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# Solved Example Problems for Physics : Properties of Matter

Physics : Properties of Matter - Book Back Numerical Problems, Book Back Conceptual Questions, Solved Example Problems for Physics

## Conceptual questions

### EXAMPLE 7.1

Within the elastic limit, the stretching strain produced in wires A, B, and C due to stress is shown in the figure. Assume the load applied are the same and discuss the elastic property of the material. Write  down  the  elastic  modulus  in ascending order.

Solution

Here, the elastic modulus is Young modulus and due to stretching, stress is tensile stress and strain is tensile strain.

Within the elastic limit, stress is proportional to strain (obey Hooke’s law). Therefore, it shows a straight line behavior. So, the modulus of elasticity (here, Young modulus) can be computed by taking slope from this straight line. Hence, calculating the slope for the straight line, we get

Slope of A > Slope of B > Slope of C

Which implies,

Young modulus of C < Young modulus of B < Young modulus of A

Notice that larger the slope, lesser the strain (fractional change in length). So, the material is much stiffer. Hence, the elasticity of wire A is greater than wire B which is greater than C. From this example, we have understood that Young’s modulus measures the resistance of solid to a change in its length.

### EXAMPLE 7.2

A wire 10 m long has a cross-sectional area 1.25 x 10-4 m2. It is subjected to a load of 5 kg. If Young’s modulus of the material is 4 x 1010 N m-2, calculate the elongation produced in the wire.

Take g = 10 ms-2.

Solution ### EXAMPLE 7.3

A metallic cube of side 100 cm is subjected to a uniform force acting normal to the whole surface of the cube. The pressure is 106 pascal. If the volume changes by 1.5 x 10-5 m3, calculate the bulk modulus of the material.

Solution ### EXAMPLE 7.4

A metal cube of side 0.20 m is subjected to a shearing force of 4000 N. The top surface is displaced through 0.50 cm with respect to the bottom. Calculate the shear modulus of elasticity of the metal.

### Solution

Here, L = 0.20 m, F = 4000 N, x = 0.50 cm

= 0.005 m and Area A = L2 = 0.04 m2

Therefore, Shear modulus ## EXAMPLE 7.5

A wire of length 2 m with the area of cross-section 10-6m2 is used to suspend a load of 980 N. Calculate i) the stress developed in the wire ii) the strain and iii) the energy stored.

Given: Y = 12 × 1010N m−2.

## Solution ### EXAMPLE 7.6

A solid sphere has a radius of 1.5 cm and a mass of 0.038 kg. Calculate the specific gravity or relative density of the sphere.

### Solution

Radius of the sphere R = 1.5 cm

mass m = 0.038 kg ### EXAMPLE 7.7

Two pistons of a hydraulic lift have diameters of 60 cm and 5 cm. What is the force exerted by the larger piston when 50 N is placed on the smaller piston?

### Solution

Since, the diameter of the pistons are given, we can calculate the radius of the piston This means, with the force of 50 N, the force of 7200 N can be lifted.

Solved Example Problems for Buoyancy

### EXAMPLE 7.8

A cube of wood floating in water supports a 300 g mass at the centre of its top face. When the mass is removed, the cube rises by 3 cm. Determine the volume of the cube.

### Solution

Let each side of the cube be l. The volume occupied by 3 cm depth of cube,

V=(3cm) × l2 = 3l2cm

According to the principle of floatation, we have

Vρg = mg  Vρ = m

ρ is density of water = 1000 kg m-3

(3l2 × 10-2m) × (1000 kgm-3)=300 × 10-3 kg l = 10 × 10-2m = 10 cm

Therefore, volume of cube V = l3 = 1000 cm3

### EXAMPLE 7.9

A metal plate of area 2.5×10-4m2 is placed on a 0.25×10-3m thick layer of castor oil. If a force of 2.5 N is needed to move the plate with a velocity 3×10-2m s-1, calculate the coefficient of viscosity of castor oil.

Given: A=2.5×10-4 m2dx = 0.25×10-3mF=2.5and dv = 3×10-2 m s-1

Solution ## EXAMPLE 7.10

Let 2 .4×104 J of work is done to increase the area of a film of soap bubble from 50 cm2 to 100 cm2. Calculate the value of surface tension of soap solution.

Solution:

A soap bubble has two free surfaces, therefore increase in surface area ∆A = A2A1 = 2(100-50) × 10-4m2 = 100 × 10-4m2.

Since, work done W = T ×ΔA T = ## EXAMPLE 7.11

If excess pressure is balanced by a column of oil (with specific gravity 0.8) 4 mm high, where R = 2.0 cm, find the surface tension of the soap bubble.

### Solution

The excess of pressure inside the soap bubble is = 15.68 ×10 2 N m1

### EXAMPLE 7.12

Water rises in a capillary tube to a height of 2.0cm. How much will the water rise through another capillary tube whose radius is one-third of the first tube?

### Solution

From equation (7.34), we have

1/r hr =constant

Consider two capillary tubes with radius r1 and r2 which on placing in a liquid, capillary rises to height h1 and h2, respectively. Then, ### EXAMPLE 7.13

Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 2 mm, made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside?. Surface tension of mercury T=0.456 N m-1; Density of mercury ρ = 13.6 × 103 kg m-3

### Solution

Capillary descent, where, negative sign indicates that there is fall of mercury (mercury is depressed) in glass tube.

## EXAMPLE 7.14

In a normal adult, the average speed of the blood through the aorta (radius r = 0.8 cm) is 0.33 ms-1. From the aorta, the blood goes into major arteries, which are 30 in number, each of radius 0.4 cm. Calculate the speed of the blood through the arteries.

## Solution:

a1v1 = 30 a2 v2  π r12v1  = 30 π r22v2 Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
11th Physics : Properties of Matter : Solved Example Problems for Physics : Properties of Matter |

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