A person of mass ‘m’ is standing on a weighing machine placed in an elevator/lift. The actual weight of Person is mg. Then

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When lift moves uniformly down word direction it’s apparent weight is m (g-a)

When lift is at rest the apparent weight is “zero”

The lift moves up ward with retardation ‘a’ it’s percentage increase in apparent weight is $\frac{- a}{g} \times 100$

The lift moves up word with acceleration ‘a’ it’s percentage increase in apparent weight is $\frac{a}{g} \times 100$

answer is A, B.

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Detailed Solution

(i) Lift moves upward with “a” (or) downward Retardation with “a’ is

$w^{1} = m (g + a) \Rightarrow w^{1} = w (1 + \frac{a}{g}) \Rightarrow \frac{w^{1}}{w} = \frac{a}{g} + 1$

$\frac{w^{1}}{w} - 1 = \frac{a}{g} \Rightarrow \frac{w^{1} - w}{w} \times 100 = \frac{a}{g} \times 100$

(ii) Lift moves down with “a” (or) upward retardation “a” is

$w^{1} = m (g - a) \Rightarrow w^{1} = m g (1 - \frac{a}{g})$

$\frac{w^{1}}{w} - 1 = - \frac{a}{g} \Rightarrow \frac{w^{1} - w}{w} \times 100 = - \frac{a}{g} \times 100$

(iii) Lift moves uniformly $w^{1} = m g$

(iv) Lift is at rest $w^{1} = m g$

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