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Test: Dimensions of Physical Quantities - NEET MCQ


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30 Questions MCQ Test Physics Class 11 - Test: Dimensions of Physical Quantities

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Test: Dimensions of Physical Quantities - Question 1

Which of the following is not the name of a physical quantity ?

Detailed Solution for Test: Dimensions of Physical Quantities - Question 1

Kilogram represent unit of physical quantity and not the physical quantity.

Test: Dimensions of Physical Quantities - Question 2

Light year is the unit of

Detailed Solution for Test: Dimensions of Physical Quantities - Question 2

The light-year is a unit of length used to express astronomical distances and measures about 9.46 trillion kilometres or 5.88 trillion miles. As defined by the International Astronomical Union, a light-year is the distance that light travels in vacuum in one Julian year. 

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Test: Dimensions of Physical Quantities - Question 3

PARSEC is a unit of

Detailed Solution for Test: Dimensions of Physical Quantities - Question 3

The parsec (symbol: pc) is a unit of length used to measure large distances to astronomical objects outside the Solar System. A parsec is defined as the distance at which one astronomical unit subtends an angle of one arcsecond, which corresponds to 648000π astronomical units.

Test: Dimensions of Physical Quantities - Question 4

Which of the following system of units is NOT based on the unit of mass, length and time alone

Detailed Solution for Test: Dimensions of Physical Quantities - Question 4

The SI system of units is a modern system and hence involves all the quantities that can't be derived using all the other quantities of the set. While the rest systems are old and local methods and hence are not scientifically accurate and explainable and thus only have three basic quantities while SI have 7.

Test: Dimensions of Physical Quantities - Question 5

In the S.I. system the unit of energy is-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 5

Joule is the SI unit of energy

Test: Dimensions of Physical Quantities - Question 6

Unit of pressure in S.I. system is-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 6

The SI unit of pressure is Pascal (represented as Pa) which is equal to one newton per square metre (N/m-2 or kg m-1s-2). Interestingly, this name was given in 1971. Before that pressure in SI was measured in newtons per square metre.

Test: Dimensions of Physical Quantities - Question 7

The mutual inductance has unit of-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 7

The unit of inductance in the SI system is the henry (H), named after American scientist Joseph Henry, which is the amount of inductance which generates a voltage of one volt when the current is changing at a rate of one ampere per second.

Test: Dimensions of Physical Quantities - Question 8

In SI unit the angular acceleration has unit of-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 8

Angular acceleration. Angular acceleration is the rate of change of angular velocity. In three dimensions, it is a pseudovector. In SI units, it is measured in radians per second squared (rad/s2), and is usually denoted by the Greek letter alpha (α).

Test: Dimensions of Physical Quantities - Question 9

The SI unit of the universal gravitational constant G is

Detailed Solution for Test: Dimensions of Physical Quantities - Question 9

We know that,       g = G. m1 m2 / r2
Where we know g has dimensions of acceleration
Thus [G] = [ g.r2 /m1 m2]
= N m2kg-2

Test: Dimensions of Physical Quantities - Question 10

Which of the following statement is wrong ?

Detailed Solution for Test: Dimensions of Physical Quantities - Question 10

The incorrect statement is:

d) Unit of surface tension is Newton metre.

The correct unit of surface tension is Newton per metre (N/m), not Newton metre.

Here's a breakdown:

  • a) Unit of K.E. is Newton-metre: True, because kinetic energy is measured in joules, and 1 joule equals 1 Newton-metre.
  • b) Unit of viscosity is poise: True, poise is indeed a unit of dynamic viscosity.
  • c) Work and energy have the same dimensions: True, both work and energy have the same dimensions and are measured in joules.
  • d) Unit of surface tension is Newton metre: Incorrect, it should be Newton per metre (N/m).
Test: Dimensions of Physical Quantities - Question 11

What are the dimensions of length in force × displacement/time

Detailed Solution for Test: Dimensions of Physical Quantities - Question 11

[F] = MLT-2
[displacement/time]  = LT-1
Thus we get [F x displacement/time] = ML2T-3
Thus the answer is 2

