The speed of a transverse wave passing through a string of length 50 cm and mass 10 g is 60 ms -1. The area of cross-section of the wire is 2.0 mm 2 and its Young's modulus is 1.2 × 1011 Nm -2. The extension of the wire over its natural length due to its tension will be x × 10-5 m. The value of x is

Question

The speed of a transverse wave passing through a string of length 50 cm and mass 10 g is 60 ms -1. The area of cross-section of the wire is 2.0 mm 2 and its Young's modulus is 1.2 × 1011 Nm -2. The extension of the wire over its natural length due to its tension will be x × 10-5 m. The value of x is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

V w =√( T /μ) 60=√( T /10 × 10-3) × 0.5 T =((60)2 × 10-2/0.5)=72 N Δ ℓ=( F ℓ/ AY )=(72 × 0.5/2 × 10-6 × 1.2 × 1011) =(72 × 5/24) × 10-5=15 × 10-5

Q. The speed of a transverse wave passing through a string of length 50cm and mass 10g is 60ms−1. The area of cross-section of the wire is 2.0mm2 and its Young's modulus is 1.2×1011Nm−2. The extension of the wire over its natural length due to its tension will be x×10−5m. The value of x is

Vw​=μT​​ 60=10×10−3T​×0.5​ T=0.5(60)2×10−2​=72N Δℓ=AYFℓ​=2×10−6×1.2×101172×0.5​ =2472×5​×10−5=15×10−5

$V_{w} = \frac{T}{μ}$
$60 = \frac{T}{10 \times 1 0 ^{- 3}} \times 0.5$
$T = \frac{( 60 ) ^{2} \times 1 0 ^{- 2}}{0.5} = 72 N$
$Δ ℓ = \frac{F ℓ}{A Y} = \frac{72 \times 0.5}{2 \times 1 0 ^{- 6} \times 1.2 \times 1 0 ^{11}}$
$= \frac{72 \times 5}{24} \times 1 0^{- 5} = 15 \times 1 0^{- 5}$