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Equipartition Theorem

The multiplie...

Question

The multiplier can be expressed algebraically as _________________.

A

K = \frac{Δ I}{Δ Y}

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B

K = \frac{Δ A E}{Δ A D}

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C

K = \frac{Δ Y}{Δ I}

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D

K = \frac{Δ Y}{Δ A D}

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Solution

The correct option is C $K = \frac{Δ Y}{Δ I}$
Investment multiplier refers to the number of time by which the increase in output or income exceeds the increase in investment. It is measured as the ratio between change in income and change in investment and it is denoted as 'k'. Algebraically,
k = ΔY/ ΔI where Y is the income and I is the investment expenditure in the economy.

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Q. PARAGRAPH
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation

z = x / y

. If the errors in

x, y

z

Δ x, Δ y

Δ z

, respectively, then

z \pm Δ z = \frac{x \pm Δ x}{y \pm Δ y} = \frac{x}{y} (1 \pm \frac{Δ x}{x}) {(1 \pm \frac{Δ y}{y})}^{- 1}

.
The series expansion for

{(1 \pm \frac{Δ y}{y})}^{- 1}

, to first power in

Δ y / y

1 \mp (Δ y / y)

. The relative errors in independent variables are always added. So the error in

z

Δ z = z (\frac{Δ x}{x} + \frac{Δ y}{y})

.
The above derivation makes the assumption that

Δ x / x ≪ 1, Δ y / y ≪ 1

. Therefore, the higher powers of these quantities are neglected.

Consider the ratio

r = \frac{(1 - a)}{(1 + a)}

to be determined by measuring a dimensionless quantity

a

. If the error in the measurement of a is

Δ a (\frac{Δ a}{a} << 1)

, then what is the error

Δ r

r

Q. PARAGRAPH
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation

z = x / y

. If the errors in

x, y

z

Δ x, Δ y

Δ z

, respectively, then

z \pm Δ z = \frac{x \pm Δ x}{y \pm Δ y} = \frac{x}{y} (1 \pm \frac{Δ x}{x}) {(1 \pm \frac{Δ y}{y})}^{- 1}

.
The series expansion for

{(1 \pm \frac{Δ y}{y})}^{- 1}

, to first power in

Δ y / y

1 \mp (Δ y / y)

. The relative errors in independent variables are always added. So the error in

z

Δ z = z (\frac{Δ x}{x} + \frac{Δ y}{y})

.
The above derivation makes the assumption that

Δ x / x ≪ 1, Δ y / y ≪ 1

. Therefore, the higher powers of these quantities are neglected.

In an experiment the initial number of radioactive nuclei is

3000

. It is found that

1000 \pm 40

nuclei decayed in the

1.0 s

| x | << 1, ln (1 + x) = x

up to first power in

x

Δ λ

, in the determination of the decay constant

λ

s^{- 1}

Q. If MPC = 0.4 and

Δ Y

Δ S

If MPC = 0.4 and $Δ Y = 1, 000,$ what will be $Δ S$ ?

y = 2 x^{3} - 2 x^{2} + 3 x - 5

x = 2

Δ x = 0.1

Δ y

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