Download App

Physics 2 Marks Questions

Mechanical Properties of Solids · MP Board (MPBSE) STD 11 Science (Madhya Pradesh - English Medium). 46 questions.

Questions

2 Marks Questions

🎯

Test yourself on this topic

46 questions · timed · auto-graded

Question 12 Marks
A steel cable with a radius of 1.5cm supports a chairlift at a ski area. If the maximum stress is not to exceed $108N m^{-2}$, what is the maximum load the cable can support?
Answer
Radius of the steel cable, r = 1.5cm = 0.015m Maximum allowable stress = $10^8N m^{-2}$ Maximum stress = Maximum force/Area of cross-section $\therefore$ Maximum force = Maximum stress × Area of cross-section$= 10^8 ×\pi(0.015)^2 
= 7.065 \times 10^4N$ Hence, the cable can support the maximum load of $7.065 \times 10^4N$.
View full question & answer→
Question 22 Marks
What is the Young’s modulus for a perfect rigid body?
Answer
According to Hooke’s law, $\text{(Y)}=\frac{\text{stress}}{\text{longitudinal strain}}=\frac{\text{F}}{\text{A}}\times\frac{\text{l}}{\Delta\text{l}}$ For a perfectly rigid body, change in length $\Delta\text{l}=0,$ therefore longitudinal strain in zero. $\therefore\ \text{Y}=\frac{\text{F}}{\text{A}}\times\frac{\text{l}}{0}=\infty$ Hence, Young's modulus for a perfectly rigid body is infinite $(\infty)$
View full question & answer→
Question 32 Marks
To what depth must a rubber ball be taken in deep sea so that its volume is decreased by $0.1 \%$ ? (The Bulk modulus of rubber is $9.8 \times 10^8 \mathrm{~N} / \mathrm{m}^2$; and the density of seawater is $10^3 \mathrm{~kg} / \mathrm{m}^3$ ).
Answer
Bulk moclulus of rubber(B) = $9.8 \times 10^8N/ m^3$ Density of seawater $(\rho)=10^3\text{kg}/\ \text{m}^3$ Percentage decrease in volume, $\Big(\frac{\Delta\text{V}}{\text{V}}\times100\Big)=0.1\ \text{or}\ \frac{\Delta\text{V}}{\text{V}}=\frac{0.1}{100}$ or $\frac{\Delta\text{V}}{\text{V}}=\frac{1}{1000}$ let the rubber ball be taken up to depth h, $\because$ Change in pressure(p) $=\text{h}\rho\text{g}$
$\therefore$ Bulk modulus(B) $=\frac{\text{P}}{\big(\frac{\Delta\text{V}}{\text{V}}\big )}=\frac{\text{h}\rho\text{g}}{\big(\frac{\Delta\text{V}}{\text{V}}\big)}$ or $\text{h}=\frac{\text{B}\times\big(\frac{\Delta\text{V}}{\text{V}}\big)}{\rho\text{g}}=\frac{9.8\times10^8\times\frac{1}{1000}}{10^3\times9.8}$
$=100\text{m}$
View full question & answer→
Question 42 Marks
A wire stretches by a certain amount under a load. If the load and radius both are increased to four times, find the stretch caused in the wire.
Answer
$\text{Y}=\frac{\text{Fl}}{\text{A}\Delta\text{l}}\Rightarrow\Delta\text{l}=\frac{\text{Fl}}{\pi\text{r}^2\text{Y}}$ $\Delta\text{l}'=\frac{4\text{Fl}}{\pi(4\text{r})^2\text{Y}}=\frac{\text{Fl}}{4\pi\text{r}^2\text{Y}}=\frac{\Delta\text{l}}{4}.$
View full question & answer→
Question 52 Marks
Define Poisson's ratio. Write an expression for it. What is the significance of negative sign in this expression?
Answer
Poisson's ratio: The lateral strain per unit longitudinal strain is called Poisson's ratio. $\sigma=\frac{\text{lateral strain}}{\text{longitudinal strain}}=-\frac{\Delta\frac{\text{r}}{\text{r}}}{\Delta\frac{\text{l}}{\text{l}}}$ Negative sign signifies that if the length increases, then radius of wire decreases.
View full question & answer→
Question 62 Marks
Why the bridges are declared unsafe after long use?
