Tag: the need for approximation

Questions Related to 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

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

  1. $0.4\% $

  2. $0.5\% $

  3. $0.2\% $

  4. $0.1\% $

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Correct Option: C
Explanation:
Given:Length of the cylinder $=l=4.28\,cm$
Error$=\Delta l=0.01\,cm$
Percentage error$=\dfrac{\Delta\,l}{l}\times 100$
$=\dfrac{0.01}{4.28}\times 100=0.234\approx\,0.2\%$

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

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

  1. $25.1 cc$

  2. $25.21 cc$

  3. $2.51 cc$

  4. $251.2 cc$

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Correct Option: D
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. $251.2\,cc$

What is the sum of the factors of 496 ? 

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

  1. 990

  2. 996

  3. 992

  4. 985

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Correct Option: C
Explanation:

The factors of $496$ are $1, 2, 4, 8, 16, 31, 62, 124, 248,496$


The sum of the factors is $1+ 2+ 4+ 8+ 16+ 31+ 62+ 124+ 248+496=992$

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

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

  1. $k$

  2. $3k$

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

  4. none of these

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Correct Option: B
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$=\frac{dV}{V}\times 100=3k$

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

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

  1. $\alpha$ %

  2. $2\alpha$ %

  3. $3\alpha$ %

  4. none of these

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Correct Option: C
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 $V =3\alpha$%

The pressure P and volume V of a gas are connected by the relation $PV^{1/4}=constant$. The percentage increase in the pressure corresponding to a deminition of $\dfrac12 \%$ in the volume is

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

  1. $\dfrac {1}{2}$ %

  2. $\dfrac {1}{4}$ %

  3. $\dfrac {1}{8}$ %

  4. none of these

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Correct Option: C
Explanation:
$PV^{1/4}=constant$
$\displaystyle \Rightarrow P=\dfrac{k}{V^{{1}/{4}}}$
$\displaystyle \Rightarrow \dfrac{dP}{dV}=-\dfrac{k}{4}V^{-{5}/{4}}$
Percentage error in V $\displaystyle= -\dfrac{1}{2}\%$
$\Rightarrow\displaystyle \dfrac{\Delta V}{V} =-\dfrac{1}{200}$
$\Rightarrow\displaystyle {\Delta V}=-\dfrac{V}{200}$
Approximate change in $P\displaystyle=dP=(\dfrac{dP}{dV}){\Delta V}$
                                           $\displaystyle  =\dfrac{1}{800} {kV^{-1/4}}=\dfrac{1}{8}\%$ of P
Percentage increase in  $V \ \displaystyle =\dfrac{1}{8}\%$

If the ratio of base radius and height of a cone is 1:2 and percentage error in radius is $\lambda$ %, then the error in its volume is

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

  1. $\lambda$ %

  2. $2\lambda$%

  3. $3\lambda$%

  4. none of these

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Correct Option: A
Explanation:
Volume of cone $V=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h $
Given, $\displaystyle \dfrac{r}{h}=\dfrac{1}{2}$
$\Rightarrow V=\dfrac { 2 }{ 3 } \pi { r }^{ 3 } $
$\displaystyle \dfrac{dV}{dr}=2\pi r^{2}$
Percentage error in measuring r $=\lambda$%
$\Rightarrow \displaystyle \dfrac{\Delta r}{r}=\dfrac{\lambda}{100}$
$\Rightarrow \displaystyle \Delta r =\dfrac{\lambda r}{100}$
Approximate error in V $\displaystyle=dV=(\dfrac{dV}{dr}) \Delta r $
                                      $\displaystyle=\dfrac{\lambda}{100} (2\pi r^{3})=\lambda\%$ of V
Percentage error in $V =\lambda\%$

If $y=x^n$, then the ratio of relative errors in $y$ and $x$ is

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

  1. $1:1$

  2. $2:1$

  3. $1:n$

  4. $n:1$

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Correct Option: D
Explanation:
$y=x^{n}$
$\Rightarrow \displaystyle \dfrac{dy}{dx}=nx^{n-1}$
Approximate error in y is $\displaystyle dy=\left (\dfrac{dy}{dx}\right) \Delta x$
                                      $=nx^{n-1} \Delta x$
Relative error in y is $\displaystyle \dfrac{dy}{y}=\dfrac{n}{x}\Delta x$
Approximate error in x is $\displaystyle dx=\left (\dfrac{dx}{dy}\right) \Delta y$
                                     $\displaystyle=\dfrac{1}{nx^{n-1}} \Delta y$
Relative error in x is $\displaystyle \dfrac{dx}{x}=\dfrac{1}{nx^{n}}\Delta y$
Required ratio $\displaystyle = \dfrac{\dfrac{n}{x}\Delta x}{\dfrac{1}{nx^{n}}\Delta y}$
                               $\displaystyle =n^{2}x^{n-1} \dfrac{\Delta x}{\Delta y}$
                               $\displaystyle =\dfrac{n}{1}$
So, the ratio of relative errors in y and x is $ n:1$.

The circumference of a circle is measured as $28 cm$ with an error of $0.01 cm$. The percentage error in the area is

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

  1. $\dfrac {1}{14}$

  2. $0.01$

  3. $\dfrac {1}{7}$

  4. none of these

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Correct Option: A
Explanation:

Circumference $C=2\pi r$
$\Rightarrow\displaystyle r=\frac{14}{\pi}$
Also, $\displaystyle \frac{dC}{dr}=2\pi$
Area of circle $A=\pi r^{2} $
$\Rightarrow\displaystyle A=\frac{{14}^{2}}{\pi}$
Also, $\displaystyle \frac{dA}{dr}=2\pi r$
$\displaystyle \Rightarrow \frac{dA}{dC}=r=\frac{14}{pi}$
Approximate error in $A$ is $\displaystyle dA=( \frac{dA}{dC}) \Delta C$
                           $\displaystyle=\frac{14}{\pi}\frac{1}{100}$
                            $\displaystyle=\frac{1}{1400}$ of A
Percentage error in $A \ \displaystyle =\frac{1}{14}\%$

If there is an error of $0.01 cm$ in the diameter of a sphere then percentage error in surface area when the radius $= 5 cm$, is

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

  1. $0.005\%$

  2. $0.05\%$

  3. $0.1\%$

  4. $0.2\%$

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Correct Option: D
Explanation:

Surface area of sphere $S=4\pi r^{2}$
$\displaystyle S=\pi D^{2}$
$\Rightarrow S=100\pi$
Also, $ \displaystyle \frac{dS}{dD}=2\pi D=20\pi$
Approximate error in S is $\displaystyle dS=(\frac{dS}{dD})\Delta D$
                                         $ =20\pi (0.01)$
                                          $=\dfrac{1}{500} S$
                                           $=0.2$% of S
Percentage error in $S=0.2%$