A Metal Wire Can Be Broken By Applying. Let l1 = l2 = l be the length of the two wires and δl1 = δl2 = δl be. Correct option is c) breaking stress is the characteristic of the material. The force required to break the wire of twice the diameter is (a) $20 \mathrm{~kg}. Web a metal wire can be broken by applying a load of 45 kg wt. Web a wire can be broken by applying a load of $20 \mathrm{~kg}$ wt. Hence, for two wires of same. The force required to break the wire of twice the diameter is: Answered apr 6, 2020 by chithrajain (84.9k points) selected apr 6, 2020 by subodhsharma. Web when the load on a wire is increased from 3 k g − w t to 8 k g − w t, the elongation increases from 0.61 m m to 1.02 m m. Thus, the ratio of longitudinal. Then the force required to break the wire of twice the diameter and the same. Express the relation for the applied force on a wire. Since the wires have same material, their modulus of elasticity must be same.
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The force required to break the wire of twice the diameter is (a) $20 \mathrm{~kg}. Hence, for two wires of same. Then the force required to break the wire of twice the diameter and the same. Express the relation for the applied force on a wire. Web a wire can be broken by applying a load of $20 \mathrm{~kg}$ wt. Since the wires have same material, their modulus of elasticity must be same. Let l1 = l2 = l be the length of the two wires and δl1 = δl2 = δl be. The force required to break the wire of twice the diameter is: Answered apr 6, 2020 by chithrajain (84.9k points) selected apr 6, 2020 by subodhsharma. Web when the load on a wire is increased from 3 k g − w t to 8 k g − w t, the elongation increases from 0.61 m m to 1.02 m m.
Metal torn wire stock image. Image of torn, string, iron 12149189
A Metal Wire Can Be Broken By Applying The force required to break the wire of twice the diameter is: Then the force required to break the wire of twice the diameter and the same. Answered apr 6, 2020 by chithrajain (84.9k points) selected apr 6, 2020 by subodhsharma. Correct option is c) breaking stress is the characteristic of the material. The force required to break the wire of twice the diameter is (a) $20 \mathrm{~kg}. Web when the load on a wire is increased from 3 k g − w t to 8 k g − w t, the elongation increases from 0.61 m m to 1.02 m m. The force required to break the wire of twice the diameter is: Let l1 = l2 = l be the length of the two wires and δl1 = δl2 = δl be. Express the relation for the applied force on a wire. Web a wire can be broken by applying a load of $20 \mathrm{~kg}$ wt. Thus, the ratio of longitudinal. Hence, for two wires of same. Web a metal wire can be broken by applying a load of 45 kg wt. Since the wires have same material, their modulus of elasticity must be same.
