Mechanical Properties of Solids Test

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Mechanical Properties of Solids Test - 64

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Weekly Quiz Competition

Question 1
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The Young's modulus of a rubber string 8 cm long and density 1.5 $$kg/m^{3}$$ is $$5\times 10^{8} N/m^{2}$$, is suspended on the ceiling in a room. The increase in length due to its own weight will be:-

A
$$9.6\times 10^{-5}m$$

B
$$9.6\times 10^{-11}m$$

C
$$9.6\times 10^{-3}m$$

D
9.6 m

Solution

$$y=5\times { 10 }^{ 8 }N/{ m }^{ 2 };a=g;l=8cm=8\times { 10 }^{ -2 }m;d=1.5kg/{ m }^{ 3 }\quad $$
$$y=\cfrac { stress }{ strain } =\cfrac {\dfrac{F}{A} }{ \dfrac{\Delta l}{l} } =\cfrac { \cfrac { F\times d }{ A\times d } }{ \dfrac{\Delta l}{l} } =\cfrac { \cfrac { F\times d }{ v } }{ \dfrac{\Delta l}{l} } =\cfrac { F\times d\times l }{ v\times \Delta l } =dg\times \cfrac { A }{ dl } \quad $$
$$=5\times { 10 }^{ 8 }=\cfrac { 1.5\times 64\times { 10 }^{ -4 }\times 10 }{ \Delta l } \quad $$
$$\Delta l=\cfrac { 1.5\times 64\times { 10 }^{ -4 }\times 10 }{ 2\times 5\times { 10 }^{ 8 } } =\cfrac { 0.3\times 64\times { 10 }^{ -12 }\times 10 }{ 2 } $$
$$\therefore \Delta l=9.6{ 10 }^{ -11 }m$$
Question 2
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In a young's double-slit experiment, the slits are separated by $$0.5 mm$$, and the screen is placed $$150 cm$$ away. A beam of light consisting of two wavelengths, $$650 nm$$, and $$520 nm$$, is used to obtain interference fringes on the screen. The least distance from the common central maxima to the point where the bright fringes due to both the wavelengths coincide is:

A
$$9.75 mm$$

B
$$15.6 mm$$

C
$$1.56 mm$$

D
$$7.8mm$$

Solution

Given,
$$d=0.5mm=0.5\times 10^{-3}m$$
$$D=150cm=1.5m$$
$$\lambda_1=650nm$$
$$\lambda_2=520nm$$
For common maxima,
$$n_1\lambda_1=n_2\lambda_2$$
$$\dfrac{n_1}{n_2}=\dfrac{\lambda_2}{\lambda_1}$$
$$\dfrac{n_1}{n_2}=\dfrac{520}{650}=\dfrac{4}{5}$$
Now, $$n_1=4, \lambda_1=650nm$$
Condition of maxima,
$$\dfrac{yd}{D}=n_1\lambda_1$$
$$y=\dfrac{n_1\lambda_1 D}{d}$$
$$y=\dfrac{4\times 650\times 10^{-9}\times 1.5}{0.5\times 10^{-3}}=7.8mm$$
The correct option is D.
Question 3
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An iron wore and copper wire having same length and cross-section are suspended from same roof Young's modulus of copper is $$\dfrac 13$$ that of iron. Then the ratio of the weights to be added at their ends so that their ends are at the same level is

A
$$1:3$$

B
$$1:9$$

C
$$3:1$$

D
$$9:1$$

Solution

Correct Answer: Option C
Step 1: Relate the change in length
Refer to figure $$1$$
The initial lengths of both the copper and iron wires are same and finally their ends are at same level.
Let the initial length of each be $$l$$
Change in length of iron be $$\Delta l_1$$ and that of copper be $$\Delta l_2$$
so,
$$l + \Delta l_1 = l + \Delta l_2$$
Therefore,
$$\Delta l_1 = \Delta l_2$$ .......... $$(I)$$

Step2: Use Young's Modulus
We know,
Young's Modulus $$Y = \dfrac{Stress}{Strain}$$

Let the weight added to Iron be $$F_1$$ and that to copper be $$F_2$$
Also, their cross section area are same.
so,
$$Y_1 = \dfrac{F_1l}{A \Delta l_1}$$ and $$Y_2 = \dfrac{F_2l}{A \Delta l_2}$$
It is given that
$$Y_1 = 3Y_2$$
therefore,
$$\dfrac{F_1}{\Delta l_1} = 3 \times \dfrac{F_2}{\Delta l_2} $$
as
$$\Delta l_1 = \Delta l_2$$
$$F_1 = 3F_2$$
so,
$$\dfrac{F_1}{F_2} = \dfrac 31$$
Hence, Option C is the correct answer
Question 4
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If the work done in stretching a wire by $$1\ mm$$ is $$2\ J$$, then work necessary for stretching another wire of same material but with double radius of cross-section and half of the length by $$1\ mm$$ is

A
$$8\ J$$

B
$$16\ J$$

C
$$4\ J$$

D
$$32\ J$$

Solution
Question 5
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An elongation of $$0.1^\circ$$ in a wire of cross sectional area $$10^{-6}m^{2}$$ causes a tension of 100n. The Young's modulus is-