Test: Dimensions of Physical Quantities - Question 12

The angular frequency is measured in rad s-1. Its dimension in length are :

Detailed Solution for Test: Dimensions of Physical Quantities - Question 12

Unit of angular frequency is rad/sec which can be said as angle/time. As angle is dimensionless and time has dimension T, we get the dimension of angular frequency as T-1

Test: Dimensions of Physical Quantities - Question 13

The dimensional formula of coefficient of viscosity is

Detailed Solution for Test: Dimensions of Physical Quantities - Question 13

Coefficient of viscosity (η)= Fr/Av   

 F= tangential Force, Area, r= distance between the layers, v= velocity.

Dimensional Formula of Force = M1L1T-2.
Dimensional Formula of Area= M0L2T0.
Dimensional Formula of distance= M0L1T0.
Dimensional Formula of velocity= M0L1T-1.

Putting these values in above equation we get,

[η]= [M1L1T-2][M0L1T0] / [M0L2T0] [M0L1T-1] = [M1L-1T-1]

Test: Dimensions of Physical Quantities - Question 14

A pair of physical quantities having the same dimensional formula is :

Detailed Solution for Test: Dimensions of Physical Quantities - Question 14

The dimensions of angular momentum are M L2T−1
That of torque is  M L2T−2
Also dimension of energy is  M L2T−2
Where as same of force is  M LT−2
And of power is  M L2T−3
Thus we get torque and energy have the same dimensional formulas.

Test: Dimensions of Physical Quantities - Question 15

Dimensions of pressure are the same as that of

Detailed Solution for Test: Dimensions of Physical Quantities - Question 15

[P] = [F/A] = MLT-2 / L2  = ML-1T-2
[F/V] = MLT-2 / L3  = ML-2T-2
[E/V] = ML2T-2 / L3  = ML-1T-2

Test: Dimensions of Physical Quantities - Question 16

 If momentum (P), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula 

Detailed Solution for Test: Dimensions of Physical Quantities - Question 16

[E] = [F][d]
= [P/T][A]½ 
[E] = P1A½T-1

Test: Dimensions of Physical Quantities - Question 17

If radian correction is not considered in specific heat measurement. The measured value of specific heat will be

Detailed Solution for Test: Dimensions of Physical Quantities - Question 17

Correct Answer :- b

Explanation : If radian correction is not considered in specific heat measurement. The measured value of specific heat will be more than its actual value.

Test: Dimensions of Physical Quantities - Question 18

The specific resistance has the unit of-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 18

We know that specific resistance, ⍴ = R A / L
Where R is net resistance, A is area and
L is length
Hence [⍴] = ohm x m.m / m
= ohm.m

Test: Dimensions of Physical Quantities - Question 19

One watt-hour is equivalent to

Detailed Solution for Test: Dimensions of Physical Quantities - Question 19

Power = energy / time.
energy = power × time.
energy = watt × hour
= 1watt × 1hour
= 1 × (60×60)
= 3600
= 3.6 × 103 joule

Test: Dimensions of Physical Quantities - Question 20

The density of mercury is 13600 kg m-3. Its value of CGS system will be :

Detailed Solution for Test: Dimensions of Physical Quantities - Question 20

We know that density of Hg = 13600 kg/m.m.m
= 13600 x 1000 g / 100cm x 100cm x 100cm
= 13.6 g/cm.cm.cm

Test: Dimensions of Physical Quantities - Question 21

What is the unit for measuring the amplitude of a sound?

