Mechanical Properties of Solids - Class 11 Physics - Chapter 8 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Mechanical Properties of Solids | NCERT | Physics | Class 11
Consider a cylindrical rod of material having a negative Poisson's ratio. A force is applied along its length which can increase or decrease its length. Then which of the following is true?
A. If the length of the rod increases, its radius decreases.
B. If the length of the rod increases, its surface area remains the same.
C. If the length of the rod decreases, its radius also decreases.
D. If the length of the rod decreases, its surface area increases.
The correct answer is C. If the length of the rod decreases, its radius also decreases.
To understand this, we should consider the Poisson's ratio, denoted as $ \mu $, which is defined as: $$ \mu = -\frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} $$ In the case of a material with a negative Poisson's ratio, when the rod is stretched longitudinally (increasing its length), the lateral dimensions (like its diameter) also expand rather than contract as they would with a normal positive Poisson's ratio. Conversely, when the rod is compressed longitudinally (decreasing its length), the lateral dimensions contract as well, which corresponds to a decrease in the radius.
Therefore, when the length of such a rod decreases, the radius also decreases, matching option C.
Choose the correct statements with respect to the property of bone china.
A. Manufacturing process involves the addition of animal bone ashes.
B. Manufacturing process involves the addition of resins.
C. It is harder than porcelain.
D. It is softer than porcelain.
The correct statements regarding the properties of bone china are:
A: Manufacturing process involves the addition of animal bone ashes.
C: It is harder than porcelain.
Bone china is indeed a type of porcelain, distinguished by its high strength and chip resistance due to the inclusion of bone ashes in its composition. The typical formula consists of bone ash fused with china clay, feldspar, and fine silica, yielding a material that surpasses regular porcelain in terms of hardness. Resins (as mentioned in option B) are not utilized in the manufacturing process, and bone china is definitely not softer than porcelain, contradicting option D.
Which is the natural hardest substance?
A) Diamond
B) Iron
C) Wurtzite Boron Nitride
D) Lonsdaleite
The correct answer is A) Diamond.
Diamond is recognized as the hardest naturally occurring substance on Earth. Diamonds are unique not only because of their hardness but also because they have the ability to bend and reflect light, breaking it into the colors of the rainbow which contributes to their captivating sparkle.
A simple spring has length (I) and force constant $k$. It is cut into two springs of lengths $I_{1}$ and $I_{2}$ such that $I_{1} = n I_{2}$ ($n =$ an integer). The force constant of the spring of length (I_{1}) is
(A) $k(1+n)$
(B) $\frac{k}{n}(1+n)$
(C) $k$
(D) $\frac{k}{n+1}$
The correct answer is (B) $\frac{k}{n}(1+n)$.
Let's denote the spring constant of the spring with length $I_1$ as $k_1$ and the spring constant of the spring with length $I_2$ as $k_2$. Given that $I_1 = n I_2$, the original spring is equivalent to $(n+1)$ parts, each having the same length as $I_2$.
From the property of springs in series, the effective spring constant $k_2$ of a section of length $I_2$ is determined by combining $(n+1)$ such sections: $$ \frac{1}{k} = \frac{n+1}{k_2} \implies k_2 = \frac{k}{n+1} $$
Since $I_1$ consists of $n$ segments, each identical to a segment of length $I_2$, the spring constant $k_1$ when springs are in series is: $$ \frac{1}{k_1} = \frac{n}{k_2} = \frac{n}{\frac{k}{n+1}} = \frac{n(n+1)}{k} $$ Therefore: $$ k_1 = \frac{k}{n}(1+n) $$
Thus, the consequent spring constant for the spring of length $I_1$ is indeed $\frac{k}{n}(1+n)$, aligning with option (B).
A piece of copper wire has twice the radius of a piece of steel wire. Young's modulus for steel is twice that of copper. One end of the copper wire is joined to one end of the steel wire so that both can be subjected to the same longitudinal force. By what fraction of its length will the steel have stretched when the length of the copper has increased by 1%?
A) 1% B) 2% C) 2.5% D) 3%
The correct option is B) 2%.
Let's denote the radius of the copper wire as $r_{cu}$ and the radius of the steel wire as $r_{st}$. Given that: $$ r_{\mathrm{cu}} = 2 r_{\mathrm{st}} $$
Since the area of a circle is proportional to the square of the radius, the cross-sectional area of the copper wire, $A_{cu}$, is four times that of the steel wire, $A_{st}$: $$ A_{cu} = 4 A_{st} $$
Given also that the Young's modulus of steel, $Y_{st}$, is twice that of copper, $Y_{cu}$: $$ Y_{st} = 2 Y_{cu} $$
When the length of the copper changes by 1%, this change is represented by: $$ \frac{\Delta l_{cu}}{l_{cu}} = 0.01 $$
Using Hooke's Law for extension under the same force $F$, for both wires, we get the following relationships for the elongation $\Delta l$:
For steel: $$ \frac{\Delta l_{st}}{l_{st}} = \frac{F}{A_{st} \cdot Y_{st}} \quad \text{(1)} $$
For copper: $$ \frac{\Delta l_{cu}}{l_{cu}} = \frac{F}{A_{cu} \cdot Y_{cu}} \quad \text{(2)} $$
Dividing equation (1) by equation (2), you simplify the expression using the area and modulus values: $$ \frac{\frac{\Delta l_{st}}{l_{st}}}{\frac{\Delta l_{cu}}{l_{cu}}} = \frac{A_{cu} \cdot Y_{cu}}{A_{st} \cdot Y_{st}} $$
Plugging in the ratios $A_{cu} = 4 A_{st}$ and $Y_{st} = 2 Y_{cu}$ into the equation above, and knowing that $\frac{\Delta l_{cu}}{l_{cu}} = 0.01$, you obtain: $$ \frac{\Delta l_{st}}{l_{st}} = \frac{4 A_{st} \cdot Y_{cu}}{A_{st} \cdot 2 Y_{cu}} \times 0.01 = \frac{4}{2} \times 0.01 = 0.02 $$
Thus, the stretch fraction of the steel wire, when the copper wire has increased by 1%, is 2%.
