Correct Answer - Option 3 : 10
Concept:
Heat generated during welding process is given by:
\(Q=VIt=I^2Rt=\frac{V^2}{R}t\)
Heat required to melt welding portion = Unit melting energy (u) × Volume
Calculation:
Given:
di = 100 mm, do = 110 mm, Voltage = 30 V, l (thickness) = 1 mm, R = 42.4 Ω, and u = 64.4 MJ/m3
Volume = \(\frac{\pi}{4}(d_o^2\;-\;d_i^2)\;×\;2l\) [∵ thickness melts in both pipes]
Volume = \(\frac{\pi}{4}(110^2\;-\;100^2)\;×\;2\;×\;1\)
Volume = \(3298.672\;mm^3=3298.672×10^{-9}\;m^3\)
Melting Energy = Unit melting energy (u) × Volume
Melting Energy = 64.4 × 106 × 3298.672 × 10-9 ⇒ 212.43 J
Heat Generated = \(\frac{{{V^2}}}{R}t\)
Since no heat goes in waste,
Heat generated = Melting energy required
\(∴ \frac{{{30^2}}}{42.4}t=212.43\;\)
∴ t = 10 sec