Units and Measurements Test

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Units and Measurements Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

A metal sample carrying a current along $$X-$$axis with density $$J_{x}$$ is subjected to a magnetic field $$B_{z}$$ (along $$z-$$axis). The electric field $$E_{y}$$ developed along $$Y-$$axis is directly proportional to $$J_{x}$$ as well as $$B_{z}$$. The constant of proportionality has $$SI$$ unit.

A
$$\dfrac{m^{2}}{A}$$

B
$$\dfrac{m^{3}}{As}$$

C
$$\dfrac{m^{2}}{As}$$

D
$$\dfrac{As}{m^{3}}$$

Solution

Given: $$E_{y} \alpha J_{x}$$ and $$E_{y} \alpha B_{Z}$$
$$E_{y} \alpha J_{x}B_{Z}$$
$$\Rightarrow E_{y}=KJ_{x}B_{Z}$$ (Where K = constant of proportionality)
By Dimensional analysis we have;
$$\left [ E_{y} \right ] = [KJ_{x}B_{Z}]$$
$$\Rightarrow \left [ K \right ] = \dfrac{\left [ E_{y} \right ]}{\left [ J_{x}B_{Z} \right ]}=\dfrac{[]M^{1}L^{1}T^{3}A^{-1}]}{[L^{-2}A^{A}][M^{1}T^{-2}A^{-1}]}$$
$$\Rightarrow [K]=\dfrac{[L^{3}]}{[A^{1}T^{1}]}\rightarrow $$ Dimensional formual of K
So, S.I unit of K will be : $$ \dfrac{m^{3}}{As}$$
Question 2
1 / -0

Give the MKS units for the following quantities.
Power of lens.

A
hertz

B
Dioptre.

C
ohms

D
unit less

Solution

The dioptre is the unit of measure for the refractive power of a lens. The power of a lens is defined as the reciprocal of its focal length in meters, or $$D = \dfrac 1f$$, where $$D$$ is the power in diopters and $$f$$ is the focal length in meters.
Question 3
1 / -0

In the formula $$X=3YZ^2$$, X and Z have dimensions of capacitance and magnetic induction respectively. The dimensions of Y in MKSQ system are ________.

A
$$-2$$ in mass, $$-2$$ in length, $$4$$ in time and $$5$$ in charge.

B
$$-3$$ in mass, $$-2$$ in length, $$4$$ in time and $$4$$ in charge.

C
$$-4$$ in mass, $$-5$$ in length, $$4$$ in time and $$4$$ in charge.

D
$$-3$$ in mass, $$-2$$ in length, $$8$$ in time and $$7$$ in charge.

Solution

Given,

$$X=3YZ^2$$

Then,

$$Y=\dfrac{X}{3Z^2}$$

$$[X]=[C]=[M^{-1}L^{-2}T^2Q^2]$$

$$[Z]=[B]=[MT^{-1}Q^{-1}]$$

Then,

$$[Y]=\dfrac{[X]}{[3Z^2]}$$

$$=\dfrac{[M^{-1}L^{-2}T^2Q^2]}{[MT^{-1}Q^{-1}]^2}=[M^{-3}L^{-2}T^4Q^4]$$

That is,

$$-3$$ in mass, $$-2$$ in length, $$4$$ in time and $$4$$ in charge.
Question 4
1 / -0

Planck's constant has the dimensions ___________.

A
$$ML^3T^{-1}$$.

B
$$ML^2T^{-1}$$.

C
$$ML^6T^{-1}$$.

D
$$ML^8T^{-1}$$.

Solution

Planck's constant, symbolized h, relates the energy in one quantum (photon) of electromagnetic radiation to the frequency of that radiation.

$$h=\dfrac Ev=\dfrac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]$$
Question 5
1 / -0

The physical quantities not having same dimensions are:

A
Torque and work

B
Momentum and Planck's constant

C
Stress and Young's modulus

D
Speed and $$(\mu_{0}\epsilon_{0})^{-1/2}$$

Solution

The torque and the work both have the same dimensions because they are given as the product of force and the distance.

$$[Momentum] = [MLT^{-1}]$$
$$[h] = \dfrac {E}{v} = ML^{2}T^{-2}T = [ML^{2}T^{-1}]$$.

On the other hand, the young's modulus is the ratio of stress and strain. The strain is a dimensionless quantity. Therefore, the unit of stress and young's modulus will be the same.

The unit of speed is $$m/s$$ and it has the dimension of $$[LT^{-1}]$$.
The speed of light is given by:
$$c=\dfrac{1}{\sqrt{\mu_0\epsilon_0}}$$
Therefore, both will have the same dimensions.

So, option $$(B)$$ is correct.
Question 6
1 / -0

If $$y$$ represents pressure and $$x$$ represents velocity gradient, then the dimensions of $$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } }$$ are

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

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

C
$$\left[ ML^{ -1 }T^{ 0 } \right] $$

D
$$\left[ M^{2}L^{ -2 }T^{ -4 } \right] $$

Solution

Dimensions of $$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } $$ = dimension of $$\dfrac { y }{ { x }^{ 2 } } $$
Dimension of y $$M{ L }^{ -1 }{ T }^{ -2 }$$
Dimension of x $${ T }^{ -1 }$$
$$\dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \quad =\dfrac { M{ L }^{ -1 }{ T }^{ -2 } }{ { \left( { T }^{ -1 } \right) }^{ 2 } } \\ =M{ L }^{ -1 }{ T }^{ 0 }$$
Question 7
1 / -0

Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii $$R_1$$and $$R_2$$, respectively. The ratio of the masses of X to that of Y is