What makes HDPE have density and strength compared to LDPE?
HDPE (High-Density Polyethylene) is generally stronger and has a higher density compared to LDPE (Low-Density Polyethylene) due to several reasons:
HDPE exhibits superior tensile strength and impact resistance, which can be attributed to its higher crystallinity and lower degree of branching in its structure. This makes the molecular chains pack together more tightly, leading to increased strength.
HDPE polymers can be easily processed using all standard thermoplastic methods, which contributes to their versatility and strength.
These plastics are remarkably resistant to chemicals, corrosion, and climate changes compared to LDPE, enhancing their durability and applicability in various environmental conditions.
Products made from HDPE materials are not only more durable but also better in terms of flexibility, strength, and they are relatively lightweight.
Overall, the structural composition and properties of HDPE confer it with characteristics that are advantageous over LDPE in several applications.
Arrange the following substances in increasing order of hardness:
A. Sponge
B. Brick
C. Iron rod
D. Diamond
To rank materials by their hardness—which is the ability of materials to resist physical changes such as deformations caused by external forces—we assess their resistance to compression and deformation. Based on these criteria, we classify the given substances in the following order:
Sponge: Soft and highly compressible.
Brick: Harder than a sponge but can still chip or break under high pressure.
Iron rod: Much more resistant to deformation, making it harder than a sponge and brick.
Diamond: Recognized as one of the hardest known natural materials on Earth, extremely resistant to scratches or deformation.
Hence, the substances in increasing order of hardness are:
$$ \text{Sponge} < \text{Brick} < \text{Iron rod} < \text{Diamond} $$
There is no change in the volume of a wire due to a change in its length on stretching. The Poisson's ratio of the material of the wire is:
(A) +0.50 (B) -0.50 (C) ? (D) -0.25
The correct answer is (B) -0.50.
To understand why, consider the formula for the fractional change in volume of the material, given by: $$ \frac{dV}{V} = \frac{dL}{L} + 2\sigma \frac{dL}{L} $$ Here, $\sigma$ is the Poisson's ratio, and $\frac{dL}{L}$ denotes the fractional change in length. The term $2\sigma \frac{dL}{L}$ accounts for changes in the transverse dimensions due to Poisson's effect.
Since we know that the volume change ($\frac{dV}{V}$) is zero (as the volume of the wire does not change): $$ \frac{dV}{V} = 0 $$ By substituting for $\frac{dV}{V}$, we have: $$ \frac{dL}{L} + 2 \sigma \frac{dL}{L} = 0 $$
Simplifying, we get: $$ (1 + 2 \sigma) \frac{dL}{L} = 0 $$
Since $\frac{dL}{L}$ cannot be zero (the wire is stretched), $(1+2\sigma)$ must be zero. Solving for $\sigma$: $$ 1 + 2 \sigma = 0 \implies \sigma = -\frac{1}{2} $$
Thus, the Poisson's ratio of the wire's material is -0.50.
Which of the following properties describe a thermosetting polymer?
A) Reusable
B) Rigid
C) Stretchable
D) Recyclable
The correct answer is B) Rigid.
Thermosetting polymers are characterized by their inability to be reformed once they have been cured or set. These types of plastics do not soften or melt when heated, which makes them rigid.
In contrast, properties such as being reusable, stretchable, and recyclable are not typically associated with thermosetting polymers. These materials undergo a chemical change that provides high resistance against heat and makes them permanently set into a rigid structure.
While building a house in highly seismic areas, the use of mud or timber is better than heavy construction materials.
A) True
B) False
The correct option is A) True.
In highly seismic areas, it is advisable to use lighter construction materials such as mud or timber for building structures, especially for the roof. These materials are preferred because, in the event of a collapse during an earthquake, the impact and damage are significantly reduced due to their light weight.
Furthermore, buildings constructed from materials like mud or bamboo not only offer good stability but are also able to withstand climate changes to a certain degree, enhancing their suitability for earthquake-prone regions.
If $\rho$ is the density of the material of a uniform rod and $\sigma$ is the breaking stress, and the length of the rod is such that the rod is just about to break due to its own weight when suspended vertically from a fixed support, then
A Length of the rod: $\frac{\sigma}{\rho g}$
B Stress at a cross section perpendicular to the length of the rod, at one fourth the length of the rod above its lowest point: $\frac{\sigma}{4}$
C Stress at all horizontal sections of the rod is the same.
