A glass slide 15 cm long, 2 cm thick is suspended from the arm of a balance with its lower edge horizontal and equibalanced. Now the slide is dipped in the soap solution and is raised such that the lower end just touches the surface of the soap solution. If now a block of 400 mg has to be placed in the other having unknown surface tension then what will be the value of surface tension of solution?

Question

A glass slide 15 cm long, 2 cm thick is suspended from the arm of a balance with its lower edge horizontal and equibalanced. Now the slide is dipped in the soap solution and is raised such that the lower end just touches the surface of the soap solution. If now a block of 400 mg has to be placed in the other having unknown surface tension then what will be the value of surface tension of solution? (A) 1.1 × 10-1 N / m (B) 1.1 N / m (C) 1.10 × 10-2 N / m (D) None of these

Tardigrade Admin · Accepted Answer

If after placing the block, force due to weight and soap solution equalize that means equilibrium has restored.

∴ F = S \times 2 (ℓ + b)

(due to glass slide being rectangle)

∴ m g = S \times 2 (ℓ + b)

\Rightarrow 400 \times 1 0^{- 3} = S \times 2 (15 + 2) \times 10

S = \frac{400 \times 1 0 ^{- 3}}{34 \times 1 0 ^{- 2}} = 11 \times 1 0^{- 1} = 1.1 N / m

$1.1 \times 1 0^{- 1} N / m$

$1.1 N / m$

$1.10 \times 1 0^{- 2} N / m$

None of these

1.1×10−1N/m

1.1N/m

1.10×10−2N/m

None of these

If after placing the block, force due to weight and soap solution equalize that means equilibrium has restored. ∴F=S×2(ℓ+b) (due to glass slide being rectangle) ∴mg=S×2(ℓ+b) ⇒400×10−3=S×2(15+2)×10 S=34×10−2400×10−3​=11×10−1=1.1N/m

$1.1 \times 1 0^{- 1} N / m$

$1.1 N / m$

$1.10 \times 1 0^{- 2} N / m$