Units and Measurements Test

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Units and Measurements Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

Planck's constant ($$h$$), speed of light in vacuum ($$c$$) and Newton's gravitational constant ($$G$$) are three fundamental constants. Which of the following combinations of these has the dimension of length?

A
$$\sqrt{\dfrac{Gc}{h^{3/2}}}$$

B
$$\dfrac{\sqrt{hG}}{c^{3/2}}$$

C
$$\dfrac{\sqrt{hG}}{c^{5/2}}$$

D
$$\sqrt{\dfrac{hc}{G}}$$

Solution

Hint : Write the Units of all given physical quantities put these values in all option to verify which expression have same unit or dimension of length

Step 1: Write the SI units of given physical quantities

The SI unit of h =J.s or N.m.s
SI unit of c =m/s
S.I unit of $$G=\dfrac{N.m^{2}}{Kg^{2}}$$

Step 2: Finding the units of all options

For A: $$\sqrt{\dfrac{(N.m^2/kg^2)(m/s)}{(N.m.s)^{3/2}}}=\dfrac{kg.m^3}{N.s^5}=\dfrac{kg.m^3}{(kg.m/s^2)(s^5)}=m^2/s^3$$

For B: $$\dfrac{\sqrt{(N.m.s)(N.m^2/kg^2)}}{(m/s)^{3/2}}=\dfrac{Ns^2}{kg}=\dfrac{(kg.m/s^2)(s^2)}{kg}=m$$, it is the unit length.

For C: $$\dfrac{\sqrt{(N.m.s)(N.m^2/kg^2)}}{(m/s)^{5/2}}=\dfrac{Ns^3}{m.kg}=\dfrac{(kg.m/s^2)(s^3)}{m.kg}=s$$

For D: $$\sqrt{\dfrac{(N.m.s)(m/s)}{N.m^2/kg^2}}=kg$$

Hence option B is correct
Question 2
1 / -0

If force $$(F)$$, velocity $$(V)$$ and time $$(T)$$ are taken as fundamental units, the dimensions of mass are

A
$$[FVT^{-1}]$$

B
$$[FVT^{-2}]$$

C
$$[FV^{-1}T^{-1}]$$

D
$$[FV^{-1}T]$$

Solution

Hint: Find the formula of the mass in terms of force, velocity, and time.

Step 1: Relate the formula of mass to the given quantities.
We know,
$$F=ma$$
Acceleration is velocity per unit time.
$$a=\dfrac{V}{T}$$
Hence,
$$F=\dfrac{mV}{T}$$

Step 2: Find out the dimensional formula of mass.
Now, from the above equation, we can write,
$$m=\dfrac{FT}{V}$$
$$m=[FV^{-1}T]$$

Hence, the dimension of mass is $$[FV^{-1}T]$$
The correct answer is option (D).
Question 3
1 / -0

The dimensions of specific resistance are:

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

B
$$[ML^3T^{-3}A^{-2}]$$

C
$$[ML^3T^{-2}A^{-1}]$$

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

Solution

Resistance of a substance=$$R=\rho\dfrac{l}{A}$$
$$\implies \rho=\dfrac{RA}{l}$$
$$=\dfrac{V}{I}\dfrac{A}{l}$$
Dimension of $$V$$ is $$[ML^2T^{-3}A^{-1}]$$
Dimension of $$I$$ is $$[A]$$
Dimension of $$A$$ is $$[L^2]$$
Dimenstion of $$l$$ is $$[L]$$
Thus dimension of $$\rho$$ is $$[ML^{3}T^{-3}A^{-2}]$$
Question 4
1 / -0

The dimensional formula for electric flux is?

A
$$[ML^3A^{-1}T^{-3}]$$

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

C
$$[ML^3A^tT^{-3}]$$

D
$$[ML^{-3}A^{-1}T^{-3}]$$

Solution

Electric flux , $$\phi= BA= \dfrac F q A= \dfrac{MLT^{-2}\times L^2}{AT}= [ML^3A^{-1}T^{-3}]$$
Question 5
1 / -0

Which of the following pairs does not have same dimensions?

A
impulse and momentum

B
moment of inertia and moment of force

C
angular momentum and Planck's constatn

D
work and torque

Solution

Moment of force $$(\tau)$$ is equal to the product of moment of inertia $$(I)$$ and the angular acceleration $$(\alpha)$$ of the rotating body.
$$\therefore$$ $$\tau = I \alpha$$
Hence the dimensions of moment of force and moment of inertia must be different.
Question 6
1 / -0

What is the dimensions of magnetic field B in terms of C (=coulomb), M, L, T?

