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 = $\frac{{\mathrm{x}}^{\mathrm{m}}{\mathrm{y}}^{\mathrm{n}}}{{\mathrm{z}}^{\mathrm{p}}}$

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

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

∴ $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{0.08}{10}$ = 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:

⇒ $\frac{\Delta \mathrm{V}}{\mathrm{V}}=3\times \frac{\Delta \mathrm{L}}{\mathrm{L}}$

⇒ $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ = 3 × 0.008

⇒ $\frac{\Delta \mathrm{V}}{\mathrm{V}}$ = 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 $\Delta \mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\Delta \mathrm{x}$

Calculation:

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

Let x + Δx = 242 = 243 - 1

Therefore, x = 243 and Δx = - 1

Assume, $\mathrm{y}={\mathrm{x}}^{1/5}$          

Differentiating with respect to x, we get

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{5}{\mathrm{x}}^{-4/5}=\frac{1}{5(\mathrm{x}{)}^{4/5}}$

At x = 243

${\left[\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right]}_{\mathrm{x}=243}=\frac{1}{405}$ and y = $(243{)}^{1/5}=3$

As we know $\Delta \mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\Delta \mathrm{x}$

So, $\Delta \mathrm{y}=\frac{1}{405}\times (-1)=0.0024$

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

Concept: 

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

Therefore $\Delta \mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\Delta \mathrm{x}$

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, $\mathrm{y}={\mathrm{x}}^{1/2}$          

Differentiating with respect to x, we get

$\Rightarrow \frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{2}{\mathrm{x}}^{-1/2}=\frac{1}{2\sqrt{\mathrm{x}}}$

At x = 36

${\left[\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right]}_{\mathrm{x}=36}=\frac{1}{12}$ and y = $(36{)}^{1/2}=6$

As we know $\Delta \mathrm{y}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\Delta \mathrm{x}$

So, $\Delta \mathrm{y}=\frac{1}{12}\times (0.01)=0.000833$

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