D The rod is about to break from its mid point.
Solution Overview
The question presents a scenario involving the stresses on a uniform rod due to its own weight. To determine the correct statements, we explore the mechanics related to tensions in the rod under gravitational forces. We focus on two primary aspects: the length of the rod and the stress at its various sections.
Key Points to Derive:
Maximum stress occurs at the topmost section (where it is fixed).
Stress varies linearly along the length from the top to bottom.
Length of the Rod ($\frac{\sigma}{\rho g}$)
Maximum stress, $\sigma$, occurs at the upper end of the rod. This maximum stress equals the total weight of the rod divided by the cross-sectional area (A). Thus: $$ \sigma = \frac{AL \rho g}{A} = L \rho g. $$ Solving for $L$, the length triggering the breaking stress, gives: $$ L = \frac{\sigma}{\rho g} $$ This confirms that the length of the rod must be $ \frac{\sigma}{\rho g} $ for it to be on the verge of breaking.
Stress at a Cross Section
For a cross-section at one fourth the length from the lower end, the weight supported is $\frac{3}{4}$ of the total rod. Calculating the stress ($\sigma_{\text{section}}$) at this point: $$ \sigma_{\text{section}} = \frac{3}{4} L \rho g = \frac{3}{4}\sigma. $$ However, the solution incorrectly states that the stress at this point is $\frac{\sigma}{4}$. Hence, option B is only half the actual calculated stress.
Stress Continuity
The stress indeed varies linearly along the length, being maximal at the top and zero at the bottom. Hence, the stress is not the same at all horizontal sections.
Mid-Point Breaking Analysis
There is no particular reasoning provided to assert that the rod would break from the midpoint. The breaking would occur predominantly where the stress first meets or exceeds $\sigma$, which, due to the setup, is at the top end of the hanging rod.
Correct Statements
A: True - The length of the rod that would cause it to break under its own weight is indeed $\frac{\sigma}{\rho g}$.
B: False - This statement underestimates the stress at one fourth the length above the lowest point (it should be $\frac{3\sigma}{4}$ not $\frac{\sigma}{4}$).
C: False - The stress is not uniform along the rod.
D: False - The assertion about the rod breaking from the midpoint is unfounded based on the given physics.
This refined analysis underlines the fact that evaluating stresses in a rod under the influence of gravity requires precise handling of the weight distribution along the rod and understanding the relationships between stress, density, and gravitational force.
The type of cartilage found in intervertebral discs of the vertebral column is:
A) hyaline cartilage
B) elastic cartilage
C) yellow cartilage
D) fibrocartilage
The correct answer is D) fibrocartilage.
Intervertebral discs are composed of fibrocartilage, which plays a crucial role in reducing friction between the vertebrae.
The elastic limit of steel is $8 \times 10^{8} , \mathrm{Nm}^{-2}$ and its Young modulus is $2 \times 10^{11} , \mathrm{Nm}^{-2}$. Find the maximum elongation of a half-meter steel wire that can be given without exceeding the elastic limit.
To find the maximum elongation of a steel wire that wouldn't exceed its elastic limit, heed the following values and formula:
Elastic limit: $8 \times 10^{8} , \mathrm{Nm}^{-2}$
Young's modulus (Y): $2 \times 10^{11} , \mathrm{Nm}^{-2}$
Original length of the wire (L): $0.5 , \mathrm{m}$
The relationship provided by Young's modulus for elastic behavior can be described as: $$ Y = \frac{F \times L}{A \times \Delta L} $$ Where:
$F$ is the force applied,
$A$ is the cross-sectional area,
$\Delta L$ is the change in length.
From Young's equation, rearrange to solve for $\Delta L$: $$ \Delta L = \frac{F \times L}{A \times Y} $$
Using the maximal force derived from the elastic limit, input the values: $$ \Delta L = \frac{8 \times 10^{8} \times 0.5}{2 \times 10^{11}} $$
Calculate $\Delta L$: $$ \Delta L = 2 \times 10^{-3} , \mathrm{m} = 2 , \mathrm{mm} $$
Thus, the maximum elongation of the steel wire, within the elastic limit, is 2 mm.
A wire is stretched by 0.01 m by a certain force $F$. Another wire of the same material whose diameter and length are double that of the original wire is stretched by the same force. Then its elongation will be
A) 0.005 m
B) 0.01 m
C) 0.02 m
D) 0.002 m
The correct option is A) 0.005 m.
The elongation $I$ of a wire under a force $F$ can be described using the formula: $$ I = \frac{F L}{\pi r^2 Y} $$ where
$L$ is the length of the wire,
$r$ is the radius of the wire,
$Y$ is the Young's modulus of the material.
Since the force $F$ and the Young's modulus $Y$ remain constant, the formula simplifies to: $$ I \propto \frac{L}{r^2} \text{ (as both } F \text{ and } Y \text{ are constants)} $$
When comparing the original wire to the new wire with double the diameter and length, we use the proportions: $$ \frac{I_{2}}{I_{1}} = \left(\frac{L_{2}}{L_{1}}\right) \left(\frac{r_{1}}{r_{2}}\right)^{2} $$
Given $L_2 = 2L_1$ (double the length) and $r_2 = 2r_1$ (double the diameter), the equation becomes: $$ \frac{I_{2}}{I_{1}} = (2) \times \left(\frac{1}{2}\right)^2 = \frac{1}{2} $$
Thus, the new elongation $I_2$ is: $$ I_2 = \frac{I_1}{2} = \frac{0.01}{2} = 0.005 \text{ m} $$
Therefore, the elongation of the new wire when stretched by the same force $F$ is 0.005 m.
