Units and Measurements Test

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Units and Measurements Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

The pair of physical quantities that have same dimensions are:
a) Reynold's number and coefficient of friction
b) Latent heat and gravitational potential
c) Curie and frequency of light wave
d) Planck constant and torque

A
b and c are correct

B
a and b are correct

C
a, b and c are correct

D
all are correct

Solution

Coefficient of friction and Reynolds no. have no dimensions.
Planck's constant $$=ML^{2}T^{-1}$$
Torque $$=ML^{2}T^{-2}$$
Latent heat $$=\dfrac{energy}{mass}=\dfrac{M^{1}L^{2}T^{-2}}{M}=L^{2}T^{-2}$$
Gravitational potential $$=\dfrac{GM}{r}=\dfrac{force}{mass}\times r$$$$=\dfrac{M^{1}L^{1}T^{-2}}{M^{1}}\times L^{1}$$$$=L^{2}T^{-2}$$
Curie $$=$$ freq $$ =M^{0}L^{0}T^{-1}$$
Question 2
1 / -0

Which of the following pairs have same dimensions.
a) Torque and work
b) Angular momentum and work
c) Energy and Youngs modulus
d) Light year and wavelength

A
a and b are correct

B
b and c are correct

C
c and d are correct

D
d and a are correct

Solution

Torque $$=$$ Work $$=$$ Energy $$=[ML^{2}T^{-2}]$$
Angular momentum $$=mvr=[ML^{2}T^{-1}]$$
Young's modulus $$=[ML^{-1}T^{-2}]$$
Light year and wavelength $$=[M^{0}L^{1}T^{0}]$$
Hence option D is correct.

Question 3

1 / -0

Names of units of some physical quantities are given in List - I and their dimensional formulae are given in List - II. Match the correct pair of the lists.

List - I	List - II
a) Pa-s	e) $$\left[ { L }^{ 2 }{ T }^{ -2 }{ K }^{ -1 } \right] $$
b) $${ NmK }^{ -1 }$$	f) $${ MLT }^{ -3 }{ K }^{ -1 }$$
c) $${ J kg }^{ -1 }{ k }^{ -1 }$$	g) $${ ML }^{ -1 }{ T }^{ -1 }$$
d) $${ Wm }^{ -1 }{ k }^{ -1 }$$	h) $$\left[ { ML }^{ 2 }{ T }^{ -2 }{ K }^{ -1 } \right] $$

a-h, b-g, c-e, d-f

a-g, b-f, c-e, d-h

a-g, b-e, c-h, d-f

a-g, b-h, c-e, d-f

Solution

$$Pa-s=M^{1}L^{-1}T^{-2}\times T^{1}=ML^{-1}T^{-1}$$
$$Nmk=M^{1}L^{1}T^{-2}\times L^{1}\times K^{1}=M^{1}L^{2}T^{-2}K^{-1}$$
$$Jkg^{-1}K^{-1}=ML^{2}T^{-2}M^{-1}K^{-1}=M^{0}L^{2}T^{-2}K^{-1}$$
$$Wm^{-1}K^{-1}=ML^{2}T^{-3}L^{-1}K^{-1}=MLT^{-3}K^{-1}$$
Hence option D is correct.

Question 4
1 / -0

Watt per meter kelvin is the unit of

A
Stress

B
Period of revolution of satellite

C
Angular displacement

D
Coefficient of thermal conductivity

Solution

$$\dfrac { Watt }{ \left( meter \right) \times \left( kelvin \right) } =\dfrac { M{ L }^{ 2 }{ T }^{ -3 } }{ LK } =M{ L }{ T }^{ -3 }{ K }^{ -1 }$$
Coefficient of thermal conductivity $$\left( k \right) =\dfrac { \left( \dfrac{q}{A} \right) \times \ t }{ dT } $$

$$\left( q/A \right) =$$ heat transfer per unit area$$\left( W/{ m }^{ 2 } \right) $$

$$dT=$$ temperature difference
$$t=$$ thickness

$$\left[ k \right] =\dfrac { \left[ M{ L }^{ 2 }{ T }^{ -3 } \right] \left[ { L }^{ -2 } \right] \left[ L \right] }{ \left[ K \right] } =M{ L }{ T }^{ -3 }{ K }^{ -1 }$$

Question 5

1 / -0

Match List I with List II and select the correct answer.

