Errors and Approximations MCQ Quiz in मल्याळम - Objective Question with Answer for Errors and Approximations - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

If an expression is A = \(\rm\frac{x^{m}y^{n}}{z^{p}}\)

The error per cent of A ,\(\rm \frac{\Delta A}{A}\) = \(\rm m \frac{\Delta x}{x}\) + \(\rm n\frac{\Delta y}{y}\) + \(\rm p\frac{\Delta z}{z}\). 

Calculation:

Given arm length of the cube is 10 cm.

The error in the length of the arm is 0.08 cm.

⇒ Δ L = 0.08

∴ \(\rm\frac{Δ L}{L}=\frac{0.08}{10}\) = 0.008.

We know that the volume of the cube is given by V = l³.

⇒ V = 10³ = 1000 cm³

Error in the volume is given by:

⇒ \(\rm \frac{Δ V}{V}=3× \frac{Δ L}{L}\)

⇒ \(\rm \frac{Δ V}{V}\) = 3 × 0.008

⇒ \(\rm \frac{Δ V}{V}\) = 0.024

⇒ Δ V = V × 0.024

⇒ Δ V = 1000 × 0.024

⇒ Δ V = 24

The approximate error in the volume of the cube is 24 cm³.

Concept: 

Let small charge in x be Δx and the corresponding change in y is Δy.

Therefore \(\rm Δ y = \rm \frac{dy}{dx}Δ x\)

Calculation:

We have to find the value of \(\rm (242)^{1/5}\)

Let x + Δx = 242 = 243 - 1

Therefore, x = 243 and Δx = - 1

Assume, \(\rm y = x^{1/5}\)          

Differentiating with respect to x, we get

\(\rm \Rightarrow \frac{dy}{dx} = \frac{1}{5}x^{-4/5} = \frac{1}{5(x)^{4/5}}\)

At x = 243

\(\rm \left[\frac{dy}{dx} \right ]_{x=243} = \frac{1}{405}\) and y = \((243)^{1/5} = 3\)

As we know \(\rm Δ y = \rm \frac{dy}{dx}Δ x\)

So, \(\rm Δ y = \frac{1}{405} \times (- 1) = 0.0024\)

Therefore, approximate value of \(\rm (242)^{1/5}\) = y + Δy = 3 - 0.0024 = 2.997

Concept: 

Let small change in x be Δx and the corresponding change in y is Δy.

Therefore \(\rm Δ y = \rm \frac{dy}{dx}Δ x\)

Calculation:

We have to find the value of \(\sqrt{36.01}\)

Let x + Δx = 36.01 = 36 + 0.01

Therefore, x = 36 and Δx = -0.01

Assume, \(\rm y = x^{1/2}\)          

Differentiating with respect to x, we get

\(\rm \Rightarrow \frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}\)

At x = 36

\(\rm \left[\frac{dy}{dx} \right ]_{x=36} = \frac{1}{12}\) and y = \(\rm (36)^{1/2} = 6\)

As we know \(\rm Δ y = \rm \frac{dy}{dx}Δ x\)

So, \(\rm Δ y = \frac{1}{12} \times (0.01) = 0.000833\)

Therefore, approximate value of \(\sqrt{36.01} = (36.01)^{1/2}\) = y + Δy = 6 + 0.00083 = 6.00083