Two persons pull a rope towards themselves. Each person exerts a force of $100 N$ on the rope. The Young’s modulus of the material of the rope, if it extends in length by $1 c m$ is $10^{n}$ . Then value of $n$ is

Original length of the rope $= 2 m$ and the area of cross-section $2 {c m}^{2}$ .

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Solution

$F = 100 N, Δ l = 1 c m = 10^{- 2} m, l = 2 m, A = 2 \times 10^{- 4} m^{2}$

$Y = \frac{F l}{A Δ l}$
$Y = \frac{100 \times 2}{2 \times 10^{- 4} \times 10^{- 2}}$
$Y = 10^{8}$
So, value of $n = 8$

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Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find the Young modulus of the material of the rope if it extends in length by 1 cm. Original length of the rope = 2m and the area of cross section = $2 c m^{2} .$

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Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. The Young’s modulus of the material of the rope, if it extends in length by 1 cm is 10n. Then value of n is Original length of the rope =2 m and the area of cross-section 2 cm2.

F=100 N,Δl=1 cm=10−2 m,l=2 m,A=2×10−4 m2 Y=FlAΔl Y=100×22×10−4×10−2 Y=108 So, value of n=8

Two persons pull a rope towards themselves. Each person exerts a force of $100 N$ on the rope. The Young’s modulus of the material of the rope, if it extends in length by $1 c m$ is $10^{n}$ . Then value of $n$ is

Original length of the rope $= 2 m$ and the area of cross-section $2 {c m}^{2}$ .

$F = 100 N, Δ l = 1 c m = 10^{- 2} m, l = 2 m, A = 2 \times 10^{- 4} m^{2}$

$Y = \frac{F l}{A Δ l}$
$Y = \frac{100 \times 2}{2 \times 10^{- 4} \times 10^{- 2}}$
$Y = 10^{8}$
So, value of $n = 8$