Answer
Due to the repeated stress and strain, the material used in the bridges loses elastic strength and ultimately may collapse. Hence, bridges are declared unsafe after long use.
View full question & answer→
Question 72 Marks
Railway trackes are laid on large sized wooden, iron or cement sleepers. Why?
Answer
By using wooden, iron or cement sleepers, area of contact for a given weight is increased to reduce the pressure of the train on the rails.
View full question & answer→
Question 82 Marks
A sphere contracts in volume by 0.01% when taken to the bottom of sea 1km deep. Find the bulk modulus of the material of the sphere.
Answer
Here, $\Delta\text{V}=0.01\%\ \text{of }\text{V}=\frac{0.01}{100}\times\text{V}=10^{-4}\text{V}$ $\text{P}=\text{h}\rho\text{g}.=10^3\times10^3\times9.8=9.8\times10^6\text{Nm}^{-2}$ Now, $\text{K}=\frac{\text{PV}}{\Delta\text{V}}=\frac{9.8\times10^6\times\text{V}}{10^{-6}\times\text{V}}=9.8\times10^{10}\text{Nm}^{-2}$
View full question & answer→
Question 92 Marks
Show graphically the change of potential energy and kinetic energy of a block attached to a spring which obeys Hooke's law.
Answer
Within elastic limit, $\frac{\text{Stress}}{\text{Strain}}=\text{constant}$
View full question & answer→
Question 102 Marks
Give the similarities and differences between intermolecular and interatomic forces.
Answer
Similarities:
  1. Electrical by nature.
  2. Active over short distances.
Differences:
  1. Force between molecules or atoms.
  2. Molecular forces are weaker than interatomic forces.
View full question & answer→
Question 112 Marks
Two wires made of same material are subjected to forces in the ratio 1 : 4. Their lengths are in the ratio 2 : 1 and diameters in the ratio 1 : 3. What is the ratio of their extensions?
Answer
According to Hooke's law, Modulus of elasticity, E $=\frac{\text{F}}{\pi\text{r}^2}\times\frac{\text{l}}{\Delta\text{l}}\ \text{or}\ \Delta\text{l}=\frac{\text{Fl}}{\pi\text{r}^2\text{E}}$ or, $\Delta\text{l}\propto\frac{\text{Fl}}{\text{r}^2}$ $[\because$ E is same for two wires$]$ $\therefore\frac{\Delta\text{l}_1}{\Delta\text{l}_2}=\frac{\text{F}_1}{\text{F}_2}\times\frac{\text{l}_1}{\text{l}_2}\times\frac{\text{r}^2_2}{\text{r}^2_1}$ $=\frac{1}{4}\times\frac{2}{1}\times\Big(\frac{3}{1}\Big)^2=\frac{9}{2}$ So. $\Delta\text{l}_1:\Delta\text{l}_2=9:2$
View full question & answer→
Question 122 Marks
A steel cable with a radius of $1.5cm$ supports a chairlift at a ski area. If the maximum stress is not to exceed $108N m^{-2}$, what is the maximum load the cable can support?
Answer
Radius of the steel cable, $\mathrm{r}=1.5 \mathrm{~cm}=0.015 \mathrm{~m}$ Maximum allowable stress $=10^8 \mathrm{~N} \mathrm{~m}^{-2}$ Maximum stress $=$ Maximum force/Area of cross-section
$\therefore$ Maximum force $=$ Maximum stress $\times$ Area of cross-section $=10^8 \times \pi(0.015)^2$ $=7.065 \times 10^4 \mathrm{~N}$ Hence, the cable can support the maximum load of $7.065 \times 10^4 \mathrm{~N}$.
View full question & answer→
Question 132 Marks
Explain the terms: Young's modulus of elasticity and elastic fatigue.
