Apparent weight at bottom most line is

2 m g

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m g

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3 m g

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4 m g

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Solution

The correct option is A $2 m g$
First, we find spring constant by equating spring force with weight, since it feels weightless at top. $k x = m g$ .
We get K as 200N/m. Now, the apparent weight at bottom is
$k x + m g = 200 \times 0.05 + 1 \times 10 = 20 N = 2 m g$ .
Option A is thus correct.

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Apparent weight at bottom most line is

The correct option is A 2 mgFirst, we find spring constant by equating spring force with weight, since it feels weightless at top. kx=mg.We get K as 200N/m. Now, the apparent weight at bottom is kx+mg=200×0.05+1×10=20N=2mg.Option A is thus correct.

The correct option is A $2 m g$
First, we find spring constant by equating spring force with weight, since it feels weightless at top. $k x = m g$ .
We get K as 200N/m. Now, the apparent weight at bottom is
$k x + m g = 200 \times 0.05 + 1 \times 10 = 20 N = 2 m g$ .
Option A is thus correct.