List - I	List - II
A) Spring constant	I) [$${ M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }$$]
B) Pascal	II) [$${ M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 }$$]
C) Hertz	III)[ $${ M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 }$$]
D) Joule	IV) [$${ M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 }$$]

A-III, B-IV, C-II, D-I

A-IV, B-III, C-I, D-II

A-IV, B-III, C-II, D-I

A-III, B-IV, C-I, D-II

Solution

$$Spring \ constant, \ k=\dfrac{F}{x}=\dfrac{MLT^{-2}}{L}=[M^1L^0T^{-2}$$]

$$Hertz=\dfrac{1}{t}=[M^0L^0T^{-1}$$]

$$Pascal=\dfrac{F}{A}=\dfrac{MLT^{-2}}{L^2}=[M^1L^{-1}T^{-2}$$]

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

Question 6

1 / -0

Study the following.

List - I	List - II
a) Same negative dimensions of mass	I) Pressure, Rydberg constant
b) Same negative dimensions of length	II) Mangnetic induction field, Potential
c) Same dimensions	III) Capacity, Universal gravitational constant
d) Same dimension of current	IV) Energy density, Surface tension

a-III, b-I, c-IV, d-II

a-III, b-IV, c-I, d-II

a-I, b-II, c-III, d-IV

a-II, b-I, c-IV, d-III

Solution

Pressure=$$\dfrac{force}{area}=\dfrac{mass\times acc^{n}}{area}=\dfrac{MLT^{-2}}{L^{2}}=ML^{-1}T^{-2}$$

Unit of Rydberg constant $$=L^{-1}$$ ( These 2 quantities have same -ve dimension of length)

Surface tension $$=M^{1}L^{0}T^{-2}$$
Energy density $$=$$ Energy per unit area $$=\dfrac{ML^{2}T^{-2}}{L^{2}}=ML^{0}T^{-2}$$
$$\rightarrow $$(same dimensional formula)

Potential$$=ML^{2}T^{-3}A^{-1}$$
Magnetic induction field $$=ML^{2}A^{-1}T^{-4}$$(same dimension for current)

Question 7

1 / -0

Match the physical quantities given in Column I with suitable dimensions expressed in Column II.

Column I	Column II
a) Angular momentum	g) $${ ML }^{ 2 }{ T }^{ -2 }$$
b) Latent heat	h) $${ ML }^{ 2 }{ Q }^{ -2 }$$
c) Torque	i) $${ ML }^{ 2 }{ T }^{ -1 }$$
d) Capacitane	j) $${ ML }^{ 3 }{ T }^{ -1 }{ Q }^{ -2 }$$
e) Inductance	k) $${ M }^{ -1 }{ L }^{ -2 }{ T }^{ 2 }{ Q }^{ 2 }$$
f) Resistivity	l) $${ L }^{ 2 }{ T }^{ -2 }$$

a-i, b-l, c-g, d-k, e-h, f-j

a-l, b-i, c-k, d-g, e-j, f-h

a-i, b-l, c-h, d-j, e-g, f-k

a-h, b-j, c-g, d-k, e-i, f-l

Solution

Angular momentum = $$mvr=ML^2T^{-1}$$

Latent Heat, $$Q=ML, \dfrac{Q}{M}=L, [L]=M^0L^2T^{-2}$$

Torque $$=F\times perpendicular \ distance= MLT^{-2}L=ML^2T^{-2}$$

Inductance $$=\dfrac{2E}{I^2}=L=\dfrac{ML^2T^{-2}}{A^2}=ML^2T^{-2}A^{-2}$$

Capacitance $$C=\dfrac{Q^2}{2E}=\dfrac{A^2T^2}{ML^2T^{-2}}$$

Resistivity$$=ML^{3}T^{-1}Q^{-2}$$

Option A is the appropriate choice.

