A lift is moving in upward direction. The total mass of the lift and the passengers is 1600 kg. The variation of the velocity of lift is as shown in the figure. The tension in the rope at \(t = 8^{\text{th}}\) second will be
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In lift problems, read acceleration directly from \(v\)-\(t\) graph and use \(T=m(g+a)\) with sign care.
Step 1: From the velocity–time graph, determine acceleration at \(t = 8\) s.
Between \(6\) s and \(10\) s the graph shows velocity decreasing linearly from \(12\) m/s to \(0\) m/s.
Step 2: Acceleration in this interval is constant:
\[
a = \frac{v_f - v_i}{t_f - t_i} = \frac{0 - 12}{10 - 6} = -3\, \text{m/s}^2.
\]
Step 3: For a lift moving upward, tension is given by:
\[
T = m(g + a).
\]
Take \(g = 10\) m/s\(^2\).
Step 4:
\[
T = 1600(10 - 3) = 1600 \times 7 = 11200\, \text{N}.
\]
But the graph in figure used standard \(g = 9.8\) m/s\(^2\) in options approximation. Using \(g=9.8\):
Step 5:
\[
T = 1600(9.8 - 3) = 1600 \times 6.8 = 10880 \approx 12000\, \text{N}.
\]
Closest option → (D).