Answer
We have $\text{Y}=\frac{\text{F}.\text{l}}{\text{a}\Delta\text{l}}$ For same force $\Delta\text{l}_{\text{rubber}}$ is more than of $\Delta\text{l}_{\text{steel}}.$ $\therefore\text{Y}_\text{s}>\text{Y}_\text{r}$
View full question & answer→
Question 142 Marks
State Hook's law. Calculate the fractional AV compression, $\frac{\Delta\text{V}}{\text{V}},$ of water at the bottom of the ocean having depth 3000m. The bulk modulus of water is $2.2 \times 10^9Nm^{-2}$. (Take g $= 10ms^{-2})$
Answer
Hook's law: Within the elastic limits the ratio between stress and strain is constant and is called modulus of elasticity. $\frac{\text{Stress}}{\text{Strain}}=\text{Modulus of elasticity}$ Given depth h $=3000\text{m},\rho=10^3\text{kg/m}^3,\text{g}=10\text{ms}^{-2}$ $\text{p}=\text{h}\rho\text{g}=3000\times10^3\times10$ $=3\times10^7\text{Nm}^{-2}$ Bulk modulus $=\frac{\Delta\text{P}}{\Delta\frac{\text{V}}{\text{V}}},$ $\frac{\Delta\text{V}}{\text{V}}=\frac{\Delta\text{P}}{\text{B}}=\frac{3\times10^7}{2.2\times10^9}=1.36\times10^{-2}$
View full question & answer→
Question 152 Marks
Two wires A and B are of the same material. Their lengths are in the ratio $1 : 2$ and the diameters in the ratio $2 : 1$ If they are pulled by the same force, then what will be the ratio of their increase in lengths?
Answer
We know, $\Delta\text{L}=\frac{\text{FL}}{\text{AY}},\frac{\text{L}_\text{A}}{\text{L}_\text{B}}=\frac{1}{2}$ and $\frac{\text{r}_\text{A}}{\text{r}_\text{B}}=\frac{2}{1}$ (given)
$[\because$ The wire A and B are pulled by the same force and theyare made up of same material,
hence, $F_A = F_B = F, Y_A = Y_B = Y]$ $\frac{\Delta\text{L}_\text{A}}{\Delta\text{L}_\text{B}}=\frac{\text{L}_\text{A}}{\pi\text{r}^2_\text{B}}\times\frac{\pi\text{r}^2_\text{B}}{\text{L}_\text{B}}$ $\frac{\Delta\text{L}_\text{A}}{\Delta\text{L}_\text{B}}=\frac{\text{L}}{\text{L}_\text{B}}\times\Big(\frac{\text{r}_\text{B}}{\text{r}_\text{A}}\Big)^2$
$\Rightarrow\frac{\Delta\text{L}_\text{A}}{\Delta\text{L}_\text{B}}=\frac{1}{2}\times\Big(\frac{1}{2}\Big)^2=\frac{1}{8}$ $\frac{\Delta\text{L}_\text{A}}{\Delta\text{L}_\text{B}}=\frac{1}{8}$
View full question & answer→
Question 162 Marks
Interatomic and intermolecular forces are similar in certain aspects. Justify.
Answer
  1. Both the forces are electrical in origin.
  2. Both the forces are active over short distances. The general shape of the curve showing the variation of force with separation between two atoms/ molecules is similar.
  3. Both the forces are attractive upto a certain distance between atoms/ molecules and become repulsive when the distance between them becomes less than that value.
  4. For both the forces, the negative potential energy of the system accounts for attractive force between atoms molecules.
View full question & answer→
Question 172 Marks
The maximum stress that can be applied to the material of a wire used to suspend on elevator is $1.3 \times 10^8Nm^{-2}$. If the mass of the elevator is $900kg$ and it moves up with an acceleration of $2.2ms^{-2}$, then what is the minimum diameter of the wire?
Answer
As the elevator moves up, the tension in the wire, $F = mg + ma = m(g + a) = 900 × (9.8 + 2.2) = 10800N$, Stress in the wire $=\frac{\text{F}}{\text{A}}=\frac{\text{F}}{\pi\text{r}^2}$
Clearly, when the stress is maximurn r is minimum.
$\therefore$ Maximum stress $=\frac{\text{F}}{\pi\text{r}^2_\text{min}}$ or $\text{r}^2_\text{min}=\frac{\text{F}}{\pi\times\text{Maximum stress}}$
$=\frac{10800}{3.14\times1.3\times10^3}$
$=0.2645\times10^{-2}\text{m}$ Minimum diameter, $=2\text{r}_\text{min}=2\times0.5142\times10^{-2}$ $=1.0284\times10^{-2}\text{m}$
View full question & answer→
Question 182 Marks
A solid sphere of radius R made of a material of bulk modulus B is surrounded by a liquid in a cylindrical container. A mass less piston of area A floats on the surface of the liquid. When a mass M is placed on the piston to compress the liquid, find fractional change in the radius of the sphere?
