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Measurement of area and volume - class-VII

Description: measurement of area and volume
Number of Questions: 36
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Tags: measurements and uncertainties physical quantities and measurement measurement making measurements learning how to measure physics measurements and units
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$1000l$ = ?

  1. $1ml$

  2. $1m^3$

  3. $1cm^3$

  4. $1000m^3$


Correct Option: B
Explanation:

$1000l=1m^3$. $l$ is used to measure the volume of a fluid and $m^3$ is generally used to express volume of a solid.

$\displaystyle 1{ m }^{ 3 }$ = ?

  1. $10^6cm^3$

  2. $\displaystyle 1{ cm }^{ 3 }$

  3. $10^{-6}cm^3$

  4. None of these


Correct Option: A
Explanation:

$1m=100cm=10^2{cm}$

So, $1m^3=({10^2})^3cm^3=10^6cm^3$

Which is the correct form of expressing SI unit of volume?

  1. $\displaystyle { m }^{ 3 }$

  2. $cubicmetre$

  3. $\displaystyle { metre }^{ 3 }$

  4. all of these


Correct Option: D
Explanation:

SI unit of volume is $m^3$ which can also be written as $cubicmetre$ and $metre^3$. It is a big unit, the smaller units are also in use.

The space occupied by a body is called its

  1. area

  2. volume

  3. mass

  4. none of these


Correct Option: B
Explanation:

The space occupied by a body is called its volume. SI unit of volume is $m^3$.

The volume of a liquid is expressed in

  1. $l$

  2. $l^3$

  3. $m^3$

  4. $kg$


Correct Option: A
Explanation:

The volume of liquids is expressed in $litre(l)$.

The volume of a stone can be determined by the displacement method using a measuring cylinder. True or false.

  1. True

  2. False


Correct Option: A
Explanation:

It is true that the volume of a stone (or any irregular shaped body) can be determined by the displacement method using a measuring cylinder. Volume of the stone is equal to the difference of final level of water in the mesuring jar and the initial level.

Name the unit in which the volume of liquid is generally measured. How can this unit is related to L.

  1. kilogram, 1 litre = 10$^{-6}$ m$^3$

  2. litre, 1 litre = 10$^{-6}$ m$^3$

  3. litre, 1 litre = 10$^{-3}$ m$^3$

  4. litre, 1 litre = 10$^{-3}$ cm$^3$


Correct Option: C
Explanation:

Volume is generally measured in liter, $1\ liter=10^{-3}m^3$

Mark the formulae which is used to measure volume of a regular shaped object.

  1. Volume of a cuboid $=$ Length $\times$ Breadth $\times$ Height

  2. Volume of a cube of side $a$ each, $V = {a}^{3}$

  3. Both A and B

  4. None of A and B


Correct Option: C
Explanation:

If the length, breadth and height are equal, a cube is formed. Let $a$ be the length of each side of the cube, then its volume $V$ is equal to $a \times a \times a$, or ${a}^{3}$.

Volume of a cube of side a each, $V  = {a}^{3}$

The volume of a cuboid of length $\left(l\right)$, breadth $\left(b\right)$ and height $\left(h\right)$, is the product of length, breadth and height.

Volume of a cuboid $=$ Length $\times$ breadth $\times$ Height

The space occupied by a substance is its

  1. Area

  2. Volume

  3. Surface area

  4. Cross section area


Correct Option: B
Explanation:

The space occupied by a substance is its volume. The $SI$ unit for volume is the cubic metre $\left({m}^{3}\right)$.

Sub-multiples of the cubic metre, the cubic centimetre and the cubic millimetre are used for smaller measurements.

In a burette, the zero mark on the graduation scale is near its mouth.

  1. True

  2. False


Correct Option: A
Explanation:

A burette is a device used in labs (chemistry) for the dispensing of variable, measured amounts of liquids or solutions. A graduated volumetric burette delivers measured volumes of liquid. Hence, a burette determines the difference between the initial and final volumes of liquid. Therefore the scale on burette is marked downwards(zero at the top), so as to measure correct amount of volume delivered.
Hence, given sentence is true. 

