# Some Mechanical Properties of Matter

## Concept Of Physics

### H C Verma

1   A load of $10$ $kg$ is suspended by a metal wire $3$ $m$ long and having a cross-sectional area $4$ $mm^2$. Find (a) the stress (b) the strain and (c) the elongation. Young's modulus of the metal is $2\cdot0$ x $10^{11}$ $N/m^2$.

##### Solution :

Stress = $\frac{F}{A}$

Y = $\frac{FL}{A\Delta{L}}$ $\Rightarrow$ $\frac{\Delta{L}}{L}$ = $\frac{F}{YA}$

F = mg

Strain $\frac{\Delta{L}}{L}$

2   A vertical metal cylinder of radius $2$ $cm$ and length $2$ $m$ is fixed at the lower end and a load of $100$ $kg$ is put on it. Find (a) the stress (b) the strain and (c) the compression of the cylinder. Young's modulus of the metal = $2$ x $10^{11}$ $N/m^2$.

##### Solution :

$e$ = strain = $\frac{\rho}{Y}$

$\rho$ = stress = $\frac{mg}{A}$

Compression $\Delta{L}$ = $eL$

3   The elastic limit of steel is $8$ x $10^8$ $N/m^2$ and its Young's modulus $2$ x $10^{11}$ $N/m^2$. Find the maximum elongation of a half-meter steel wire that can be given without exceeding the elastic limit.

##### Solution :

Y = $\frac{F}{A}$ . $\frac{L}{\Delta{L}}$ $\Rightarrow$ $\Delta{L}$ = $\frac{FL}{AY}$

4   A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of (a) the stresses developed in the two wires and (b) the strains developed. $Y$ of steel = $2$ x $10^{11}$ $N/m^2$. $Y$ of copper = $1\cdot3$ X $10^{11}$ $N/m^2$.

##### Solution :

a) $\frac{Stress\ of\ cu}{Stress\ of\ st}$ = $\frac{F_{cu}}{A_{}cu}\frac{A_{g}}{F_{g}}$ = $\frac{F_{cu}}{F_{st}}$ = 1

$L_{steel}$ = $L_{cu}$ and $A_{steel}$ = $A_{cu}$

b) Strain = $\frac{\Delta{Lst}}{\Delta{lcu}}$ = $\frac{F_{st}L_{st}}{A_{st}Y_{st}}$ . $\frac{A_{cu}Y_{cu}}{F_{cu}I_{cu}}$ $\\$ ($\because$ $L_{cu}$ = $I_{st}$ ; $A_{cu}$ = $A_{st}$)

5   In figure ($14-E1$) the upper wire is made of steel and the lower of copper. The wires have equal cross-section. Find the ratio of the longitudinal strains developed in the two wires.

##### Solution :

$\big(\frac{\Delta{L}}{L}\big)_{cu}$ = $\frac{F}{AY_{cu}}$

($\because$ $A_{cu}$ = $A_{st}$) = $\frac{Y_{cu}}{Y_{st}}$

$\big(\frac{\Delta{L}}{L}\big)_{st}$ = $\frac{F}{AY_{st}}$

$\frac{Strain\ steel\ wire}{Strain\ om\ copper\ wire}$ = $\frac{F}{AY_{st}}$ x $\frac{AY_{cu}}{F}$

6   The two wires shown in figure ($14-E2$) are made of the same material which has a breaking stress of $8$ x $10^8$ $N/m^2$. The area of cross-section of the upper wire is $0\cdot006$ $cm^2$ and that of the lower wire is $0\cdot003$ $cm^2$. The mass $m_{1}$ = $10$ $kg$, $m_{2}$ = $20$ $kg$ and the hanger is light. (a) Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased ? (b) Repeat the above part if $m_{1}$ = $10$ $kg$ and $m_{2}$ = $36$ $kg$.