A $5$ m long aluminum wire $\left(Y = 7 \times 10^{10} N/m^{2}\right)$ of diameter $3$ mm supports a $40$ kg mass. In order to have the same elongation in a copper wire $\left(Y = 12 \times 10^{10} N/m^{2}\right)$ of the same length under the same weight, the diameter should now be, in mm:$5$
To determine the appropriate diameter for the copper wire, follow these steps.
Given:
Length ($L$) of both wires: 5 m
Diameter ($d_{Al}$) of aluminum wire: 3 mm
Young's modulus ($Y_{Al}$) for aluminum: $7 \times 10^{10}$ N/m²
Young's modulus ($Y_{Cu}$) for copper: $12 \times 10^{10}$ N/m²
Mass ($m$) supported by the wire: 40 kg
First, let's consider the elongation ($\Delta L$) formula: $$ \Delta L = \frac{F \cdot L}{\pi r^2 \cdot Y} = \frac{4 \cdot F \cdot L}{\pi d^2 \cdot Y} \text{ (Since } r = \frac{d}{2} \text{)} $$
Given that the elongation is the same for both wires under the same load, we can equate the expressions for the aluminum and copper wires as follows: |$$ \Delta L_{Al} = \Delta L_{Cu} $$| Thus, $$ \frac{4 \cdot F \cdot L}{\pi d_{Al}^2 \cdot Y_{Al}} = \frac{4 \cdot F \cdot L}{\pi d_{Cu}^2 \cdot Y_{Cu}} $$
By simplifying, we get: $$ \frac{1}{d_{Al}^2 \cdot Y_{Al}} = \frac{1}{d_{Cu}^2 \cdot Y_{Cu}} $$
Rearranging for $d_{Cu}^2$: $$ d_{Cu}^2 \cdot Y_{Cu} = d_{Al}^2 \cdot Y_{Al} $$
Solving for $d_{Cu}$: $$ d_{Cu} = d_{Al} \times \sqrt{\frac{Y_{Al}}{Y_{Cu}}} $$
Substituting the given values: $$ d_{Cu} = 3 \times \sqrt{\frac{7 \times 10^{10}}{12 \times 10^{10}}} $$
Calculating the value: $$ d_{Cu} = 3 \times \sqrt{\frac{7}{12}} \approx 2.29 \text{ mm} $$
Therefore, the appropriate diameter for the copper wire to achieve the same elongation as the aluminum wire under the same weight is approximately 2.3 mm.
So, the correct option is: $\mathbf{2.3 : mm}$
A wire is stretched 1 mm by a force of 1 kN. How far would a wire of the same material and length but of four times that diameter be stretched by the same force? (A) $\frac{1}{2}$ mm B $\frac{1}{2}$ mm (C) $\frac{1}{8}$ mm D $\frac{1}{16}$ mm
The correct option is D: $\frac{1}{16} \mathrm{~mm}$
To find how far a wire of the same material and length, but with four times the diameter, would be stretched by the same force, we start by using Young's modulus formula:
$$ Y = \frac{F L}{A \Delta L} $$
where:
$Y$ is Young's modulus,
$F$ is the force applied,
$L$ is the length of the wire,
$A$ is the cross-sectional area,
$\Delta L$ is the change in length.
Since the material and the applied force remain the same, Young's modulus ($Y$), the initial length ($L$), and the force ($F$) are constants. This implies:
$$ \Delta L \propto \frac{1}{A} $$
For a wire with a circular cross-section, the area ($A$) is proportional to the square of the diameter ($D$):
$$ A \propto D^2 $$
Thus, the elongation ($\Delta L$) will be inversely proportional to the square of the diameter:
$$ \Delta L \propto \frac{1}{D^2} $$
Given that the new diameter is four times the original diameter:
$$ D_2 = 4 \times D_1 $$
The ratio of the elongations of the two wires is:
$$ \frac{\Delta L_2}{\Delta L_1} = \frac{D_1^2}{D_2^2} = \frac{D_1^2}{(4D_1)^2} = \frac{D_1^2}{16D_1^2} = \frac{1}{16} $$
Therefore, the elongation of the new wire $\Delta L_2$ is:
$$ \Delta L_2 = \frac{1}{16} \Delta L_1 $$
Since the original wire is stretched $1 , \text{mm}$ by the force, the elongation of the wire with four times the diameter will be:
$$ \Delta L_2 = \frac{1}{16} , \text{mm} $$
Thus, the correct answer is $\frac{1}{16} , \text{mm}$.
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A steel wire of length $4.7 \mathrm{~m}$ and cross-sectional area $3.0 \times 10^{-5} \mathrm{~m}^{2}$ stretches by the same amount as a copper wire of length $3.5 \mathrm{~m}$ and cross-sectional area of $4.0 \times 10^{-5} \mathrm{~m}^{2}$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?
