Question:
Two wires \(A\) and \(B\) of same cross-section are connected end to end. When same tension is created in both wires, the elongation in \(B\) wire is twice the elongation in \(A\) wire. If \(L_A\) and \(L_B\) are the initial lengths of the wires \(A\) and \(B\) respectively, then \((\text{Young's modulus of material of wire }A=2\times 10^{11}\,\text{N m}^{-2}\text{ and Young's modulus of material of wire }B=1.1\times 10^{11}\,\text{N m}^{-2})\)
Two wires \(A\) and \(B\) of same cross-section are connected end to end. When same tension is created in both wires, the elongation in \(B\) wire is twice the elongation in \(A\) wire. If \(L_A\) and \(L_B\) are the initial lengths of the wires \(A\) and \(B\) respectively, then \((\text{Young's modulus of material of wire }A=2\times 10^{11}\,\text{N m}^{-2}\text{ and Young's modulus of material of wire }B=1.1\times 10^{11}\,\text{N m}^{-2})\)
Show Hint
For wires under the same tension and having the same cross-sectional area,
\[
\Delta L\propto \frac{L}{Y}
\]
So, compare elongations using the ratio of length to Young's modulus.
Updated On: Jun 22, 2026
- \(\dfrac{L_A}{L_B}=\dfrac{10}{11}\)
- \(\dfrac{L_A}{L_B}=\dfrac{4}{5}\)
- \(\dfrac{L_A}{L_B}=\dfrac{9}{11}\)
- \(\dfrac{L_A}{L_B}=\dfrac{3}{7}\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Use the formula for elongation of a wire.
The elongation of a wire is given by \[ \Delta L=\frac{FL}{AY} \] where, \[ F=\text{tension in the wire} \] \[ L=\text{initial length of the wire} \] \[ A=\text{area of cross-section} \] \[ Y=\text{Young's modulus} \]
Step 2: Write elongations for wires \(A\) and \(B\).
For wire \(A\), \[ \Delta L_A=\frac{FL_A}{AY_A} \] For wire \(B\), \[ \Delta L_B=\frac{FL_B}{AY_B} \] Since both wires have the same cross-section and the same tension, \(F\) and \(A\) are common.
Step 3: Use the given relation between elongations.
It is given that elongation in wire \(B\) is twice the elongation in wire \(A\).
Therefore, \[ \Delta L_B=2\Delta L_A \] Substituting the expressions, \[ \frac{FL_B}{AY_B}=2\left(\frac{FL_A}{AY_A}\right) \] Cancelling \(F\) and \(A\), \[ \frac{L_B}{Y_B}=2\frac{L_A}{Y_A} \] Rearranging, \[ \frac{L_A}{L_B}=\frac{Y_A}{2Y_B} \]
Step 4: Substitute the values of Young's moduli.
Given, \[ Y_A=2\times 10^{11}\,\text{N m}^{-2} \] \[ Y_B=1.1\times 10^{11}\,\text{N m}^{-2} \] Therefore, \[ \frac{L_A}{L_B} = \frac{2\times 10^{11}}{2(1.1\times 10^{11})} \] \[ \frac{L_A}{L_B} = \frac{2}{2.2} \] \[ \frac{L_A}{L_B} = \frac{20}{22} \] \[ \frac{L_A}{L_B} = \frac{10}{11} \]
Step 5: Final conclusion.
Hence, \[ \boxed{\frac{L_A}{L_B}=\frac{10}{11}} \]
The elongation of a wire is given by \[ \Delta L=\frac{FL}{AY} \] where, \[ F=\text{tension in the wire} \] \[ L=\text{initial length of the wire} \] \[ A=\text{area of cross-section} \] \[ Y=\text{Young's modulus} \]
Step 2: Write elongations for wires \(A\) and \(B\).
For wire \(A\), \[ \Delta L_A=\frac{FL_A}{AY_A} \] For wire \(B\), \[ \Delta L_B=\frac{FL_B}{AY_B} \] Since both wires have the same cross-section and the same tension, \(F\) and \(A\) are common.
Step 3: Use the given relation between elongations.
It is given that elongation in wire \(B\) is twice the elongation in wire \(A\).
Therefore, \[ \Delta L_B=2\Delta L_A \] Substituting the expressions, \[ \frac{FL_B}{AY_B}=2\left(\frac{FL_A}{AY_A}\right) \] Cancelling \(F\) and \(A\), \[ \frac{L_B}{Y_B}=2\frac{L_A}{Y_A} \] Rearranging, \[ \frac{L_A}{L_B}=\frac{Y_A}{2Y_B} \]
Step 4: Substitute the values of Young's moduli.
Given, \[ Y_A=2\times 10^{11}\,\text{N m}^{-2} \] \[ Y_B=1.1\times 10^{11}\,\text{N m}^{-2} \] Therefore, \[ \frac{L_A}{L_B} = \frac{2\times 10^{11}}{2(1.1\times 10^{11})} \] \[ \frac{L_A}{L_B} = \frac{2}{2.2} \] \[ \frac{L_A}{L_B} = \frac{20}{22} \] \[ \frac{L_A}{L_B} = \frac{10}{11} \]
Step 5: Final conclusion.
Hence, \[ \boxed{\frac{L_A}{L_B}=\frac{10}{11}} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Elasticity Questions
- A 4 kg stone is attached to a steel wire being whirled at a constant speed of \( 12 \) m/s in a horizontal circle. The wire is 4 m long with a diameter of 2 mm, and Young’s modulus is \( 2 \times 10^{11} \) Nm\(^2\). The strain in the wire is:
If Young’s modulus of elasticity is $Y = \dfrac{2mg l^2}{5b t e}$, where ‘g’ is the acceleration due to gravity, ‘m’ is the mass, ‘l’ is the length, ‘b’ is the breadth, ‘t’ is the thickness and ‘e’ is the elongation, then the value of $k$ is
- The bulk modulus \( B \) and compressibility \( k \) are related by:
- The sag developed in a bar of length $l$, breadth $b$, and thickness $d$ when subjected to a load $w$ at the nter is (Y -- Young's modulus of the material of the bar)
- A copper wire of length 2.4 m and an aluminum wire of length 0.7 m, both having diameter 2 mm, are connected end to end. When stretched by a load, the obtained elongation is found to be 0.6 mm. The applied load is
(Given: Young’s modulus of copper \( Y_C = 1.2 \times 10^{11} \, \text{Nm}^{-2} \), Young’s modulus of aluminum \( Y_A = 0.7 \times 10^{11} \, \text{Nm}^{-2} \))
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions