Question

an elongation of 0.1% in a wire of cross section 10^-6 M2; causes a tension of 100N Y for the wire is​

Answer 1

Given :

length of the wire = ∆l/l × 100 = 0.1

Area of cross section = 10¯⁶ m²

Tension = 100 N.

To Find :

The Young's modulus of the wire

Solution :

We know that,

$\boxed{\bf Y = \dfrac{F}{A} \times \dfrac{ l}{e}}$

Here,

• Y denotes Young's modulus
• F denotes force
• A denotes area of cross section
• l denotes length
• e denotes elasticity

by substituting all the given values in the formula,

$\dashrightarrow\sf Y = \dfrac{F}{A} \times \dfrac{ l}{e}$

$\dashrightarrow\sf Y = \dfrac{100}{ {10}^{ - 6} } \times \dfrac{1}{ {10}^{ - 3} }$

$\dashrightarrow\sf Y = \dfrac{100}{ {10}^{ - 9}}$

$\dashrightarrow\sf Y = 100 \times {10}^{9}$

$\dashrightarrow\sf Y = {10}^{11} \: Nm^{-2}$

Thus, the Young's modulus of the wire is 10¹¹ N/m².

Answer 2

Answer:

The Correct Answer Is 10 to the power 11

Explanation:

Hope It was Helpful