The ratio of the Young's modulus of steel to that of copper is approximately:
[ \frac{Y_{\text{steel}}}{Y_{\text{copper}}} \approx 0.559 ]
Hence, the Young's modulus of steel is approximately 0.559 times that of copper.
Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?
Fig. 8.9
(a) Young's Modulus
The calculation for Young's modulus yields:
[ Y = \frac{200 \times 10^6 , \text{N/m}^2}{0.002} = 100 \times 10^{9} , \text{N/m}^2 = 100 , \text{GPa} ]
So, Young's modulus for the material is 100 GPa.
(b) Approximate Yield Strength
From the stress-strain curve, the approximate yield strength at the point where the curve starts deviating from the linear region is:
[ \sigma_{y} \approx 250 \times 10^6 , \text{N/m}^2 ]
So, the approximate yield strength for the material is 250 MPa.
Summary
Young's modulus (Y): 100 GPa
Approximate yield strength (σ_y): 250 MPa
These values are essential for understanding the elastic properties and mechanical limits of the given material.
The stress-strain graphs for materials $A$ and $B$ are shown in Fig. 8.10.
Fig. 8.10
The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young's modulus?
(b) Which of the two is the stronger material?
To determine the properties of materials (A) and (B) from their stress-strain graphs, let's analyze them based on Young's modulus and tensile strength.
(a) Greater Young's Modulus
Young's Modulus (E) is defined as the ratio of stress (σ) to strain (ε) within the proportional limit (linear region) of the stress-strain curve. [ E = \frac{\sigma}{\epsilon} ]
Greater Young's Modulus: The slope of the linear region of the stress-strain graph gives the Young's modulus. A steeper slope indicates a greater Young's modulus.
From the graphs, material B has a steeper initial linear slope compared to material A.
Therefore, material B has a greater Young's modulus.
(b) Stronger Material
Strength of a material, often referred to as the ultimate tensile strength (σ_u), is the maximum stress the material can withstand before failure (fracture).
Stronger Material: Look for the higher value of stress before the material breaks.
From the graphs, the stress value at the highest point of the curve (just before the failure) for material A is higher compared to that of material B.
Therefore, material A is the stronger material.
Read the following two statements below carefully and state, with reasons, if it is true or false.
(a) The Young's modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.
Statement (a)
False.
The Young's modulus of rubber is much smaller than that of steel. Young's modulus is a measure of the stiffness of a material. Steel, being a much stiffer material, has a higher Young's modulus which means it requires a larger force to produce the same amount of strain compared to rubber. In contrast, rubber is more flexible and can be stretched more easily, indicating a lower Young's modulus.
Statement (b)
True.
The stretching of a coil is indeed determined by its shear modulus. A coil (such as a helical spring) typically experiences a deformation due to twisting or shearing forces. The shear modulus measures the material's response to such shear stress and indicates how easily it can be twisted or deformed sideways. Hence, it is the appropriate modulus to describe the stretching behavior of coils.
Two wires of diameter $0.25 \mathrm{~cm}$, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is $1.5 \mathrm{~m}$ and that of brass wire is $1.0 \mathrm{~m}$. Compute the elongations of the steel and the brass wires.
Fig. 8.11
Let's summarize the results:
Cross-sectional Area (A):$$ A = \pi \left( \frac{0.0025}{2} \right)^2 = 4.90874 \times 10^{-6} \ \text{m}^2 $$
Elongation of Steel Wire:
Using, $$ \Delta L_{\text{steel}} = \frac{F_{\text{steel}} \times L_{\text{steel}}}{A \times Y_{\text{steel}}} $$ Substituting, $$ \Delta L_{\text{steel}} = \frac{(98 \ \text{N}) \times (1.5 \ \text{m})}{(4.90874 \times 10^{-6} \ \text{m}^2) \times (2 \times 10^{11} \ \text{N} \ \text{m}^{-2})} = 0.000149733 \ \text{m} = 0.1497 \ \text{mm} $$
Elongation of Brass Wire:
Using, $$ \Delta L_{\text{brass}} = \frac{F_{\text{brass}} \times L_{\text{brass}}}{A \times Y_{\text{brass}}} $$ Substituting, $$ \Delta L_{\text{brass}} = \frac{(58.8 \ \text{N}) \times (1.0 \ \text{m})}{(4.90874 \times 10^{-6} \ \text{m}^2) \times (1 \times 10^{11} \ \text{N} \ \text{m}^{-2})} = 0.000119786 \ \text{m} = 0.1198 \ \text{mm} $$
Therefore, the elongations are:
Steel Wire: ($0.1497 \ \text{mm}$
Brass Wire: $0.1198 \ \text{mm}$
These elongations give us an idea of how much each material will stretch under the given load.
