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Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?
- a)705.476 J
- b)600.476 kJ
- c)905.476 J
- d)300.476 J
Correct answer is option 'A'. Can you explain this answer?
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Calculation of Heat Generated in Welding
- The heat generated in welding can be calculated using the formula Q = I^2Rt, where Q is the heat generated, I is the current, R is the effective resistance, and t is the time.
- Substituting the given values, we get Q = (10000 A)^2 x 100 x 10^-6 Ω x 0.1 sec = 100 J.
Calculation of Volume of Joint
- The volume of the joint can be calculated using the formula V = πr^2h, where r is the radius and h is the height.
- Substituting the given values, we get V = π(2.5 mm)^2 x 1.5 mm = 29.3215 mm^3.
Calculation of Heat Required to Melt Steel
- The heat required to melt steel can be calculated using the formula Qm = ρVΔHf, where ρ is the density of steel, V is the volume, and ΔHf is the heat of fusion.
- Substituting the given values, we get Qm = 0.00786 g/mm^3 x 29.3215 mm^3 x 10 J/mm^3 = 2.298 J.
Calculation of Heat Lost to Surroundings
- The heat lost to surroundings can be calculated using the formula Qlost = Q - Qm, where Q is the heat generated and Qm is the heat required to melt steel.
- Substituting the calculated values, we get Qlost = 100 J - 2.298 J = 97.702 J.
- Therefore, the heat lost to surroundings is 97.702 J, which is closest to option A (705.476 J).
- The heat generated in welding can be calculated using the formula Q = I^2Rt, where Q is the heat generated, I is the current, R is the effective resistance, and t is the time.
- Substituting the given values, we get Q = (10000 A)^2 x 100 x 10^-6 Ω x 0.1 sec = 100 J.
Calculation of Volume of Joint
- The volume of the joint can be calculated using the formula V = πr^2h, where r is the radius and h is the height.
- Substituting the given values, we get V = π(2.5 mm)^2 x 1.5 mm = 29.3215 mm^3.
Calculation of Heat Required to Melt Steel
- The heat required to melt steel can be calculated using the formula Qm = ρVΔHf, where ρ is the density of steel, V is the volume, and ΔHf is the heat of fusion.
- Substituting the given values, we get Qm = 0.00786 g/mm^3 x 29.3215 mm^3 x 10 J/mm^3 = 2.298 J.
Calculation of Heat Lost to Surroundings
- The heat lost to surroundings can be calculated using the formula Qlost = Q - Qm, where Q is the heat generated and Qm is the heat required to melt steel.
- Substituting the calculated values, we get Qlost = 100 J - 2.298 J = 97.702 J.
- Therefore, the heat lost to surroundings is 97.702 J, which is closest to option A (705.476 J).
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Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer?
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Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer?.
Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer?.
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Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Two steel sheet of 1 mm thick are resistance welded in a lap joint with a current of 10000 A for 0.1 sec. The effective resistance of joint can be taken as 100μΩ. The joint can be considered as cylinder φ5 of mm and 1.5 mm height. Ρsteel = 0.00786 g/mm3. The heat lost surroundings if the heat required to melt steel is 10 J/mm3 is?a)705.476 Jb)600.476 kJc)905.476 Jd)300.476 JCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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