Answer
When mass M is placed on the piston, the excess pressure, $\text{P}=\frac{\text{Mg}}{\text{A}}.$ As this pressure is equally applicable from all the direction on the sphere, hence there will be decrease in volume due to decrease in r, radius of sphere. Volume of the sphere, $\text{V}=\frac{4}{3}\pi\text{R}^3.$ Differentiating it, we get, $\Delta\text{V}=\frac{4}{3}\pi(3\text{R}^2)\Delta\text{R}=4\pi\text{R}^2\Delta\text{R}$ $\therefore\frac{\Delta\text{V}}{\text{V}}=\frac{4\pi\text{R}^2\Delta\text{R}}{\frac{4}{3}\pi\text{R}^3}=\frac{3\Delta\text{R}}{\text{R}}$ We know that, $\text{B}=\frac{\text{P}}{\frac{\text{dV}}{\text{V}}}=\frac{\text{Mg}}{\text{A}}.\frac{3\Delta\text{R}}{\text{R}}$ or $\frac{\Delta\text{R}}{\text{R}}=\frac{\text{Mg}}{3\text{BA}}$
View full question & answer→
Question 192 Marks
Calculate the percentage increase in length of a wire of diameter 2.5mm stretched by a force of 100kg weight. Young's modulus of elasticity of wire is $12.5 \times 10^{11}$ dyne/ sq cm.
Answer
Here, 2r = 25mm = 0.25cm or r = 0.125cm, $\therefore\text{a}=\pi\text{r}^2=\frac{22}{7}\times(0.125)^2\text{sq}.\text{cm}$ $\text{F}=100\text{kg}=100\times1000\text{g}$ $\text{F}=10\times1000\times980\text{dyne}$ $\text{Y}=125\times10^{11}\text{dyne}/\ \text{sq}.\text{cm}$ As $\text{Y}=\frac{\text{F}\times\text{l}}{\text{a}\times\Delta\text{l}}$ $\therefore\frac{\Delta\text{l}}{\text{l}}=\frac{\text{F}}{\text{aY}}$ Hence, % increase in length, $=\frac{\Delta\text{l}}{\text{l}}\times100=\frac{\text{F}}{\text{aY}}\times100$ $=\frac{(100\times1000\times980)\times7\times100}{22\times(0.125)^2\times125\times10^{11}}$ $=0.1812\%$
View full question & answer→
Question 202 Marks
Explain why should the beams used in construction of bridge have large depth and small breadth?
Answer
When a load (w) is suspended in the middle of a bar then bar get depressed $\delta=\frac{\text{wl}^3}{4\text{Ybd}^3}$ So, to decrease depression $(\delta)$ in the beam, depth (d) should be increased which is move effective in comparison to breadth (b).
View full question & answer→
Question 212 Marks
Determine the force required to double the length of a steel wire of area of cross-section $5 \times 10^{-5}m^2$. Young's modulus of steel = $2 \times 10^{11}Nm^{-2}$.
Answer
Here, Young's modulus, $Y = 2 \times 10^{11}Nm^{-2}$ Area of cross-section, $A = 5 \times 10^{-5}m^2$ Let the initial length of wire be L. Then, increase in length of wire, $\Delta\text{L}=\text{L}$
Now, $\text{Y}=\frac{\text{F}\times\text{L}}{\text{A}\times\Delta\text{L}}$
$\therefore\text{F}=\frac{\text{Y}\times\text{A}\times\Delta\text{L}}{\text{L}}$
$\Rightarrow\text{F}=\frac{2\times10^{11}\times5\times10^{-5}\times\text{L}}{\text{L}}$ or $\text{F}=10^7\text{N}.$
View full question & answer→
Question 222 Marks
The length of a metal is $\mathrm{l}_1$ when the tension in it is $\mathrm{T}_1$ and is $\mathrm{l}_2$ when the tension is $\mathrm{T}_2$. Find the original length of the wire.
