\mathrm{\,Nm\,}}^{-2}``
In order to decrease the volume (V) of a water sample by 0.01%, let the increase in pressure be P.
`` \frac{V\times 0.01}{100}=∆V``
`` \Rightarrow \frac{∆V}{V}={10}^{-4}``
`` \,\mathrm{\,From\,}B=\frac{PV}{∆V},\,\mathrm{\,we\,}\,\mathrm{\,have\,}:``
`` \Rightarrow P=B\left(\frac{∆V}{V}\right)``
`` =2.1\times {10}^{9}\times {10}^{-4}``
`` =2.1\times {10}^{5}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
Hence, the required increase in pressure is `` 2.1\times {10}^{5}{\,\mathrm{\,Nm\,}}^{-2}``.
Page No 301:

Qstn #14
Estimate the change in the density of water in ocean at a depth of 400 m below the surface. The density of water at the surface = 1030 kg m^-3 and the bulk modulus of water = 2 × 10⁹ N m^-2.
Ans : Given:
`` \,\mathrm{\,Bulk\,}\,\mathrm{\,modulus\,}\,\mathrm{\,of\,}waterB=2\times {10}^{9}\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
`` ``
Depth (d) = 400 m
Density of water at the surface (ρ₀) = 1030 kg/m³
We know that:
`` \,\mathrm{\,Density\,}\,\mathrm{\,at\,}\,\mathrm{\,surface\,}{\rho }_{0}=\frac{m}{{\,\mathrm{\,V\,}}_{0}}``
`` \,\mathrm{\,Density\,}\,\mathrm{\,at\,}\,\mathrm{\,depth\,}{\rho }_{\,\mathrm{\,d\,}}=\frac{m}{{\,\mathrm{\,V\,}}_{\,\mathrm{\,d\,}}}``
`` \Rightarrow \frac{{\rho }_{\,\mathrm{\,d\,}}}{{\rho }_{0}}=\frac{{V}_{0}}{{V}_{d}}...\left(\,\mathrm{\,i\,}\right)``
Here: ρ_d= density of water at a depth
m = mass
V₀ = volume at the surface
V_d = volume at a depth
`` \,\mathrm{\,Pressure\,}\,\mathrm{\,at\,}\,\mathrm{\,a\,}\,\mathrm{\,depth\,}d={\,\mathrm{\,\rho \,}}_{0}gd``
`` \,\mathrm{\,Acceleration\,}\,\mathrm{\,due\,}\,\mathrm{\,to\,}\,\mathrm{\,gravity\,}g=10{\,\mathrm{\,ms\,}}^{2}``
`` \,\mathrm{\,Volume\,}\,\mathrm{\,strain\,}=\frac{{V}_{0}-{V}_{d}}{{V}_{0}}``
`` B=\frac{\,\mathrm{\,Pressure\,}}{\,\mathrm{\,Volume\,}\,\mathrm{\,strain\,}}``
`` \Rightarrow B=\frac{{\,\mathrm{\,\rho \,}}_{0}gd}{\left(\frac{{V}_{0}-{V}_{d}}{{V}_{0}}\right)}``
`` \Rightarrow 1-\frac{{V}_{d}}{{V}_{0}}=\frac{{\,\mathrm{\,\rho \,}}_{0}gd}{B}``
`` \Rightarrow \frac{{V}_{d}}{{V}_{0}}=\left(1-\frac{{p}_{0}gd}{B}\right)...\left(\,\mathrm{\,ii\,}\right)``
Using equations (i) and (ii), we get:
`` \frac{{\rho }_{\,\mathrm{\,d\,}}}{{\rho }_{0}}=\frac{1}{\left(1-{\displaystyle \frac{{\rho }_{0}gd}{B}}\right)}``
`` \Rightarrow {\rho }_{d}=\frac{1}{\left(1-{\displaystyle \frac{{\rho }_{0}gh}{B}}\right)}{\rho }_{0}``
`` \Rightarrow {\rho }_{d}=\frac{1030}{\left(1-{\displaystyle \frac{1030\times 10\times 400}{2\times {10}^{9}}}\right)}\approx 1032\,\mathrm{\,kg\,}/{\,\mathrm{\,m\,}}^{3}``
`` \,\mathrm{\,Change\,}\,\mathrm{\,in\,}\,\mathrm{\,density\,}={\rho }_{\,\mathrm{\,d\,}}-{\rho }_{0}``
`` =1032-1030=2\,\mathrm{\,kg\,}/{\,\mathrm{\,m\,}}^{3}``
Hence, the required density at a depth of 400 m below the surface is 2 kg/m³.
Page No 301:

