A uniform heavy rod of weight 10 kg ms -2, crosssectional area 100 cm 2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 1011 Nm -2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.

Question

A uniform heavy rod of weight 10 kg ms -2, crosssectional area 100 cm 2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 1011 Nm -2. Neglecting the lateral contraction, find the elongation of rod due to its own weight. (A) 2 × 10-9 m (B) 5 × 10-8 m (C) 4 × 10-8 m (D) 5 × 10-10 m

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4e62086ff5544c00b25d713c1cf7cdfe-.png" /> We know, Δ ℓ=(W L/2 A Y) Δ ℓ=(10 × 1/2 × 5) × 100 × 10-4 × 2 × 1011 Δ ℓ=(1/2) × 10-9 =5 × 10-10 m

$2 \times 1 0^{- 9} m$

$5 \times 1 0^{- 8} m$

$4 \times 1 0^{- 8} m$

$5 \times 1 0^{- 10} m$

We know,
$Δ ℓ = \frac{W L}{2 A Y}$
$Δ ℓ = \frac{10 \times 1}{2 \times 5} \times 100 \times 1 0^{- 4} \times 2 \times 1 0^{11}$
$Δ ℓ = \frac{1}{2} \times 1 0^{- 9}$
$= 5 \times 1 0^{- 10} m$

Q. A uniform heavy rod of weight 10kgms−2, crosssectional area 100cm2 and length 20cm is hanging from a fixed support. Young modulus of the material of the rod is 2×1011Nm−2. Neglecting the lateral contraction, find the elongation of rod due to its own weight.

2×10−9m

5×10−8m

4×10−8m

5×10−10m

We know, Δℓ=2AYWL​ Δℓ=2×510×1​×100×10−4×2×1011 Δℓ=21​×10−9 =5×10−10m

$2 \times 1 0^{- 9} m$

$5 \times 1 0^{- 8} m$

$4 \times 1 0^{- 8} m$

$5 \times 1 0^{- 10} m$

We know,
$Δ ℓ = \frac{W L}{2 A Y}$
$Δ ℓ = \frac{10 \times 1}{2 \times 5} \times 100 \times 1 0^{- 4} \times 2 \times 1 0^{11}$
$Δ ℓ = \frac{1}{2} \times 1 0^{- 9}$
$= 5 \times 1 0^{- 10} m$