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Physics 2 Marks Questions

PART - 2 CH - 8 Mechanical Properties of Solids · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 20 questions.

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Question 12 Marks
The breaking stress for a metal is $2 \times$ $10^6 N / m ^2$. Find the maximum length of this metal wire, by which it can suspended under its own weight without breaking. (Density of metal $=8 \times 10^3 kg / m ^3$ and $g=10 m / s ^2$ )
Answer
Suppose the maximum length of the wire is L and area of the cross-section is A then the own weight of the wire
$W = ALd g$ where $d=$ density
Stress due to this weight $=\frac{ AL d g}{A}= L d g$
If this is the breaking stress then according to the questions :
$\begin{aligned}L d g & =2 \times 10^6 N / m^2 \\L & =\frac{2 \times 10^6}{d g}=\frac{2 \times 10^6}{8 \times 10^3 \times 10} \\L & =\frac{2 \times 10^6}{8 \times 10^4}=\frac{2 \times 10^2}{8}=\frac{2 \times 100}{8} \\L & =25 m\end{aligned}$
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Question 22 Marks
A solid ball of rubber is taken from the surface of a 400 m deep lake to its bottom, then its volume decreases by 0.2 percent. If the density of water is $1.0 \times 10^3 kg / m ^3$, calculate the Bulk's Modulus of elasticity of rubber.
Answer
Bulk modulus of elasticity (K)
$\begin{aligned}& =\frac{\text { Change in pressure }}{\text { Volume strain }} \$K) & =-\frac{d P}{\frac{\Delta V}{V}}\end{aligned}$
Change in pressure $=h d g$
$\begin{array}{l}=400 \times 10^3 \times 10 \\=4 \times 10^6 N / m^2\end{array}$
Volume strain $=\frac{0.2}{100}=\frac{2}{1000}=2 \times 10^{-3}$
Hence, Bulk modulus of elasticity $K =\frac{4 \times 10^6}{2 \times 10^{-3}}$
$=2 \times 10^9 N / m^2$
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Question 32 Marks
When 100 atm pressure is applied on a sphere in the normal and uniform direction its volume decreases by 0.01%. Calculate the Bulk's Modulus of elasticity of this material.
Answer
$K=\frac{\Delta P}{\frac{-\Delta V}{V}}$
Given : $P =100 \times 1.013 \times 10^5 N / m ^2$
$\begin{aligned}-\Delta V & =\frac{V \times .01}{100}=\frac{1}{10000} \\-\frac{\Delta V}{V} & =\frac{1}{10000}=10^{-4} \\K=\frac{P}{\frac{-\Delta V}{V}} & =\frac{100 \times 1.013 \times 10^5 N / m^2}{10^{-4}} \\& =1.013 \times 10^{11} N / m^2\end{aligned}$
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Question 42 Marks
The area of cross - section of a steel wire is $1 cm^2$. How much force would be required to double its length? $\left(Y=2 \times 10^{12}\right.$ $\text {dyne}$ $\left./ cm ^2\right)$.
Answer
Given : $\quad A=1 cm^2$
Let length $= L$, Increase $=2 L- L = L$
$\begin{aligned}
Y & =\frac{MgL}{A \times \Delta L} \\
2 \times 10^{12} & =\frac{MgL}{1 \times L} \\Mg & =2 \times 10^{12}\end{aligned}$
Hence, the necessary force required is $2 \times 10^{12}$ dyne.
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Question 52 Marks
On a wire whose diameter is 0.4 cm, a weight of 25 kg is suspended due to this, the length of 50 cm wire becomes 51 cm then find the Young's Modulus of elasticity?
Answer
Given,
$ \begin{aligned} 2 r & =0.4 cm \\ r & =0.2 cm=0.2 \times 10^{-2} m \\ r & =2 \times 10^{-3} m \\ Mg & =25 kg=250 \text { Newton } \\ L & =50 cm=50 \times 10^{-2} m \\ l & =51-50=1 cm=10^{-2} m \\ Y & =? \end{aligned} $
We know that,
$ \begin{aligned} Y & =\frac{MgL}{A \times \Delta L} \\ Y & =\frac{MgL}{\pi r^2 \Delta L} \\ Y & =\frac{25 \times 10 \times 50 \times 10^{-2}}{3.14 \times\left(2 \times 10^{-3}\right)^2 \times 10^{-2}} \\ Y & =\frac{12500 \times 10^{-2}}{3.14 \times 4 \times 10^{-8}} \\ Y & =\frac{125}{12.56} \times 10^8 \\ & =9.95 \times 10^8 Newton / m^2 \end{aligned} $
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Question 62 Marks
From two wires of a metal equal weights are suspended. One of the wire is 2 m long and has diameter of 1 mm and the second has length 1m and diameter 0.5 mm. Find the ratio of increases in length of these.
