Units and Measurements Test

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Units and Measurements Test - 48

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

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Units and Measurements Test - 48

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Question 1

[ML2T−2A−1]\displaystyle \left [ ML^2T^{-2}A^{-1} \right ][ML2T−2A−1]

[ML3T−4A−2]\displaystyle \left [ ML^{3}T^{-4}A^{-2} \right ][ML3T−4A−2]

[M0L−2T−2A−2]\displaystyle \left [ M^{0}L^{-2}T^{-2}A^{-2} \right ][M0L−2T−2A−2]

[ML2T−1A2]\displaystyle \left [ ML^{2}T^{-1}A^{2} \right ][ML2T−1A2]

Solution

Question 2

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

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

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

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

Solution

Question 3

[ML2T−2A−2][ML^2T^{-2}A^{-2}][ML2T−2A−2]

[ML2T3A−1][ML^2T^{3}A^{-1}][ML2T3A−1]

[ML2T−3A1][ML^2T^{-3}A^{1}][ML2T−3A1]

[ML2T−3A−1][ML^2T^{-3}A^{-1}][ML2T−3A−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

10−110^{-1}10−1

10−210^{-2}10−2

10110^1101

10210^2102

Solution

Question 7

[ML2T−2]and[MLT−1]\displaystyle \left [ ML^{2}T^{-2} \right ]and\left [ MLT^{-1} \right ][ML2T−2]and[MLT−1]

[ML2T−1]and[ML2T−1]\displaystyle \left [ ML^{2}T^{-1} \right ]and\left [ ML^{2}T^{-1} \right ][ML2T−1]and[ML2T−1]

[ML3T1]and[ML2T−2]\displaystyle \left [ ML^{3}T^{1} \right ]and\left [ ML^{2}T^{-2} \right ][ML3T1]and[ML2T−2]

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

Solution

Question 8

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

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

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

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

Solution

Question 9

[M2L2T]\displaystyle \left [ M^{2}L^{2}T \right ][M2L2T]

[M0L0T0]\displaystyle \left [ M^{0}L^{0}T^{0} \right ][M0L0T0]

[ML2T−2]\displaystyle \left [ ML^{2}T^{-2} \right ][ML2T−2]

[M2L2T−2]\displaystyle \left [ M^{2}L^{2}T^{-2} \right ][M2L2T−2]

Solution

Question 10

[M0L0T−1]\displaystyle \left [ M^{0}L^{0}T^{-1} \right ][M0L0T−1]

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

[M0LT−1]\displaystyle \left [ M^{0}LT^{-1} \right ][M0LT−1]

[ML0L−1]\displaystyle \left [ ML^{0}L^{-1} \right ][ML0L−1]

Solution

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

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

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

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

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

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

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

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

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

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

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

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

$10^{-1}$

$10^{-2}$

$10^1$

$10^2$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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