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When an empty lift is moving down with an acceleration of $\frac{g}{4} m s^{- 2}$ , the tension in the cable is $6000 N$ . When the lift is moving up with an acceleration of $\frac{g}{2} m s^{- 2}$ , the tension in the cable is,

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$12000 N$

$18000 N$

$15000 N$

$16000 N$

answer is C.

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Detailed Solution

When lift is going down,

$6000 = m (g - \frac{g}{4})$

$\Rightarrow \frac{3}{4} m g = 6000$

$\Rightarrow m g = 8000 N$

When lift is going up,

$T = m (g + \frac{g}{2})$ $= \frac{3}{2} m g = 12000 N$

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