A
$$\displaystyle \left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }C \right] $$

B
$$\displaystyle \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -1 }{ C }^{ -1 } \right] $$

C
$$\displaystyle \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 }C \right] $$

D
$$\displaystyle \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -1 }C \right] $$

Solution

We know that force on a particle moving perpendicular to magnetic field is
$$F=qvB$$
$$\Rightarrow B=\dfrac{F}{qv}$$
Dimension of $$F$$ is $$[M^1L^1T^{-2}]$$
Dimension of $$q$$ is $$[C^1]$$
Dimension of $$v$$ is $$[L^1T^{-1}]$$
Hence, the dimensions of $$B$$ is $$[M^1L^{0}T^{-1}C^{-1}]$$
Question 7
1 / -0

Dimensional formula of angular momentum is

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

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

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

D
$$MLT^{-1}$$

Solution
Question 8
1 / -0

Dimensional formula of $$\Delta Q$$ heat supplied to the system is given by

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

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

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

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

Solution

Heat has the same dimesnsions as energy which has the same dimensions as force times distance. Hence the answer is $$ {[{M}{L}{T}^{-2}]}{[{L}]}={[{M}{L}^{2}{T}^{-2}]} $$.
Question 9
1 / -0

The dimensional formula of Planck's constant is:

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

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

C
$$[ML^0T^{2}]$$

D
$$[MLT^{2}]$$

Solution

$$\textbf{Hint:}$$ Use the dimensional formula of energy and frequency.
$$\textbf{Step 1:}$$ Energy of photon is given as
$$E=h\nu$$
$$h=\dfrac{E}{\nu}$$
$$\textbf{Step 2:}$$ Calculating the dimension of the Planck's constant,
Dimension of $$E=[ML^2T^{-2}]$$
Dimension of $$\nu$$ $$=$$ $$[T^{-1}]$$
So, the dimension of plank's constant $$h$$ is $$[ML^2T^{-1}]$$
$$\textbf{Option A is correct.}$$
Question 10
1 / -0

A cube has a side of length $$\displaystyle 1.2\times { 10 }^{ -2 }m$$. Calculate its volume.

A
$$\displaystyle 1.7\times { 10 }^{ -6 }{ m }^{ 3 }$$

B
$$\displaystyle 1.73\times { 10 }^{ -6 }{ m }^{ 3 }$$

C
$$\displaystyle 1.70\times { 10 }^{ -6 }{ m }^{ 3 }$$

D
$$\displaystyle 1.732\times { 10 }^{ -6 }{ m }^{ 3 }$$

Solution

Volume, $$\displaystyle V={ l }^{ 3 }={ \left( 1.2\times { 10 }^{ -2 }m \right) }^{ 3 }$$
$$\displaystyle =1.728\times { 10 }^{ -6 }{ m }^{ 3 }$$
Since length (l) has two significant figure, the volume (V) will also have two significant figure.
Therefore, the correct answer is $$\displaystyle V=1.7\times { 10 }^{ -6 }{ m }^{ 3 }$$

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Re-Attempt Weekly Quiz Competition

Units and Measurements Test - 18

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Units and Measurements Test - 18

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Question 1

$$\sqrt{\dfrac{Gc}{h^{3/2}}}$$

$$\dfrac{\sqrt{hG}}{c^{3/2}}$$

$$\dfrac{\sqrt{hG}}{c^{5/2}}$$

$$\sqrt{\dfrac{hc}{G}}$$

Solution

Question 2

$$[FVT^{-1}]$$

$$[FVT^{-2}]$$

$$[FV^{-1}T^{-1}]$$

$$[FV^{-1}T]$$

Solution

Question 3

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

$$[ML^3T^{-3}A^{-2}]$$

$$[ML^3T^{-2}A^{-1}]$$

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

Solution

Question 4

$$[ML^3A^{-1}T^{-3}]$$

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

$$[ML^3A^tT^{-3}]$$

$$[ML^{-3}A^{-1}T^{-3}]$$

Solution

Question 5

impulse and momentum

moment of inertia and moment of force

angular momentum and Planck's constatn

work and torque

Solution

Question 6

$$\displaystyle \left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }C \right] $$

$$\displaystyle \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -1 }{ C }^{ -1 } \right] $$

$$\displaystyle \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -2 }C \right] $$

$$\displaystyle \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -1 }C \right] $$

Solution

Question 7

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

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

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

$$MLT^{-1}$$

Solution

Question 8

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

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

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

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

Solution

Question 9

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

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

$$[ML^0T^{2}]$$

$$[MLT^{2}]$$

Solution

Question 10

$$\displaystyle 1.7\times { 10 }^{ -6 }{ m }^{ 3 }$$

$$\displaystyle 1.73\times { 10 }^{ -6 }{ m }^{ 3 }$$

$$\displaystyle 1.70\times { 10 }^{ -6 }{ m }^{ 3 }$$

$$\displaystyle 1.732\times { 10 }^{ -6 }{ m }^{ 3 }$$

Solution

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