Quantitative Aptitude

Mensuration

2,507 Questions
All Direct Formula Application 1067 Shape Transformations 343 Mensuration Ratio Equations 277 Combined & Hollow Solids 247 Successive Percentage Changes 178 Inscribed & Circumscribed Figures 165 Cost Estimation Problems 100 Liquid Pour & Submergence 67 Unit Conversions 35 Path & Border Calculations 17 Time & Work Integration 7 Scale & Model Proportions 2

Mensuration Questions

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The error in the measurement of the radius of a sphere is $0.5$ %. Find the permissible error in the measurement of surface area?

  1. $0.1$ %
  2. $10$ %
  3. $5$ %
  4. $1$ %
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The percentage error in measurement of radius is given, $\dfrac{\Delta r}{r}\times 100 =0.5$%
Thus, $\dfrac{\Delta r}{r}=0.5/100=0.005$
The surface area of sphere is $S=4\pi r^2$
Take ln and differentiate, $\dfrac{\Delta S}{S}=2\dfrac{\Delta r}{r}=2\times 0.005=0.01 $
The permissible or % error in the measurement of surface area $=\dfrac{\Delta S}{S}\times 100=0.01\times 100=1$%

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

The radius and height of a cone are measured as $6cms$ each by scale in which there is an error of $0.01cm$ in each cm. then the approximate error in its volume is.

  1. $.14$
  2. $.12$
  3. $.36$
  4. $0.16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Formula,

$V=\pi r^2 \dfrac{h}{3}$

$=\pi\times 6^2 \dfrac{6}{3}=226.19$

$\dfrac{\Delta V}{V}=\dfrac{\pi}{3}6 \times 0.01 \times 0.01=0.000628$

The change in volume is,

$V=0.000628\times 226.19=0.14$%

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

Length of a thin cylinder as measured by vernier callipers having least cound $0.01\ cm$ is $3.25\ cm$ and its radius of cross-section is measured by a screw gauge having least count $0.01\ mm$ as $2.75\ mm$. The percentage error in the measurement of volume of the cylinder will be

  1. $2\%$
  2. $3\%$
  3. $1\%$
  4. $1.5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

If error in measurement of radius of a sphere is 1%, what will be the error in measurement of volume?

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

  3. 3%

  4. 10%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The volume is given by $V=\dfrac{4}{3} \pi R^3$, where $R$ is radius of sphere.


$\dfrac{\delta V}{V}= 3\dfrac{ \delta R}{R}$

$\dfrac{\delta V}{V}= 3 \times 1 $ % $=3 $ %

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

A uniform wire of radius $r=0.5 \pm 0.005 cm$ length $l =5\pm 0.05 cm $. The maximum percentage error in its volume is

  1. $30 \%$
  2. $3 \%$
  3. $2 \%$
  4. $1.5 \%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The volume of the wire is $V=\pi r^2 l$
Thus, the relative error, $\dfrac{\Delta V}{V}=\pm \left(2\dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)$
The maximum % error in V  $=\dfrac{\Delta V}{V}\times 100= \left(2\dfrac{\Delta r}{r}+\dfrac{\Delta l}{l}\right)\times 100=\left(2\dfrac{0.005}{0.5}+\dfrac{0.05}{5}\right)100=3\%$

Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

A cube has a side of length 1.2 x $10^{-2}$. Its volume upto correct significant figures is

  1. 1.7 x $10^{-6}m^3$
  2. 1.73 x $10^{-6}m^3$
  3. 1.78 x $10^{-6}m^3$
  4. 1.732 x $10^{-6}m^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Here
Length of the cube, $L =12 x10^{-2} m$
Volume of the cube, $V= (1.2 \times10^{-2} m)^3= 1.728 \times 10^{-6}m^3$
As the result can have only two significant figures, therefore, on rounding off, we get, $V. 1.7 \times 10^{-6} m^3$
Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

Calculate area enclosed by a circle of diameter $1.06\ m$ to correct number of significant figures.

  1. $ 0.883 m^2$
  2. $ 0.0883 m^2$
  3. $ 0.88333 m^2$
  4. $ 0.8830 m^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area = $ \pi (\dfrac{d}{2})^2  $

$ \Rightarrow Area = \dfrac{22}{7}\times 0.53^2 = 0.882828 $
But, no of significant in diameter should be equal to the no. of significant in Area, so Ans is $ 0.883 m^2$

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The radius of a sphere is 1.41 cm. Its volume to an appropriate number of significant figures is then

  1. 11.73 $cm^3$
  2. 11.736 $cm^3$
  3. 11.7 $cm^3$
  4. 117 $cm^3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Radius of the sphere, $r = 1.41 cm$ ($3$ significant figures)
Volume of the sphere,
$\displaystyle V = \dfrac {4}{3} \pi r^3 = \dfrac{4}{3} \times 3.14 \times (1.41)^3\,cm^3 = 11.736\, cm^3$
Rounded off upto $3$ significant figures $= 11.7 cm^3$.
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

the length and breath of a metal sheet are $2.325\ m$ and $3.142\ m$ respectively. What is the area of this sheet using proper number of significant figures

  1. $7.30515\ m^{2}$
  2. $7.3051\ m^{2}$
  3. $7.305\ m^{2}$
  4. $7.31\ m^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The area is calculated by multiplying length and breadth: 2.325 * 3.142 = 7.30515 m^2. Since both given measurements have four significant figures, the result must be rounded to four significant figures, giving 7.305 m^2.

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The radius of a sphere is $5$ cm. Its volume will be given by (according to the theory of significant figures) :

  1. $523.33\ { cm }^{ 3 }$
  2. $5.23\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$
  3. $5.0\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$
  4. $5\ { \times\ { 10 }^{ 2 }cm }^{ 3 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume$=\dfrac{4}{3}{\pi r}^{3}=\dfrac{4}{3}\times \pi \left ( 5 \right )^{3}=523.33\ {cm}^{3}$$=5.2333\times 10^{2}\ cm^{3}$
                                                                                   $\downarrow $
                                                                          5 significant figures
Since radius has single significant figure, so, volume should also have single significant figure.

For single significant figure, we have to drop all after decimal.
$\Rightarrow$ Volume $= 5\times 10^{2}\ cm^{2}$ (if the digit to be dropped is less than 5, preceding digit is left unchanged)

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The diameter of a sphere is $4.24\ m$. Its surface area with due regard to significant figures is :

  1. 5.65 ${ m }^{ 2 }$
  2. 56.5 ${ m }^{ 2 }$
  3. 565 ${ m }^{ 2 }$
  4. 5650 ${ m }^{ 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$d=4.24\;m$
$SA=$ Surface area $=4\pi r^{2}=4\pi \left ( \dfrac{d}{2} \right )^{2}=\pi d^{2}$
$=\pi \left ( 4.24 \right )^{2}$
$=56.47\ m^{2}$
Since significant figure in 4.24 is 3, so we express the answer in 3 significant figures.
$SA=56.47\ m^2 \approx 56.5\ m^2$
(Rounding off to 3 significant figures - if digit to be dropped is more than 5, the preceding digit is raised by 1)

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The volume of a sphere is $ 1.76\;{cm }^{ 3 }$. The volume of 25 such spheres according to the idea of significant figures in ${cm}^{ 3 }$ is 

  1. 44.00

  2. 44.0

  3. 44

  4. 4.4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of 1 sphere $=1.76 cm^{3}$
Volume of 25 spheres $=25\times 1.76$
                                      $=44 {cm}^{3}$
                                      $=44.0 {cm}^{3}$
                                 (Since volume is reported in 3 significant figure)