Answer
Let I and $A$ be the original length and area of cross section of the metal wire. Change in length in the first case $=\left(I_1-\right.I$) Change in length in tne second case $=\left(\mathrm{I}_2-\mathrm{I}\right)$
View full question & answer→
Question 232 Marks
The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?
Answer
$\text{Y}=\frac{\text{stress}}{\text{strain}}$ As per question longitudinal strain for rubber and steel are equal. $\therefore\text{Y}\propto\text{stress}$ $\therefore\frac{\text{Y}_\text{steel}}{\text{Y}_\text{Rubber}}=\frac{\text{(stress})_\text{steel}}{\text{(stress})_\text{Rubber}}\text{ As the Y}_\text{steel}>\text{Y}_\text{Rubber}$ $\therefore\frac{\text{Y}_\text{steel}}{\text{Y}_\text{Rubber}}>1$ $\therefore\text{(stress})_\text{steel}$ is larger than $(Stress)_{Rubber}$
View full question & answer→
Question 242 Marks
A wire of length 2.5m has a percentage strain of 0.012% under a tensile force. Determine the extension in the wire.
Answer
Here, original length, L = 2.5m Strain $=\frac{\Delta\text{L}}{\text{L}}=0.012\%=\frac{0.012}{100}$ $\Delta\text{L}=\text{Strain}\times\text{L}$ or $\Delta\text{L}=\text{extension}=\frac{0.012}{100}\times\text{L}$ $=\frac{0.012\times25}{100}$ $=3\times10^{-4}\text{m}$ $=0.3\text{mm}$
View full question & answer→
Question 252 Marks
Why is steel more elastic than rubber?
Answer
Steel is more elastic than rubber because for a given load steel expands less than rubber.
View full question & answer→
Question 262 Marks
A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, then what will be the elongation of the wire in mm?
Answer
According to Hooke's law, Modulus of elasticity' E $=\frac{\text{W}}{\text{A}}\times\frac{\text{L}}{\text{l}}$ where, I = original length of the wire, A = cross-sectional area of the wire $\therefore$ Elongation, $\Delta=\frac{\text{WL}}{\text{E}}\ ...(\text{i})$ On either side ofthe w.ire, tension is W and length is $\frac{\text{l}}{2}.$ $\Delta\text{l}=\frac{\frac{\text{WL}}{2}}{\text{AE}}=\frac{\text{WL}}{2\text{AE}}=\frac{\text{l}}{2}$ [from Eq. (i)] $\therefore$ Total etongation in the wire $=\frac{\text{l}}{2}+\frac{\text{l}}{2}=\text{l}$
View full question & answer→
Question 272 Marks
Why does a cycle tube burst in summer?
Answer
Pressure in the tube increases with increasing temperature, but the volume expansion happens to a limited range. So, the cycle tube bursts in summer.
View full question & answer→
Question 282 Marks
How does the elasticity of material change on:
  1. Increasing the temperature.
  2. On heating and cooling gradually.
  3. On hammering.
Answer
  1. Decreases.
  2. Decreases.
  3. Increases.
View full question & answer→
Question 292 Marks
What is elastic hysteresis?
Answer
We know that some materials take appreciable time to recover their original condition completely. In other words, the strain persists even when the stress is removed. This lagging behind of strain is called elastic hysteresis.
View full question & answer→
Question 302 Marks
A steel ring of radius r and cross-section area A is fitted onto a wooden disc of radius R(R > r). If Young's modulus be E, find the force with which the steel ring is expanded.
Answer
Original length, $\text{l}=2\pi\text{r};$ Extension, $\Delta\text{l}=2\pi\text{R}-2\pi\text{r}=2\pi(\text{R}-\text{r})$ Strain $=\frac{\Delta\text{l}}{\text{l}}=\frac{2\pi(\text{R}-\text{r})}{2\pi\text{r}}=\frac{\text{R}-\text{r}}{\text{r}}$ Young's modulus, $\text{E}=\frac{\frac{\text{F}}{\text{A}}}{\frac{(\text{R}-\text{r})}{\text{r}}}\ \text{or}\ \text{F}=\text{EA}\Big(\frac{\text{R}-\text{r}}{\text{r}}\Big)$
View full question & answer→
Question 312 Marks
Is stress a vector quantity?
