Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

The force required to stretch a steel wire of $$1 cm^{2}$$ cross-section to $$1.1$$ times its length would be ($$Y =2 \times 10^{11} Nm^{-2}$$

A
$$2 \times 10^{6}N$$

B
$$2 \times 10^{3}N$$

C
$$2 \times 10^{-6}N$$

D
$$2 \times 10^{-7}N$$

Solution

$$F =A \times Y \times \text{Strain} = 1 \times 10^{-4} \times 2 \times 10^{11} \times 0.1 = 2\times 10^6N$$
Question 2
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Which statement is true for a metal

A
$$Y <\eta$$

B
$$Y=\eta$$

C
$$Y>\eta$$

D
$$Y< \dfrac{1}{\eta}$$

Solution

$$Y = 2\eta (1+\sigma)$$
Question 3
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The length of a rod is $$20 cm$$ and area of cross-section $$2 cm^2$$. The Young's modulus of the material of wire is $$1.4 \times 10^{11} N/m^2$$ . If the rod is compressed by $$5 kg-wt$$ along its length, then increase in the energy of the rod in joules will be

A
$$8.57 \times 10^{-6}$$

B
$$22.5 \times 10^{-4}$$

C
$$9.8 \times 10^{-5}$$

D
$$45.0 \times 10^(-5)$$

Solution

Energy =$$ \dfrac{1}{2}Fl=\dfrac{1}{2}\times F\left( \dfrac{Fl}{AY}\right) = \dfrac{1}{2}\times \dfrac{F^{2}l}{AY}= \dfrac{\left( 50\right) ^{2}\times 20\times 10^{-2}}{2\times 10^{-4} \times 1\cdot 4\times 10^{11}} = 8.57 \times 10^{-6}$$
Question 4
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In solids, inter-atomic forces are

A
Totally repulsive

B
Totally attractive

C
Combination of (a) and (b)

D
None of these

Solution

In a solid, each atom or molecules surrounded by neighboring atoms or molecules. These are bonded together by inter atomic or intermolecular forces and stay in a stable equilibrium position. When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causing a change in the inter atomic (or intermolecular) distances. When the deforming force is removed, the inter atomic forces tend to drive them back to their original positions. Thus the body regains its original shape and size.
So, when the force stends to strech the solid the inter-atomic forces are attractive. And when the force tends to compress the solid the inter-atomic forces tend to repel. Hence, they are a combination of both attractive and repulsive.
Question 5
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If a spring is extended to length $$l$$ then according to Hook's law

A
$$F=kl$$

B
$$F=\dfrac{k}{l}$$

C
$$F=k^2l$$

D
$$F=\dfrac{k^2}{l}$$

Solution

According to Hook's law within the elastic limit, stress is directly proportional to strain. So, we can say that,
$$Stress \propto Strain$$
or. $$\frac{Stress}{Strain}=K$$
This constant of proportionality is called the modulus of elasticity.
Stress is given by force per area and strain is given by the change in length by total length. Let's assume the cross sectional area as $$A$$ and the total length as $$L$$. Hence,
$$\frac{\frac{F}{A}}{\frac{l}{L}}=K$$
or, $$\frac{F L}{Al}=K$$
or, $$F=\frac{AK}{L}l$$
The area total length and the modulus of elasticity are all constants. Let's replace them with $$k=\frac{AK}{L}$$
Hence, we have
$$F=kl$$.
Question 6
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A spring of force constant $$k = 100 N/m$$ is compressed to $$x = 0.5 m$$ by a block of mass $$100 g$$ and released. There should be a box placed at a distance $$d$$ on the ground so that the block falls in it. Find the distance $$d.$$

A
$$5 m$$

B
$$10 m$$

C
$$15 m$$

D
$$20 m$$

Solution

Given,
Mass of block $$100\ gm=0.1\ kg$$
Spring constant $$k=100\ N/m$$
Vertical height $$h=2\ m$$
Spring compressed by$$=0.5\ m$$

Here, we can find the horizontal velocity $$v$$ with which the block will leave the surface.
So, using principle of conservation of energy we have,
$$\dfrac {1}{2} kx^{2} = \dfrac {1}{2} mv^{2}$$
$$\Rightarrow \dfrac {1}{2} \times 100\times (5\times 10^{-1})^{2} = \dfrac {1}{2} \times 0.1\times v^{2}$$
$$\Rightarrow v = 5\sqrt {10} m/s$$

Now, the horizontal distance moved by the block is given by $$d = vt$$, where $$t$$ is the time taken by the block to fall the distance $$2 m$$

