Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 72

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Weekly Quiz Competition

Question 1
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A uniform rod of mass m, length L, area of cross-section A and Youngs modulus Y hangs from a rigid support. Its elongation under its own weight will be:

A
zero

B
mgL/2YA

C
mgL/YA

D
2mgL/YA.

Solution

$$\begin{array}{l}M=\text { Mass } \\A=\text { Area of cross Sechon } \\L=\text { Length of Rod }\end{array}$$
$$Y=\text { Yourgls modulus }$$
$$\text { mass of } \operatorname{leng}{th } \ L=M$$

$$\text { Mass of length } x=\left(\frac{M}{L}\right) x$$

$$\begin{array}{l}\text { Tension at point } B=\left(\frac{M}{L} x\right) g \\Y=\frac{\text { Tension/Area }}{\frac{\Delta L}{L}}=\text { strain }\end{array}$$

$$\begin{array}{l}\therefore Y=\frac{\frac{I}{A}}{\frac{d c}{d x}} \quad\left(d c=\begin{array}{l}\text { incressed } \\\text { length }\end{array}\right) \\d c=\int \frac{T}{y_{A}} d x\end{array}$$

$$\begin{array}{l}\text { integrating botu sides } \\\Delta L=\int_{0}^{L} \frac{\left(\frac{M}{L} x\right) g}{Y_{A}} d x\end{array}$$

$$\Delta L=\frac{M g}{A Y L} \int_{0}^{L } x d\quad\therefore \Delta L=\frac{M g L}{2Y A}$$
Hence option B is correct
Question 2
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Directions For Questions

A light rigid bar is suspended horizontally from two vertical wires, one of steel and one of brass, as shown in figure. Each wire is $$2.00\:m$$ long. The diameter of the steel wire is $$0.60\:mm$$ and the length of the bar $$AB$$ is $$0.20\:m$$. When a mass of $$10\:kg$$ is suspended from the centre of the $$AB$$ bar remains horizontal. The Young modulus for steel $$=2.0 \times 10^{11}\:Pa$$, the Young modulus for brass $$=1.0 \times 10^{11}\:Pa$$

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Calculate the extension of the steel wire and the energy stored in it.

A
$$45\ J$$

B
$$4.5\ J$$

C
$$0.45\ J$$

D
$$0.045\ J$$

Solution

Work done on the wire during stretching = average force x extension
Energy Stored$$=\dfrac{1}{2}F.\Delta\,l$$
Put $$\Delta\,l=\dfrac{F l}{A Y_{s}}$$
Energy Stored$$=\dfrac{1}{2}\dfrac{l}{A}\dfrac{F^2}{Y_s}$$
$$=\dfrac{1}{2}\dfrac{2\times(50)^2} {\pi \times(0.6 \times 10^{-3}/2)^2\times2\times10^{11}}$$
$$=0.045J$$
Question 3
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A steel rope has length $$L$$, area of cross-section $$A$$, Young's modulus $$Y$$. [$$Density = d$$]. If the steel rope is vertical and moving with the force acting vertically up at the upper end, find the strain at a point $$\displaystyle \frac{L}{3}$$ from lower end.

A
$$(dgL)/2Y$$

B
$$(dgL)/4Y$$

C
$$(dgL)/6Y$$

D
$$(dgL)/8Y$$

Solution

Since steel rope has mass therefore, tension in it will not remain constant
and it will proportional to the length of it from the end with which it is tied up.
$$T\alpha \ l$$

Therefore tension at $$ l=L/3 $$ will be .

$$ T_{l=L/3}=T/3 $$

$$ T_{l=L/3}=[dALg]/6 $$

We know that,

$$stress=F/A=T/A$$

$$stress= [dALg]/6A$$

$$strain=stress/Y$$

$$strain= [dALg]/6AY$$

Therefore, option(C) is correct
Question 4
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A slightly conical wire of length L and end radii $$r_1$$ and $$r_2$$ is stretched by two forces F and F applied parallel to the length in opposite directions and normal to the end faces. If Y denotes Young's modulus, then the extension produced is:

