A soap film of surface tension 3 × 10-2 Nm -1 formed in a rectangular frame can support a straw as shown in figure. If g=10 ms -2, the mass of the straw is

Question

A soap film of surface tension 3 × 10-2 Nm -1 formed in a rectangular frame can support a straw as shown in figure. If g=10 ms -2, the mass of the straw is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9dca8f13e4ba9ea5888f0573efe27a40-.png" /> (A) 0.006 g (B) 0.06 g (C) 0.6 g (D) 6 g

Tardigrade Admin · Accepted Answer

As soap film can support the straw, m g=2 σ l m=(2 σ l/g)=(2 ×(3 × 10-2 ( N / m )) ×(0.1 m )/10 ( m ) s 2) ( uparrow2 σ l/ downarrowm g Start ) =6 × 10-4 kg =0.6 gram. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/10deb953b179eddf06739d64f10f4a86-.png" />

Q. A soap film of surface tension $3 \times 1 0^{- 2} N m^{- 1}$ formed in a rectangular frame can support a straw as shown in figure. If $g = 10$ $m s^{- 2},$ the mass of the straw is

0.006 g

0.06 g

0.6 g

6 g

As soap film can support the straw,
$m g = 2 σ l$
$m = \frac{2 σ l}{g} = \frac{2 \times ( 3 \times 1 0 ^{- 2} \frac{N}{m} ) \times ( 0.1 m )}{10 \frac{m}{s ^{2}}} \frac{↑ ^{2 σ l}}{↓ _{m g}^{St a r t}}$
$= 6 \times 1 0^{- 4} k g = 0.6$ gram.

Q. A soap film of surface tension 3×10−2Nm−1 formed in a rectangular frame can support a straw as shown in figure. If g=10 ms−2, the mass of the straw is

0.006 g

0.06 g

0.6 g

6 g

As soap film can support the straw, mg=2σl m=g2σl​=10s2m​2×(3×10−2mN​)×(0.1m)​↓mgStart​↑2σl​ =6×10−4kg=0.6 gram.

Q. A soap film of surface tension $3 \times 1 0^{- 2} N m^{- 1}$ formed in a rectangular frame can support a straw as shown in figure. If $g = 10$ $m s^{- 2},$ the mass of the straw is

As soap film can support the straw,
$m g = 2 σ l$
$m = \frac{2 σ l}{g} = \frac{2 \times ( 3 \times 1 0 ^{- 2} \frac{N}{m} ) \times ( 0.1 m )}{10 \frac{m}{s ^{2}}} \frac{↑ ^{2 σ l}}{↓ _{m g}^{St a r t}}$
$= 6 \times 1 0^{- 4} k g = 0.6$ gram.