140855
A wire of length $1 \mathrm{~m}$ and radius $2 \mathrm{~mm}$ is vertically clamped. The lower end is twisted through an angle of $45^{\circ}$. The angle of shear is
1 $0.09^{\circ}$
2 $0.9^{\circ}$
3 $9^{\circ}$
4 $90^{\circ}$
Explanation:
A Given, Radius $(\mathrm{r})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$, Twisted angle $(\theta)=45^{\circ}$ $\text { Length }(l)=1 \mathrm{~m}$ According to relation, $\mathrm{r} \theta=l \phi$ Angle of shear $(\phi)=\frac{\mathrm{r} \theta}{l}=\frac{2 \times 10^{-3} \times 45}{1}$ $\phi=0.09^{\circ}$
AP EAMCET (23.09.2020) Shift-I
Mechanical Properties of Solids
140857
A material has Poisson's ratio 0.50. If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is
1 0.6
2 0.4
3 0.2
4 0
Explanation:
D Given, Poisson ratio $(\sigma)=0.5$ Longitudinal strain $\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, Poisson ratio $=-\frac{\text { Lateral change }(\text { strain })}{\text { Longitudnal change }(\text { strain })}$ $0.5=\frac{-\frac{\mathrm{dr}}{\mathrm{r}}}{2 \times 10^{-3}}$ $\frac{\mathrm{dr}}{\mathrm{r}}=-1 \times 10^{-3}$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{~L}$ Taking $\log$ of both side and differentiating $\frac{\mathrm{dV}}{\mathrm{V}} =\frac{2 \mathrm{dr}}{\mathrm{r}}+\frac{\mathrm{dL}}{\mathrm{L}}$ $=\left(-2 \times 1 \times 10^{-3}\right)+\left(2 \times 10^{-3}\right)=0$ $\%$ Change in volume $(\Delta \mathrm{V})=\frac{\mathrm{dV}}{\mathrm{V}} \times 100=0 \times 100=0 \%$
AP EAMCET (22.09.2020) Shift-I
Mechanical Properties of Solids
140859
$\quad$ A force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2 R$ will be
1 $\frac{F}{2}$
2 $2 F$
3 $4 F$
4 $\frac{F}{4}$
Explanation:
C We know, Breaking stress $=\frac{\text { Breaking force }}{\text { Area }}$ Breaking force $=$ breaking stress $\times$ Area $\mathrm{F}=\sigma \times \pi \mathrm{R}^{2}$ Given, $\mathrm{R}$ become 2 times, then breaking force, $\mathrm{F}^{\prime}=\sigma \times \pi(2 \mathrm{R})^{2}$ $\mathrm{~F}^{\prime}=4 \sigma \pi \mathrm{R}^{2}$ $\mathrm{~F}^{\prime}=4 \mathrm{~F}$
AP EAMCET (18.09.2020) Shift-II
Mechanical Properties of Solids
140860
The diagram shows stress versus strain curve for the materials $A$ and $B$. from the curves, we can infer that
1 A is brittle but $\mathrm{B}$ is ductile
2 A is ductile but $\mathrm{B}$ is brittle
3 Both A and B are ductile
4 Both A and B are brittle
Explanation:
B Curve A is ductile and curve B is Brittle.