Answer
Equal weight are suspended
For same metal $\begin{aligned} F _1 & = F _2 \\ Y _1 & = Y _2\end{aligned}$
We know that, $\quad Y =\frac{ F \times L }{ A \times \Delta L }$
$\therefore$ For the first wire
$Y_1=\frac{F_1 \times L_1}{\pi r_1^2 \times l_1}\ldots\ldots (1)$
For the second wire
$Y _2=\frac{ F _2 \times L _2}{\pi r_2^2 \times l_2}\dots\ldots (2)$
By making equal the eq. ($1)$ & $$(2),
$\because \quad F _1= F _2$ and $Y _1= Y _2$
$\therefore \quad \frac{ L _1}{\pi r_1^2 \times l_1}=\frac{ L _2}{\pi r_2^2 \times l_2}$
$\begin{aligned} \frac{l_1}{l_2} & =\left(\frac{ L _1}{L_2}\right) \times\left(\frac{r_2}{r_1}\right)^2 \\ \frac{l_1}{l_2} & =\left(\frac{ L _1}{L_2}\right) \times\left(\frac{2 r_2}{2 r_1}\right)^2=\left(\frac{2}{1}\right) \times\left(\frac{0.5}{1}\right)^2 \\ & =2 \times 0.25 \\ \frac{l_1}{l_2} & =0.5=\frac{1}{2}\end{aligned}$
$l_1: l_2=1: 2$
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Question 72 Marks
A steel girder can bear a load of 25 tonnes. If the depth of the girdes is doubled, how much load can it bear?
Answer
If a girder can bear $F _1$ load at depth $d_1$ and $F _2$ load at depth $d_2$, then
$\frac{F_2}{F_1}=\frac{d_2^3}{d_1^3}$
In question, $F _1=25$ tonnes, $d_2=2 d_1$
$\begin{array}{l}
\frac{F_2}{25}=\frac{\left(2 d_1\right)^3}{\left(d_1\right)^3} \\\frac{F_2}{25}=\frac{8 d_1^3}{d_1^3}=8 \\F_2=25 \times 8=200 \text { tonnes }\end{array}$
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Question 82 Marks
For a wire of a material radius 0.5 mm , Young's modulus of elasticity of the material of the wire $2 \times 10^{11} N / m ^2$ and if strain of the wire cannot be more than $10^{-3}$ then how much weight can be suspended from the wire?
Answer
Radius of wire $r=0.5 mm=5 \times 10^{-4} m$
Cross-section area of the wire $=\pi r^2$
$\begin{array}{l}
=3.14 \times\left(5 \times 10^{-4}\right)^2 \\
=3.14 \times 25 \times 10^{-8} m^2 \\
=78.5 \times 10^{-8} m^2
\end{array}$
Let maximum force F Newton is being applied on the wire
$\begin{aligned}
\text { Maximum force stress } & =F / A \\
& =\frac{F}{78.5 \times 10^{-8}} N / m^2 \end{aligned}$
Given,
$\text { Strain }=10^{-3}$
We know that, $\quad Y =\frac{\text { Stress }}{\text { Strain }}$$\begin{array}{l} \frac{F}{78.5 \times 10^{-8}}=2 \times 10^{11} \times 10^{-3} \\
=2 \times 10^8 \\ \because Y=2 \times 10^{11} N / m^2 \text { and Strain }=10^{-3} \\
F=2 \times 10^8 \times 78.5 \times 10^{-8} \\ =157 \text { Newton } \end{array}$
Hence, the maximum Force applied o the wire is 157 Newton.
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Question 92 Marks
When a stress of $2 \times 10^6 N / m ^2$ is applied on a copper wire ( $Y =1^{11} N / m ^2$ ), what will be the percentage increase in its length?
Answer
Given :
$\begin{aligned}
Y & =10^{11} N / m^2 \\
\text { Stress } & =\frac{F}{A}=2 \times 10^6 N / m^2 \\Y & =\frac{FL}{Al} \\
10^{11} & =2 \times 10^6 \times \frac{L}{l}
\end{aligned}$
$\therefore \quad \frac{l}{L}=\frac{2 \times 10^6}{10^{11}}=2 \times 10^{-5}$
For the percentage multiplying by 100 both sides
$\frac{1}{L} \times 100=2 \times 10^{-5} \times 100$
Percentage increase in length $\%$
$=2 \times 10^{-3}=0.002 \%$
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Question 102 Marks
The area of cross - section of a steel wire is $1 ~ c m ^2$. How much force would be required to double its length? If Young's Modulus of elasticity of steel is $20 \times 10^{10} N / m ^2$.
Answer
Given : $A =1 ~cm^2=10^{-4} ~m^2$
$
Y=20 \times 10^{10} N / m^2
$
Let length is L
$\therefore$ Increase in length $l=2 L- L = L$
We know that $Y =\frac{ F / A }{l / L }=\frac{ F \times L }{ A \times l}$
$
20 \times 10^{10}=\frac{F \times L}{10^4 \times L} \quad(\because l=L)
$
$
\begin{array}{l}
F=20 \times 10^{10} \times 10^{-4} \\
F=20 \times 10^6 N
\end{array}
$
Therefore, the force required $=20 \times 10^6 N$.