Which of the following is used to measure the volume of a regular shaped object?

  1. Metre scale

  2. Vernier callipers

  3. Screw guage

  4. All of these


Correct Option: D
Explanation:

To find the volume of a regular shaped object ,first its dimensions are measured using either a metre scale, a vernier callipers or a screw gauge and then volume is calculated using corresponding relations.

What is best way to measure the volume of an irregularly shaped solid?

  1. By using a beam balance

  2. By using water displacement

  3. By using a ruler to measure the length of each side of the object

  4. By using formula : length x width x height


Correct Option: B

Available capacity(s) of standard measuring flask is / are :

  1. 50 ml only

  2. 100 ml only

  3. 200 ml only

  4. all of these


Correct Option: D
Explanation:

Measuring flask consists of a long narrow neck provided with a stopper. Different capacities (50 ml, 100 ml, 250 ml etc) of standard measuring flasks are available.

Choose the INCORRECT statement from the following.

  1. Volume is a chemical property of matter

  2. Volume is a physical property of matter

  3. Volume is needed to find density

  4. Volume is the amount of space something takes up


Correct Option: A
Explanation:

Option D is the definition of Volume. It is only a physical property of any object and has nothing to do with the chemical properties. Since density is mass/ volume, it is necessary to find density.

Which of the following is used to measure the volume of a liquid or an irregular shaped object?

  1. A burette

  2. A measuring flask

  3. A measuring cylinder

  4. All of these


Correct Option: D
Explanation:

Volume of a liquid or an irregular solid object is measured with the help of burettes, measuring flasks and measuring cylinders.

Which of the following is not used to measure the volume of a liquid or an irregular shaped object?

  1. A pipette

  2. A measuring flask

  3. A graduated cylinder

  4. A meter scale


Correct Option: D
Explanation:

A liquid or an irregular shaped object does not have a geometric shape. So, by measuring its certain dimensions, volume may not be calculated. Hence a meter scale can not be used to measure the volume of a liquid or an irregular shaped object.

Find the volume of a ring immersed in water ,increasing the level from 55.2 ml to 75.3 ml.

  1. $20.1\ ml$

  2. $29\ ml$

  3. $10.1\ ml$

  4. $14.1\ ml$


Correct Option: A
Explanation:

Initial volume level   $V _i = 55.2 \ ml$
Final volume level   $V _f = 75.3 \ ml$
Thus, volume of ring    $V = V _f-V _i$
$\implies \ V = 75.3-55.2 = 20.1 \ ml$

The volume enclosed by the cylinder of diameter $1.06\ m$  and height 7.2 m to the correct no. of the significant figure is:

  1. $6.350\ m^3$

  2. $6.35\ m^3$

  3. $6.3\ m^3$

  4. $6\ m^3$


Correct Option: C
Explanation:

Diameter of cylindrical vessel  $d = 1.06 \ m$
Height of cylindrical vessel  $h = 7.2 \ m$
Volume   $V = \dfrac{\pi d^2 h}{4}$
$\implies \ V = \dfrac{\pi\times (1.06)^2\times 7.2}{4} = 6.354 \ m^3$
Since the number $7.2$ has two significant figures and $1.06$ has three significant figures.
So the final answer must have 2 significant figures.
Thus volume of cylinder   $V = 6.3 \ m^3$

Volume enclosed by the sphere of diameter $1.06\ m$ to the correct no. of significant figure is:

  1. $0.62329 \ m^3$

  2. $0.623 \ m^3$

  3. $0.6232 \ m^3$

  4. None of these


Correct Option: B
Explanation:

The significant figures are the number of digits in a value, that are significant. In $1.06m$, the number of significant figures is 3. In the answer also we must maintain 3 significant figures only. Hence option B is correct.

The density of steel rod is $7850\ kg/m^3$, its mass is $5\ kg$.Find its volume?

