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SS1: MATHEMATICS - 2ND TERM

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  1. The Set Theory I | Week 1
    4 Topics
  2. The Set Theory II | Week 2 & 3
    2 Topics
  3. The Set Theory III | Week 4
    2 Topics
    |
    1 Quiz
  4. Simple Equations & Change of Subject of Formula | Week 5
    1 Topic
    |
    1 Quiz
  5. Algebraic - Variations | Week 6
    4 Topics
    |
    1 Quiz
  6. Quadratic Equations I | Week 7
    5 Topics
    |
    1 Quiz
  7. Logical Reasoning | Week 8
    6 Topics
    |
    1 Quiz
  8. Construction & Locus I | Week 9
    5 Topics
    |
    1 Quiz



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What is Joint Variation

A joint variation exists when a quantity varies directly and / or inversely with two or more other quantities. E.g.

G varies directly with M and t and inversely with the square of r i.e.

\( \scriptsize G \propto \normalsize \frac{Mt}{r^2}\scriptsize \; or \; G = \normalsize \frac{KMT}{r^2}\)

Where k is a constant.

Example 4

The weight W kg of a metal bar varies jointly as its length L metres and the square of its diameter d mm. If W = 140 when d = 4 and L = 54, find d in terms of W and L (JAMB)

Solution:

\( \scriptsize W \propto Ld^2 \)

\( \scriptsize W = K Ld^2 \)

When W = 140, d = 4, L = 54

\( \scriptsize 140 = K \: \times \: 54 \: \times \: 4^2 \)

\( \scriptsize K = \normalsize \frac {140}{54 \: \times \: 16} \\ = \frac {140}{864} \\ = \frac {35}{216}\)

K = 0.16

If \( \scriptsize W = K Ld^2 \)

Then, \( \scriptsize d^2 = \normalsize \frac {W}{KL} \)

Then, \( \scriptsize d = \normalsize \sqrt { \frac {W}{KL}} \)

so d in terms of W and L

⇒ \( \scriptsize d = \normalsize \sqrt { \frac {W}{0.16L}} \)

or

\( \scriptsize d = \normalsize \sqrt { \frac {216W}{35L}} \)

Note: The various units of measurement have not been considered in solving this problem

Example 5

The universal gas law states that the volume V(m3) of a given mass of an ideal gas varies directly with its absolute temperature T(K) and inversely with its pressure P (n/m2 )A certain mass of gas at an absolute temperature 275K and pressure \( \scriptsize 10^5 \normalsize \frac {N}{m^2}\) has a volume 0.0225m3

(a) Find the formula that connects P, V, and T.

(b) Hence find the pressure of the gas when its absolute temperature is 374K and its volume is 0.018m3

Solution

(a) \( \scriptsize V \propto \normalsize \frac{T}{P} \)

i.e \( \scriptsize V = K \normalsize \frac{T}{P} \)

(b) when V = 0.0225, T = 275 and P = \( \scriptsize 10^5 \)

\( \scriptsize K = \normalsize \frac{VP}{T} \)

\( \scriptsize K = \normalsize \frac{0.0225 \; \times \; 10^5}{275} \)

\( \scriptsize K = \normalsize \frac{2250}{275} = \frac{90}{11} \)

∴ the formula conecting P, V, and T is given as

\( \scriptsize V = \normalsize \frac{90}{11} \: \times \: \frac{T}{P}\)

\( \scriptsize V = K \normalsize \frac{T}{P} \)

make P the subject of the formula

\( \scriptsize PV = K T \)

∴ \( \scriptsize P = \normalsize \frac{KT}{V} \)

⇒ \( \scriptsize P = \normalsize \frac{90}{11}\frac{T}{V} \)

⇒ When T = 374 and V = 0.018

\( \scriptsize P = \normalsize \frac{90}{11} \scriptsize \; \times \; \normalsize \frac{374}{0.018} \)

⇒ \( \frac{33,600}{0.198} \)

 i.e. P = 170,000

\( \scriptsize P = 1.7 \times 10^5 \normalsize \frac {N}{m^2} \)

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