The area of cross-section of a wire of length 1.1 m is 1mm2 . It is leaded with 1 kg of Youngs modulus of copper is 1.1× 1011Nm-2, then the increase in length will be (if g=10ms-2 )

Question

The area of cross-section of a wire of length 1.1 m is 1mm2 . It is leaded with 1 kg of Youngs modulus of copper is 1.1× 1011Nm-2, then the increase in length will be (if g=10ms-2 ) (A) 0.01 mm (B) 0.075 mm (C) 0.1 mm (D) 0.15mm

Tardigrade Admin · Accepted Answer

Increase in length l=(mgL/AY) =(1× 10× 1.1/1.1× 1011× 10-6)m=0.1 mm ?(i)

Q. The area of cross-section of a wire of length 1.1 m is $1 m m^{2}$ . It is leaded with 1 kg of Youngs modulus of copper is $1.1 \times 10^{11} N m^{- 2},$ then the increase in length will be (if $g = 10 m s^{- 2}$ )

0.01 mm

0.075 mm

0.1 mm

0.15mm

Increase in length $l = \frac{m gL}{A Y}$ $= \frac{1 \times 10 \times 1.1}{1.1 \times 10 ^{11} \times 10 ^{- 6}} m = 0.1 mm$ ?(i)

Q. The area of cross-section of a wire of length 1.1 m is 1mm2 . It is leaded with 1 kg of Youngs modulus of copper is 1.1×1011Nm−2, then the increase in length will be (if g=10ms−2 )