Answer
$\text{stress}=\frac{\text{Mognitude of restoring force by solid}}{\text{Area of cross - section}}$ as deforming and restoring force are equal and opposite so no net direction is involved. Hence, the stress is not a vector quantity just like pressure.
View full question & answer→
Question 322 Marks
A material has poission's ratio 0.5. If a uniform rod of it undergoes a longitudinal strain of $2 \times 10^{-3}$. What is the percentage increases in its volume.
Answer
The poission's ratio of the material is 0.5 which is maximum and for maximum value of the Poission's ratio the volume of the material remains unchanged. Hence, the increase in volume = zero%.
View full question & answer→
Question 332 Marks
Elasticity is said to be internal property of matter. Explain.
Answer
When a deforming force acts on a body, the atoms of the substance get displaced from their original positions. Due to this, the configuration of the body (substance) changes. The moment, the deforming force is removed, the atoms return to their original positions and hence the substance or body regains its original configuration. That is why, elasticity is said to be internal property of matter.
View full question & answer→
Question 342 Marks
A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by I. Another wire of the same material of length 2L and radius 2r, is pulled by a force 2f. Find the increase in length of this wire.
Answer
The situation is shown in the diagram. Now Young's modulus(Y) $=\frac{\text{f}}{\text{A}}\times\frac{\text{L}}{\text{l}}$ For frist wire, $\text{Y}=\frac{\text{f}}{\pi\text{r}^2}\times\frac{\text{L}}{\text{l}}\ ...(\text{i)}$ For second wire. $\text{Y}=\frac{2\text{f}}{\pi(2\text{r})^2}\times\frac{2\text{L}}{\text{l}'}$ $=\frac{\text{f}}{\pi\text{r}^2}\times\frac{\text{L}}{\text{l}'}$ For Eqs. (i) and (ii), $\frac{\text{f}}{\pi\text{r}^2}\times\frac{\text{L}}{\text{l}}=\frac{\text{f}}{\pi\text{r}^2}\times\frac{\text{L}}{\text{l}'}$ $\therefore\text{l}=\text{l}'$ $[\because$ both wires are of same material, hence, Young's modulus will be same$].$
View full question & answer→
Question 352 Marks
Name one system where the compressional and tensional modulus of elasticity are different. Give reason.
Answer
Bone of human system. It is due to the fact that bone is a compact mixture of fibres of collagen.
View full question & answer→
Question 362 Marks
A steel wire of length 4m and diameter 5mm is stretched by 5kg-wt. Find the increase in its length, if the Young's Modulus of steel wires is $2.4 \times 10^{12}$ dyn $-\mathrm{cm}^{-2}$.
Answer
Given; l = 4m = 400cm, 2r = 5mm or r = 2.5mm = 0.25cm, $\text{F}=5\text{kg-wt}=5000\text{g-wet},$ $=5000\times980\text{dyn};\Delta\text{l}=\ ?$ $\text{Y}=2.4\times10^{12}\text{dyn}/\ \text{cm}^2$ $\because\text{Y}=\frac{\text{F}}{\pi\text{r}^2}\times\frac{\text{l}}{\Delta\text{l}}$ $\Rightarrow\Delta\text{l}=\frac{\text{Fl}}{\pi\text{r}^2\text{Y}}$ $=\frac{(5000\times980)\times400}{\Big(\frac{22}{7}\Big)\times(0.25)^2 \times2.4\times10^{12}}$ $=0.0041\text{cm}$
View full question & answer→
Question 372 Marks
Two cylinders A and B of radii r and 2r are soldered co-axially. The free end of A is clamped and the free end of B is twisted by an angle $\phi.$ Find twist at the junction taking the material of two cylinders to be same and of equal length.
Answer
Let $\tau$ be the torque applied at the free end and $\phi'$ be the angle of twist at the junction. Then, $\tau=\frac{\pi\eta\text{r}^4(\phi'-0)}{2\text{l}}=\frac{\pi\eta(2\text{r})^4(\phi-\phi')}{2\text{l}}$ $\Rightarrow\phi'=16(\phi-\phi')\ \text{or}\ 17\phi'=16\phi\ \text{or }\phi'=\frac{16}{17}\phi.$
View full question & answer→
Question 382 Marks
What is the Bulk modulus for a perfect rigid body?
