Units and Measurements Test

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Units and Measurements Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following pair does not have same dimensions ?

A
Pressure, Modulus of Elasticity

B
Angular velocity, Velocity gradient

C
Surface tension, Force constant

D
Impulse, Torque

Solution

$${\textbf{Explanation of options A, B and C:}}$$
A. Pressure $$=$$ Modulus of Elasticity $$= \dfrac{F}{A} = M^{1} L^{-1} T^{-2}$$
B. Angular velocity $$= \omega = T^{-1}$$
Velocity gradient $$= \dfrac{dv}{dz} = \dfrac{LT^{-1}}{L} = T^{-1}$$
C. Surface Tenstion $$= \dfrac{F}{l} = $$ Force constant $$= \dfrac{M^{1} L^{1} T^{-2}}{L^{1}} = M T^{-2}$$
$${\textbf{Explanation of option D:}}$$
Impulse $$= F\times t = M^{1} L^{1} T^{-2} \times T^{1} = M^{1} L^{1} T^{-1}$$
Torque $$= \vec{r} \times \vec{F} = L^{1} \times M^{1} L^{1} T^{-2} = M^{1} L^{2} T^{-2}$$

$${\textbf{Correct option: D}}$$
Question 2
1 / -0

Arrange the following physical quantities in the decreasing order of dimension of length.
I. Density
II. Pressure
III. Power
IV. Impulse

A
I, II, III, IV

B
III, II, I, IV

C
IV, I, II, III

D
III, IV, II, I

Solution

$$Density = \dfrac{Mass}{length} = M^{1} L^{-3} T^{0}$$
$$ Pressure = \dfrac{force}{Area} = \dfrac{M^{1}L^{1}t^{-2}}{L^{2}} = M^{1}L^{-1}T^{-2}$$
$$Power = F\times v = M^{1}L^{1}T^{-2} \times LT^{-1} = M^{1}L^{2}T^{-3}$$
$$Impulse = f\times t = M^{1}L^{1}T^{-2}\times T^{1} = M^{1}L^{1}T^{-1}$$

Hence, option D is correct.
Question 3
1 / -0

Dimensional formula for capacitance is:

A
$${ M }^{ -1 }{ L }^{ -2 }{ T }^{ 4 }{ I }^{ 2 }$$

B
$${ M }^{ 1 }{ L }^{ 2 }{ T }^{ 4 }{ I }^{ -2 }$$

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

D
$${ MLT }^{ -1 }$$

Solution

$$E=\dfrac{Q^2}{2C}$$
Dimensions of C will be dimension of $$\dfrac{Q^2}{E}$$$$=\dfrac{I^2T^2}{ML^2T^{-2}}$$$$=M^{-1}L^{-2}T^4I^2$$
Hence, option A is correct.
Question 4
1 / -0

The dimensional formula for coefficient of kinematic viscosity is :

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

B
$${ M }^{ 0 }{ L }^{ 2 }{ T }^{ -1 }$$

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

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

Solution

Kinematic viscocity $$=\dfrac{coeff \ of \ viscocity}{ density}$$
$$=\dfrac{ML^{-1}T^{-1}}{ML^{-3}}$$

$$=M^{0}L^{2}T^{-1}$$
Question 5
1 / -0

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

A
Stress

B
Young's Modulus

C
Pressure

D
All the above

Solution

Stress and pressure both are defined as $$\dfrac{F}{A}=\dfrac{MLT^{-2}}{L^2}=ML^{-1}T^{-2}$$
Young's modulus is calculated by formula $$\dfrac{stress}{strain}$$. As strain is dimensionless the Young's modulus will also have dimensions of stress only.
Hence, option D is correct.
Question 6
1 / -0

Dimensions of impulse are :

A
$${ MLT }^{ -2 }$$

B
$${ M }^{ 2 }{ LT }^{ -1 }$$

C
$${ MLT }^{ -1 }$$

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

Solution

$$Impulse=F\times

t=M^{1}L^{1}T^{-2}\times T^{1}=M^{1}L^{1}T^{-1}$$
Question 7
1 / -0

The dimensional formula for Magnetic induction is :

A
$${ MT }^{ -1 }{ A }^{ -1 }$$

B
$${ MT }^{ -2 }{ A }^{ -1 }$$

C
$${ MLA }^{ -1 }$$

D
$${ MT }^{ -2 }A$$

Solution

$$F=BQV$$
Dimension of B will be $$\dfrac{MLT^{-2}}{ATLT^{-1}}=ML^0T^{-2}A^{-1}$$
Question 8
1 / -0

Dimensions of $$C\times R$$ (Capacity x Resistance) is:

A
frequency

B
energy

C
time period

D
current

Solution

C = Capacity $$=M^{-1}L^{-2}I^{2}T^{4}$$
R = Resistance $$=ML^{2}I^{-2}T^{-3}$$
So, $$C\times R=T^{1}M^{0}L^{0}I^{0}$$
Question 9
1 / -0

Dimensional formula for Angular momentum is:

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

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

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

D
$${ ML }^{ 3 }{ T }^{ -2 }$$

Solution

$$\textbf {Hint :}$$ Use angular momentum $$L=mvr$$
$${\textbf{Explanation:}}$$
Angular Momentum $$L=mvr$$
m is mass, v is velocity and r is radius
We know that dimension formula of velocity is $$ LT^{-1}$$
Dimensional formula of $$L=M \times LT^{-1}\times L=ML^2T^{-1}$$

$${\textbf{Correct option: A}}$$
Question 10
1 / -0

The pair of physical quantities not having the same dimensional formula are:

A
Acceleration, Gravitational field strength

B
Torque, Angular momentum

C
Pressure, Modulus of Elasticity

D
All the above

Solution

Acceleration $$=LT^{-2}=$$Gravitational field strength
Torque $$=ML^{2}T^{-2}$$
Angular momentum $$=ML^{2}T^{-1}$$
Modulus of elasticity $$=ML^{-1}T^{-2}=$$ Pressure

Hence, option B is correct.

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Units and Measurements Test - 21

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Units and Measurements Test - 21

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Question 1

Pressure, Modulus of Elasticity

Angular velocity, Velocity gradient

Surface tension, Force constant

Impulse, Torque

Solution

Question 2

I, II, III, IV

III, II, I, IV

IV, I, II, III

III, IV, II, I

Solution

Question 3

$${ M }^{ -1 }{ L }^{ -2 }{ T }^{ 4 }{ I }^{ 2 }$$

$${ M }^{ 1 }{ L }^{ 2 }{ T }^{ 4 }{ I }^{ -2 }$$

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

$${ MLT }^{ -1 }$$

Solution

Question 4

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

$${ M }^{ 0 }{ L }^{ 2 }{ T }^{ -1 }$$

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

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

Solution

Question 5

Stress

Young's Modulus

Pressure

All the above

Solution

Question 6

$${ MLT }^{ -2 }$$

$${ M }^{ 2 }{ LT }^{ -1 }$$

$${ MLT }^{ -1 }$$

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

Solution

Question 7

$${ MT }^{ -1 }{ A }^{ -1 }$$

$${ MT }^{ -2 }{ A }^{ -1 }$$

$${ MLA }^{ -1 }$$

$${ MT }^{ -2 }A$$

Solution

Question 8

frequency

energy

time period

current

Solution

Question 9

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

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

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

$${ ML }^{ 3 }{ T }^{ -2 }$$

Solution

Question 10

Acceleration, Gravitational field strength

Torque, Angular momentum

Pressure, Modulus of Elasticity

All the above

Solution

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