A
$$\displaystyle \frac{FL}{\pi r_1 r_2Y}$$

B
$$\displaystyle \frac{FLY}{\pi r_1 r_2}$$

C
$$\displaystyle \frac{FL}{\pi r_1 Y}$$

D
$$\displaystyle \frac{FL}{\pi r_1^{2}Y}$$

Solution

If $$R$$ be the rate of increase of radius per length, $$R=\dfrac{r_2-r_1}{L}$$
Consider a element of thickness $$dx$$ at a distance $$x$$ from narrow end of the wire.
The radius of the element is $$y=r_1+Rx$$.
The extension in the element is $$\displaystyle dl=\dfrac{F dx}{(\pi y^2)Y}$$
thus, $$\displaystyle l=\frac{F}{\pi Y}\int_0^L\frac{dx}{r_1+Rx}$$
let $$r_1+Rx=p, Rdx=dp$$
or $$\displaystyle l=\dfrac{F}{\pi Y}\int^{r_1+RL}_0\dfrac{dp}{R p^2}=\dfrac{F}{\pi YR}\left[\dfrac{1}{r_1}-\dfrac{1}{r_1+RL}\right]=\dfrac{F}{\pi YR}\left[\dfrac{r_1+RL-r_1}{r_1(r_1+RL)}\right]$$
or $$\displaystyle l=\dfrac{FL}{\pi Y(r_2-r_1)}.\frac{r_2-r_1}{r_1(r_1+r_2-r_1)}=\dfrac{FL}{\pi r_1r_2 Y}$$
Question 5
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The speed of a traverse wave travelling on a wire having a length $$50\space cm$$ and mass $$50\space g$$ is $$80\space ms^{-1}$$. The area of cross-section of the wire is $$1\space mm^2$$ and its Young's modulus is $$16\times10^{11}\space Nm^{-2}$$. Find the extension of the wire over natural length.

A
$$2\space cm$$

B
$$2\space mm$$

C
$$0.2\space mm$$

D
$$0.02\space mm$$

Solution

$$\quad v = \sqrt{\displaystyle\frac{T}{\mu}}$$

$$or \quad T = v^2\mu = (80)^2\left(\displaystyle\frac{50}{0.5}\times10^{-3}\right) = 640\space N$$

$$and \quad Y = \displaystyle\frac{Fl}{A\space\triangle l}$$

$$or \quad \triangle l = \displaystyle\frac{Fl}{AY} = \displaystyle\frac{640\times0.5}{10^{-6}\times16\times10^{11}} = 20\times10^{-5}\space m$$
Question 6
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A ball of radius R and with bulk modulus of elasticity K is kept in a liquid inside a cylindrical container. It is pressed by putting a mass m on a massless piston of cross-sectional area A, then the fractional decrease in the radius of ball will be

A
$$\displaystyle \frac{Mg}{3KR}$$

B
$$\displaystyle \frac{Mg}{3KA}$$

C
$$\displaystyle \frac{Mg}{KA}$$

D
$$\displaystyle \frac{MgK}{3AR}$$

Solution

$$K=\dfrac{VdP}{dV}$$
or $$\dfrac{dV}{V}=\dfrac{P}{K}$$.....(1) also P=Mg/A
$$V=\dfrac{4\pi R^3}{3}$$ on differentiating wrt R
we get $$\dfrac{dV}{V}=\dfrac{3dR}{R}$$ ...(2)
from 1 and 2
we get $$\dfrac{dR}{R}=\dfrac{Mg}{3KA}$$
Question 7
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A small solid sphere of radius R made of a material of bulk modulus B is surrounded by an incompressible liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. The magnitude of fractional change in the radius of the sphere $$\left( \dfrac{dR}{R} \right)$$when a mass M is placed slowly on the piston to compress the liquid is:

A
$$\dfrac{Mg}{3AB}$$

B
$$\dfrac{Mg}{AB}$$

C
$$\dfrac{3Mg}{AB}$$

D
None of these

Solution

Pressure formula,
$$ P = P_{0} + \rho gh $$
and $$P' = P_{0} + \rho gh + \dfrac{Mg}{A} $$
So, $$ \Delta P = \dfrac{Mg}{A}$$
Bulk modulus, $$ B = \dfrac{ \Delta P}{ \Delta V/V} $$
or, $$ \dfrac{ \Delta V}{V} = \dfrac{ \Delta P}{B}$$
Now, $$ V = \dfrac{4}{3} \pi r^{3}$$
So, $$ ln V = ln \dfrac{4}{3}+ln\pi +3lnR$$
or, $$ \dfrac{ \Delta V}{V} = \dfrac{3 \Delta R}{R}$$
or, $$ \dfrac{ \Delta R}{R} = \dfrac{ \Delta V}{3V}= \dfrac{ \Delta P}{3B} = \dfrac{Mg}{3AB}$$
Question 8
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Six identical uniform rods $$PQ, QR, RS, ST, TU$$ and $$UP$$ each weighing W are freely joined at their ends to form a hexagon. The rod $$PQ$$ is fixed in a horizontal position and middle points of $$PQ$$ and $$ST$$ are connected by a vertical string. The tension in string is

A
$$W$$

B
$$3W$$

C
$$2W$$

D
$$4W$$

Solution

Let a small displacement be given to the system in vertical plane in frame such that ST remains horizontal then let vertical displacement of centres of rods UP and QR be y then vertical displacement of centres of UT and RS will be 3y and that of TS will by 4y. Equating the total virtual work to zero we get.
$$(W + W) \delta y + (W + W) 3 \delta y + W (4 \delta y) - T (4 \delta y) = 0$$
(where T is the tension in thread)
$$\Rightarrow 2W + 6 W + 4W = 4 T$$
$$\Rightarrow T = 3 W$$
Question 9
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A small solid sphere of radius $$R$$ made of a material of bulk modulus $$B$$ is surrounded by an incompressible liquid in a cylindrical container. A massless piston of area $$A$$ floats on the surface of the liquid. The magnitude of fractional change in the radius of the sphere $$(\dfrac {dR}{R})$$ when a mass $$M$$ is placed slowly on the piston to compress the liquid is:

A
$$\dfrac {Mg}{3AB}$$

B
$$\dfrac {Mg}{AB}$$

C
$$\dfrac {3Mg}{AB}$$

D
none of these

Solution

We know that $$P=\dfrac{F}{A}$$
So, $$ \Delta P = \dfrac{Mg}{A}$$

We know, Bulk Modulus, $$ B = \dfrac{ \Delta P}{ \Delta V/V} $$

Or, $$ \dfrac{ \Delta V}{V} = \dfrac{ \Delta P}{B}$$

Now, for a sphere, $$ V = \dfrac{4}{3} \pi r^{3}$$

So, $$ ln V = ln \dfrac{4 \pi}{3} +3lnR$$

Or, $$ \dfrac{ \Delta V}{V} = \dfrac{3 \Delta R}{R}$$

Or, $$ \dfrac{ \Delta R}{R} = \dfrac{ \Delta V}{3V}= \dfrac{ \Delta P}{3B} = \dfrac{Mg}{3AB}$$
Question 10
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Two opposite forces $$F_1 = 120N \:and \:F_2 = 80N$$ act on an heavy elastic plank of modulus of elasticity $$y=2\times 10^{11}N/m^2$$ and length $$L = 1m$$ placed over a smooth horizontal surface. The cross-sectional area of plank is $$A = 0.5m^2$$. If the change in the length of plank is (in nm )

A
1

B
0.5

C
5

D
2

Solution

We use the formula
$$ \Delta L= \dfrac{FL}{Ay}$$
The average force acting on the plank is $$\dfrac{120+80}{2}=100 N$$, Thus
$$\Delta L=\dfrac{100\times 1}{0.5\times 2\times 10^11}$$
or
$$\Delta L=1\times 10^{-9}=1 nm$$