Therefore, $$d = 5\sqrt {10} \times \sqrt {(2h)/g}$$
$$\Rightarrow d = 5\sqrt {10} \times \sqrt {4/10} = 10 m$$.
Question 7
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The equation of a stationary wave in a metal rod is given by $$y=0.002\displaystyle \sin\frac{\pi x}{3}\sin 1000t$$, where $$\mathrm{x}$$ is in cm and $$\mathrm{t}$$ is in second. The maximum tensile stress at a point $$\mathrm{x}=2\mathrm{c}\mathrm{m}$$ (Young's modulus $$\mathrm{Y}$$ of material of rod $$=8\times 10^{11}dyne/cm^{2})$$ will be

A
$$\displaystyle \frac{\pi}{3}\times 10^{8}dyne/cm^{2}$$

B
$$\displaystyle \frac{4\pi}{3}\times 10^{8}dyne/cm^{2}$$

C
$$\displaystyle \frac{8\pi}{3}\times 10^{8}dyne/cm^{2}$$

D
$$\displaystyle \frac{2\pi}{3}\times 10^{8}dyne/cm^{2}$$

Solution

The equation of stationary wave in the metal rod is: $$y=0.002Sin\dfrac{\pi x}{3} sin 1000t$$
The maximum tensile stress at the point $$x=2\ cm$$ will be:
$$T= Y \times (\dfrac{dy}{dx})_{max}=Y \times 2 \times 10^{-3} \times \dfrac{\pi}{3}Cos \dfrac{2 \pi}{3}= \dfrac{8 \pi}{3} \times 10^8 dyne/cm^2$$
Question 8
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A copper wire is held at the two ends between two rigid supports. At $$30^{\mathrm{o}}\mathrm{C}$$, the wire is just taut,with negligible tension. If $$Y=13\times 10^{11}Nm^{-2}, \alpha =1.7\times 10^{-5}(^{\circ}C)^{-1}$$ and density $$ \rho=9\times 10^{3}kgm^{-3}$$, then the speed of transverse wave in this wire at $$10^{o}$$C is:

A
$$ 90 ms^{-1} $$

B
$$70 ms^{-1}$$

C
$$60 ms^{-1}$$

D
$$100 ms^{-1}$$

Solution

$$V=\sqrt{\dfrac{T}{m}}=\sqrt{\dfrac{Y\alpha \Delta t}{\rho}}$$
where Thermal force $$=$$ Tension in wire $$=T =YA\alpha\Delta t$$
and linear density $$= m =A\rho$$
So, now putting in the derived expression,
$$V=\sqrt{\dfrac{13\times10^{11}\times1.7\times10^{-5}\times10}{9\times10^3}}=70 m/s$$
Question 9
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In performing an experiment to determine the Young's modulus Y of steel, a student can record the following values:
length of wire l$$=(\ell_{0}\pm\Delta$$l$$){m}$$
diameter of wire $${d}=({d}_{0}\pm\Delta {d})$$ mm
force applied to wire $${F}$$=$$({F}_{0}\pm\Delta {F}){N}$$
extension of wire $${e}=({e}_{0}\neq\Delta {e})$$ mm
In order to obtain more reliable value for Y, the followlng three techniques are suggested.
Technique (i) A shorter wire ls to be used.
Technique (ii) The diameter shall be measured at several places with a micrometer screw gauge.
Technique (iii) Two wires are made irom the same ntaterial and of same length. One is loaded at a fixed weight and acts as a reference for the extension of the other which is load- tested
Which of the above techniques is/are useful?

A
i and ii only

B
ii and iii only

C
i only

D
iii only

Solution

for (1) the wire should be large to increase the elasticity.Strain is inversely proportional to young's modulus.
for (2) micrometer is used for diameter reading to reduce the error.its important point
for (3) using wire of same material,it reduces the error due to thermal expansion.
hence,point 2 & 3 is correct for accurate reading
Question 10
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In the figure above, a wire is subjected to a gradually increasing force $${F}_{i}$$ which causes an extension e. The way in which e depends on F is shown by the line OP. Then, the force is gradually decreased and relation between e and F in this case is shown by the line PQ. The mechanical energy that is recovered from the wire in this process when the force is removed, is given by the shaded area in

A

B

C

D

Solution

Work done W by the applied force F on the wire for an increment in length $$\Delta y$$ is $$W=F. \Delta y$$, where y is measured from equilibrium point of wire at $$F=0$$
$$\Rightarrow$$ The elastic potential energy (U) stored in the wire as it is stretched from $$y=0$$ to $$y=e$$ is given by $$U=\int_{0}^{e}F.dy=$$. area bounded by curve, the vertical axis and the horizontal line passing through P,
Here, in graph (A), the area is bounded by horizontal axis.
$$\Rightarrow$$ (A) is not correct.
The area in (B) gives the energy gained by the stretched wire and the wire suffers a permanent strain when F is removed at P.(it is called plastic deformation and the energy gained is dissipated as heat in the wire)
The area in (C) is the energy stored in the wire when it is stretched from y=0 to y=e at point P.
$$\Rightarrow$$ Subtracting area in graph (B) from area in graph (C) gives the area in (D) which is required mechanical energy.