Units and Measurements Test

Result

Units and Measurements Test - 48

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The dimensional formula for magnetic flux is

A
$$\displaystyle \left [ ML^2T^{-2}A^{-1} \right ]$$

B
$$\displaystyle \left [ ML^{3}T^{-4}A^{-2} \right ]$$

C
$$\displaystyle \left [ M^{0}L^{-2}T^{-2}A^{-2} \right ]$$

D
$$\displaystyle \left [ ML^{2}T^{-1}A^{2} \right ]$$

Solution

Formula for magnetic flux will be $$B \times A$$
$$B$$ =magnetic field. Dimensions will be $$[MT^{-2}I^{-1}]$$
$$A$$ =Area. Dimensions will be $$[L^2]$$
Dimension for magnetic flux will be $$[ML^2T^{-2}I^{-1}]$$
Question 2
1 / -0

Which one of the following has the dimensions of pressure?

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

B
$$\displaystyle \left [ M^{-1}L^{-1} \right ]$$

C
$$\displaystyle \left [ MLT^{-2} \right ]$$

D
$$\displaystyle \left [ ML^{-1}T^{-2} \right ]$$

Solution

Dimension of pressure is given as $$\displaystyle \frac{|F|}{|A|}=\frac{|ma|}{|A|}=\frac{M.L{T}^{-2}}{{L}^{2}}=M{L}^{-1}{T}^{-2}$$
Question 3
1 / -0

The dimensions of voltage in terms of mass $$(M)$$, length $$(L)$$ and time $$(T)$$ and ampere $$(A)$$ are

A
$$[ML^2T^{-2}A^{-2}]$$

B
$$[ML^2T^{3}A^{-1}]$$

C
$$[ML^2T^{-3}A^{1}]$$

D
$$[ML^2T^{-3}A^{-1}]$$

Solution

Voltage(V) is a representation of the electric potential energy(W) per unit charge(Q).
So, $$[V]=\left [\dfrac {W}{Q}\right ]=\dfrac {[ML^2T^{-2}]}{[AT]}=[ML^2A^{-1}T^{-3}]$$
Question 4
1 / -0

Consider the following pairs of quantities:
1. Young's modulus; pressure
2. Torque; energy
3. Linear momentum; work
4. Solar day; light year
In which cases are the dimensions, within a pair, same?

A
1 and 3

B
1 and 4

C
1 and 2

D
2 and 4

Solution

Young's Modulus $$=$$ Pressure $$=\left[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } \right] $$
[Torque] $$=$$ [Energy] $$=\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right] $$
Linear momentum $$ =\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -1 } \right] $$, work $$ =\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 } \right] $$
Solar day $$ =[T]$$ and Light year$$ =[L]$$
Question 5
1 / -0

The physical quantity having the dimensions $$\displaystyle \left [ M^{-1}T^3L^{-3}A^{2} \right ]$$ is:

A
resistance

B
resistivity

C
electrical conductivity

D
electromotive force

Solution

Electrical conductivity is given as,
$$\sigma = \dfrac{Gl}{a}$$
where $$G$$ is the conductance i.e. $$G = \dfrac{I}{V}$$
So, $$[\sigma]=[\dfrac{Gl}{a}]=[\dfrac{Il}{Va}]=[\dfrac{AL}{ML^2T^{-2}L^2/AT}]$$
$$\implies [\sigma]=[M^{-1}T^3L^{-3}A^2]$$
Question 6
1 / -0

Directions For Questions

The thickness of a metal sheet is measured to be $$326 mm$$. Express its order of magnitude in

...view full instructions

metres

A
$$10^{-1}$$

B
$$10^{-2}$$

C
$$10^1$$

D
$$10^2$$

Solution

1 mm is $$1\times { 10 }^{ -3 }m$$.
So, $$326mm = 3.26\times1\times { 10 }^{ 2 }\times { 10 }^{ -3 }m$$.
Hence its order of magnitude in metres is $${ 10 }^{ -1 }m$$
Question 7
1 / -0

The dimensional formula for Planck's constant and angular momentum are

A
$$\displaystyle \left [ ML^{2}T^{-2} \right ]and\left [ MLT^{-1} \right ]$$

B
$$\displaystyle \left [ ML^{2}T^{-1} \right ]and\left [ ML^{2}T^{-1} \right ]$$

C
$$\displaystyle \left [ ML^{3}T^{1} \right ]and\left [ ML^{2}T^{-2} \right ]$$

D
$$\displaystyle \left [ MLT^{-1} \right ]and\left [ MLT^{-2} \right ]$$

Solution

We know that $$E=h\nu$$
where dimensions of Energy, E are $$M{L}^{2}{T}^{-2}$$ and dimensions of frequency are inverse of Time Period, i.e. $${T}^{-1}$$
Thus, $$M{L}^{2}{T}^{-2}=[h]{T}^{-1}$$
or, $$[h]=M{L}^{2}{T}^{-1}$$
Now, angular momentum, $$L=I\omega$$
where units of omega are inverse of time (rad/sec, where radian is dimensionless).
Units of I are $$M{L}^{2}$$ (because I is of the form $$kM{x}^{2}$$)
Thus $$[L]=M{L}^{2}{T}^{-1}$$
Question 8
1 / -0

Dimensions of linear impulse are

A
$$\displaystyle \left [ ML^{-2}T^{-3} \right ]$$

B
$$\displaystyle \left [ ML^{-2} \right ]$$

C
$$\displaystyle \left [ MLT^{-1} \right ]$$

D
$$\displaystyle \left [ MLT^{-2} \right ]$$

Solution

Linear impulse is given as the change in linear momentum, $$I=m\Delta v$$
Thus, $$[I]=ML{T}^{-1}$$
Question 9
1 / -0

Which of the following is the dimension of the coefficient of friction?

A
$$\displaystyle \left [ M^{2}L^{2}T \right ]$$

B
$$\displaystyle \left [ M^{0}L^{0}T^{0} \right ]$$

C
$$\displaystyle \left [ ML^{2}T^{-2} \right ]$$

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

Solution

Coefficient of friction is unitless and dimensionless.
Thus, $$\displaystyle \left [ M^{0}L^{0}T^{0} \right ]$$
option B is correct.
Question 10
1 / -0

Dimension of velocity gradient is

A
$$\displaystyle \left [ M^{0}L^{0}T^{-1} \right ]$$

B
$$\displaystyle \left [ ML^{-1}T^{-1} \right ]$$

C
$$\displaystyle \left [ M^{0}LT^{-1} \right ]$$

D
$$\displaystyle \left [ ML^{0}L^{-1} \right ]$$

Solution

Velocity gradient = $$\dfrac{Velocity}{Distance}$$
Thus $$[Velocity \ Gradient]=\dfrac{L{T}^{-1}}{L}={M}^{0}{L}^{0}{T}^{-1}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Units and Measurements Test - 48

Result

Units and Measurements Test - 48

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle \left [ ML^2T^{-2}A^{-1} \right ]$$

$$\displaystyle \left [ ML^{3}T^{-4}A^{-2} \right ]$$

$$\displaystyle \left [ M^{0}L^{-2}T^{-2}A^{-2} \right ]$$

$$\displaystyle \left [ ML^{2}T^{-1}A^{2} \right ]$$

Solution

Question 2

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

$$\displaystyle \left [ M^{-1}L^{-1} \right ]$$

$$\displaystyle \left [ MLT^{-2} \right ]$$

$$\displaystyle \left [ ML^{-1}T^{-2} \right ]$$

Solution

Question 3

$$[ML^2T^{-2}A^{-2}]$$

$$[ML^2T^{3}A^{-1}]$$

$$[ML^2T^{-3}A^{1}]$$

$$[ML^2T^{-3}A^{-1}]$$

Solution

Question 4

1 and 3

1 and 4

1 and 2

2 and 4

Solution

Question 5

resistance

resistivity

electrical conductivity

electromotive force

Solution

Question 6

$$10^{-1}$$

$$10^{-2}$$

$$10^1$$

$$10^2$$

Solution

Question 7

$$\displaystyle \left [ ML^{2}T^{-2} \right ]and\left [ MLT^{-1} \right ]$$

$$\displaystyle \left [ ML^{2}T^{-1} \right ]and\left [ ML^{2}T^{-1} \right ]$$

$$\displaystyle \left [ ML^{3}T^{1} \right ]and\left [ ML^{2}T^{-2} \right ]$$

$$\displaystyle \left [ MLT^{-1} \right ]and\left [ MLT^{-2} \right ]$$

Solution

Question 8

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

$$\displaystyle \left [ ML^{-2} \right ]$$

$$\displaystyle \left [ MLT^{-1} \right ]$$

$$\displaystyle \left [ MLT^{-2} \right ]$$

Solution

Question 9

$$\displaystyle \left [ M^{2}L^{2}T \right ]$$

$$\displaystyle \left [ M^{0}L^{0}T^{0} \right ]$$

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

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

Solution

Question 10

$$\displaystyle \left [ M^{0}L^{0}T^{-1} \right ]$$

$$\displaystyle \left [ ML^{-1}T^{-1} \right ]$$

$$\displaystyle \left [ M^{0}LT^{-1} \right ]$$

$$\displaystyle \left [ ML^{0}L^{-1} \right ]$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.