  1. $7.36\times 10^{-4}\ m^3$

  2. $36\times 10^{-4}\ m^3$

  3. $6.36\times 10^{-4}\ m^3$

  4. $6\times 10^{-4}\ m^3$


Correct Option: C
Explanation:

Mass of steel rod $m = 5 \ kg$
Density of steel rod   $d = 7850 \ kg/m^3$
Volume of steel rod  $V = \dfrac{m}{d}$
$\implies \ V = \dfrac{5}{7850} = 6.36\times 10^{-4} \ m^3$

Each side of a cube is measured to be $7.203\ m$.The volume of the cube to appropriate significant figure is:

  1. $31.3\ m^3$

  2. $313\ m^3$

  3. $373.7\ m^3$

  4. $37.3\ m^3$


Correct Option: C
Explanation:

Side of cube  $a = 7.203 \ m$
Volume of cube  $V = a^3$
$\implies \ V = (7.203)^3 = 373.714 \ m^3$
Since the value of side of cube has 4 significant figures, thus the value of volume of cube must also have 4 significant figure.
Thus volume of cube  $V = 373.7 \ m^3$

Density and mass of the solid block is $5.2\  kg/m^3$ and $2\ kg$.Find its volume?

  1. $0.3846\ m^3$

  2. $7846\ cm^3$

  3. $68.46\ cm^3$

  4. $98.46\ cm^3$


Correct Option: A
Explanation:

Mass of solid block  $m = 2 \ kg$
Density of solid block   $d = 5.2 \ kg/m^3$
Volume of solid block  $V = \dfrac{m}{d}$
$\implies \ V = \dfrac{2}{5.2} = 0.3846 \ m^3$

Which of the following instruments can be used to measure the volume of a liquid?

  1. Measuring cylinder

  2. Vernier callipers

  3. Pipette

  4. Measuring flask


Correct Option: A,C,D
Explanation:

Measuring cylinder, pipette and measuring flask are used to measure the volume of the given liquid whereas vernier calliper is used measure the length of the object.

A rectangular vessel $ 1 m \times 0.6 m \times 0.5 m $ filled with maize grains. Treat the grains spheres of diameter 0.3 cm each.The number of grains in the vessel is the nearest to 

  1. $ 5 \times 10^7 $

  2. $ 1 \times 10^7 $

  3. $ 2 \times 10^7 $

  4. $ 2 \times 10^5 $


Correct Option: C
Explanation:

Let n number of grains are in the vessel.

So, volume of cuboid vessel is equal to n times volume of spheres. i.e.

$ 1\times 0.6\times 0.5=n\left( \dfrac{4}{3}\times 3.14\times {{\left( \dfrac{3}{2000} \right)}^{3}} \right) $

$ 0.3=n\dfrac{4}{3}\times 3.14\times 3.375\times {{10}^{-9}} $

$ 0.3=n\times 14.13\times {{10}^{-9}} $

$ n=\dfrac{0.3}{14.13\times {{10}^{-9}}} $

$ n=0.021\times {{10}^{9}}\, $

$ n=2\times {{10}^{7}}$

A single sheet of aluminium foil is folded twice to produce a stack of four sheets. The total thickness of the stack of sheets is measured to be $(0.80\pm 0.02)mm$. This measurement is made using a digital caliper with a zero error or $(-0.20\pm 0.02)mm$
What is the percentage uncertaintyin calculated thickness of a single sheet?
  1. $1.0$%

  2. $2.0$%

  3. $4.0$%

  4. $6.7$%


Correct Option: C
Explanation:

The zero error (−0.20) can be removed but its uncertainty (± 0.02) must be added to the measurement uncertainty.

True value of total thickness =$ 0.80 + 0.20 = 1.0 mm  $

Total uncertainty in measurement = $(0.02+0.02)=0.04\ mm$

So the four sheets have a true thickness of $ (1.00 ± 0.04) mm.$
 A single sheet would have a thickness of $(0.25 ± 0.01) mm.$

Percentage error= $\dfrac{0.01}{0.25}\times 100 = 4.0$%

What is the S.I. unit of volume ? How is it related to litre ?

  1. $m^3, 1 \ m^3 = 1000$ litre

  2. $cm^3, 1 \ cm^3 = 10^{-3}$ litre

  3. $m^3, 1 \ m^3 = 10^6$ litre

  4. None of the above


Correct Option: A
Explanation:

S.I. unit of volume is $m^3$. 

