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

ㅤ$\;\;\underline{\textbf{\textsf{ Given :-}}}$

•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

ㅤ$\;\;\underline{\textbf{\textsf{ To Find :-}}}$

•The change in length of the string.

ㅤ$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Let the change in length of the string be 'l'

We know that :-

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

Substitute the given values

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

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

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

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

$\sf{\dashrightarrow A=0.0000785}$

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

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Again, we know that :-

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

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

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

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

$\sf{\dashrightarrow 5l\times 10^3\times392.5=1}$

$\sf{ \dashrightarrow 5l \times 392500=1}$

$\sf{\dashrightarrow l=5.09 \times 10^{-7}\ m}$   (approx.)

ㅤ$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ The change in length of the string is \textbf{ 5.09×10⁷ }}}.$

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Answer 2

Answer:

stepbby step 5×10^-7.........