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A needle of length 5cm is floating on water of surface tension 0.07 Jm^-2 and is not wetted by water. Then the weight of the needle will be

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7 x 10^-3 N

0.7 x 10^-3 N

5.2 x 10^-2 N

2 x 10^-2 N

answer is A.

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Detailed Solution

Weight of the neddle is balanced by surface tensional force

T x 2l = mg

0.07 x 2 x 5 x 10^-2 = mg

mg = 7 x 10^-3 N

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