NEET odisha - 2019
Mechanical Properties of Solids
140861
There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be
1 $1: 1$
2 $2: 1$
3 $1: 2$
4 $4: 1$
Explanation:
D Given, $\mathrm{L}_{1}=\mathrm{L}, \mathrm{d}_{1}=\mathrm{d}$ Then, $\delta \mathrm{L}_{1}=\frac{\mathrm{PL}}{\mathrm{AE}}$ $\mathrm{L}_{2}=\mathrm{L}, \mathrm{d}_{2}=2 \mathrm{~d}_{1}=2 \mathrm{~d}$ $\delta \mathrm{L}_{2}=\frac{\mathrm{PL}_{2}}{\mathrm{AE}}$ $\mathrm{P}, \mathrm{L}$, are same for both wire $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{\frac{\pi}{4} \mathrm{~d}_{2}^{2}}{\frac{\pi}{4} \mathrm{~d}_{1}^{2}}$ $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{(2 \mathrm{~d})^{2}}{(\mathrm{~d})^{2}}=\frac{4}{1}$ $\delta \mathrm{L}_{1}: \delta \mathrm{L}_{2}=4: 1$
140855
A wire of length $1 \mathrm{~m}$ and radius $2 \mathrm{~mm}$ is vertically clamped. The lower end is twisted through an angle of $45^{\circ}$. The angle of shear is
1 $0.09^{\circ}$
2 $0.9^{\circ}$
3 $9^{\circ}$
4 $90^{\circ}$
Explanation:
A Given, Radius $(\mathrm{r})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$, Twisted angle $(\theta)=45^{\circ}$ $\text { Length }(l)=1 \mathrm{~m}$ According to relation, $\mathrm{r} \theta=l \phi$ Angle of shear $(\phi)=\frac{\mathrm{r} \theta}{l}=\frac{2 \times 10^{-3} \times 45}{1}$ $\phi=0.09^{\circ}$
AP EAMCET (23.09.2020) Shift-I
Mechanical Properties of Solids
140857
A material has Poisson's ratio 0.50. If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is
1 0.6
2 0.4
3 0.2
4 0
Explanation:
D Given, Poisson ratio $(\sigma)=0.5$ Longitudinal strain $\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, Poisson ratio $=-\frac{\text { Lateral change }(\text { strain })}{\text { Longitudnal change }(\text { strain })}$ $0.5=\frac{-\frac{\mathrm{dr}}{\mathrm{r}}}{2 \times 10^{-3}}$ $\frac{\mathrm{dr}}{\mathrm{r}}=-1 \times 10^{-3}$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{~L}$ Taking $\log$ of both side and differentiating $\frac{\mathrm{dV}}{\mathrm{V}} =\frac{2 \mathrm{dr}}{\mathrm{r}}+\frac{\mathrm{dL}}{\mathrm{L}}$ $=\left(-2 \times 1 \times 10^{-3}\right)+\left(2 \times 10^{-3}\right)=0$ $\%$ Change in volume $(\Delta \mathrm{V})=\frac{\mathrm{dV}}{\mathrm{V}} \times 100=0 \times 100=0 \%$
AP EAMCET (22.09.2020) Shift-I
Mechanical Properties of Solids
140859
$\quad$ A force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2 R$ will be
1 $\frac{F}{2}$
2 $2 F$
3 $4 F$
4 $\frac{F}{4}$
Explanation:
C We know, Breaking stress $=\frac{\text { Breaking force }}{\text { Area }}$ Breaking force $=$ breaking stress $\times$ Area $\mathrm{F}=\sigma \times \pi \mathrm{R}^{2}$ Given, $\mathrm{R}$ become 2 times, then breaking force, $\mathrm{F}^{\prime}=\sigma \times \pi(2 \mathrm{R})^{2}$ $\mathrm{~F}^{\prime}=4 \sigma \pi \mathrm{R}^{2}$ $\mathrm{~F}^{\prime}=4 \mathrm{~F}$
AP EAMCET (18.09.2020) Shift-II
Mechanical Properties of Solids
140860
The diagram shows stress versus strain curve for the materials $A$ and $B$. from the curves, we can infer that
1 A is brittle but $\mathrm{B}$ is ductile
2 A is ductile but $\mathrm{B}$ is brittle
3 Both A and B are ductile
4 Both A and B are brittle
Explanation:
B Curve A is ductile and curve B is Brittle.
