A steel wire is 1 m long and 1mm^2 in area of cross-section. If it takes 200N to stretch its wire by 1 mm, how much force will be required to stretch a wire of the same material as well as diameter from its normal length of 10 m to a length of 1002 cm?
10 thanks= i.b
Answers
The force required will be 400 N
Explanation :
In the first case
Length L = 1 m
Area A = 1 mm² = 10⁻⁶ m²
Force F = 200 N
Change in length , L' = 1 mm = 10⁻³ m
we know that young's modulus is given by
⇒Y = FL/AL'
⇒ 200 x 1/(10⁻⁶ x 10⁻³)
⇒ 2 x 10¹¹
Similarly in the second case,
L = 10 m
Area A = 1 mm² = 10⁻⁶ m²
New length = 1002 cm = 10.02 m
Change in length , L' = 10.02 - 10 = 0.02 m = 2 x 10⁻² m
Hence,
⇒Y = FL/AL'
⇒ F = YAL'/L
⇒ (2 x 10¹¹ x 10⁻⁶ x 2 x 10⁻²)/10
⇒ 400 N
Hence the force required will be 400 N
The force required will be 400 N
Explanation :
In the first case
Length L = 1 m
Area A = 1 mm² = 10⁻⁶ m²
Force F = 200 N
Change in length , L' = 1 mm = 10⁻³ m
we know that young's modulus is given by
Y = FL/AL'
= 200 x 1/(10⁻⁶ x 10⁻³)
= 2 x 10¹¹
Similarly in the second case,
L = 10 m
Area A = 1 mm² = 10⁻⁶ m²
New length = 1002 cm = 10.02 m
Change in length , L' = 10.02 - 10 = 0.02 m = 2 x 10⁻² m
Hence,
Y = FL/AL'
=> F = YAL'/L
= (2 x 10¹¹ x 10⁻⁶ x 2 x 10⁻²)/10
= 400 N
Hence the force required will be 400 N