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Question 112 Marks
The stress-strain curves of two different types of rubbers are shown in the following figure. How are these stress-strain curves of rubber different from those of metals?
Answer
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Question 122 Marks
A wire is cut to half its length. For a given load what would be the effect on increase in its length? What would be the effect on the maximum load that it can bear?
Answer
(i) Increase in length would remain half.
(ii) Would remain unaffected.
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Question 132 Marks
If the length of a wire increases by 1 mm when a 1 kg object is hung, then how much will the length of that wire increases when a 2 kg object is hung? How much does it hung 50 kg of weight?
Answer
$\because I \propto Mg$. Therefore when a weight of 2 kg is hung, the length of the wire doubles.
i.e. $
2 \times 1=2
$
Similarly when a weight of 50 kg is hung, the length of the wire $50 \times 1 = 50$ mm increases. Will increase provided that 50 kg the stress produce by the weight must remain within the limit of elasticity otherwise the wire will break.
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Question 142 Marks
The Young's Modulus of elasticity of brass is half than that of iron. The length of a brass is equal to the length of another iron wire. There is equal stress on both the wires. What will be the value of the ratio of increase in length of these two wires?
Answer
The length L of the wires of both the material is same and on both the stress (F/A) is the same. Therefore,
$
l \propto \frac{1}{Y}
$
Thus, $\quad \frac{l_1}{l_2}=\frac{ Y _2}{ Y _1}=\frac{ Y }{\frac{1}{2} Y }=\frac{2}{1}$
i.e. $\quad$ Ratio $=2: 1$
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Question 152 Marks
Why does a wire break when stretched beyond a limit?
Answer
When the wire is stretched with more force, the distance between its atoms increases. When this distance becomes so much that the force of attraction between the atoms remains negligible. In this situation, when the force is removed, the atoms of the material are not able to return to theis original position and the wire is not able to assume its previous shape. This is the limit of elasticity. By applying more force, the permanent position of the atoms gets disturbed and if the force is increased, further on the wire, it breaks.
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Question 162 Marks
Why is spring made of steel and not of copper?
Answer
The elasticity of steel is greater than that of copper. After applying and removing equal deforming force steel regains its original form sooner as compared to copper because Young's Modulus of elasticity of steel is more. For this reason the spring is made of steel instead of copper.
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Question 172 Marks
When a load of 1 kg is suspended from a wire its length increases by 12 mm. Keeping other conditions the same, if the radius of the wire is doubled, what will be the increase in length?
Answer
$ Y =\frac{ F / A }{l / L }=\frac{ FL }{ A l}$
$\therefore  l=\frac{ FL }{ AY }=\frac{ FL }{\pi r^2 Y }$
Thus, $ l \propto \frac{1}{r^2}$
$\therefore \frac{l_2}{l_1}=\left(\frac{r_1}{r_2}\right)^2$
$\begin{aligned}\text {or} \quad l_2 & =l_1\left(\frac{r_1}{r_2}\right)^2=12 \text {mm}\left(\frac{r_1}{2 r_1}\right)^2 \\
& =12 \times 1 / 4=3 \text {mm} .
\end{aligned}$
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Question 182 Marks
Steel is more elastic compared to rubber. Show how?
Answer
If steel and rubber wires of same length and thickness are taken then on suspending equal loads the increase in the length of rubber wire would be more compared to steel wire. Since the value of strain in steel is less than that of rubber. Since it is clear from the elasticity $=\frac{\text { Stress }}{\text { Strain }}$ that when the strain is less the value of elasticity is higher. This show that steel is more elastic compared to rubber.
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Question 192 Marks
If the length of a wire is cut to half then
(i) What would be the effect on its increase in length within the given weight?
(ii) What would be the effect on the maximum weight which it bears?
Answer
It is clear from the increase in the length of the wire $l \propto L$ that if the length is half, the increase in length will be half. Maximum weight bear capacity does not depend on length.
Breaking stress $=\frac{\text { Maximum force }}{\text { Cross-sectional area }}$
Therefore, there is no effect on the maximum weight bear capacity.
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Question 202 Marks
Write the definitions of longitudinal strain and longitudinal stress.
Answer
Longitudinal strain : The ratio of change in length and original length is called longitudinal strain. Due to tensile stress, the original length L of the wire or rod increased by $\Delta L$. Hence,
Longitudinal strain $=\frac{\Delta L }{ L }$
Longitudinal stress : When force is applied on a wire and its length increases in the direction of longitudinal deforming forces. The restoring force acting per unit area of the body is called longitudinal stress.
Longitudinal stress $=\frac{\text { Force }}{\text { Area }}=\frac{ F }{ A }$
Where, $F =$ Applied force and $A =\pi r^2$ Area of the wire or rod.
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