Q.33. [Figure: Extension vs Load graph] Figure represents the extension of a wire of length 1 meter

Q: What is the correct answer for this exam Physics Q33?

The correct answer is C. Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} = \frac{Fl}{A\Delta l}$. From the graph, we can pick a point, for example, when $W = 60N$, $\Delta l = 6 \times 10^{-4} m$. Given $l = 1m$, $A = 10^{-6} m^2$. $Y = \frac{60 N \times 1 m}{10^{-6} m^2 \times 6 \times 10^{-4} m} = \frac{60}{6 \times 10^{-10}} = 10 \times 10^{10} = 1.0 \times 10^{11} N/m^2$. Alternativel

Q: Q.33. [Figure: Extension vs Load graph] Figure represents the extension of a wire of length 1 meter, suspended from the ceiling of the room at one end with load W connected to the other end. If the…

The correct answer is C. Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta l/l} = \frac{Fl}{A\Delta l}$. From the graph, we can pick a point, for example, when $W = 60N$, $\Delta l = 6 \times 10^{-4} m$. Given $l = 1m$, $A = 10^{-6} m^2$. $Y = \frac{60 N \times 1 m}{10^{-6} m^2 \times 6 \times 10^{-4} m} = \frac{60}{6 \times 10^{-10}} = 10 \times 10^{10} = 1.0 \times 10^{11} N/m^2$. Alternativel

Q: What are the options for this this exam Physics question?

A. $1.0 \times 10^{11}$ · B. $2.0 \times 10^{10}$ · C. $1.0 \times 10^{10}$ · D. $2.0 \times 10^{11}$

Question

Q.33. [Figure: Extension vs Load graph] Figure represents the extension  of a wire of length 1 meter, suspended from the ceiling of the room at one end with load W connected to the other end. If the cross-sectional area of the wire is  then the Young's modulus of the wire is _______ N/m.

FirstInTest · Accepted Answer

Young's modulus $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l} = \frac{Fl}{A\Delta l}$. From the graph, we can pick a point, for example, when $W = 60N$, $\Delta l = 6 	imes 10^{-4} m$. Given $l = 1m$, $A = 10^{-6} m^2$. $Y = \frac{60 N 	imes 1 m}{10^{-6} m^2 	imes 6 	imes 10^{-4} m} = \frac{60}{6 	imes 10^{-10}} = 10 	imes 10^{10} = 1.0 	imes 10^{11} N/m^2$. Alternatively, the slope of the $\Delta l$ vs $W$ graph is $\frac{\Delta l}{W}$. From the graph, $\frac{\Delta l}{W} = \frac{6 	imes 10^{-4} m}{60 N} = 10^{-5} m/N$. So, $Y = \frac{l}{A 	imes (\Delta l/W)} = \frac{1 m}{10^{-6} m^2 	imes 10^{-5} m/N} = \frac{1}{10^{-11}} = 10^{11} N/m^2$. This matches option (3) $1.0 	imes 10^{11}$.

Answer

$1.0 	imes 10^{11}$

Answer

$2.0 	imes 10^{10}$

Answer

$1.0 	imes 10^{10}$

Answer

$2.0 	imes 10^{11}$

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