Question

A spring has a diameter of 1cm and the original length of 2cm, the string is pulled by a force of 200N, determine the change in length of the string. Young's modulus of the string =5×10^9N/m^2

Answer 1

✯ The change in length of the string = $\sf{1.019\times\:10^{-5}}$ m. ✯

Explanation:

Given:

• Diameter of the string= 1 cm
• Original length(L) = 2 cm
• Force(F) = 200 N
• Young's modulus of the string(Y) = 5 × 10 ^9 N/m²

To find:

• The change in length of the string.

Solution:

• Diameter = 1 cm = 0.01 m
• Original length = 2 cm = 0.02 m

Let the the change in length of the string be l m.

• Radius of the string (r) = 0.01/2 = 0.005 m

Then,

Area of cross section (A) of the string will be,

= πr²

=( 3.14 × 0.005 × 0.005 ) m²

= 0.0000785 m²

= $\sf{7.85\times\:10^{-5}\:m^2}$

We know that,

${\boxed{\sf{Y=\dfrac{FL}{Al}}}}$

• [Put values]

$\implies\sf{5\times\:10^9=\dfrac{200\times\:0.02}{7.85\times\:10^{-5}\times\:l}}$

$\implies\sf{5\times\:10^9=\dfrac{4}{7.85\times\:10^{-5}\times\:l}}$

$\implies\sf{5\times\:10^9\times\:7.85\times\:10^{-5}\times\:l=4}$

$\implies\sf{392.5\times\:10^3\times\:l=4}$

$\implies\sf{l=\dfrac{4}{392.5\times\:10^3}}$

$\implies\sf{l=1.019\times\:10^{-5}}$

Therefore, the change in length of the string is $\sf{1.019\times\:10^{-5}}$ m.

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Answer 2

$\underline{\underline{\sf{\color{lime}{GIVEN :-}}}}$

For a spring,

• Diameter, d = 1 cm = 0.01 m
• Original length, L = 2 cm = 0.02 cm
• Force applied, F = 200 N
• Young's modulus of the string, Y = 5 × 10⁹ N/m²

$\underline{\underline{\sf{\color{lime}{TO\ DETERMINE :-}}}}$

The change in length of the string.

$\underline{\underline{\sf{\color{lime}{SOLUTION :-}}}}$

Let the change in length of the string be l.

◘ Area of cross section of the string →

$\sf{A=\pi r ^2}$

$\sf{\longrightarrow A=\dfrac{22}{7} \times \bigg( \dfrac{d}{2} \bigg) ^2}$

$\sf{\longrightarrow A=\dfrac{22}{7}\times\bigg( \dfrac{0.01}{2} \bigg)^2}$

$\sf{\longrightarrow A=\dfrac{22}{7}\times (0.005)^2 }$

$\sf{\longrightarrow A=\dfrac{22}{7} \times 0.000025}$

$\sf{\longrightarrow A=0.0000785}$

$\sf{\longrightarrow A=785\times 10^{-6}\ m^2}$

__________________________

As we know that :

$\boxed{\bf{Y=\dfrac{FL}{Al} }}$

Substituting the values and calculating further :

$\sf{\longrightarrow 5\times 10^9=\dfrac{200\times0.02}{785\times 10^{-6}\times l} }$

$\sf{\longrightarrow 5l \times 785 \times 10^9 \times 10^{-6}=200\times 0.02}$

$\sf{\longrightarrow 5l\times785\times 10^{9-6}=4}$

$\sf{\longrightarrow 5l\times 10^3\times196.25=1}$

$\sf{\longrightarrow 5l \times 196250=1}$

$\sf{\longrightarrow 5l=0.0000050}$

$\sf{\longrightarrow l=0.000001}$

$\sf{\longrightarrow l=1 \times 10^{-5}\ m}$   (approx.)

Therefore, the change in length of the string is 1 × 10⁻⁵ m (approx) .