a uniform steel rod of 1m in length and area of cross section 20cm² is hanging from a fixed support find the increase in the length of the rod (Ysteel=2×10^11N/m²) (density of steel=7.85×10^3kg/m³)
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Young’s modulus (Y) of wire of length (L) and area of cross section (A) is given as
Y = FL / (Ae)
e = FL / (AY)
= mgL / (AY)
= AL²dg / (AY)
= L²dg / Y
= (1 m)² × 7.85 × 10³ kg/m³ × 9.8 m/s² / (2 × 10¹¹ N/m²)
= 384.65 × 10⁻⁹ m
= 384.65 nm
Elongation in wire is 384.65 nm
Y = FL / (Ae)
e = FL / (AY)
= mgL / (AY)
= AL²dg / (AY)
= L²dg / Y
= (1 m)² × 7.85 × 10³ kg/m³ × 9.8 m/s² / (2 × 10¹¹ N/m²)
= 384.65 × 10⁻⁹ m
= 384.65 nm
Elongation in wire is 384.65 nm
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0
Explanation:
Answer is 1.923 *10^-5
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