Units and Measurements Test

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Units and Measurements Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The deBroglie wavelength associated with a particle of mass $$m$$ and energy $$E$$ is $$\dfrac{h}{\sqrt {2mE}}$$. The dimensional formula of Planck's constant h is

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

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

C
$$[MLT^{-2}]$$

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

Solution

Planck constant $$=h=\lambda \sqrt {2mE}=L\sqrt {M(ML^2T^{-2)}}=[ML^2T^{-1}]$$
Question 2
1 / -0

Let $$[\epsilon_0]$$ denote the dimensional formula of the permittivity of vacuum. If $$M=mass, L=length, T=time$$ and $$A=electric$$ current, then

A
$$\epsilon_0 = [M^{-1}L^{-3}T^2A^2]$$

B
$$\epsilon_0 = [M^{-2}L^{-3}T^4A^2]$$

C
$$\epsilon_0 = [M^{1}L^{2}T^1A^2]$$

D
$$\epsilon_0 = [M^{1}L^{2}T^1A]$$

Solution

Using Coulomb law, $$F=\dfrac {1}{4\pi \epsilon_0}\dfrac {q_1q_2}{R^2}$$

$$\Rightarrow \epsilon_0=\dfrac {q_1q_2}{4\pi FR^2}$$

Substituting the units in the equation:

So, $$\epsilon_0=\dfrac {C^2}{N.m^2}$$

$$=\dfrac {[AT]^2}{MLT^{-2}.L^2}=[M^{-1}L^{-3}T^4A^2]$$
Question 3
1 / -0

Match List I with List II and select the correct answer using the codes given below the list
List I List II
P. Boltzmann constant 1. $$[ML^2T^{-1}]$$
Q. Coefficient of viscosity 2. $$[ML^{-1}T^{-1}]$$
R. Planck constant 3. $$[MLT^{-3}K^{-1}]$$
S. Thermal conductivity 4. $$[ML^2T^{-2}K^{-1}]$$

A
3 1 2 4

B
3 2 1 4

C
4 2 1 3

D
4 1 2 3

Solution

Boltzmann constant $$=\displaystyle \frac {R}{N}=\frac {PV}{nTN}=\frac {ML^{-1}T^{-2}L^3}{K}=ML^2T^{-2}K^{-1}$$

Coefficient of viscosity $$ =\displaystyle \frac {F}{6\pi rv}=\frac {MLT^{-2}}{L.LT^{-1}}=ML^{-1}T^{-1}$$

Planck constant $$=\displaystyle \frac {E}{v}=\frac {ML^2T^{-2}}{T^{-1}}=ML^{2}T^{-1}$$

Thermal conductivity $$=\displaystyle \frac {Hl}{tA\Delta T}=\frac {ML^2T^{-2}L}{TL^2K}=MLT^{-3}K^{-1}$$
Question 4
1 / -0

If g is the acceleration due to gravity and $$\lambda$$ is wavelength then which physical quantity does $$\sqrt {\lambda g}$$ represent?

A
Velocity

B
Length

C
Time

D
Charge

Solution

Since $$\lambda$$ is wavelength, hence its dimension is that of length = $$L$$.
Dimension of g is $$LT^{-2}$$
Thus, dimension of $$\sqrt{\lambda g}=\sqrt{L^2T^{-2}}=LT^{-1}$$, which represents velocity.
Question 5
1 / -0

The dimensions of $$\dfrac {1}{\epsilon_0}\dfrac {e^2}{hc}$$ are

A
$$M^{-1}L^{-3}T^4A^2$$

B
$$ML^{3}T^{-4}A^{-2}$$

C
$$M^0L^0T^0A^0$$

D
$$M^{-1}L^{-3}T^2A$$

Solution

Here, $$h$$ =planck's constant, $$c$$ = light velocity
Using Coulomb's law we have $$F=\dfrac {1}{4\pi \epsilon_0}\dfrac {e^2}{r^2}$$
$$\therefore \dfrac {e^2}{\epsilon_0}=4\pi Fr^2$$ (dimensionally)
$$[\dfrac {e^2}{\epsilon_0hc}]=[\dfrac {4\pi Fr^2}{hc}]=\dfrac {(MLT^{-2})L^2}{ML^2T^{-1}[LT^{-1}]}=[M^0L^0T^0A^0]$$
The term $$\dfrac {e^2}{\epsilon_ohc}$$ is called fine structure constant & has the value $$=\dfrac {1}{137}$$.
Question 6
1 / -0

If $$e$$ is the charge, $$V$$ the potential difference, $$T$$ the temperature, then the units of $$\dfrac {eV}{T}$$ are the same as that of