A
$$(R_1/R_2)^{1/2}$$

B
$$R_2/R_1$$

C
$$(R_1/R_2)^2$$

D
$$R_1/R_2$$

Solution

We know,

$$R=\dfrac{\sqrt{2mqV}}{qB}$$

As q,B,V are same

$$R\propto \sqrt m$$

$$\dfrac{R_1}{R_2}=\sqrt{\dfrac{m_1}{m_2}}$$

$$\dfrac{m_1}{m_2}=\dfrac{(R_1)^2}{(R_2)^2}$$

Option $$\textbf C$$ is the correct answer.
Question 8
1 / -0

The physical quantities which do not have same dimensions are

A
Energy and Torque

B
Pressure and Stress

C
Work and Energy

D
Strain and Youngs Modulus

Solution

The energy and torque have the same dimensions as both can be represented as the product of force and distance.

The pressure and stress are the quantities that have the same dimensions. They both are defined as force per unit area.

The work and energy are the quantities that have the same dimensions. They are the product of force and distance.

The young's modulus is defined as the ratio of stress and strain. The strain is a dimensionless quantity. So, Young's modulus will have the same unit as of stress.
Strain =$$\dfrac { change\quad in\quad dimension }{ orginal\quad dimension } $$
$$={ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }$$

Youngs Modulus $$={ M }{ L }^{ -1 }{ T }^{ -2 }$$
Question 9
1 / -0

The velocity of a body moving viscous medium is given by $$V=\dfrac { A }{ B } [1-{ e }^{ { -t }/{ B } }]$$ where $$t$$ is time, $$A$$ and $$B$$ are constants. Then the dimensional formula of $$A$$ is:

A
$${ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }$$

B
$${ M }^{ 0 }{ L }^{ 1 }{ T }^{ 0 }$$

C
$${ M }^{ 0 }{ L }^{ 1 }{ T }^{ -1 }$$

D
$${ M }^{ 1 }{ L }^{ 1 }{ T }^{ -1 }$$

Solution

In $$e^\frac{-t}{B}$$ is the exponential term. So, the $$\dfrac{t}{B}$$ is constant.
So, B has the dimensions as:
$$B=[T]$$

On the other hand, the dimension of $$\dfrac AB$$ will be the same as that of the velocity.
$$\dfrac{A}{B}$$ has dimensions as velocity

$$[LT^{-1}]=\dfrac{a}{[T]}$$

dimensional formula of A is $$M^0L^1 T^0$$
Question 10
1 / -0

The equation of the state of some gases can be expressed as $$(P + \dfrac{a}{V^2} = \dfrac{R\theta}{V})$$, where P is the pressure. V the volume, $$\theta$$ the absolute temperature and a and b are constant. The dimensional formula of a is

A
$$[ML^5T^{-2}]$$

B
$$[M^{-1}L^5T^{-2}]$$

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

D
$$[ML^{-5}T^{-2}]$$

Solution

Dimensional formula of Pressure is $$[M^{1} L^{-1} T^{-2}]$$
Dimensional formula of Volume is $$[M^{0} L^{3} T^{0}]$$

From the given equation:
$$P+\dfrac{a}{V^2}=\dfrac{R\theta}{V}$$

Dimensional formula of $$\dfrac{a}{V^{2}}$$ is same as pressure.
$$\dfrac{a}{L^6}=[M^1L^{-1}T^{-2}]$$

Therefore dimensional formula of a is $$[M^{1} L^{5} T^{-2}]$$

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Units and Measurements Test - 66

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Units and Measurements Test - 66

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Question 1

$$\dfrac{m^{2}}{A}$$

$$\dfrac{m^{3}}{As}$$

$$\dfrac{m^{2}}{As}$$

$$\dfrac{As}{m^{3}}$$

Solution

Question 2

hertz

Dioptre.

ohms

unit less

Solution

Question 3

$$-2$$ in mass, $$-2$$ in length, $$4$$ in time and $$5$$ in charge.

$$-3$$ in mass, $$-2$$ in length, $$4$$ in time and $$4$$ in charge.

$$-4$$ in mass, $$-5$$ in length, $$4$$ in time and $$4$$ in charge.

$$-3$$ in mass, $$-2$$ in length, $$8$$ in time and $$7$$ in charge.

Solution

Question 4

$$ML^3T^{-1}$$.

$$ML^2T^{-1}$$.

$$ML^6T^{-1}$$.

$$ML^8T^{-1}$$.

Solution

Question 5

Torque and work

Momentum and Planck's constant

Stress and Young's modulus

Speed and $$(\mu_{0}\epsilon_{0})^{-1/2}$$

Solution

Question 6

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

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

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

$$\left[ M^{2}L^{ -2 }T^{ -4 } \right] $$

Solution

Question 7

$$(R_1/R_2)^{1/2}$$

$$R_2/R_1$$

$$(R_1/R_2)^2$$

$$R_1/R_2$$

Solution

Question 8

Energy and Torque

Pressure and Stress

Work and Energy

Strain and Youngs Modulus

Solution

Question 9

$${ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }$$

$${ M }^{ 0 }{ L }^{ 1 }{ T }^{ 0 }$$

$${ M }^{ 0 }{ L }^{ 1 }{ T }^{ -1 }$$

$${ M }^{ 1 }{ L }^{ 1 }{ T }^{ -1 }$$

Solution

Question 10

$$[ML^5T^{-2}]$$

$$[M^{-1}L^5T^{-2}]$$

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

$$[ML^{-5}T^{-2}]$$

Solution

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