##### Solution :

$\frac{T_{2}}{A_{u}}$ $\Rightarrow$ $\frac{m_{2}g\ +\ m_{1} g\ +\ \omega{g}}{A_{u}}$ = $8$ x $10^8$ $\Rightarrow$ $\omega_{0}$ = 2 kg

Stress in upo rod = $\frac{T_{2}}{A_{u}}$ $\Rightarrow$ $\frac{m_{2}g\ +\ m_{1}g\ +\ wg}{A_{u}}$ $\Rightarrow$ $w$ = 0.18 kg

$\frac{T_{1}}{A_{1}}$ = $\frac{m_{1}g\ +\ \omega{g}}{A_{1}}$ = $8$ x $10^8$ $\Rightarrow$ $w$ = 14 kg

Stress in the lower rod = $\frac{T_{1}}{A_{1}}$ $\Rightarrow$ $\frac{m_{1}g\ +\ \omega{g}}{A_{1}}$ $\Rightarrow$ $w$ = 14 kg

The maximum load that can be put is 2kg. Upper wire will break first if load is increased

For same stress, the max load that can be put bis 14 kg. If the load is increased the lower wire will break first.

7   Two persons pull a rope towards themselves. Each person exerts a force of $100$ $N$ on the rope. Find the Young's modulus of the material of the rope if it extends in length by $1$ $cm$. Original length of the rope = $2$ $m$ and the area of cross-section = $2$ $cm^2$.

##### Solution :

Y = $\frac{F}{A}$ . $\frac{L}{\Delta{L}}$

8   A steel rod of cross-sectional area $4$ $cm^2$ and length $2$ $m$ shrinks by $0\cdot1$ $cm$ 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's modulus of steel = $1\cdot9$ x $10^{11}$ $N/m^2$.

##### Solution :

$Y$ = $\frac{F}{A}$ $\frac{L}{\Delta{L}}$ $\Rightarrow$ $F$ = $\frac{YA\Delta{L}}{L}$

9   Consider the situation shown in figure ($14-E3$). The force $F$ is equal to the $m_{2}$ $g/2$. If the area of cross-section of the string is $A$ and its Young's modulus $Y$, find the strain developed in it. The string is light and there is no friction anywhere.

##### Solution :

Again , $T$ = $F$ + $m_{1}a$

and $T$ - $F$ = $m_{1}a$ $...(2)$

$\Rightarrow$ $\frac{\Delta{L}}{L}$ = $\frac{(m_{2}^{2}\ +\ 2m_{1}m_{2})g}{2(m_{1}\ +\ m_{2})AY}$ = $\frac{m_{2}g(m_{2}\ +\ 2m_{1})}{2AY(m_{1}\ +\ m_{2})}$

From equation (1) and (2) , we get $\frac{m_{2}g}{2(m_{1}\ +\ m_{2})}$

$m_{2}g$ - $T$ = $m_{2}a$ $....(1)$

Now, $Y$ = $\frac{FL}{A\Delta{L}}$ $\Rightarrow$ $\frac{\Delta{L}}{L}$ = $\frac{F}{AY}$

$\Rightarrow$ $a$ = $\frac{m_{2}g\ -\ F }{m_{1}\ +\ m_{2}}$

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 $\theta$ with the vertical and is released. Find the maximum value of $\theta$ so that the sphere does not rub the floor. Young's modulus of the metal of the wire is $2\cdot0$ x $10^{11}$ $N/m^2$. Make appropriate approximations.

