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Two rods, one made of brass and the other of lead, have the same dimensions. When a particular stress is applied to the brass rod, it stretches by 0.18 mm. How much does the lead rod stretch under the same stress? Use a value of 9.0×10 10 Pa for the Young’s modulus of brass and use a value of 1.6×10 10 Pa for the Young’s modulus of lead. |
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Answer» tion:The DIAMETER of a brass rod is 4 mm and YOUNG's MODULUS of brass is 9×10 10 N/m 2 . The force REQUIRED to stretch by 0.1% of its length is :- ANSWERANSWERY= ANSWERY= AΔlANSWERY= AΔlFLANSWERY= AΔlFL ANSWERY= AΔlFL ANSWERY= AΔlFL F= ANSWERY= AΔlFL F= LANSWERY= AΔlFL F= LYΔlAANSWERY= AΔlFL F= LYΔlA ANSWERY= AΔlFL F= LYΔlA = ANSWERY= AΔlFL F= LYΔlA = LANSWERY= AΔlFL F= LYΔlA = LYA×0.1LANSWERY= AΔlFL F= LYΔlA = LYA×0.1L ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 2ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3 ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3 ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3 ) ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3 ) 2ANSWERY= AΔlFL F= LYΔlA = LYA×0.1L =0.1YA=0.1×9×10 10 ×π×( 24×10 −3 ) 2 =360πmark as brainlist plz. |
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