A
planck's constant

B
stefan's constant

C
boltzmann constant

D
gravitational constant

Solution

$$\displaystyle \frac {eV}{T}=\frac {Workdone}{T}=\frac {PV}{T}=\frac { R }{ N } =k=Boltzmann \ constant$$
(By using ideal gas equation $$PV=\dfrac { RT }{ N } $$)
Question 7
1 / -0

The length, breadth and thickness of a block are given by $$l = 12\ cm$$, $$b = 6\ cm$$ and $$t = 2.45 cm$$. The volume of the block according to the idea of significant figures should be :

A
$$\displaystyle 1\times 10^{2}\:cm^{3}$$

B
$$\displaystyle 2\times 10^{2}\:cm^{3}$$

C
$$\displaystyle 1.763\times 10^{2}\:cm^{3}$$

D
None of these

Solution

Since question contains atleast 1 significant number in one of the value, therefore, we should write the answer in such a form that it contains only one significant figure. $$V=lbt=12\times 6\times 2.45=2\times 10^2\ cm^3$$
Question 8
1 / -0

If time $$T$$, acceleration $$A$$ and force $$F$$ are regarded as base units, then the dimensional formula of work is

A
$$[FA]$$

B
$$[FAT]$$

C
$$[FAT^2]$$

D
$$[FA^2T]$$

Solution

$$Work=Force \times distance=F\times L$$
We need distance in terms of acceleration and time.
$$A=LT^{-2}$$
$$L=AT^2$$
$$W=FAT^2$$
Question 9
1 / -0

Using mass $$(M)$$, length $$(L)$$, time $$(T)$$ and electric current $$(A)$$ as fundamental quantities the dimensions of permittivity will be

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

B
$$[MLT^{-2}A^{-2}]$$

C
$$[M^{-1}L^{-3}T^{+4}A^{2}]$$

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

Solution

Using Coulomb's law, Force $$F=\dfrac {1}{4\pi \epsilon_0}\dfrac {q_1q_2}{r^2}\Rightarrow \epsilon_2=\dfrac {q_1q_2}{4\pi Fr^2}$$
So dimension of $$[\epsilon_0]$$$$=\dfrac {[AT]^2}{[MLT^{-2}][L^2]}=[M^{-1}L^{-3}T^4A^2]$$
Question 10
1 / -0

The dimensional formula of farad is

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

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

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

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

Solution

One farad is defined as the capacitance $$C$$ of a capacitor across which, when charged with one coulomb $$Q$$ of electricity, there is a potential difference $$V$$ of one volt.
So, $$\displaystyle [C]=\left [\frac {Q}{V}\right ]=\left [\frac {Q^2}{W}\right ]=\left [\frac {Q^2}{M^{1}L^{2}T^{-2}}\right ]=[M^{-1}L^{-2}T^2Q^2]$$

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Re-Attempt Weekly Quiz Competition

List I	List II
P. Boltzmann constant	1. $$[ML^2T^{-1}]$$
Q. Coefficient of viscosity	2. $$[ML^{-1}T^{-1}]$$
R. Planck constant	3. $$[MLT^{-3}K^{-1}]$$
S. Thermal conductivity	4. $$[ML^2T^{-2}K^{-1}]$$

Units and Measurements Test - 47

Result

Units and Measurements Test - 47

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Rank

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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

$$\epsilon_0 = [M^{-1}L^{-3}T^2A^2]$$

$$\epsilon_0 = [M^{-2}L^{-3}T^4A^2]$$

$$\epsilon_0 = [M^{1}L^{2}T^1A^2]$$

$$\epsilon_0 = [M^{1}L^{2}T^1A]$$

Solution

Question 3

3 1 2 4

3 2 1 4

4 2 1 3

4 1 2 3

Solution

Question 4

Velocity

Length

Time

Charge

Solution

Question 5

$$M^{-1}L^{-3}T^4A^2$$

$$ML^{3}T^{-4}A^{-2}$$

$$M^0L^0T^0A^0$$

$$M^{-1}L^{-3}T^2A$$

Solution

Question 6

planck's constant

stefan's constant

boltzmann constant

gravitational constant

Solution

Question 7

$$\displaystyle 1\times 10^{2}\:cm^{3}$$

$$\displaystyle 2\times 10^{2}\:cm^{3}$$

$$\displaystyle 1.763\times 10^{2}\:cm^{3}$$

None of these

Solution

Question 8

$$[FA]$$

$$[FAT]$$

$$[FAT^2]$$

$$[FA^2T]$$

Solution

Question 9

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

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

$$[M^{-1}L^{-3}T^{+4}A^{2}]$$

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

Solution

Question 10

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

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

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

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

Solution

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