A straight metal wire of length 10 cm rests horizontally on the...

For a straight wire resting on a liquid surface, the liquid is in contact with both sides of the wire's length.

Effective length of contact, $L_{\text{eff}} = 2L$

Upward force due to surface tension, $F = T \times L_{\text{eff}} = T \times 2L$

For the wire to be supported without sinking, this upward force must balance the downward weight of the wire ($W = mg$).

$mg = 2TL$

$m = \dfrac{2TL}{g}$

Given:
$T = 0.07 \text{ N m}^{-1}$
$L = 10 \text{ cm} = 0.1 \text{ m}$
$g = 10 \text{ m s}^{-2}$

$m = \dfrac{2 \times 0.07 \times 0.1}{10} = \dfrac{0.014}{10} = 0.0014 \text{ kg}$

Converting to grams:
$m = 0.0014 \times 1000 \text{ g} = 1.4 \text{ g}$

Answer: $1.4 \text{ g}$