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SS1: PHYSICS – 3RD TERM

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  1. Production of Electric Current | Week 1
    6 Topics
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    1 Quiz
  2. Electric Current | Week 2
    5 Topics
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    1 Quiz
  3. Electrical Resistance of a Conductor | Week 3
    5 Topics
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    1 Quiz
  4. Particulate Nature of Matter | Week 4
    5 Topics
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  5. Crystalline and Non-crystalline Substances | Week 5
    3 Topics
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    1 Quiz
  6. Elastic Properties of Solids | Week 6 & 7
    4 Topics
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    1 Quiz
  7. Fluids at Rest & in Motion | Week 8 & 9
    6 Topics
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    1 Quiz
  8. Solar Collector
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Stress:

The stress of a material is defined as the ratio of force to the area. It is the ratio of force that acts on a material to the area of the material.

\( \scriptsize Stress, \: \sigma \) = \( \frac{Force}{Area} \\ = \frac{F}{A} \)

Stress is frequently represented by a lowercase Greek letter sigma (σ)

Its unit is Nm-2 or N/m2

Example 1:

A metal rod is being compressed by a machine with a force of 750N. The cross-sectional area of the metal rod is 30cm2. What is the compressive stress on the metal rod?

Solution:

Force = 750 N

Area = 30cm2
convert to m2 (divide by 10,000)
= \( \frac{30}{10000} = \scriptsize 0.003m^2\)

\( \scriptsize Stress, \: \sigma = \normalsize \frac{force}{area} \\ = \normalsize \frac{750}{0.003} \\ = \scriptsize 250000 Nm^{-2} \\ = \scriptsize 2.5 \: \times \: 10^{5} Nm^{-2}\)

Strain:

Strain is the ratio of extension produced in a solid/wire to the original length.

\(\scriptsize Strain, \: \epsilon = \normalsize \frac{Extension}{Original \: Length}\\= \normalsize \frac{e}{L}\\= \normalsize \frac{metre}{metre}\)

Strain has no unit.

Example 2:

A hot liquid enters through a copper pipe 15.00 m long. This causes an increase in length to 15.17 m. Calculate the longitudinal strain?

Solution:

Let the original length be l1 and the extended length l2

∴ e = l2 – l1

e = 15.17 – 15 = 0.17m

⇒ \(\scriptsize Strain = \normalsize \frac{Extension}{Original \: Length}\)

⇒ \(\scriptsize Strain = \normalsize \frac{e}{l_1}\)

⇒ \(\scriptsize Strain = \normalsize \frac{0.17}{15}\)

⇒ \(\scriptsize Strain = 0.0113\)

Hence, the longitudinal strain is 0.01

Young Modulus:

If a force F (Newton) is applied to a wire of original length L (in metres) and cross-sectional area A (m2) and produces an extension, e, in (m).

youngs modulus

⇒ \( \scriptsize Tensile\:strength = \normalsize \frac{Tensile \: Force}{Area} \\ = \normalsize \frac{F}{A}\)

⇒ \( \scriptsize Tensile\:stress = \normalsize \frac{extension}{Original \: Length} \\= \frac{e}{L} \)

∴ Stress & Strain, according to Hooke’s law can be written as:

Stress is proportional to strain provided elastic limit is not exceeded.

The constant of proportionality is known as Young’s Modulus. It is denoted as E or Y

⇒ \(\scriptsize Y = \normalsize \frac{Stress}{Strain}\)

⇒ \(\normalsize \large \frac{\frac{F}{A}}{\frac{e}{L}}\)

= \( \frac{FL}{Ae}\)

Y = Young modulus- constant.

⇒ \(\scriptsize Y = \normalsize \frac{Tensile \:Stress}{Tensile\:Strain}\)

Example 3:

A force of 40N applied at the end of a wire of length 4m and diameter 2.00mm produces an extension of 0.24mm.

Calculate the;
(a) stress on the wire.
(b) the strain on the wire.
(c) Young’s Modulus material of the wire.
(π = 3.14)

Solution:

(a) Stress = \( \frac{Force}{Area} \)

Radius, r = \( \frac{diameter}{2} \)

r = \( \frac {2.00mm}{2} \)

= 1 mm or 0.001m (convert mm to m)

∴ For F = 40N and r = 0.001m

Area = πr2 = 3.14 x (0.001)2

\(\scriptsize Stress = \normalsize \frac{40}{3.14 \: \times \: (0.001)^2}\\ = \scriptsize 12.74 \: \times \: 10^6 \: Nm^{-2} \)

(b) \(\scriptsize Strain = \normalsize \frac{extension}{original \: length}\\ = \normalsize \frac{e}{l} \)

l = 4m;
\( \scriptsize e = 0.24mm \\ = \frac{0.24}{1000} \\ = \scriptsize 2.4 \: \times \: 10^{-4} m \)

\(\scriptsize \therefore \: Strain = \normalsize \frac{2.4 \: \times \: 10^{-4}}{4} \\ = \scriptsize 6.0 \: \times \: 10^{-5}\)

(c) Young’s Modulus, Y = \( \normalsize \frac{stress}{strain}\\ = \frac{12.74 \: \times \: 10^6}{6 \: \times \: 10^{-5}} \\ \scriptsize = 2.12 \: \times \: 10^{11} Nm^{-2} \)

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