Question

1. A drop of olive oil of radius 0.25 mm spreads into a circular
film of radius 10 cm on the water surface. Estimate the
molecular size of olive oil.

ans question 2 is in attached image.​

Answer 1

SOLUTION:

(Question 1.)

Given:

• Radius of olive oil (r) = 0.25mm

Converting it into cm,

0.25mm = 0.025cm

∴ 10mm = 1cm

• The radius of the circular film, (R) = 10cm

To find:

• The molecular size (t)

As we know that

$\boxed{\sf{t =\frac{volume\ of\ oil\ drop}{Area\ of\ film}}}$

$\sf{t =\dfrac{(\frac{4}{3}\pi r^{3})}{(\pi R^{2})}}$

$\sf{t =\dfrac{(\frac{4}{3} * \pi * (0.025)^{3})}{(\pi * (10)^{2})}}$

(π) will be cancelled in both numerator and denominator.

$\sf{t =\dfrac{(\frac{4}{3} * (0.025)^{3})}{ (10)^{2}}}$

$\sf t = \frac{4}{3} * 25^{3} *10^{-11}$

$\boxed{\sf t = 2.08 \times 10^{-7}}$

Hence,

The molecular size of olive oil is $\boxed{\sf t = 2.08 \times 10^{-7}}$ cm

- -

QUESTION 2.

If the size of a nucleus (≈ 10⁻¹⁵ m) is scaled up to the tip os a sharp pin (≈ 10⁻⁵ m), what roughly is the size of the atom?

SOLUTION:

$\tt Magnification\ in\ size = \frac{Size\ of\ the\ tip\ of\ sharp\ pin}{Size\ of\ nucleus}$

$\tt Magnification\ in\ size = \frac{10^{-15}}{10^{-5}}$

Magnification in size = 10¹⁰

So,

The actual size of an atm is 10⁻¹⁰m = 1 Angstrom, which is scaled up by a factor 10¹⁰.

Therefore,

The apparent size of an atom = 10⁻¹⁰ × 10¹⁰ = 1m

Hence,

A nucleus in an atom is a small in size as the tip of a sharp pin located at the centre of a sphere of the radius of one metre.

Answer 2

Qᴜᴇsᴛɪᴏɴ : 1

A drop of olive oil of radius 0.25 mm spreads into a circular film of radius 10 cm on the water surface. Estimate the molecular size of olive oil.

Aɴsᴡᴇʀ :

Given,

• Radius of a drop of olive oil is 0.25 mm.

$\longmapsto\:\:\bf{Radius\:(R)\:=\:0.25\:mm\:=\:0.25\times{10^{-3}}\:m}$

• Radius of the oil film on the water surface is 10 cm.

$\longmapsto\:\:\bf{Radius\:(r)\:=\:10\:cm\:=\:10^{-1}\:m}$

To Find,

• The molecular size of olive oil.

Calculation,

↝ The thickness of the oil film formed represents the size of the oil molecule.

$\red\bigstar\:{\underline{\orange{\boxed{\bf{\green{\dfrac{4}{3}\:\pi\:R^3\:=\:\pi\:r^2\times{Size\:of\:a\:molecule}\:}}}}}}$

$\implies\bf{Size\:of\:a\:molecule\:=\:\dfrac{4}{3}\times{\dfrac{R^3}{r^2}}\:}$

$\implies\bf{Size\:of\:a\:molecule\:=\:\dfrac{4}{3}\times{\dfrac{(0.25\times{10^{-3}})^3}{(10^{-1})^2}}\:}$

$\implies\bf{Size\:of\:a\:molecule\:=\:\dfrac{4}{3}\times{0.015625}\times{\dfrac{10^{-9}}{10^{-2}}}\:}$

$\implies\bf\pink{Size\:of\:a\:molecule\:=\:0.0208\times{10^{-7}}\:m}$

$\Large\bf{Therefore,}$

The molecular size of olive oil is 0.0208 × 10⁻⁷ m.

__________________________

Qᴜᴇsᴛɪᴏɴ : 2

If the size of a nucleus (≃ 10⁻¹⁵ m) is scaled upto the top of a sharp pin (≃ 10⁻⁵ m), what roughly is the size of the atom ?

Aɴsᴡᴇʀ :

Given,

• The size of a nucleus is 10⁻¹⁵ m.

• The size of a sharp pin is 10⁻⁵ m.

To Find,

• The size of the atom.

Calculation,

✯ If we magnify the size of nucleus from 10⁻¹⁵ m to 10⁻⁵ m, then the magnification is

$\pink\bigstar{\underline{\green{\boxed{\bf{\blue{Magnification\:=\:\dfrac{Size\:of\:pin}{Size\:of\:nucleus}\:}}}}}}$

$:\implies\:\bf{Magnification\:=\:\dfrac{10^{-5}}{10^{-15}}\:}$

$:\implies\:\bf{Magnification\:=\:10^{-5}\times{10^{15}}\:}$

$:\implies\:\bf{Magnification\:=\:10^{10}\:}$

Thus,

★ We have to magnify the size of the atom by 10¹⁰.

➻ If same magnification 10¹⁰ is made on atom, it's dimension will br,

➙ 10⁻¹⁰ × 10¹⁰

➙ 1 m

$\Large\bf{Therefore,}$

The size of the atom is 1 m.