Units and Measurements Test

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Units and Measurements Test - 52

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Question 1
1 / -0

A cube has a side of length $$ 1.2 \times 10^{-2}m $$. Calculate its volume.

A
$$ 1.7 \times 10^{-6} m^3 $$

B
$$ 1.73 \times 10^{-6} m^3 $$

C
$$ 1.70 \times 10^{-6} m^3 $$

D
$$ 1.728 \times 10^{-6} m^3 $$

Solution

Length of the cube, $$L = 1.2 \times 10^{-2} m$$
$$\therefore$$ Volume of the cube, $$V = L^3 = (1.2 \times 10^{-2})$$ $$\times (1.2 \times 10^{-2})$$ $$\times (1.2 \times 10^{-2}) m^3$$
$$\implies V = 1.728 \times 10^{-6} m^3$$
But as the length contains only $$2$$ significant digits, thus volume must also contain $$2$$ significant digits.
Hence, volume of the cube is $$1.7 \times 10^{-6} m^3$$
Question 2
1 / -0

Count total number of S.F. in $$6.020\, \times\, 10^{23}$$

A
4

B
3

C
6

D
7

Solution

S.F. = Four; 6, 0, 2, 0; remaining 23 zeros are not significant.
Question 3
1 / -0

Count total number of S.F. in 300.00

A
5

B
6

C
8

D
9

Solution

S.F. = Five, trailing zeros after decimal point are significant.
Question 4
1 / -0

What are the dimensions of electrical resistance ?

A
[ML$$\displaystyle ^{2}$$T$$\displaystyle ^{-2}$$I$$\displaystyle ^{2}]$$

B
[ML$$\displaystyle ^{2}$$T$$\displaystyle ^{-3}$$I$$\displaystyle ^{-2}]$$

C
[ML$$\displaystyle ^{2}$$T$$\displaystyle ^{-3}$$I$$\displaystyle ^{2}]$$

D
[ML$$\displaystyle ^{2}$$T$$\displaystyle ^{-2}$$I$$\displaystyle ^{-2}]$$

Solution

Hint:
Resistance is defined as the opposition offered by any conductor to the flow of current.

Step 1: Write formula for Electrical Resistance.
By, Ohm's Law, we know that
$$R = \dfrac{V}{I}$$
where $$V$$ is potential difference, and $$I$$ is current.

Step 2: Find Dimensions of Potential and Current.
Dimention of current is given as,
$$[I] = I^1$$

Since we know that Potential is given as,
$$V = \dfrac{W}{q}$$
where, $$W$$ is Work and $$q$$ is charge
Dimention of potential is given as,
$$[V] = \dfrac{[W]}{[q]}$$

$$\Rightarrow[V] = \dfrac{[ML^2T^{-2}]}{[IT]}$$

$$\Rightarrow[V] = [ML^2T^{-3}I^{-1}]$$

Step 3: Calculate Dimension for Resistance.
Dimension for Resistance is given as,
$$[R] = \dfrac{[V]}{[I]}$$

$$\Rightarrow [R] = \dfrac{[ML^2 T^{-3}I^{-1}]}{[I]}$$

$$\Rightarrow [R] = [ML^2 T^{-3}I^{-2}]$$

Thus, Dimension for Electrical Resistance is, $$ [R] = [ML^2 T^{-3}I^{-2}]$$.
Option B is correct.
Question 5
1 / -0

If a conductance of a conductor (G) is $$\displaystyle \frac { { I }^{ 2 }t }{ W } $$ , where I is current, t is time and W is work done then write the unit of conductance expressed in terms of fundamental units.

A
$$\dfrac { { A }^{ 2 }{ s }^{ 3 } }{ { kg\quad m }^{ 2 } } $$

B
$$\dfrac { { A }^{ 3 }{ s }^{ 2 } }{ { kg\quad m }^{ 2 } } $$

C
$$\dfrac { { A }^{ 2 }{ s }^{ 3 } }{ { kg\quad m }^{ 1 } } $$

D
$$\dfrac { { A }^{ 3}{ s }^{ 2 } }{ { kg\quad m }^{1 } } $$

Solution

Fundamental unit of $$I$$ is $$A$$ and that of $$t$$ is $$s$$.

The fundamental unit of work done $$W $$ is $$Joules$$ and it is given as $$kg \ m^2/s^2$$ in the SI system.

Thus unit of conductance $$ = \dfrac{I^2 t}{W} = \dfrac{A^2 \ s}{kg \ m^2/s^2} = \dfrac{A^2 \ s^3}{kg \ m^2}$$
Question 6
1 / -0

An ice cube has a stone of 500 g placed on its top, is floating in water with its lateral sides placed vertically. It displaces 5 kg of water. Suddenly, the stone slips into water. Because of this ice cube rises by $$\displaystyle \frac{1}{10}th$$ of its length above the water level. What is the density of the ice cube (in $$kg/m^3$$)?