Detailed Solution for Test: Dimensions of Physical Quantities - Question 21
Unit for measuring the amplitude of a sound:
The unit for measuring the amplitude of a sound is the decibel (dB).
Explanation:
- The amplitude of a sound wave refers to the maximum displacement of particles in the medium through which the sound wave is traveling.
- The decibel is a logarithmic unit used to express the ratio between two values, in this case, the sound pressure level.
- The decibel scale is based on a logarithmic scale because the human perception of sound intensity is not linear. It allows for a wider range of values to be expressed in a more manageable format.
- The decibel scale is relative, meaning it compares the sound pressure level to a reference level, typically the threshold of hearing.
- The decibel scale ranges from 0 dB, which represents the threshold of hearing, to around 120-130 dB, which is the threshold of pain for most individuals.
- The decibel scale is commonly used in various fields, including physics, engineering, and acoustics, to quantify the intensity or amplitude of sound waves.
- By using decibels, it becomes easier to compare the amplitudes of different sounds and evaluate their relative loudness.
Therefore, the correct answer is A: Decibel (dB).
Test: Dimensions of Physical Quantities - Question 22

The unit of intensity of magnetisation is-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 22

Intensity of magnetisation : It represents the extent to which a specimen is magnetised when placed in a magnetising field. Or in other words the intensity of magnetisation of a magnetic material is defined as the magnetic moment per unit volume of the material. 

It indicates how the sample affected by magnetic field when placed in it.

M = Magnetic moment/volume = μM / V

unit SI : M= Amp.metre-1

Test: Dimensions of Physical Quantities - Question 23

The M.K.S. units of coefficient of viscosity is-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 23

We know that coefficient of viscosity (η)= Fr/Av where F = tangential force, r = distance between the layers , v = velocity and A is the area of the surface.
 Thus we get [η] = MLT-2.L / L2. (L/T)
= M / LT
Thus its unit is kg / m sec

Test: Dimensions of Physical Quantities - Question 24

For 10(at+3) , the dimension of a is-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 24

As 'at' and 3 are added in the equation, we get at and 3 have same dimensions i.e.1
Thus a has dimensions same as 1/t.

Test: Dimensions of Physical Quantities - Question 25

The pressure of 106 dyne/cm2 is equivalent to

Detailed Solution for Test: Dimensions of Physical Quantities - Question 25

We know that 105 dyne = 1N
And 104cm2  = 1 m2
Thus we get 10 dyne / cm2 = N / m2
Hence 106 dyne / cm2 =105 N / m2

Test: Dimensions of Physical Quantities - Question 26

The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is

Detailed Solution for Test: Dimensions of Physical Quantities - Question 26

The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as present in the number having the least number of decimal places.

The final result should, therefore, be rounded off to one decimal place, i.e. 663.8

Test: Dimensions of Physical Quantities - Question 27

 The SI unit of length is the meter. Suppose we adopt a new unit of length which equals to x meters. The area 1m2 expressed in terms of the new unit has a magnitude-

Detailed Solution for Test: Dimensions of Physical Quantities - Question 27

We have  1 unit = x meters
So  1 unit2 = x2 meter2
Hence, we get  1 meter2 = 1/x2 unit2

Test: Dimensions of Physical Quantities - Question 28

 r = 2 g/cm3 convert it into MKS system -

Detailed Solution for Test: Dimensions of Physical Quantities - Question 28

multiply the mass / volume value by 1000

Test: Dimensions of Physical Quantities - Question 29

Given that v is the speed, r is radius and g is acceleration due to gravity. Which of the following is dimension less

Detailed Solution for Test: Dimensions of Physical Quantities - Question 29

v2r/g= (L1T-1)2L1/LT-2=L2
v2/rg= (LT-1)2 L1(L1 T-2) ​=M0L0T0
v2g/r​=(LT-1)2 LT-2/L​=L2T-4
v2rg=(LT-1)2(L1) (LT-1) =L4T-3
So, option D is correct.

Test: Dimensions of Physical Quantities - Question 30

 The value of G = 6.67 × 10_11 N m2 (kg)_2. Its numerical value in CGS system will be :

Detailed Solution for Test: Dimensions of Physical Quantities - Question 30


G = 6.67 x 10-11 Nm2/kg2

so, when expressed in VGS units we shall convert N to dynes, m to cm and kg to g, thus

G = 6.67 x 10-11 x [(105dynes x (102)2cm2 ) / (103)2 g2]

=  6.67 x 10-11 x [ (105 x 104 ) / 106 ] 

Thus,

in CGS system

G = 6.67 x 10-8 dyne.cm2/g2

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