$1m^3=1000\ litre$

Volume of a single liquid drop can be measured using :

  1. Burette

  2. Pipette

  3. Flask

  4. Measuring cone


Correct Option: A
Explanation:

A Burette  is a device used in analytical chemistry for the dispensing of variable, measured amounts of a chemical solution. A volumetric burette delivers measured volumes of liquid.

1 litre is equal to

  1. $10^3\,cm^3$

  2. $10\,dm^3$

  3. $10^{-3}\,cm^3$

  4. all the above


Correct Option: A
Explanation:

We know that $1$ litre $=10^3 cm^3$

$1 cm=10^{-1} dm$ or $1 cm^3=10^{-3} dm^3$
So, $10^3 cm^3=10^3\times 10^{-3} dm^3=1 dm^3$
Hence, 1 litre $=1 dm^3$
Thus, option A will be correct. 

State whether true or false.

A measuring jar is graduated in ml from bottom to top.

  1. True

  2. False


Correct Option: A
Explanation:

The given statement is true that the measuring jar is graduated in ml from bottom to top.

The volume of water in a measuring cylinder is 55 mL When a stone tied to a string is immersed in the water the water level rises to 83 mL. Find the volume of the stone

  1. $28ml^2$

  2. $28ml$

  3. $38ml$

  4. $28m^2l$


Correct Option: B
Explanation:
Initial level of water  $v _i = 55 \ ml$
Final level of water  $v _f = 83  ml$
Volume of stone  $V _{stone} = V _f - V _i = 83 - 55 = 28 \ ml$

 In a given system of units, the ratio of unit of volume to that of its area gives 

  1. Density

  2. Length

  3. Mass

  4. Decibel


Correct Option: B
Explanation:

SI Unit of volume is $m^3$ 

Unit of area is $m^2$
So $\dfrac{Volume}{area}=m^3/m^2=m$, that's the unit of length 

A cubic metre of water at $0^o$C is solidified into ice, density of ice is $0.96$ of water at $0^o$C. Which of the following deductions are true?
$1$. Water expands when solidified.
$2$. Ice will float in water at $0^o$C with half its volume above the surface.
$3$. Density of all liquids are higher than its density when solidified.

  1. $3$ only

  2. $2$ only

  3. $1$ only

  4. $1$ and $2$


Correct Option: C

A cubical block of wood of specific gravity $0.5$ and a chunk of concrete of specific gravity $2.5$ are fastened together. The ratio of mass of wood to the mass of concert which makes the combination to float with its entire volume submerged in water is 

  1. $1/5$

  2. $1/3$

  3. $3/5$

  4. $2/5$


Correct Option: A

There are $n\ sets$ of real numbers. In every set there are four real numbers in AP. Such that the square of the last term is equal to the sum of the square of the first three terms. Then: -

  1. $N = 4$

  2. $N = 1000$

  3. $N = 10$

  4. $N \rightarrow \infty$


Correct Option: C

Four  positive charges $ ( 2 \sqrt 2-1 )Q $ are arranged at corner of square . another charge q is placed at the centre of the square. resultant force acting on each corner is zero if q is :

  1. $-7Q/4$

  2. $-4Q/ 7$

  3. $-Q$

  4. $none$


Correct Option: C

A public park, in the form of a square, has an area of $(100 \pm 0.2 )m^2 $ .The side of park is :

  1. $ (10 \pm 0.01 ) m $

  2. $ (10 \pm 0.1 ) m $

  3. $ (10 \pm 0.02 ) m $

  4. $ (10 \pm 0.2 ) m $


Correct Option: A
Explanation:

Let the side length of the square park is $l$.

The area of a square is given as,

$A = {l^2}$

$100 = {l^2}$

$l = 10\;{\rm{m}}$

The error in length is given as,

$\dfrac{{\Delta A}}{A} = 2\dfrac{{\Delta l}}{l}$

$\dfrac{{0.2}}{{100}} = 2\dfrac{{\Delta l}}{{10}}$

$\Delta l = 0.01\;{\rm{m}}$

Thus, the side of the park is $\left( {10 \pm 0.01} \right)\;{\rm{m}}$.

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