Question

Find the maximum weight of needle which can float
on water having surface tension 0.073 N/m if length
of needle is 4 cm.
(1) 0.58 kg
(2) 0.584 g
(3) 1.3 g
(4) 1.3 kg​

Answer 1

The maximum weight of needle "(2) 0.584 g is correct".

Explanation:

Given,

Water having surface tension = 0.073 $\dfrac{N}{m}$ and

Length  of needle = 4 cm = $\dfrac{4}{100}$ m

To find, the maximum weight of needle = ?

The maximum weight of needle = T × 2l

= 0.073 × 2 × 0.04 kg

= 0.00584 kg

= 0.00584 × $10^{3}$ g

= 0.584 g

Hence, the maximum weight of needle "(2) 0.584 g is correct".

Answer 2

Answer:

(2) 0.584 g

Explanation:

Max. weight = Surface tension × 2 length

= 0.073 × 2 × 4 × 10 ⁻² Kg

=

0.584 × 10 ⁻² × 10 ³ g

= 0.584 g

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