##### Solution :

$\Rightarrow$ Again, by work energy principle,

So, $\Delta{T}$ = $\frac{m\ [\ 2gr(1 - cos\theta)\ ]}{r}$ = $2mg(1 - cos\theta)$

When it moves to angle $\theta$, and released, the tension the $T$ at lowest point is

$\Rightarrow$ $v^2$ = 2$gr(1 - cos\theta)$ $......(2)$

$\Rightarrow$ $F$ =$\frac{YA\Delta{L}}{L}$ = 2$mg$ - 2$mg$ $cos\theta$ $\Rightarrow$ 2$mg$ $cos\theta$ = 2$mg$ - $\frac{YA\Delta{L}}{L}$ = $cos\theta$ = 1 - $\frac{YA\Delta{L}}{L(2mg)}$

The cahneg in tension is due to centrifugal force $\Delta{T}$ = $\frac{mv^2}{r}$ $.......(1)$

At equilibrium $\Rightarrow$ $T$ = $mg$

$\Rightarrow$ $\frac{1}{2}$ $mv^2$ - 0 = $mgr(1- cos\theta)$

$\Rightarrow$ $F$ = $\Delta{T}$

$\Rightarrow$ $T'$ = $mg$ + $\frac{mv^2}{r}$

11   A steel wire of original length $1$ $m$ and cross-sectional area $4\cdot00$ $mm^2$ is clamped at the two ends so that it lies horizontally and without tension. If a load of $2\cdot16$ $kg$ is suspended from the middle point of the wire, what would be its vertical depression ? $\\$ $Y$ of the steel = $2.0$ x $10^{11}$ $N/m^2$. Take $g$ = $10$ $m/s^2 ##### Solution : From equation (1), (2) and (3) and the frebody diagram,$\\2Icos\theta$=$mg$. Increase in length$\Delta{L}$=$(AC\ +\ CB)$-$AB$So,$\Delta{L}$=$2(I^2\ +\ x^2)^\frac{1}{2}$-$100.....(2)Y$=$\frac{F}{A}\frac{I}{\Delta{I}}....(3)$Here,$AC$=$(I^2\ +\ x^2)^\frac{1}{2}$12 A copper wire of cross-sectional area$0\cdot01cm^2$is under a tension of$20N$. Find the decrease in the cross-sectional area. Young's modulus of copper =$1\cdot1$x$10^{11} 4N/m^2$and Poisson's ratio =$0\cdot32$. [Hint :$\frac{\Delta{A}}{A}$=$2\frac{\Delta{r}}{r}$] ##### Solution :$\sigma$=$\frac{\Delta{D}/D}{\Delta{L}/L}\Rightarrow\frac{\Delta{D}}{D}$=$\frac{\Delta{L}}{L}\Rightarrow\Delta{A}$=$\frac{2\Delta{r}}{r}y$=$\frac{FL}{A\Delta{L}}\Rightarrow\frac{\Delta{L}}{L}$=$\frac{F}{Ay}$Again,$\frac{\Delta{A}}{A}$=$\frac{2\Delta{r}}{r}$13 Find the increase in pressure required to decrease the volume of a water sample by$0\cdot01$%. Bulk modulus of water =$2\cdot1$x$10^9N/m^2$. ##### Solution :$B$=$\frac{Pv}{\Delta{v}}\RightarrowP$=$B\big(\frac{\Delta{v}}{v}\big)$14 Estimate the change in the density of water in ocean at a depth of$400m$below the surface. The density of water at the surface =$1030kg/m^3$and the bulk modulus of water =$2$x$10^9N/m^2$. ##### Solution :$\Rightarrow\frac{vD}{v_{0}}$=$\big($1$ - $\frac{\rho_{0}gh}{B}\big)$

So, $\frac{\rho_{d}}{\rho_{0}}$ = $\frac{V_{0}}{V_{d}}$ $.......(1)$

$B$ = $\frac{\rho_{0}gh}{(V_{0}\ -\ V_{d})}$ $\Rightarrow$ $1$ - $\frac{V_{d}}{v_{0}}$ = $\frac{\rho_{0}gh}{B}$

$\rho_{0}$ = $\frac{m}{v_{0}}$ = $\frac{m}{v_{d}}$

Vol.strain = $\frac{V_{0}\ -\ V_{d}}{V_{0}}$

15   A steel plate of face-area $4$ $cm^2$ 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 = $8\cdot4$ X $10^{10}$ $N/m^2$.