Question 8
1 / -0

The correct order in which the dimensions of length increases in the following physical quantities is:
a) permittivity
b) resistance
c) magnetic permeability
d) stress

A
a, b, c, d

B
d, c, b, a

C
a, d, c, b

D
c, b, d, a

Solution

a) Permittivity $$=M^{-1}L^{-3}T^{4}I^{2}={\epsilon}_{0}$$
b) Resistance$$=ML^{2}T^{-1}Q^{-2}$$
c) Magnetic permeability$$=\mu_{0}=MLT^{-2}I^{-2}$$
d) Stress$$=F/A=ML^{-1}T^{-2}$$
Hence, dimensions of length in a<d<c<b.
Question 9
1 / -0

The dimensional formula for pressure gradient is :

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

B
$${ M }^{ 1 }{ L }^{ -2 }{ T }^{ -2 }$$

C
$${ M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }$$

D
$${ M }^{ 1 }{ L }^{ -1 }{ T }^{ -3 }$$

Solution

Pressure gradient $$=\dfrac{dp}{dx}$$

$$\displaystyle p=\frac{force}{Area}=\left ( kg\frac{m}{s^{2}} \right )\left ( \frac{1}{m^{2}} \right )=ML^{-1}T^{-2}$$

Where, $$x=L\left ( M^{0} L^{1}T^{0}\right )$$

So, $$\displaystyle \frac{dp}{dx}=\frac{ML^{-1}T^{-2}}{M^{0}L^{1}T^{0}}=ML^{-2}T^{-2}$$
Question 10
1 / -0

L, C and R represents inductance, capacitance and resistance respectively. The combinations that have the dimensions of frequency are :

A
$$\dfrac { 1 }{ CR } $$

B
$$\dfrac { R }{ L } $$

C
$$\dfrac { 1 }{ \sqrt { LC } } $$

D
All of the above

Solution

We know that unit for the frequency is per second i.e.$$ [T]^{ -1 }$$
Now let us consider the dimensions of the given physical quantities.
We know that Capacitance i.e.
$$C=Q/V$$
where Q is electric charge and V is electric potential.
Now we know that Q= It where I is current and t is time.
Also that V=IR where R is resistance.
So,
$$C= It/IR$$
Thus,
$$C= t/R$$
also $$1/t=1/RC$$
Thus dimension of $$1/RC$$ will be $$[T^{-1}]$$ that is same as that of frequency.

Now Inductance(L) can be given as
$$L= Vt/I$$
And also $$V=IR$$
Therefore $$L=IRt/I$$
Thus, $$R/L = 1/t$$

It means both A and B are correct now checking for the third option
$$LC= Rt * t/R = t^2$$
That gives $$1/\sqrt { LC } = 1/t$$
Thus dimension of $$1/\sqrt{LC}$$ will be $$T^{-1}$$ that is same as that of frequency.

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Units and Measurements Test - 43

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Units and Measurements Test - 43

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Question 1

b and c are correct

a and b are correct

a, b and c are correct

all are correct

Solution

Question 2

a and b are correct

b and c are correct

c and d are correct

d and a are correct

Solution

Question 3

a-h, b-g, c-e, d-f

a-g, b-f, c-e, d-h

a-g, b-e, c-h, d-f

a-g, b-h, c-e, d-f

Solution

Question 4

Stress

Period of revolution of satellite

Angular displacement

Coefficient of thermal conductivity

Solution

Question 5

A-III, B-IV, C-II, D-I

A-IV, B-III, C-I, D-II

A-IV, B-III, C-II, D-I

A-III, B-IV, C-I, D-II

Solution

Question 6

a-III, b-I, c-IV, d-II

a-III, b-IV, c-I, d-II

a-I, b-II, c-III, d-IV

a-II, b-I, c-IV, d-III

Solution

Question 7

a-i, b-l, c-g, d-k, e-h, f-j

a-l, b-i, c-k, d-g, e-j, f-h

a-i, b-l, c-h, d-j, e-g, f-k

a-h, b-j, c-g, d-k, e-i, f-l

Solution

Question 8

a, b, c, d

d, c, b, a

a, d, c, b

c, b, d, a

Solution

Question 9

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

$${ M }^{ 1 }{ L }^{ -2 }{ T }^{ -2 }$$

$${ M }^{ 1 }{ L }^{ 2 }{ T }^{ -2 }$$

$${ M }^{ 1 }{ L }^{ -1 }{ T }^{ -3 }$$

Solution

Question 10

$$\dfrac { 1 }{ CR } $$

$$\dfrac { R }{ L } $$

$$\dfrac { 1 }{ \sqrt { LC } } $$

All of the above

Solution

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