A
$$10^{12}N/M^{2}$$

B
$$10^{11}N/M^{2}$$

C
$$10^{10}N/M^{2}$$

D
$$10^{2}N/M^{2}$$

Solution

Given,
$$\dfrac{\Delta l}{l}\times 100=0.1$$%
$$\dfrac{\Delta l}{l}=\dfrac{0.1}{100}=1\times 10^{-3}$$
Cross-sectional area, $$A=10^{-6}m^2$$
$$T=100N$$
Young's modulus,
$$Y=\dfrac{T/A}{\Delta l/l}=\dfrac{100/10^{-6}}{10^{-3}}=\dfrac{100}{10^{-9}}$$
$$Y=10^{11}N/m^2$$
The correct option is B.
Question 6
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A $$40\ cm$$ wire having a mass $$3.2\ gm$$ and area of cross-section $$1\ mm^{2}$$ is stretched between two supports $$40.05\ cm$$ apart. In its fundamental note, the wire vibrates with a frequency $$220\ Hz$$. The Young's molulus is :

A
$$1.98\times 10^{11}\ N/m^{2}$$

B
$$2.2\times 10^{11}\ N/m^{2}$$

C
$$3.96\times 10^{11}\ N/m^{2}$$

D
$$3.2\times 10^{11}\ N/m^{2}$$

Solution

$$l = 40\,cm$$
$$\Delta l = 0.05\,cm$$

$$f = \dfrac{v}{{2l}}$$

$$ \Rightarrow v = 22 \times 8 = 176m/s$$
Now$$,$$ $$v = \sqrt {\dfrac{T}{\mu }} $$

$$ \Rightarrow T = 176 \times 176 \times \dfrac{{3.2 \times {{10}^{ - 3}}}}{{40 \times {{10}^{ - 2}}}}$$

$$ = \dfrac{{176 \times 176 \times 32}}{{40 \times 10 \times 10}}$$

$$ = \dfrac{{176 \times 176 \times 32}}{{4 \times {{10}^3}}} = 247.8\,N$$

$$\therefore \gamma = \dfrac{{247.8 \times 40}}{{1 \times {{10}^{ - 6}} \times 0.05}}$$

$$ = 1.98 \times {10^{11}}\,N/{m^2}$$
Hence,
option $$(A)$$ is correct answer.
Question 7
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A steel wire (original length = 2m, diameter = 1mm) and a copper wire ( original length =1m, diameter =2mm) are loaded as shown in the figure . Find the ratio of extension of steel wire to that of copper wire. Given , Young's modulus of steel = $$2\times 10^{11} Nm^{-2}$$ and that of copper is $$ 10^{11} Nm^{-2}$$.

A
$$\frac{10}{3}$$

B
$$5$$

C
$$\frac{20}{3}$$

D
$$\frac{14}{3}$$

Solution
Question 8
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A body of mass 1 Kg is fastended to one end of a steel wire of cross - sectional area 3$$\times 10^{-6}$$ m$$^{2}$$ and it rotated in horizontal circle of radius 20 cm with a constant speed of 2 m/s .The elongation in the wire is (Y = 2$$\times 10^{11}$$N / m$$^{2}$$)

A
$$0 . 33\times 10^{-5}$$ m

B
$$0 . 67 \times 10^{-5}$$ m

C
$$2 \times 10^{-5}$$ m

D
$$4\times10^{-5}$$ m

Solution
Question 9
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The force required to punch a square hole 2 cm side in steel sheet 2 mm thick is:-
(shearing stress of steel sheet $$=3.5 \times10^8 N/m^2)$$

A
$$5.6\times10^4 N$$

B
$$3.4\times10^4 N$$

C
$$9.1\times10^4 N$$

D
$$6.8\times10^4 N$$

Solution

$$\begin{array}{l}\text { Breaking area }=2 \pi r\times L \\=2 \pi x \\\text { Breaking area = (Side length) }\times \text { (thickness) } \times 4\\=2\times 10^{-2} \times 2 \times 10^{-3} \times 4\\=4 \times 10^{-5} \times 4 \\=16 \times 10^{-5}\end{array}$$

$$\begin{aligned}F &=(\text { Breaking stress })\times \text { Area } \\&=16 \times 10^{-5} \times 3.5 \times 10^{8} \\F &=5.6 \times 10^{4}\mathrm{~N}\end{aligned}$$
Question 10
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A man grows into a giant such that his linear dimensions increase by a factor of $$9$$. Assuming that his density remains same, the stress in the leg will change by a factor of?

A
$$81$$

B
$$\cfrac{1}{81}$$

C
$$9$$

D
$$\cfrac{1}{9}$$

Solution

$$\begin{aligned}\text { Stress } &=\frac{\text { Force }}{\text { Area }}=\frac{(\text { height })\times \text { Area }\times \text { density }\times g}{\text { Area }} \\&=\text { (height) }(g\times\text { density })\end{aligned}$$

$$\begin{array}{l}\text { Stress a height } \\\text { So stress in leg will also change by a factor of } 9 .\end{array}$$