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Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Find the approximate error in the volume of a cube with edge $x$ cm, when the edge is increased by $2\%$

  1. $4\%$
  2. $2\%$
  3. $6\%$
  4. $8\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

in a mutiplication while multiplying we have to add the percentage errors 


so if the edge of the cube is $x$cm then the volume will be ${ x }^{ 3 }$

but for errors we have to add the percentage 

therefore the total error  percentage now becomes $6\%$

therefore the error is increased by $4\%$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If the length of cylinder is measured to be $4.28 cm$ with an error of $0.01 cm$, the percentage error in the measured length is nearly

  1. $0.4\% $
  2. $0.5\% $
  3. $0.2\% $
  4. $0.1\% $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given:Length of the cylinder $=l=4.28\,cm$
Error$=\Delta l=0.01\,cm$
Percentage error$=\dfrac{\Delta\,l}{l}\times 100$
$=\dfrac{0.01}{4.28}\times 100=0.234\approx\,0.2\%$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The radius of the sphere is measured as $ \left( {10 \pm 0.02} \right)cm$. The error in the measurement of its volume is 

  1. $25.1 cc$
  2. $25.21 cc$
  3. $2.51 cc$
  4. $251.2 cc$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $r$ be the radius of the sphere.


$\Rightarrow$  $r=10$


Error in the measurement of radius $=\Delta r$

$\therefore$  $\Delta r=0.02\,m$

$\Rightarrow$  Volume of the sphere $(V)=\dfrac{4}{3}\pi r^3$

We need to find error in calculating the volume that is $\Delta V$

$\Delta V=\dfrac{dv}{dr}\times \Delta r$

         $=\dfrac{d\left(\dfrac{4}{3}\pi r^3\right)}{dr}\times \Delta r$

         $=\dfrac{4}{3}\pi\dfrac{d(r^3)}{dr}\times \Delta r$

         $=\dfrac{4}{3}\pi(3r^2)\times (0.0.2)$

         $=4\pi r^2\times 0.02$

         $=4\times 3.14\times (10)^3\times 0.02$

         $=251.2\,cm^3$ i.e. $251.2\,cc$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If there is an error of $k%$ in measuring the edge of a cube, then the percent error in estimating its volume is

  1. $k$
  2. $3k$
  3. $\displaystyle \frac{k}{3}$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of cube$V=x^{3}$
$\displaystyle \frac{dV}{dx}=3x^{2}$
Percentage error in measuring side $= k%$
$\displaystyle \Rightarrow \frac{\delta x}{x}=\frac{k}{100}$
$\displaystyle {\delta{x}}=\frac{xk}{100}$
Approximate error in estimating volume $=dV=(\frac{dV}{dx}){\delta x}=3x^{2}\frac{xk}{100}$
Percentage error in estimating volume$=\frac{dV}{V}\times 100=3k$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The height of a cylinder is equal to the radius. If an error of $\alpha$ % is made in the height, then percentage error in its volume is

  1. $\alpha$ %
  2. $2\alpha$ %
  3. $3\alpha$ %

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Volume of cylinder $V= \pi { r }^{ 2 }h$
Since, $h=r$
$V=\pi {h}^{3}$ 
$\displaystyle \dfrac{dV}{dh}=3\pi h^{2}$
Given, percentage error in measuring height $=\alpha$%
$\Rightarrow \displaystyle \dfrac { \Delta h }{ h } =\dfrac { \alpha }{ 100 } $
$\Rightarrow \displaystyle  { \Delta h }=\dfrac {\alpha h }{ 100 } $
Now, approximate error in measuring V$\displaystyle =dV= (\dfrac{dV}{dh}){ \Delta h}$
                                          $\displaystyle = \dfrac{3\alpha }{100} {\pi h^{3}} =3\alpha$% of V
Percentage error in measuring $V =3\alpha$%