**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 $10^{11}$ $N/m^2$.

F = mg

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

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

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

**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 $10^{11}$ $N/m^2$.

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

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

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.

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 $10^{11}$ $N/m^2$.

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

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

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.

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

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

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

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

**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 $m_{2}$ = $36$ $kg$.

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

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

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

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

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

**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 = $2$ $cm^2$.

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 $10^{11}$ $N/m^2$.

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

**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.

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

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

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

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

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

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

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

**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.

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

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

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

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

$\Rightarrow$ Again, by work energy principle,

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

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

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

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

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

**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

$Y$= $\frac{F}{A}$ $\frac{I}{\Delta{I}}$ $....(3)$

Increase in length $\Delta{L}$ = $(AC\ +\ CB)$ - $AB$

Here, $AC$ = $(I^2\ +\ x^2)^\frac{1}{2}$

So, $\Delta{L}$ = $2(I^2\ +\ x^2)^\frac{1}{2}$ - $100$ $.....(2)$

From equation (1), (2) and (3) and the frebody diagram, $\\$ $2I$ $cos\theta$ = $mg$.

**12.** A copper wire of cross-sectional area $0\cdot01$ $cm^2$ is under
a tension of $20$ $N$. 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 : $\frac{\Delta{A}}{A}$ = $2\frac{\Delta{r}}{r}$]

$y$ = $\frac{FL}{A\Delta{L}}$ $\Rightarrow$ $\frac{\Delta{L}}{L}$ = $\frac{F}{Ay}$

$\sigma$ = $\frac{\Delta{D}/D}{\Delta{L}/L}$ $\Rightarrow$ $\frac{\Delta{D}}{D}$ = $\frac{\Delta{L}}{L}$

Again, $\frac{\Delta{A}}{A}$ = $\frac{2\Delta{r}}{r}$

$\Rightarrow$ $\Delta{A}$ = $\frac{2\Delta{r}}{r}$

**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 $10^9$ $N/m^2$.

$B$ = $\frac{Pv}{\Delta{v}}$ $\Rightarrow$ $P$ = $B$ $\big(\frac{\Delta{v}}{v}\big)$

**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 = $2$ x $10^9$ $N/m^2$.

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

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

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

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

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

**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 $10^{10}$ $N/m^2$.

$\eta$ = $\frac{F}{A\theta}$ $\\$ 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 = $0\cdot076$ $N/m$.

$F$ = $T\ I$

**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.

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

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

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.

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

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

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

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

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

**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 $10^{-2}$ $N/m$.

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

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

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

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

$\frac{h_{\omega}}{h_{Hg}}$ = $\frac{T _{\omega}}{T_{Hg}}$ 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 = $13600\ kg/m^3$.

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

**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 = $0\cdot075\ N/m$.

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

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

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

$A = \pi r^2$

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

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

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

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.

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

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

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
= $135°$.

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

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

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

**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
= $0\cdot075\ N/m$.

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

**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.

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
= $0\cdot030\ N/m$.

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

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

Considering symmetric elements,

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

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

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

**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

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

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

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

$ 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}$

**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 = $1000\ kg/m^3$.

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

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

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

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

Thus,

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

or

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

**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 ?

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