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Question: A brass rod of cross-sectional area \(1 \mathrm {~cm} ^ { 2 }\) and length 0.2 m is compressed lengt...

A brass rod of cross-sectional area 1 cm21 \mathrm {~cm} ^ { 2 } and length 0.2 m is compressed lengthwise by a weight of 5 kg. If Young's modulus of elasticity of brass is 1×1011 N/m21 \times 10 ^ { 11 } \mathrm {~N} / \mathrm { m } ^ { 2 } and g=10 m/sec2g = 10 \mathrm {~m} / \mathrm { sec } ^ { 2 }, then increase in the energy of the rod will be

A

10−510 ^ { - 5 }J

B

2.5×10−52.5 \times 10 ^ { - 5 }J

C

5×10−55 \times 10 ^ { - 5 }J

D

2.5×10−42.5 \times 10 ^ { - 4 }J

Answer

2.5×10−52.5 \times 10 ^ { - 5 }J

Explanation

Solution

U=12×( stress )2Y×U = \frac { 1 } { 2 } \times \frac { ( \text { stress } ) ^ { 2 } } { Y } \times volume= 12×F2×A×LA2×Y\frac { 1 } { 2 } \times \frac { F ^ { 2 } \times A \times L } { A ^ { 2 } \times Y }

=12×F2LAY=12×(50)2×0.21×10−4×1×1011\frac { 1 } { 2 } \times \frac { F ^ { 2 } L } { A Y } = \frac { 1 } { 2 } \times \frac { ( 50 ) ^ { 2 } \times 0.2 } { 1 \times 10 ^ { - 4 } \times 1 \times 10 ^ { 11 } }=