A
960

B
840

C
900

D
820

Solution
Question 7
1 / -0

What is the dimensional formula of universal gravitational 'G'? The gravitational force of attraction between two objects of masses $$\displaystyle { m }_{ 1 }\quad and\quad { m }_{ 2 }$$ seperated by a distance d is given by $$\displaystyle F=\frac { G{ m }_{ 1 }{ m }_{ 2 } }{ { d }_{ 2 } } $$. Where 'G' is the universal gravitational constant.

A
$$\displaystyle [M\quad { LT }^{ -2 }]$$

B
$$\displaystyle [{ M }^{ -1 }{ L }^{ 2 }{ T }^{ 2 }]$$

C
$$\displaystyle [{ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }]$$

D
$$\displaystyle\ [{ M }^{ 2 }{ L }^{ 2 }{ T }^{ -2 }$$

Solution

Gravitational constant can be written as $$G = \dfrac{Fr^2}{m_1m_2}$$
So, its dimensions are $$[G] = \dfrac{[F][r^2]}{[m_1][m_2]}$$

$$\implies \ [G] = \dfrac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1}L^3T^{-2}]$$
Question 8
1 / -0

Which one of the following defines the mass of a body?

A
Total weight of the body

B
Gravitational force exerted by the body

C
Matter contained in the body

D
Gravitational force exerted on the body

E
None of these

Solution
Question 9
1 / -0

The pair of quantities that do not have the same dimensions is:

A
Latent heat, specific heat

B
Gravitational force, Coulomb force

C
Kinetic energy of a freely falling body, potential energy of a compressed spring

D
Coefficient of friction, number of molecules in a container

Solution

$$\textbf{Explanation:}$$
$$\bullet$$The quantity that has same unit will have same dimension representation in terms of MLTAK where M is mass, L is length, T is time , A is ampere and K is for kelvin (temperature).

$$\bullet$$Gravitational force and coulomb force are types of force which can be measured in Newton. Both have same unit. So, there dimensions will be same as well.

$$\bullet$$Similarly, coefficient of friction and number of molecules in a container both are unitless and kinetic energy of freely falling body and potential energy of compressed spring both are measure in joule. So, they have same dimensions.

$$\bullet$$But latent heat is energy required to change the state of a unit mass while specific heat is energy required to change to change unit mass by 1°. So, unit of latent heat is joule/kg while unit of specific heat is joule/°c. As unit of both quantity are different, their dimensions will be different.

$$Answer:$$

Hence, option A is the correct answer.
Question 10
1 / -0

The distance between two places is 25 km. Write the magnitude of the measurement.

A
km

B
distance

C
25

D
two

Solution

The distance between the two places is $$25$$ km where $$25$$ represents the magnitude of measurement and km represents the unit.

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Units and Measurements Test - 52

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Units and Measurements Test - 52

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Question 1

$$ 1.7 \times 10^{-6} m^3 $$

$$ 1.73 \times 10^{-6} m^3 $$

$$ 1.70 \times 10^{-6} m^3 $$

$$ 1.728 \times 10^{-6} m^3 $$

Solution

Question 2

4

3

6

7

Solution

Question 3

5

6

8

9

Solution

Question 4

[ML$$\displaystyle ^{2}$$T$$\displaystyle ^{-2}$$I$$\displaystyle ^{2}]$$

[ML$$\displaystyle ^{2}$$T$$\displaystyle ^{-3}$$I$$\displaystyle ^{-2}]$$

[ML$$\displaystyle ^{2}$$T$$\displaystyle ^{-3}$$I$$\displaystyle ^{2}]$$

[ML$$\displaystyle ^{2}$$T$$\displaystyle ^{-2}$$I$$\displaystyle ^{-2}]$$

Solution

Question 5

$$\dfrac { { A }^{ 2 }{ s }^{ 3 } }{ { kg\quad m }^{ 2 } } $$

$$\dfrac { { A }^{ 3 }{ s }^{ 2 } }{ { kg\quad m }^{ 2 } } $$

$$\dfrac { { A }^{ 2 }{ s }^{ 3 } }{ { kg\quad m }^{ 1 } } $$

$$\dfrac { { A }^{ 3}{ s }^{ 2 } }{ { kg\quad m }^{1 } } $$

Solution

Question 6

960

840

900

820

Solution

Question 7

$$\displaystyle [M\quad { LT }^{ -2 }]$$

$$\displaystyle [{ M }^{ -1 }{ L }^{ 2 }{ T }^{ 2 }]$$

$$\displaystyle [{ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }]$$

$$\displaystyle\ [{ M }^{ 2 }{ L }^{ 2 }{ T }^{ -2 }$$

Solution

Question 8

Total weight of the body

Gravitational force exerted by the body

Matter contained in the body

Gravitational force exerted on the body

None of these

Solution

Question 9

Latent heat, specific heat

Gravitational force, Coulomb force

Kinetic energy of a freely falling body, potential energy of a compressed spring

Coefficient of friction, number of molecules in a container

Solution

Question 10

km

distance

25

two

Solution

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