Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Properties of Matter question

2021 · 17 Mar · Shift 2 · Q61
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Properties of Matter
  5. /2021 · 17 Mar · Shift 2 · Q61

Properties of Matter question

2021 · 17 Mar · Shift 2 · Q61

JEE MainPhysicsProperties of MatterNumerical+4 / −1
Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm3 of oleic acid per cm3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm2 by considering 100 spherical drops of radius (340π)13×10−3{\left( {{3 \over {40\pi }}} \right)^{{1 \over 3}}} \times {10^{ - 3}}(40π3​)31​×10−3 cm. Then the thickness of oleic acid layer will be x ×\times× 10 −-− 14 m. Where x is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 25

  1. Given data
  • Concentration of oleic acid in solution: 0.01 cm3 of oleic acid per cm3 of solution0.01\ \text{cm}^3 \text{ of oleic acid per cm}^3 \text{ of solution}0.01 cm3 of oleic acid per cm3 of solution
  • Number of drops used: 100100100
  • Radius of each spherical drop: r=(340π)1/3×10−3 cmr=\left(\frac{3}{40\pi}\right)^{1/3}\times 10^{-3}\ \text{cm}r=(40π3​)1/3×10−3 cm
  • Area of monomolecular film formed: A=4 cm2A=4\ \text{cm}^2A=4 cm2

We need thickness of oleic acid layer.

  1. Volume of one drop of solution

Since each drop is spherical, Vdrop=43πr3V_{\text{drop}}=\frac{4}{3}\pi r^3Vdrop​=34​πr3

Now, r3=340π×10−9 cm3r^3=\frac{3}{40\pi}\times 10^{-9}\ \text{cm}^3r3=40π3​×10−9 cm3

So, Vdrop=43π(340π×10−9)V_{\text{drop}}=\frac{4}{3}\pi\left(\frac{3}{40\pi}\times 10^{-9}\right)Vdrop​=34​π(40π3​×10−9)

Canceling terms, Vdrop=440×10−9=10−10 cm3V_{\text{drop}}=\frac{4}{40}\times 10^{-9}=10^{-10}\ \text{cm}^3Vdrop​=404​×10−9=10−10 cm3

  1. Total volume of 100 drops of solution

Vsolution=100×10−10=10−8 cm3V_{\text{solution}}=100\times 10^{-10}=10^{-8}\ \text{cm}^3Vsolution​=100×10−10=10−8 cm3

  1. Volume of oleic acid present in this solution

Only 0.01 cm30.01\ \text{cm}^30.01 cm3 of oleic acid is present per 1 cm31\ \text{cm}^31 cm3 of solution. Thus oleic acid fraction is 0.01=10−20.01=10^{-2}0.01=10−2

Therefore, Vacid=10−2×10−8=10−10 cm3V_{\text{acid}}=10^{-2}\times 10^{-8}=10^{-10}\ \text{cm}^3Vacid​=10−2×10−8=10−10 cm3

  1. Thickness of monomolecular film

For a thin film, thickness=volumearea\text{thickness} = \frac{\text{volume}}{\text{area}}thickness=areavolume​

Hence, t=10−104 cmt=\frac{10^{-10}}{4}\ \text{cm}t=410−10​ cm t=14×10−10 cm=2.5×10−11 cmt=\frac{1}{4}\times 10^{-10}\ \text{cm}=2.5\times 10^{-11}\ \text{cm}t=41​×10−10 cm=2.5×10−11 cm

Now convert cm to m: 1 cm=10−2 m1\ \text{cm}=10^{-2}\ \text{m}1 cm=10−2 m

So, t=2.5×10−11×10−2 mt=2.5\times 10^{-11}\times 10^{-2}\ \text{m}t=2.5×10−11×10−2 m t=2.5×10−13 mt=2.5\times 10^{-13}\ \text{m}t=2.5×10−13 m

Write as x×10−14 mx\times 10^{-14}\ \text{m}x×10−14 m: 2.5×10−13=25×10−142.5\times 10^{-13}=25\times 10^{-14}2.5×10−13=25×10−14

Thus, x=25x=25x=25

  1. Comparison with stored answer

Derived answer: 252525

Stored correct answer: 252525

They match.

PreviousNext

More from Properties of Matter

  • Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when they are subjected to a force of 2 N. Assume that both the wires are made up of same material and the radius of wire B is 4 times that of the radius of wire A.…2021 · Numerical
  • Consider a water tank as shown in the figure. It's cross-sectional area is 0.4 m2. The tank has an opening B near the bottom whose cross-section area is 1 cm2. A load of 24 kg is applied on the water at the top when the height of the water… Includes diagram2021 · Numerical
  • The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are l1 and l2 respectively. The actual length of the metal wire is :2021 · MCQ
  • The length of a metal wire is l1, when the tension in it is T1 and is l2 when the tension is T2. The natural length of the wire is :2021 · MCQ
  • Two small drops of mercury each of radius R coalesce to form a single large drop. The ratio of total surface energy before and after the change is :2021 · MCQ
  • If Y, K and η are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.2021 · MCQ
  • A hydraulic press can lift 100 kg when a mass 'm' is placed on the smaller piston. It can lift ​ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4…2021 · Numerical
  • A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be ​ cm.2021 · Numerical