The edge of an aluminium cube is $10 \mathrm{~cm}$ long. One face of the cube is firmly fixed to a vertical wall. A mass of $100 \mathrm{~kg}$ is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Given Data:
Mass (m): 100 kg
Gravity (g): $10 m/s(^2)$ (approximation)
Shear Modulus (G): 25 GPa
Side length of the aluminium cube (L): 10 cm = 0.1 m
Conversions:
1 GPa = $ 1 * 10^9 ) N/m(^2$
Calculations:
1. Calculate the Force (F):
[ F = m \cdot g ] [ F = 100 , \text{kg} \cdot 10 , \text{m/s}^2 ] [ F = 1000 , \text{N} ]
2. Shear Modulus (G) in N/m(^2):
[ G = 25 , \text{GPa} = 25 \times 10^9 , \text{N/m}^2 ]
3. Area of one face (A):
[ A = L^2 = (0.1 , \text{m})^2 ] [ A = 0.01 , \text{m}^2 ]
4. Shearing Stress (( \sigma_s )):
[ \sigma_s = \frac{F}{A} ] [ \sigma_s = \frac{1000 , \text{N}}{0.01 , \text{m}^2} ] [ \sigma_s = 100000 , \text{N/m}^2 ]
5. Shearing Strain (( \epsilon_s )):
[ \epsilon_s = \frac{\sigma_s}{G} ] [ \epsilon_s = \frac{100000 , \text{N/m}^2}{25 \times 10^9 , \text{N/m}^2} ] [ \epsilon_s = 4 \times 10^{-6} ]
6. Vertical Deflection (( \Delta x )):
[ \Delta x = \epsilon_s \cdot L ] [ \Delta x = 4 \times 10^{-6} \cdot 0.1 , \text{m} ] [ \Delta x = 4 \times 10^{-7} , \text{m} ] [ \Delta x = 0.4 , \text{μm} ]
Therefore, the vertical deflection of the face is $ 0.4 , \text{μm} $.
Four identical hollow cylindrical columns of mild steel support a big structure of mass $50,000 \mathrm{~kg}$. The inner and outer radii of each column are 30 and $60 \mathrm{~cm}$ respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
The compressional strain of each column is approximately $\mathbf{7.25 \times 10^{-7}} $.
This means that each column undergoes a very small change in length relative to its original length under the applied load.
A piece of copper having a rectangular cross-section of $15.2 \mathrm{~mm} \times 19.1 \mathrm{~mm}$ is pulled in tension with $44,500 \mathrm{~N}$ force, producing only elastic deformation. Calculate the resulting strain?
Area of cross-section:[ A = 15.2 , \text{mm} \times 19.1 , \text{mm} = 2.9032 \times 10^{-4} , \text{m}^2 ]
Stress $( \sigma )$:[ \sigma = \frac{F}{A} = 1.533 \times 10^8 , \text{N/m}^2 ]
Young's modulus of copper (( Y )):[ Y = 1.3 \times 10^{11} , \text{Pa} ]
Finally, we can calculate the strain $( \epsilon )$: [ \epsilon = \frac{\sigma}{Y} ]
Substituting in the values: [ \epsilon = \frac{1.533 \times 10^8 , \text{N/m}^2}{1.3 \times 10^{11} , \text{Pa}} ]
Therefore, the resulting strain is: [ \epsilon = 1.179 \times 10^{-3} , \text{(unitless)} ]
Result:The resulting strain is $1.179 \times 10^{-3}$.
A steel cable with a radius of $1.5 \mathrm{~cm}$ supports a chairlift at a ski area. If the maximum stress is not to exceed $10^{8} \mathrm{~N} \mathrm{~m}^{-2}$, what is the maximum load the cable can support?
The maximum load that the steel cable can support is given by:
Cross-sectional area $( A )$: [ A = \pi (0.015, \text{m})^2 \approx 7.1 \times 10^{-4}, \text{m}^2 ]
Maximum load $( F )$ using the stress formula: $$F = (10^{8} , \text{N} , \text{m}^{-2}) \times (7.1 \times 10^{-4}, \text{m}^2) \approx 71000, \text{N} $$
Therefore, the maximum load that the steel cable can support is 71,000 N or 71 kN.
A rigid bar of mass $15 \mathrm{~kg}$ is supported symmetrically by three wires each $2.0 \mathrm{~m}$ long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
The ratio of the diameters of the copper wires to the iron wire, given that each wire is to have the same tension, is:
$$ \frac{d_{\text{Cu}}}{d_{\text{Fe}}} = \sqrt{\frac{Y_{\text{Fe}}}{Y_{\text{Cu}}}} \approx 1.31 $$
So, the diameters of the copper wires should be approximately 1.31 times the diameter of the iron wire.
A $14.5 \mathrm{~kg}$ mass, fastened to the end of a steel wire of unstretched length $1.0 \mathrm{~m}$, is whirled in a vertical circle with an angular velocity of $2 \mathrm{rev} / \mathrm{s}$ at the bottom of the circle. The cross-sectional area of the wire is $0.065 \mathrm{~cm}^{2}$. Calculate the elongation of the wire when the mass is at the lowest point of its path.
Using: [ \Delta L = \frac{2432 , \text{N} \times 1.0 , \text{m}}{0.065 \times 10^{-4} , \text{m}^2 \times 2.0 \times 10^{11} , \text{N/m}^2} ]
The elongation ( \Delta L ) of the wire is:
[ \Delta L = 0.00187 , \text{m} ] or [ \Delta L = 1.87 , \text{mm} ]
Conclusion:
The elongation of the wire when the mass is at the lowest point of its path is 1.87 mm.
