One end of a nylon rope of length 4.5 m and diameter 6 mmis fixed to a free limb. A monkey weighing 100 N jumps to catch the free end and stays there. Find the elongation of the rope (Given Young’s modulus of nylon=4.8×1011 N m-2 and Poisson’s ratio of nylon = 0.2.), what will be the change in the diameter of the rope?

Question

One end of a nylon rope of length 4.5 m and diameter 6 mmis fixed to a free limb. A monkey weighing 100 N jumps to catch the free end and stays there. Find the elongation of the rope (Given Young’s modulus of nylon=4.8×1011 N m-2 and Poisson’s ratio of nylon = 0.2.), what will be the change in the diameter of the rope? (A) 8.8×10-9 m (B) 7.4×10-9 m (C) 6.4×10-8 m (D) 5.6×10-9 m

Tardigrade Admin · Accepted Answer

Poissonâs ratio, σ =(Δ D/ D/Δ l /l) =(Δ D/D)⋅ (l/Δ l) ∴ Δ D= (σ D Δ l/l) =(0.2×6×10 -3×3.32×10-5/4.5) =8.8×10-9 m

$8.8 \times 1 0^{- 9} m$

$7.4 \times 1 0^{- 9} m$

$6.4 \times 1 0^{- 8} m$

$5.6 \times 1 0^{- 9} m$

Poisson’s ratio, $σ$ $= \frac{Δ D / D}{Δ l / l}$ $= \frac{Δ D}{D} \cdot$ $\frac{l}{Δ l}$
$∴$ $Δ D =$ $\frac{σ D Δ l}{l}$ $= \frac{0.2 \times 6 \times 1 0 ^{- 3} \times 3.32 \times 1 0 ^{- 5}}{4.5}$
$= 8.8 \times 1 0^{- 9} m$

8.8×10−9m

7.4×10−9m

6.4×10−8m

5.6×10−9m

Poisson’s ratio, σ =Δl/lΔD/D​ =DΔD​⋅ Δll​ ∴ ΔD= lσDΔl​ =4.50.2×6×10−3×3.32×10−5​ =8.8×10−9m

$8.8 \times 1 0^{- 9} m$

$7.4 \times 1 0^{- 9} m$

$6.4 \times 1 0^{- 8} m$

$5.6 \times 1 0^{- 9} m$

Poisson’s ratio, $σ$ $= \frac{Δ D / D}{Δ l / l}$ $= \frac{Δ D}{D} \cdot$ $\frac{l}{Δ l}$
$∴$ $Δ D =$ $\frac{σ D Δ l}{l}$ $= \frac{0.2 \times 6 \times 1 0 ^{- 3} \times 3.32 \times 1 0 ^{- 5}}{4.5}$
$= 8.8 \times 1 0^{- 9} m$