##### Solution :

$\eta$ = $\frac{F}{A\theta}$ $\\$ Lateral displacement = $I\theta$

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 = $0\cdot076$ $N/m$.

##### Solution :

$F$ = $T\ I$

17   Find the excess pressure inside (a) a drop of mercury of radius $2$ $mm$ (b) a soap bubble of radius $4$ $mm$ and (c) an air bubble of radius $4$ $mm$ formed inside a tank of water. Surface tension of mercury, soap solution and water are $0\cdot465$ $N/m$, $0\cdot03$ $N/m$ and $0\cdot076$ $N/m$ respectively.

##### Solution :

b) $P$ = $\frac{42T_{g}}{r}$

a) $P$ = $\frac{2T_{Hg}}{r}$

c) $P$ = $\frac{2T_{g}}{r}$

18   Consider a small surface area of $1$ $mm^2$ at the top of a mercury drop of radius $4.0$ $mm$. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = $1\cdot0$ x $10^5$ $Pa$ and surface tension of mercury = $0\cdot465$ $N/m$. Neglect the effect of gravity. Assume all numbers to be exact.

##### Solution :

$F = PA = \frac{2T}{r}A$

b) pressure = $P_{0}$ + $(\frac{2T}{r})$

$P = \frac{2T}{r}$

a) $F$ = $P_{0}A$

$F$ = $P'A$ = $(P_{0} + \frac{2T}{r})A$

19   The capillaries shown in figure ($14-E4$) have inner radii $0\cdot5$ $mm$, $1\cdot0$ $mm$ 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 $7\cdot5$ x $10^{-2}$ $N/m$.

##### Solution :

b) $h_{B} = \frac{2Tcos\theta}{r_{B} \rho g}$

a) $h_{A} = \frac{2Tcos\theta}{r_{A}\ -\ \rho g}$

c) $h_{c} = \frac{2Tcos\theta}{r_{c} \rho g}$

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, upto what height will the water rise in the capillary ?

##### Solution :

$h_{\omega}$ = $\frac{2t_{\omega}Cos\theta_{\omega}}{r\rho_{\omega}g}$ where, the symbols have their usual meanbings.

$h-{Hg}$ = $\frac{2T_{Hg}Cos\theta_{Hg}}{r\rho_{Hg}g}$

$\frac{h_{\omega}}{h_{Hg}}$ = $\frac{T _{\omega}}{T_{Hg}}$ x $\frac{\rho_{Hg}}{\rho_{\omega}}$ x $\frac{Cos\theta_{\omega}}{Cos\theta_{Hg}}$

21   A barometer is constructed with its tube having radius $1\cdot0\ mm$. 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 = $0\cdot465\ N/m$. Density of mercury = $13600\ kg/m^3$.

##### Solution :

$h = \frac{2Tcos\theta}{r_{\rho}g}$

22   A capillary tube of radius $0\cdot50\ mm$ is dipped vertically in a pot of water. Find the difference between the pressure of the water in the tube $5\cdot0\ cm$ below the surface and the atmospheric pressure. Surface tension of water = $0\cdot075\ N/m$.

##### Solution :

$P = \frac{F}{r}$

$P = \frac{2T}{r}$

23   Find the surface energy of water kept in a cylindrical vessel of radius $6\cdot0\ cm$. Surface tension of water = $0\cdot075\ J/m^2$.

##### Solution :

$A = \pi r^2$

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 = $0\cdot465\ J/m^2$.

##### Solution :

$\Rightarrow r = \frac{R}{2} = 2$

$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi r^3 \times 8$

Increase in surface energy = TA' - TA

25   A capillary tube of radius $1\ mm$ is kept vertical with the lower end in water. (a) Find the height of water raised in the capillary. (b) If the length of the capillary tube is half the answer of part (a), find the angle $\theta$ made by the water surface in the capillary with the wall.