NEET odisha - 2019
Mechanical Properties of Solids
140861
There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be
1 $1: 1$
2 $2: 1$
3 $1: 2$
4 $4: 1$
Explanation:
D Given, $\mathrm{L}_{1}=\mathrm{L}, \mathrm{d}_{1}=\mathrm{d}$ Then, $\delta \mathrm{L}_{1}=\frac{\mathrm{PL}}{\mathrm{AE}}$ $\mathrm{L}_{2}=\mathrm{L}, \mathrm{d}_{2}=2 \mathrm{~d}_{1}=2 \mathrm{~d}$ $\delta \mathrm{L}_{2}=\frac{\mathrm{PL}_{2}}{\mathrm{AE}}$ $\mathrm{P}, \mathrm{L}$, are same for both wire $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{\frac{\pi}{4} \mathrm{~d}_{2}^{2}}{\frac{\pi}{4} \mathrm{~d}_{1}^{2}}$ $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{(2 \mathrm{~d})^{2}}{(\mathrm{~d})^{2}}=\frac{4}{1}$ $\delta \mathrm{L}_{1}: \delta \mathrm{L}_{2}=4: 1$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Mechanical Properties of Solids
140855
A wire of length $1 \mathrm{~m}$ and radius $2 \mathrm{~mm}$ is vertically clamped. The lower end is twisted through an angle of $45^{\circ}$. The angle of shear is
1 $0.09^{\circ}$
2 $0.9^{\circ}$
3 $9^{\circ}$
4 $90^{\circ}$
Explanation:
A Given, Radius $(\mathrm{r})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$, Twisted angle $(\theta)=45^{\circ}$ $\text { Length }(l)=1 \mathrm{~m}$ According to relation, $\mathrm{r} \theta=l \phi$ Angle of shear $(\phi)=\frac{\mathrm{r} \theta}{l}=\frac{2 \times 10^{-3} \times 45}{1}$ $\phi=0.09^{\circ}$
AP EAMCET (23.09.2020) Shift-I
Mechanical Properties of Solids
140857
A material has Poisson's ratio 0.50. If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is
1 0.6
2 0.4
3 0.2
4 0
Explanation:
D Given, Poisson ratio $(\sigma)=0.5$ Longitudinal strain $\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, Poisson ratio $=-\frac{\text { Lateral change }(\text { strain })}{\text { Longitudnal change }(\text { strain })}$ $0.5=\frac{-\frac{\mathrm{dr}}{\mathrm{r}}}{2 \times 10^{-3}}$ $\frac{\mathrm{dr}}{\mathrm{r}}=-1 \times 10^{-3}$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{~L}$ Taking $\log$ of both side and differentiating $\frac{\mathrm{dV}}{\mathrm{V}} =\frac{2 \mathrm{dr}}{\mathrm{r}}+\frac{\mathrm{dL}}{\mathrm{L}}$ $=\left(-2 \times 1 \times 10^{-3}\right)+\left(2 \times 10^{-3}\right)=0$ $\%$ Change in volume $(\Delta \mathrm{V})=\frac{\mathrm{dV}}{\mathrm{V}} \times 100=0 \times 100=0 \%$
AP EAMCET (22.09.2020) Shift-I
Mechanical Properties of Solids
140859
$\quad$ A force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2 R$ will be
1 $\frac{F}{2}$
2 $2 F$
3 $4 F$
4 $\frac{F}{4}$
Explanation:
C We know, Breaking stress $=\frac{\text { Breaking force }}{\text { Area }}$ Breaking force $=$ breaking stress $\times$ Area $\mathrm{F}=\sigma \times \pi \mathrm{R}^{2}$ Given, $\mathrm{R}$ become 2 times, then breaking force, $\mathrm{F}^{\prime}=\sigma \times \pi(2 \mathrm{R})^{2}$ $\mathrm{~F}^{\prime}=4 \sigma \pi \mathrm{R}^{2}$ $\mathrm{~F}^{\prime}=4 \mathrm{~F}$
AP EAMCET (18.09.2020) Shift-II
Mechanical Properties of Solids
140860
The diagram shows stress versus strain curve for the materials $A$ and $B$. from the curves, we can infer that
1 A is brittle but $\mathrm{B}$ is ductile
2 A is ductile but $\mathrm{B}$ is brittle
3 Both A and B are ductile
4 Both A and B are brittle
Explanation:
B Curve A is ductile and curve B is Brittle.
NEET odisha - 2019
Mechanical Properties of Solids
140861
There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be
1 $1: 1$
2 $2: 1$
3 $1: 2$
4 $4: 1$
Explanation:
D Given, $\mathrm{L}_{1}=\mathrm{L}, \mathrm{d}_{1}=\mathrm{d}$ Then, $\delta \mathrm{L}_{1}=\frac{\mathrm{PL}}{\mathrm{AE}}$ $\mathrm{L}_{2}=\mathrm{L}, \mathrm{d}_{2}=2 \mathrm{~d}_{1}=2 \mathrm{~d}$ $\delta \mathrm{L}_{2}=\frac{\mathrm{PL}_{2}}{\mathrm{AE}}$ $\mathrm{P}, \mathrm{L}$, are same for both wire $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{\frac{\pi}{4} \mathrm{~d}_{2}^{2}}{\frac{\pi}{4} \mathrm{~d}_{1}^{2}}$ $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{(2 \mathrm{~d})^{2}}{(\mathrm{~d})^{2}}=\frac{4}{1}$ $\delta \mathrm{L}_{1}: \delta \mathrm{L}_{2}=4: 1$
140855
A wire of length $1 \mathrm{~m}$ and radius $2 \mathrm{~mm}$ is vertically clamped. The lower end is twisted through an angle of $45^{\circ}$. The angle of shear is
1 $0.09^{\circ}$
2 $0.9^{\circ}$
3 $9^{\circ}$
4 $90^{\circ}$
Explanation:
A Given, Radius $(\mathrm{r})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$, Twisted angle $(\theta)=45^{\circ}$ $\text { Length }(l)=1 \mathrm{~m}$ According to relation, $\mathrm{r} \theta=l \phi$ Angle of shear $(\phi)=\frac{\mathrm{r} \theta}{l}=\frac{2 \times 10^{-3} \times 45}{1}$ $\phi=0.09^{\circ}$
AP EAMCET (23.09.2020) Shift-I
Mechanical Properties of Solids
140857
A material has Poisson's ratio 0.50. If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is
1 0.6
2 0.4
3 0.2
4 0
Explanation:
D Given, Poisson ratio $(\sigma)=0.5$ Longitudinal strain $\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, Poisson ratio $=-\frac{\text { Lateral change }(\text { strain })}{\text { Longitudnal change }(\text { strain })}$ $0.5=\frac{-\frac{\mathrm{dr}}{\mathrm{r}}}{2 \times 10^{-3}}$ $\frac{\mathrm{dr}}{\mathrm{r}}=-1 \times 10^{-3}$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{~L}$ Taking $\log$ of both side and differentiating $\frac{\mathrm{dV}}{\mathrm{V}} =\frac{2 \mathrm{dr}}{\mathrm{r}}+\frac{\mathrm{dL}}{\mathrm{L}}$ $=\left(-2 \times 1 \times 10^{-3}\right)+\left(2 \times 10^{-3}\right)=0$ $\%$ Change in volume $(\Delta \mathrm{V})=\frac{\mathrm{dV}}{\mathrm{V}} \times 100=0 \times 100=0 \%$
AP EAMCET (22.09.2020) Shift-I
Mechanical Properties of Solids
140859
$\quad$ A force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2 R$ will be
1 $\frac{F}{2}$
2 $2 F$
3 $4 F$
4 $\frac{F}{4}$
Explanation:
C We know, Breaking stress $=\frac{\text { Breaking force }}{\text { Area }}$ Breaking force $=$ breaking stress $\times$ Area $\mathrm{F}=\sigma \times \pi \mathrm{R}^{2}$ Given, $\mathrm{R}$ become 2 times, then breaking force, $\mathrm{F}^{\prime}=\sigma \times \pi(2 \mathrm{R})^{2}$ $\mathrm{~F}^{\prime}=4 \sigma \pi \mathrm{R}^{2}$ $\mathrm{~F}^{\prime}=4 \mathrm{~F}$
AP EAMCET (18.09.2020) Shift-II
Mechanical Properties of Solids
140860
The diagram shows stress versus strain curve for the materials $A$ and $B$. from the curves, we can infer that
1 A is brittle but $\mathrm{B}$ is ductile
2 A is ductile but $\mathrm{B}$ is brittle
3 Both A and B are ductile
4 Both A and B are brittle
Explanation:
B Curve A is ductile and curve B is Brittle.
NEET odisha - 2019
Mechanical Properties of Solids
140861
There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be
1 $1: 1$
2 $2: 1$
3 $1: 2$
4 $4: 1$
Explanation:
D Given, $\mathrm{L}_{1}=\mathrm{L}, \mathrm{d}_{1}=\mathrm{d}$ Then, $\delta \mathrm{L}_{1}=\frac{\mathrm{PL}}{\mathrm{AE}}$ $\mathrm{L}_{2}=\mathrm{L}, \mathrm{d}_{2}=2 \mathrm{~d}_{1}=2 \mathrm{~d}$ $\delta \mathrm{L}_{2}=\frac{\mathrm{PL}_{2}}{\mathrm{AE}}$ $\mathrm{P}, \mathrm{L}$, are same for both wire $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{\frac{\pi}{4} \mathrm{~d}_{2}^{2}}{\frac{\pi}{4} \mathrm{~d}_{1}^{2}}$ $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{(2 \mathrm{~d})^{2}}{(\mathrm{~d})^{2}}=\frac{4}{1}$ $\delta \mathrm{L}_{1}: \delta \mathrm{L}_{2}=4: 1$