Compute the bulk modulus of water from the following data: Initial volume $=100.0$ litre, Pressure increase $=100.0 \mathrm{~atm}\left(1 \mathrm{~atm}=1.013 \times 10^{5} \mathrm{~Pa}\right)$, Final volume $=100.5$ litre Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
The bulk modulus of water is approximately:
$$ B = 2.026 \times 10^9 , \text{Pa} $$
Comparing with the Bulk Modulus of Air
The bulk modulus for air (at constant temperature) is about:
$$ B_{\text{air}} \approx 1.0 \times 10^5 , \text{Pa} $$
To compare the two:
$$ \text{Ratio} = \frac{B_{\text{water}}}{B_{\text{air}}} = \frac{2.026 \times 10^9}{1.0 \times 10^5} = 2.026 \times 10^4 $$
Explanation of the Large Ratio
The ratio is so large because water is much less compressible than air.
Water molecules are very closely packed due to strong intermolecular forces, making it difficult to compress. This results in a high bulk modulus.
Air, on the other hand, consists of molecules that are much farther apart with weaker intermolecular forces, making it much more compressible and giving it a significantly lower bulk modulus.
Thus, solids and liquids generally have much higher bulk moduli compared to gases due to the differences in intermolecular forces and spacing between molecules.
What is the density of water at a depth where pressure is $80.0 \mathrm{~atm}$, given that its density at the surface is $1.03 \times 103 \mathrm{~kg} \mathrm{~m}^{-3}$ ?
To find the density of water at a depth where the pressure is $80.0 , \text{atm}$, we can use the bulk modulus ($B$) of water and the fact that the initial density ($\rho_0$) at the surface is $1.03 \times 10^3 , \text{kg/m}^3$.
The bulk modulus ( B ) is given as $13 , \text{GPa} = 13 \times 10^9 , \text{Pa}$.
Convert the pressure from atm to pascals:
$ 80 , \text{atm} = 80 \times 101325 , \text{Pa} = 8106000 , \text{Pa} $
Use the formula for bulk modulus:
[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} ]
Rearrange for (\frac{\Delta V}{V}):
[ \frac{\Delta V}{V} = -\frac{\Delta P}{B} ]
Where ( \Delta P ) is the change in pressure, and ( B ) is the bulk modulus.
Substitute into the formula:
[ \frac{\Delta V}{V} = -\frac{8106000 , \text{Pa}}{13 \times 10^9 , \text{Pa}} \approx -6.23 \times 10^{-4} ]
The change in density ((\Delta \rho)) due to the change in volume$(\Delta V)$ can be found by:
[ \rho = \rho_0 (1 + \frac{\Delta \rho}{\rho_0}) ]
Since $\frac{\Delta V}{V} \approx -\frac{\Delta \rho}{\rho_0}$:
[ \frac{\Delta \rho}{\rho_0} \approx 6.23 \times 10^{-4} ]
Therefore the new density (\rho) at 80 atm is:
[ \rho = \rho_0 (1 + \frac{\Delta \rho}{\rho_0}) = (1.03 \times 10^3 , \text{kg/m}^3)(1 + 6.23 \times 10^{-4}) ]
[ \rho \approx 1.03 \times 10^3 , \text{kg/m}^3 \times 1.000623 ]
[ \rho \approx 1.032 \times 10^3 , \text{kg/m}^3 ]
So, the density of water at a depth where the pressure is$80.0 , \text{atm}$ is approximately $1.032 \times 10^3 , \text{kg/m}^3$.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of $10 \mathrm{~atm}$.
The fractional change in volume of the glass slab when subjected to a hydraulic pressure of $10 \text{ atm}$ is:
[ \frac{\Delta V}{V} = 0.000027378 ]
or approximately:
[ \frac{\Delta V}{V} \approx 2.74 \times 10^{-5} ]
This means the volume of the glass slab would compress by a small fraction when subjected to this pressure.
Determine the volume contraction of a solid copper cube, $10 \mathrm{~cm}$ on an edge, when subjected to a hydraulic pressure of $7.0 \times 10^{6} \mathrm{~Pa}$.
The volume contraction of the solid copper cube, when subjected to a hydraulic pressure of $7.0 \times 10^{6} \mathrm{~Pa}$, is:
[ \Delta V = 5 \times 10^{-8} \mathrm{~m}^3 ]
Therefore, the volume contraction of the solid copper cube is $5 \times 10^{-8} \mathrm{~m}^3$.
How much should the pressure on a litre of water be changed to compress it by $0.10 \%$ ? carry one quarter of the load.
The bulk modulus of water is already provided in your chapter:
$$ B = 2.2 \times 10^9 , \text{N m}^{-2} $$
To find the pressure change required to compress water by $0.10%$:
The volume strain ( \Delta V / V ) is given as $0.10%$.
The relation between pressure change ($\Delta P$) and volume strain for bulk modulus (B) is:
$$ B = -\Delta P / (\Delta V / V) $$
Given:
$$ \Delta V / V = 0.10% = 0.001 $$
Let's compute the pressure change ( \Delta P ) needed:
[ \Delta P = - B \cdot \left( \Delta V / V \right) ]
Plugging in the values:
[ \Delta P = -(2.2 \times 10^9 , \text{N m}^{-2}) \cdot 0.001 ]
[ \Delta P = -2.2 \times 10^6 , \text{N m}^{-2} ]
So, the pressure change required to compress a litre of water by $0.10%$ is
[ \Delta P = 2.2 \times 10^6 , \text{N m}^{-2} \text{ or } 2.2 , \text{MPa}. ]
Hence, the pressure should be increased by $2.2$ MPa to compress a litre of water by $0.10%$.