##### Solution :

$\Rightarrow cos\theta = \frac{h'r_{\rho}g}{2T}$

$h = \frac{2Tcos\theta}{r_{\rho}g} , h' = \frac{2Tcos\theta}{r_{\rho}g}$

So, $\theta = cos^{-1} \frac{1}{2} = 60^o$

26   The lower end of a capillary tube of radius $1\ mm$ is dipped vertically into mercury. (a) Find 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 = $0\cdot465\ N/m$ and the contact angle of mercury with glass = $135°$.

##### Solution :

b) $T\times 2\pi r cos\theta = \pi r^2 h\times \rho\times g$

a) $h = \frac{2Tcos\theta}{r_{\rho}g}$

$\therefore\ cos\theta = \frac{hr_{\rho}g}{2T}$

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 = $0\cdot075\ N/m$.

##### Solution :

$T(2I) = [ 1 \times (10^{-3}) \times h]\rho g$

28   Consider an ice cube of edge $1\cdot0\ cm$ kept in a gravity free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.

##### Solution :

Surface area = $4\pi r^2$

29   A wire forming a loop is dipped into soap solution and taken out so that a film of soap solution is formed. A loop of $6\cdot28\ cm$ long thread is gently put on the film and the film is pricked with a needle inside the loop. The thread loop takes the shape of a circle. Find the tension in the thread. Surface tension of soap solution = $0\cdot030\ N/m$.

##### Solution :

So, $F = 2Tr \int_{0}^{\pi/2} sin\theta d\theta = 2Tr [cos\theta]_{0}^{\pi/2} = T\times 2r$

$dF_{y} = T \times r\ d\ \theta$

$dF_{y} = 2T\ r d\theta , sin\theta\ [dF_{x} = 0]$

The length of small element = $r\ d\ \theta$

Tension $\Rightarrow 2T_{t} = T \times 2r \Rightarrow T_{1} = Tr$

Considering symmetric elements,

30   A metal sphere of radius $1\ mm$ and mass $50\ mg$ falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is $1\ cm/s$, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine = $1260\ kg/m^3$ and its coefficient of viscosity at room temperature = $8\cdot0$ poise

##### Solution :

b) Hydrostatic force = B = $\big(\frac{4}{3}\big) \pi r^3 \sigma g$

$v = \frac{2}{9} \frac{r^2(\rho\ -\ \sigma)g}{\eta} \Rightarrow \frac{2}{9}\ r^2 \frac{\Large(\frac{m}{(4/3)\pi r^3} -\ \sigma\Large) g}{n}$

a) Viscous force = $6 \pi \eta r v$

c) $6\pi \eta\ r v + \big(\frac{4}{3}\big) \pi r^3 \sigma g$ = mg

31   Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = $0.02\ cm$, viscosity of air = $1\cdot8$ x $10^{-4}$ poise, $g = 9.9\ m/s^2$ and density of water = $1000\ kg/m^3$.

##### Solution :

Thus,

v = $\frac{2r^2 \rho g}{9 \eta}$

i) The weight $(4/3)\pi r^3\ \rho g$ downward.

or

Because, $\sigma$ of air is very small, the force of buoyancy may be neglected.

To find terminal velocity of rain drops, the forces acting on the drop are,

$6\pi \eta\ r v + \big(\frac{4}{3}\big) \pi r^3 \sigma g$

ii) Force of buoyancy $(4/3)\pi r^3\ \sigma g$ upward.

32   Water flows at a speed of $6\ cm/s$ through a tube of radius $41\ cm$. Coefficient of viscosity of water at room temperature is $0\cdot01$ poise. Calculate the Reynolds number. Is it a steady flow ?

##### Solution :

$v = \frac{R\eta}{\rho D} \Rightarrow R = \frac{v \rho D}{\eta}$