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QUESTION: 1

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

Solution:

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

QUESTION: 2

Light year is the unit of

Solution:

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.

QUESTION: 3

PARSEC is a unit of

Solution:

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**.

QUESTION: 4

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

Solution:

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.

QUESTION: 5

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

Solution:

Joule is the SI unit of energy

QUESTION: 6

Unit of pressure in S.I. system is-

Solution:

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^{-1}s^{-2}). Interestingly, this name was given in 1971. Before that pressure in SI was measured in newtons per square metre.

QUESTION: 7

The mutual inductance has unit of-

Solution:

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.

QUESTION: 8

In SI unit the angular acceleration has unit of-

Solution:

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/s^{2}), and is usually denoted by the Greek letter alpha (α).

QUESTION: 9

The SI unit of the universal gravitational constant G is

Solution:

We know that, g = G. m_{1} m_{2} / r_{2}

Where we know g has dimensions of acceleration

Thus [G] = [ g.r^{2} /m_{1} m_{2}]

= N m^{2}kg^{-2}

QUESTION: 10

Which of the following statement is wrong ?

Solution:

Unit of surface tension is Newton per metre

QUESTION: 11

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

Solution:

[F] = MLT^{-2}

[displacement/time] = LT^{-1}

Thus we get [F x displacement/time] = ML^{2}T^{-3}

Thus the answer is 2

QUESTION: 12

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

Solution:

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}

QUESTION: 13

The dimensional formula of coefficient of viscosity is

Solution:

**Coefficient of viscosity (η)= Fr/Av **

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

Dimensional Formula of Force = M^{1}L^{1}T^{-2}.

Dimensional Formula of Area= M^{0}L^{2}T^{0}.

Dimensional Formula of distance= M^{0}L^{1}T^{0}.

Dimensional Formula of velocity= M^{0}L^{1}T^{-1}.

Putting these values in above equation we get,

**[η]= [M ^{1}L^{1}T^{-2}][M^{0}L^{1}T^{0}] / [M^{0}L^{2}T^{0}] [M^{0}L^{1}T^{-1}] = [M^{1}L^{-1}T^{-1}]**

QUESTION: 14

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

Solution:

The dimensions of angular momentum are M L^{2}T^{−1}

That of torque is M L^{2}T^{−2}

Also dimension of energy is M L^{2}T^{−2}

Where as same of force is M LT^{−2}

And of power is M L^{2}T^{−3}

Thus we get torque and energy have the same dimensional formulas.

QUESTION: 15

Dimensions of pressure are the same as that of

Solution:

[P] = [F/A] = MLT^{-2} / L^{2} = ML^{-1}T^{-2}

[F/V] = MLT^{-2} / L3 = ML^{-2}T^{-2}

[E/V] = ML^{2}T^{-2} / L^{3} = ML^{-1}T^{-2}

QUESTION: 16

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

Solution:

[E] = [F][d]

= [P/T][A]^{½}

[E] = P^{1}A^{½}T^{-1}

QUESTION: 17

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

Solution:

**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.

QUESTION: 18

The specific resistance has the unit of-

Solution:

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

QUESTION: 19

One watt-hour is equivalent to

Solution:

Power = energy / time.

energy = power × time.

energy = watt × hour

= 1watt × 1hour

= 1 × (60×60)

= 3600

= 3.6 × 10^{3} joule

QUESTION: 20

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

Solution:

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

QUESTION: 21

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

Solution:

QUESTION: 22

The unit of intensity of magnetisation is-

Solution:

**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}

QUESTION: 23

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

Solution:

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 / L^{2}. (L/T)

= M / LT

Thus its unit is kg / m sec

QUESTION: 24

For 10^{(at+3)} , the dimension of a is-

Solution:

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.

QUESTION: 25

The pressure of 10^{6} dyne/cm^{2} is equivalent to

Solution:

We know that 10^{5} dyne = 1N

And 10^{4}cm^{2} = 1 m^{2}

Thus we get 10 dyne / cm^{2} = N / m^{2}

Hence 10^{6} dyne / cm^{2} =10^{5} N / m^{2}

QUESTION: 26

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

Solution:

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

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 1m^{2} expressed in terms of the new unit has a magnitude-

Solution:

We have 1 unit = x meters

So 1 unit^{2} = x^{2} meter^{2}

Hence, we get 1 meter^{2} = 1/x^{2} unit^{2}

QUESTION: 28

r = 2 g/cm^{3} convert it into MKS system -

Solution:

multiply the mass / volume value by 1000

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

Solution:

v^{2}r/g= (L^{1}T^{-1})2L1/LT^{-2}=L^{2}

v^{2}/rg= (LT^{-1})^{2} L1(L^{1} T^{-2}) =M^{0}L^{0}T^{0}

v^{2}g/r=(LT^{-1})2 LT^{-2}/L=L^{2}T^{-4}

v^{2}rg=(LT^{-1})^{2}(L1) (LT^{-1}) =L^{4}T^{-3}

So, option D is correct.

QUESTION: 30

The value of G = 6.67 × 10^{_11} N m^{2} (kg)^{_2}. Its numerical value in CGS system will be :

Solution:

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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