A film of soap solution is trapped between a vertical frame and a light wire ab of length 0-1 m- If g -10 ms -2 and -25 - 10-3 N m -2- then the load W that should be suspended from the wire to keep it in equilibrium isA- 0-4 gB- 0-3 gC- 0-5 gD- 0-2 g

The correct option is C 0.5 g 25× 10^ 3× 2 × 0.1 =m× 10 or m=5× 10^ 3/10kg=0.5g ...

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Standard XII

Chemistry

Heat and Work

A film of soa...

Question

A film of soap solution is trapped between a vertical frame and a light wire ab of length 0.1 m. If $g = 10 m s^{- 2}$ and $σ = 25 \times 10^{- 3} N m^{- 2}$ , then the load W that should be suspended from the wire to keep it in equilibrium is

0.2 g

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0.3 g

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0.4 g

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0.5 g

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Solution

The correct option is D
0.5 g

$25 \times 10^{- 3} \times 2 \times 0.1 = m \times 10$

or $m = \frac{5 \times 10^{- 3}}{10} k g = 0.5 g$

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A film of soap solution is trapped between a vertical frame and a light wire ab of length 0.1 m. If g=10 ms−2 and σ=25×10−3 Nm−2, then the load W that should be suspended from the wire to keep it in equilibrium is

The correct option is D 0.5 g 25×10−3×2×0.1=m×10 or m=5×10−310kg=0.5g

A film of soap solution is trapped between a vertical frame and a light wire ab of length 0.1 m. If $g = 10 m s^{- 2}$ and $σ = 25 \times 10^{- 3} N m^{- 2}$ , then the load W that should be suspended from the wire to keep it in equilibrium is

The correct option is D
0.5 g

$25 \times 10^{- 3} \times 2 \times 0.1 = m \times 10$

or $m = \frac{5 \times 10^{- 3}}{10} k g = 0.5 g$