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Concepts of Physics

Some Mechanical Properties of Matter

A drop of mercury of radius 2 {~mm} is split into 8 identical

Concepts of Physics

Class 11

HC Verma Part I

Chapter 1: Introduction to Physics

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Chapter 3: Rest and Motion: Kinematics

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Chapter 4: The Forces

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Chapter 5: Newton's Law of Motion

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Chapter 8: Work and Energy

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Chapter 9: Centre of Mass, Linear momentum Collision

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Chapter 10: Rotational Mechanics

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Chapter 11: Gravitation

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Chapter 12: Simple Harmonic Motion

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Chapter 13: Fluid Mechanics

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Chapter 14: Some Mechanical Properties of Matter

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Find the surface energy of water kept in a cylindrical vessel of radius

6.0 cm

. Surface tension of water

= 0.075 J m^{- 2}

A drop of mercury of radius

2 mm

is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury

= 0.465 J m^{- 2}

A capillary tube of radius 1 mm is kept vertical with the lower end in water.
(a) Find the height of water raised in the capillary.
(b) If the length of the capillary tube is half the answer of part , find the angle θ made by the water surface in the capillary with the wall.

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Chapter 15: Wave Motion and Waves on a String

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Chapter 20: Dispersion and Spectra

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Chapter 21: Speed of Light

14 questions

Chapter 22: Photometry

38 questions

Question

Hard

Solving time: 10 mins

A drop of mercury of radius $2 mm$ is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury $= 0.465 J m^{- 2}$ .

Name: Video solution 1: [Solved] A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury =0.465 J m−2 . - Concepts of Physics
Uploaded: 2023-01-24T05:54:33.000Z
Duration: 6 min
Description: Video solution 1: [Solved] A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury =0.465 J m−2 . - Concepts of Physics

Views: 6,482 students

Updated on: Jan 24, 2023

Book:

Concepts of Physics (HC Verma Part I)

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A drop of mercury of radius

2 mm

is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury

= 0.465 J m^{- 2}

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

Given:
Initial radius of mercury drop

R = 2 mm = 2 \times 1 0^{- 3} m

Surface tension of mercury

T = 0.465 J / m^{2}

Let the radius of a small drop of mercury be

r

.

As one big drop is split into 8 identical droplets:

volume of initial drop

= 8 \times

(volume of a small drop)

(\frac{4}{3}) π R^{3} = (\frac{4}{3}) π r^{3} \times 8

Taking cube root on both sides of the above equation:

r = \frac{R}{2} = 1 0^{- 3} m

Surface energy

= T \times

surface area

∴

Increase in surface energy

= T A^{'} - T A

= (8 \times 4 □ π r^{2} - 4 π R^{2}) T = 4 π T [8 \times (\frac{R ^{2}}{4}) - R^{2}] = 4 π T R^{2} = 4 \times (3.14) \times (0.465) \times (4 \times 1 0^{- 6}) = 23.36 \times 1 0^{- 6} = 23.4 μ J

Hence, the required increase in the surface energy of the mercury droplets is

23.4 μ J

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Question Text	A drop of mercury of radius $2 mm$ is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury $= 0.465 J m^{- 2}$ .
Updated On	Jan 24, 2023
Topic	Mechanical Properties of Solids
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 1
Upvotes	223
Avg. Video Duration	6 min

A drop of mercury of radius $2 mm$ is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury $= 0.465 J m^{- 2}$ .

Text solutionVerified

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A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury =0.465 J m−2 .

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A drop of mercury of radius $2 mm$ is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury $= 0.465 J m^{- 2}$ .