Qstn #15
A steel plate of face area 4 cm² and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel = 8.4 × 10¹⁰ N m^-2.
Ans : Given:
Face area of steel plate A = 4 cm² = 4 × 10^-4 m²
Thickness of steel plate d = 0.5 cm = 0.5 × 10^-2 m
Applied force on the upper surface F = 10 N
Rigidity modulus of steel = 8.4 × 10¹⁰ N m^-2
Let θ be the angular displacement.
Rigidity modulus `` m=\frac{F}{A\theta }``
`` \Rightarrow m=\left(\frac{10}{4\times {10}^{-4}\theta }\right)``
`` \Rightarrow \theta =\frac{10}{4\times {10}^{-4}\times 8.4\times {10}^{10}}``
`` =0.297\times {10}^{-6}``
∴ Lateral displacement of the upper surface with respect to the lower surface = θ × d
⇒ (0.297) × 10^-6 × (0.5) × 10^-2
⇒ 1.5 × 10^-9 m
Hence, the required lateral displacement of the steel plate is 1.5 × 10^-9 m.
Page No 301:

Qstn #16
A 5.0 cm long straight piece of thread is kept on the surface of water. Find the force with which the surface on one side of the thread pulls it. Surface tension of water = 0.076 N m^-1.
Ans : Given:
Length of thread l = 5 cm = 5 × 10^-2 m
Surface tension of water T = 0.76 N/m
We know that:
F = T × l = 0.76 × 5 × 10^-2
= 3.8 × 10^-3 N
Therefore, the water surface on one side of the thread pulls it with a force of 3.8 × 10^-3 N.
Page No 301:

Qstn #17
Find the excess pressure inside
Ans : Given:
`` \,\mathrm{\,Radius\,}\,\mathrm{\,of\,}\,\mathrm{\,mercury\,}\,\mathrm{\,drop\,}r=2\,\mathrm{\,mm\,}=2\times {10}^{-3}\,\mathrm{\,m\,}``
`` \,\mathrm{\,Radius\,}\,\mathrm{\,of\,}\,\mathrm{\,soap\,}\,\mathrm{\,bubble\,}r=4\,\mathrm{\,mm\,}=4\times {10}^{-3}\,\mathrm{\,m\,}``
`` \,\mathrm{\,Radius\,}\,\mathrm{\,of\,}\,\mathrm{\,air\,}\,\mathrm{\,bubble\,}r=4\,\mathrm{\,mm\,}=4\times {10}^{-3}\,\mathrm{\,m\,}``
`` \,\mathrm{\,Surface\,}\,\mathrm{\,tension\,}\,\mathrm{\,of\,}\,\mathrm{\,mercury\,}{T}_{\,\mathrm{\,Hg\,}}=0.465\,\mathrm{\,N\,}/\,\mathrm{\,m\,}``
`` \,\mathrm{\,Surface\,}\,\mathrm{\,tension\,}\,\mathrm{\,of\,}\,\mathrm{\,soap\,}\,\mathrm{\,solution\,}{T}_{s}=0.03\,\mathrm{\,N\,}/\,\mathrm{\,m\,}``
`` \,\mathrm{\,Surface\,}\,\mathrm{\,tension\,}\,\mathrm{\,of\,}\,\mathrm{\,water\,}{T}_{a}=0.076\,\mathrm{\,N\,}/\,\mathrm{\,m\,}``
`` ``

#17-a
a drop of mercury of radius 2 mm
Ans : Excess pressure inside mercury drop:
`` P=\frac{2{T}_{Hg}}{r}``
`` =\frac{0.465\times 2}{2\times {10}^{-3}}=465\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``

#17-b
a soap bubble of radius 4 mm and
Ans : Excess pressure inside the soap bubble:
`` P=\frac{4{T}_{s}}{r}``
`` =\frac{4\times 0.03}{4\times {10}^{-3}}=30\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
`` ``

#17-c
an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury, soap solution and water are 0.465 N m^-1, 0.03 N m^-1 and 0.076 N m^-1 respectively.
Ans : Excess pressure inside the air bubble:
`` P=\frac{2{T}_{a}}{r}``
`` =\frac{2\times 0.076}{4\times {10}^{-3}}=38\,\mathrm{\,N\,}/{\,\mathrm{\,m\,}}^{2}``
Page No 301:

Qstn #18
Consider a small surface area of 1 mm² at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area
Ans : Given:
Surface area of mercury drop, A = 1 mm² = 10^-6 m²
Radius of mercury drop, r = 4 mm = 4 × 10^-3 m
Atmospheric pressure, P₀ = 1.0 × 10⁵ P_a
Surface tension of mercury, T = 0.465 N/m

#18-a
by the air above it
Ans : Force exerted by air on the surface area:
F = P₀A
⇒ F= 1.0 × 10⁵ × 10^-6 = 0.1 N