Question

solve this...class 11

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Answer 1

Given:

Radius of wire = $\frac{2.5}{2}$ = 1.25 mm = 1.25 x $10^{-3}$ m

Force = 100 N

Youngs modulus = 12.5 x $10^{11}$ dyne/sq.cm = 12.5 x $10^6$ N/sq.cm

= 12.5 x $10^{10$ N/sq.m

To Find:

Percentage increase in length.

Or, it can also be written as (ΔL x 100)/L , where ΔL is the chande/increase in length and L is initial length.

Process:

By formula, we know that:

Y= (FxL)/(AxΔL)

Here :

F is the force

L is the initial length

ΔL is the increase in length

A is the area of the wire

Area = $\pi r^2 = 3.14$ x $1.25$ x $1.25$ x $10^{-6}$ = $4.94$ x $10^{-6}$ $m^2$

By solving the youngs modulus equation, we get:

ΔL/L = F/ (A x Y)

Substituting the values,we get:

ΔL/L = 1/617.5

Percentage change will be  (ΔL x 100)/L = 100/617.5

= 0.162 %

Hence, percentage increase will be 0.162%

Answer 2

★ Concept :-

Here the concept of Young's Modulus has been used. We see that we need to calculate the percentage increase in the length of the wire. So firstly we can convert all the given units into the standard units. Then we by applying the formula of Young's Modulus we can find the required quantities. Then we can change them into percentage.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{Y\;=\;\dfrac{F/A}{\Delta L/L}}}}$

$\;\boxed{\sf{\pink{\%\;increase\;in\;Length\;=\;\dfrac{\Delta L}{L}\:\times\:100}}}$

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★ Solution :-

Given,

» Young's Modulus of Wire = Y = 12.5 × 10¹¹ dyne/cm²

» Diameter of the wire = 2.5 mm

» Radius of the wire = r = 2.5/2 = 1.25 mm

» Force on the wire = F = 100 Kg Weight

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~ For the values in standard units ::

For conversion of units we must understand basic conversion rules .

• 10000 cm² = 1 m²
• 1 mm = 10¯³ m

• Young's Modulus = N/m²

Here we have to convert the given value in the required form. So,

• Y = 12.5 × 10¹⁰ N/m²

• r = 1.25 mm = 1.25 × 10¯³ m

• Force = Newton

Here we have 100 Kg Weight. So we have to multiply by gravitational force.

• F = 100 × 9.8 = 980 N

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~ For the Area of the wire (A) ::

According to area of circle, we know that

→ Area of circle = πr²

→ Area of circular wire = π(1.25 × 10¯³)² m²

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~ For the ratio of length of the wire ::

• Let the increased length of the wire be ∆L

• Let the original length of the wire be L

By the formula of Young's Modulus, we know that

$\;\sf{\rightarrow\;\;Y\;=\;\dfrac{F/A}{\Delta L/L}}$

On replacing the value, we get

$\;\sf{\rightarrow\;\;\dfrac{\Delta L}{L}\;=\;\bf{\dfrac{F/A}{Y}}}$

$\;\sf{\rightarrow\;\;\dfrac{\Delta L}{L}\;=\;\bf{\dfrac{F}{YA}}}$

By applying values, we get

$\;\sf{\rightarrow\;\;\dfrac{\Delta L}{L}\;=\;\bf{\dfrac{980}{12.5\:\times\:10^{10}\:\times\:\pi\:\times\:(1.25\:\times\:10^{3})^{2}}}}$

By applying the value of π as 3.14 we get,

$\;\sf{\rightarrow\;\;\dfrac{\Delta L}{L}\;=\;\bf{\dfrac{980}{12.5\:\times\:10^{10}\:\times\:3.14\:\times\:(1.25\:\times\:10^{3})^{2}}}}$

$\;\bf{\rightarrow\;\;\red{\dfrac{\Delta L}{L}\;=\;\bf{0.0016}}}$

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~ For the % increase in length of wire ::

We know that,

$\;\bf{\Longrightarrow\;\;\%\;increase\;in\;Length\;=\;\dfrac{\Delta L}{L}\:\times\:100}$

By applying values, we get

$\;\bf{\Longrightarrow\;\;\%\;increase\;in\;Length\;=\;0.0016\:\times\:100}$

$\;\bf{\Longrightarrow\;\;\green{\%\;increase\;in\;Length\;=\;0.16\;\;\%}}$

This is the required answer.

$\;\underline{\boxed{\tt{Required\;\:Increase\;\:in\;\:\%\;=\;\bf{\purple{0.16\;\;\%}}}}}$

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★ More to know :-

$\;\tt{\leadsto\;\;Y\;=\;\dfrac{Stress}{Strain}}$

$\;\tt{\leadsto\;\;Stress\;=\;\dfrac{Force}{Area}}$

$\;\tt{\leadsto\;\;Stress\;=\;\dfrac{Change\;in\; Dimension}{Original\; Dimension}}$