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Notes - Mechanical Properties of Solids | Class 11 NCERT | Physics
Understanding the Mechanical Properties of Solids: Class 11 Notes
Studying the mechanical properties of solids is crucial for understanding how materials behave under various forces. These principles are fundamental in engineering, construction, and everyday applications. This article explores the key concepts of mechanical properties of solids, making it easier for Class 11 students to grasp the topic.
Introduction to Mechanical Properties of Solids
Definition and Importance
Mechanical properties of solids describe how materials respond to external forces. Understanding these properties is essential in selecting the right materials for construction, manufacturing, and numerous other applications.
Applications in Engineering and Real-Life Scenarios
Knowledge of mechanical properties is vital in design and engineering. For instance, designing a bridge, building, or even a simple wrench requires a thorough understanding of how materials will behave under stress and strain.
Stress and Strain
Definitions and Concepts
Stress is the force exerted per unit area within materials. It's measured in Pascals (Pa) and can be represented by the formula: [ \text{Stress} = \frac{F}{A} ]
Strain is the deformation experienced by the material per unit length due to applied stress. It is a dimensionless quantity given by: [ \text{Strain} = \frac{\Delta L}{L} ]
Types of Stress
-
Tensile Stress: When a material is stretched.
- Compressive Stress: When a material is compressed.
- Shear Stress: When layers within the material are pushed in opposite directions.
Types of Strain
- Longitudinal Strain: Change in length due to tensile or compressive stress.
- Shearing Strain: Deformation due to shear stress.
- Volume Strain: Change in volume due to hydraulic stress.
Hooke’s Law
Explanation of Hooke's Law
Hooke's Law states that, for small deformations, stress is directly proportional to strain: [ \text{Stress} \propto \text{Strain} ] [ \text{Stress} = k \times \text{Strain} ] Here, ( k ) is the proportionality constant, known as the modulus of elasticity.
Limitations of Hooke’s Law
Hooke’s Law applies only to the elastic region where the material returns to its original shape after the removal of stress.
Stress-Strain Curve
How to Plot a Stress-Strain Curve
A stress-strain curve plots stress on the vertical axis and strain on the horizontal axis. This graphical representation helps to understand the behaviour of materials under different loads.
Regions of the Stress-Strain Curve
- Proportional Limit: The linear region where Hooke's Law is valid.
- Elastic Limit: The maximum stress a material can withstand without permanent deformation.
- Yield Point: The point beyond which the material deforms permanently.
- Ultimate Strength: Maximum stress the material can withstand.
- Fracture Point: The point at which the material breaks.
Interpretation of the Curve
- Elastic Region: The material returns to its original shape when the stress is removed.
- Plastic Region: Permanent deformation occurs beyond the yield point.
- Fracture: The material breaks at the fracture point.
Elastic Moduli
Young’s Modulus
Definition and Importance: Young’s Modulus measures the ability of a material to withstand changes in length: [ Y = \frac{\text{Stress}}{\text{Strain}} ]
Calculation Formula:
[ Y = \frac{F \times L}{A \times \Delta L} ] Examples and Uses: Essential for materials that require high tensile strength like steel in construction.
Shear Modulus
Definition and Importance: Shear Modulus measures the deformation due to shear stress. [ G = \frac{\text{Shearing Stress}}{\text{Shearing Strain}} ]
Calculation Formula:
[ G = \frac{F \times L}{A \times \Delta x} ]
Examples and Uses: Important in materials used for making shafts and gears.
Bulk Modulus
Definition and Importance: Bulk Modulus measures the material’s response to uniform pressure. [ B = -\frac{p}{\Delta V / V} ]
Calculation Formula:
[ B = -\frac{\text{Hydraulic Stress}}{\text{Volume Strain}} ]
Examples and Uses: Crucial for materials used in hydraulic systems.
Poisson’s Ratio
Definition
Poisson’s Ratio is the ratio of lateral strain to longitudinal strain. [ \text{Poisson’s Ratio} = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}} ]
Calculation Formula:
[ \mu = -\frac{\Delta d / d}{\Delta L / L} ]
Applications and Examples
Used in understanding the deformation characteristics of materials like rubber and metals.
Applications of Mechanical Properties
Material Selection in Engineering
Selecting materials with appropriate mechanical properties ensures reliability and longevity in engineering projects.
Design of Structures
Knowledge of mechanical properties is crucial in the design of structures like bridges, buildings, and vehicles to ensure they can withstand various loads and stresses.
Everyday Examples
Examples include using flexible materials for packaging, selecting metals for construction, and choosing rubber for tyres.
Summary and Key Takeaways
Recap of Main Points
- Mechanical properties describe how a material responds to forces.
- Key properties include stress, strain, Young’s Modulus, Shear Modulus, and Bulk Modulus.
- Understanding these properties is crucial in selecting materials for engineering and construction.
Importance of Mechanical Properties Understanding
A deep understanding of mechanical properties allows for better material selection and design, ensuring safety, efficiency, and durability of structures and products.
Understanding the mechanical properties of solids not only enriches your knowledge but also equips you with the understanding necessary for advanced studies and practical applications in fields like engineering and material science.
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