Question

A liquid drop of diameter 4mm breaks into 1000 droplets of equal size. calculate the Work done. (Surface tension of liquid = 0.07 N/m)​

Answer 1

Answer: 3.165 × 10^-5 J/m^2

Explanation:

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Answer 2

$\huge{ \blue{\underline{ \overline{ \boxed{ \orange{ \mathbb{ANSWER}}}}}}}$

GIVEN :

☆ Diameter of liquid drop (D) = 4mm

☆ Surface Tension of liquid = 0.07 N/m

TO FIND :

☆ Workdone (W) for the process.

SOLUTION :

D = 4 mm

R = 2 mm

R = $\sf \: 2 \: \times \: {10}^{ - 3} m$

Let radius of the larger drop be 'R' and as smaller drops are of equal size, so let their radius be 'r'.

☆ According to the Question,

$\boxed{ \sf{ \green{Vol^{m}. \: of \: larger \: drop = 1000 \times Vol^{m}. \times of \: each \: smaller \: drop.}}}$

$\implies \: \sf{\frac{4}{3} \pi {R}^{3} = 1000 \times \frac{4}{3} \pi {r}^{3} }$

$\implies \: \sf{ \cancel\frac{4}{3} \cancel\pi \: {R}^{3} = 1000 \times \cancel\frac{4}{3}\cancel\pi \: {r}^{3} }$

$\implies \: \sf{ {R}^{3} = 1000 \times {r}^{3} }$

$\implies \: \sf{ {(R)}^{3} = {(10r)}^{3} }$

$\implies \: \sf{ R = 10r}$

[Putting the value of R from given]

$\implies \: \sf{ \sf \: 2 \: \times \: {10}^{ - 3} m = 10r}$

$\boxed{\therefore \: { \sf r = 2 \: \times \: {10}^{ - 4} m}}$

☆ Now, Initial Area of larger drop = $\sf4\pi {R}^{2}$

☆ Final area of 1000 droplets = $1000 \times \sf4\pi {r}^{2}$

$\boxed{ \red{\sf{ Change \: in \: Area, \Delta A \: = Final \: Area - Initial \: Area }}}$

$\sf{\Delta A \: = 1000 \times 4\pi {r}^{2} - 4\pi {R}^{2} }$

$\sf{\Delta A \: = 4\pi (1000{r}^{2} - {R}^{2})}$

$\sf{\Delta A \: = 4\pi[ 1000( { \frac{R}{10} )}^{2} - {R}^{2} ) } ]$

$\sf{\Delta A \: = 4\pi[ 10{R}^{2} - {R}^{2}}]$

$\sf{\Delta A \: = 4\pi[ 9{R}^{2}}]$

$\sf{\Delta A \: = 4 \times 3.14 \times 9 \times( {2 \times {10}^{ - 3} )}^{2}}$

$\sf{\Delta A \: = 36 \times 3.14 \times {4 \times {10}^{ - 6}}}$

$\sf{\Delta A \: = 144\times 3.14 \times {10}^{ - 6}}$

$\boxed{ \blue {\sf{ Now, \: We \: have\: : \Delta W = S \times \Delta A}}}$

[On putting the values we get, ]

$\sf{ \Delta W = 0.07\times 144\times 3.14 \times {10}^{ -6}}$

$\sf{ \Delta W = 31.65 \times {10}^{ -6}}$

$\sf{ \Delta W = 3.165 \times {10}^{ -5}}$

$\blue{\boxed {\sf{Hence \: our \: required \: Work \: done \: is \: \: \: 3.165 \times {10}^{ -5} J}}}$

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