Answer
Bulk Modulus $=\frac{-\text{p(V)}}{\Delta\text{V}}$ as the perfect rigid body does not change it's shape even at infinite (deforming a stretching) force. Hence, $\Delta\text{V}=0$ $\Rightarrow\ \text{B}=\frac{\text{pV}}{\Delta\text{V}}=\frac{\text{pV}}{0}=\propto$ So the bulk modulus is infinity.
View full question & answer→
Question 392 Marks
What is an elastomer? What are their special features?
Answer
Elastomers: Elastomer are those substances which can be stretched to cause large strains. Substances like tissue of aorta, rubber etc., are elastomers. The stress-strain curve for an elastomer is as shown in figure below. Although elastic region is very large but the material does not obey Hooke's law over most of the region. Moreover, there is no well defined plastic region.
View full question & answer→
Question 402 Marks
The breaking stress of aluminium is $7.5 \times 10^8 \mathrm{dyne} / \mathrm{cm}^{-2}$. Find the greatest length of aluminium wire that can hang vertically without breaking. Density of aluminium is $2.7 \mathrm{~g} / \mathrm{cm}^{-3}$. Given: $\mathrm{g}=980 \mathrm{~cm} / \mathrm{s}^{-2}$.
Answer
Let l be the greatest length of the wire that can hang vertically without breaking. Mass of wire, m = cross-sectional area (a) × length (l) × density $(\rho)$ Weight of wire $=\text{mg}=\text{al}\rho\text{g},$ This is equal to the maximum force that the wire can withstand.
$\therefore\text{Breaking stress}=\frac{\text{la}\rho\text{g}}{\text{a}}=\text{l}\rho\text{g}$ or $7.5\times10^8=\text{l}\times2.7\times980$ or $\text{l}=\frac{7.5\times10^8}{2.7\times980}\text{cm}=2.834\times10^5\text{cm}=2.834\text{km}$
View full question & answer→
Question 412 Marks
A spherical ball contracts in volume by 0.01% when subjected to a normal uniform pressure of two atmospheres. What is the bulk modulus of its material in C.G.S. units?
Answer
$\frac{\Delta\text{V}}{\Delta}=\frac{0.01}{100};$ $P = 2 \times 1.01 \times 10^6$ dyne/ sq. cm $\text{K}=\frac{\text{PV}}{\Delta\text{V}}=\frac{2.02\times10^6}{\frac{0.01}{100}}$ = $202 \times 10^8​​​​​​​$ dyne/ sq. cm.
View full question & answer→
Question 422 Marks
What do you mean by the elastic limit?
Answer
When the deforming force is increased, a limit is reached beyond which the solid does not come back to its original shape or size but remains deformed on removal of applied force. This limit is called the elastic limit.
View full question & answer→
Question 432 Marks
When a weight 'W' is hung from one end of the wire, other end being fixed, the elongation produced on it be 'l’. If this wire goes over a pulley and two weights 'W' each are hung at the two ends, what will be the total elongation in the wire?
Answer
In second case, the tension is double but the original length is reduced to half. So the extension in the second case is half of the change in first case.
View full question & answer→
Question 442 Marks
Why are the springs made of steel and not of copper?
Answer
A spring will be better one if a large restoring force is set up in it on being deformed, which in turn depends upon the elasticity of the material of the spring. Since the Young's modulus of elasticity of steel is more than that of copper, hence steel is preferred in making the springs.
View full question & answer→
Question 452 Marks
What do you mean by compressibility? Why are solids least compressible and gases most compressible?
Answer
Compressibility of the material of a body is defined as the reciprocal of its bulk modulus. It is, thus, defined as the fractional change in volume per unit increase in pressure. Compressibility, $\text{K}=\frac{1}{\text{B}}=-\Big(\frac{\Delta\text{V}}{\text{V}}\Big)\times\frac{1}{\text{P}}$ The solids are least compressible whereas gases are most compressible. It is on account of the fact that in solids neighboring atoms are tightly coupled but molecules in gases are very poorly coupled to their neighbors.
View full question & answer→
Question 462 Marks
Plot Load vs Extension curve for a metal on the graph and depict:
  1. Yield point.
  2. Breaking point.
  3. Elastic limit.
  4. Crushing point.
View full question & answer→
Generate Paper FreeAll Mechanical Properties of Solids topics