140855
A wire of length $1 \mathrm{~m}$ and radius $2 \mathrm{~mm}$ is vertically clamped. The lower end is twisted through an angle of $45^{\circ}$. The angle of shear is
1 $0.09^{\circ}$
2 $0.9^{\circ}$
3 $9^{\circ}$
4 $90^{\circ}$
Explanation:
A Given, Radius $(\mathrm{r})=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$, Twisted angle $(\theta)=45^{\circ}$ $\text { Length }(l)=1 \mathrm{~m}$ According to relation, $\mathrm{r} \theta=l \phi$ Angle of shear $(\phi)=\frac{\mathrm{r} \theta}{l}=\frac{2 \times 10^{-3} \times 45}{1}$ $\phi=0.09^{\circ}$
AP EAMCET (23.09.2020) Shift-I
Mechanical Properties of Solids
140857
A material has Poisson's ratio 0.50. If a uniform rod made of this material suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is
1 0.6
2 0.4
3 0.2
4 0
Explanation:
D Given, Poisson ratio $(\sigma)=0.5$ Longitudinal strain $\left(\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)=2 \times 10^{-3}$ We know, Poisson ratio $=-\frac{\text { Lateral change }(\text { strain })}{\text { Longitudnal change }(\text { strain })}$ $0.5=\frac{-\frac{\mathrm{dr}}{\mathrm{r}}}{2 \times 10^{-3}}$ $\frac{\mathrm{dr}}{\mathrm{r}}=-1 \times 10^{-3}$ Volume $(\mathrm{V})=\pi \mathrm{r}^{2} \mathrm{~L}$ Taking $\log$ of both side and differentiating $\frac{\mathrm{dV}}{\mathrm{V}} =\frac{2 \mathrm{dr}}{\mathrm{r}}+\frac{\mathrm{dL}}{\mathrm{L}}$ $=\left(-2 \times 1 \times 10^{-3}\right)+\left(2 \times 10^{-3}\right)=0$ $\%$ Change in volume $(\Delta \mathrm{V})=\frac{\mathrm{dV}}{\mathrm{V}} \times 100=0 \times 100=0 \%$
AP EAMCET (22.09.2020) Shift-I
Mechanical Properties of Solids
140859
$\quad$ A force $F$ is needed to break a copper wire having radius $R$. The force needed to break a copper wire of radius $2 R$ will be
1 $\frac{F}{2}$
2 $2 F$
3 $4 F$
4 $\frac{F}{4}$
Explanation:
C We know, Breaking stress $=\frac{\text { Breaking force }}{\text { Area }}$ Breaking force $=$ breaking stress $\times$ Area $\mathrm{F}=\sigma \times \pi \mathrm{R}^{2}$ Given, $\mathrm{R}$ become 2 times, then breaking force, $\mathrm{F}^{\prime}=\sigma \times \pi(2 \mathrm{R})^{2}$ $\mathrm{~F}^{\prime}=4 \sigma \pi \mathrm{R}^{2}$ $\mathrm{~F}^{\prime}=4 \mathrm{~F}$
AP EAMCET (18.09.2020) Shift-II
Mechanical Properties of Solids
140860
The diagram shows stress versus strain curve for the materials $A$ and $B$. from the curves, we can infer that
1 A is brittle but $\mathrm{B}$ is ductile
2 A is ductile but $\mathrm{B}$ is brittle
3 Both A and B are ductile
4 Both A and B are brittle
Explanation:
B Curve A is ductile and curve B is Brittle.
NEET odisha - 2019
Mechanical Properties of Solids
140861
There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be
1 $1: 1$
2 $2: 1$
3 $1: 2$
4 $4: 1$
Explanation:
D Given, $\mathrm{L}_{1}=\mathrm{L}, \mathrm{d}_{1}=\mathrm{d}$ Then, $\delta \mathrm{L}_{1}=\frac{\mathrm{PL}}{\mathrm{AE}}$ $\mathrm{L}_{2}=\mathrm{L}, \mathrm{d}_{2}=2 \mathrm{~d}_{1}=2 \mathrm{~d}$ $\delta \mathrm{L}_{2}=\frac{\mathrm{PL}_{2}}{\mathrm{AE}}$ $\mathrm{P}, \mathrm{L}$, are same for both wire $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{\frac{\pi}{4} \mathrm{~d}_{2}^{2}}{\frac{\pi}{4} \mathrm{~d}_{1}^{2}}$ $\frac{\delta \mathrm{L}_{1}}{\delta \mathrm{L}_{2}}=\frac{(2 \mathrm{~d})^{2}}{(\mathrm{~d})^{2}}=\frac{4}{1}$ $\delta \mathrm{L}_{1}: \delta \mathrm{L}_{2}=4: 1$
