Units and Measurements Test

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Units and Measurements Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

The equation of state of a gas is given by $$\begin{pmatrix}P+\dfrac{a}{v^3}\end{pmatrix} (V-b^2) = cT$$, where P, V, T are pressure, volume and temperature respectively, and a, b, c are constants. The dimensions of a and b are respectively

A
$$ML^8T^{-2}$$ and $$L^{3/2}$$

B
$$ML^5T^{-2}$$ and $$L^{3}$$

C
$$ML^5T^{-2}$$ and $$L$$

D
$$ML^6T^{-2}$$ and $$L^{3/2}$$

Solution

The dimensions of $$\dfrac{a}{v^3}$$ must be that of pressure.
Dimension of pressure are $$[ML^{-1}T^{-2}]$$
Hence dimension of $$a$$ is $$[ML^{-1+3*3}T^{-2}]=[ML^8T^{-2}]$$
The dimension of $$b^2$$ must be that of volume.
Hence $$[b^2]=[L^3]$$
$$\implies [b]=[L^{3/2}]$$
Question 2
1 / -0

$$ML^{-1}T^{-2}$$ is the dimensional formula of

A
force

B
coefficient of friction

C
modulus of elasticity

D
energy

Solution

Modulus of elasticity is given by $$Y = \dfrac{F}{A}.\dfrac{L}{\Delta L}$$
Dimension of force $$F = [MLT^{-2}]$$
Dimension of Area $$A = [L^2]$$
Dimension of length $$L = [L]$$
Thus dimension of modulus of elasticity $$Y = \dfrac{[MLT^{-2}][L]}{[L^2][L]} = [ML^{-1}T^{-2}]$$
Question 3
1 / -0

Two quantities X and Y have different dimension. Which mathematical operation given below is physically meaningful?

A
$$X+Y$$

B
$$X-Y$$

C
$$\dfrac{X}{Y}$$

D
None of these

Solution

Since, X and Y have different dimension, they can only be divided.
Question 4
1 / -0

Four students measure the height of a tower. Each student uses different methods and each measures the height many times. The data for each are plotted. The measurement with the highest precision is:

A
I

B
II

C
III

D
IV

Solution

Precision refers to the closeness of measurements to each other, while accuracy refers to closeness of measurements to a known value.

In the 4 different measurements, the data in figure (I) are close to each other compared to (II), (III) and (IV).

Thus, the measurements with highest precision are made in (I)
Question 5
1 / -0

The dimensions formula of ( velocity $$)^2 /$$ radius are the same of that of :

A
Planck's constant

B
Gravitational constant

C
Dielectric constant

D
None of these

Solution

Dimensions formula of ( velocity $$)^2 /$$ radius

$$ = \dfrac {[M^0 LT^{-1}]^2 }{[M^0 LT^0]} = [LT^{-2}]$$

Thus, it has the dimensions of acceleration
Question 6
1 / -0

If the dimensions of length are expressed as a $$G^x \cdot C^y \cdot h^z$$, where G, C and h are the universal gravitational constant, speed of light and plank constant respectively, then value of x,y,z will be :

A
$$x=1, y=-\dfrac{3}{2}, z=\dfrac{1}{2}$$

B
$$x=\dfrac{1}{2}, y=\dfrac{1}{2}, z=\dfrac{3}{2}$$

C
$$x=\dfrac{1}{2}, y=-\dfrac{3}{2}, z=\dfrac{1}{2}$$

D
$$x=-\dfrac{3}{2}, y=\dfrac{1}{2}, z=\dfrac{1}{2}$$

Solution

Length $$\propto G^xC^yh^z$$
$$[L]=[M^{-1}L^3T^{-2}]^x[LT^{-1}]^y[ML^2T^{-1}]^z$$
On comparing the power of M, L and T in both sides, we get
$$-x+z=0$$ ....(i)
$$3x+y+2z= 1$$ ....(ii)
and $$-2x-y-z=0$$ ....(iii)
By solving Eqs. (i), (ii) and (iii) we get
$$x=\dfrac{1}{2}; y=-\dfrac{3}{2}$$ and $$z=\dfrac{1}{2}$$
Question 7
1 / -0

The dimension of $$\displaystyle \frac{a}{b}$$ in the equation $$\displaystyle p = \frac{a-t^2}{bx}$$ where p is pressure, x is distance and t is time is

A
$$\displaystyle [LT^{-3}]$$

B
$$\displaystyle [ML^3 T^{-1}]$$

C
$$\displaystyle [M^2 L T^{-3}]$$

D
$$\displaystyle [MT^{-2}]$$

Solution

Given equation $$\displaystyle P= \frac{a-t^2}{bx} = \frac{a}{bx} - \frac{t^2}{bx}$$
Applying the principal of homogeneity, we have
$$\displaystyle [pressure]= \left[ \frac{force}{area} \right] = \left[ \frac{a}{bx} \right]$$
or $$\displaystyle [ML^{-1}T^{-2}] = \left[ \frac{a}{b} \right] \left[ \frac{1}{L} \right]$$
or $$\displaystyle \left[ \frac{a}{b} \right] = [MT^{-2}]$$
Question 8
1 / -0

The Quantum Hall Resistance $$R_{H}$$ is a fundamental constant with dimensions of resistance. If $$h$$ is Planck's constant and $$e$$ the electron charge, then the dimension of $$R_{H}$$ is the same as.

A
$$e^{2}/h$$

B
$$h/e^{2}$$

C
$$h^{2}/e$$

D
$$e/h^{2}$$

Solution

From the fundamental relation of energy of a photon, we have $$E=h\nu$$,
$$[h] = \textrm{[Energy][Time]}$$ and $$[e] = [Charge]$$
$$\textrm{[Energy] = [Charge][Volt]}$$

Thus, $$\textrm{[Volt]} =\frac{[Energy]}{\textrm{[Charge]}}=\frac{[h]}{[e][Time]}$$

Similarly, $$\textrm{[Current]} = \frac{\textrm{[Charge]}}{\textrm{[Time]}}=\frac{[e]}{\textrm{[Time]}}$$

From Ohm's Law, $$R = \frac{V}{I}$$, we have
$$[R_H]=\frac{\textrm{[Volt]}}{\textrm{[Current]}}=\frac{[h]}{[e]\textrm{[Time]}} \times \frac{\textrm{[Time]}}{[e]} = [\frac{h}{e^2}]$$

Thus, the dimensions of $$R_H$$ match with the dimensions of $$\frac{h}{e^2}$$
Question 9
1 / -0

Express 0.006006 into scientific notation in three significant digits:

A
$$6.01 \times 10^{-3}$$

B
$$6.0006 \times 10^{-3}$$

C
$$6.00 \times 10^{-3}$$

D
$$6.0 \times 10^{-3}$$

Solution

The number $$0.006006$$ has 6 significant figures.
We need to round off the given number to convert into 3 significant figure. Thus, the correct answer is $$6.01\times 10^{-3}$$ as it has 3 significant figure including rounding off.
Question 10
1 / -0

The dimensions of capacitance are :

A
$$\displaystyle \left[ { ML }^{ -2 }{ Q }^{ -2 }{ T }^{ 2 } \right] $$

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

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

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

Solution

$$\textbf{Step 1-Formula of Capacitance}$$
If charge on capacitor is C and potential on it is V.
then capacitance $$C=\dfrac{Q}{V}$$

$$\textbf{Step 2- Dimensions of charge and potential}$$
Dimension of charge [Q]={AT]
Potential $$V=\dfrac{W}{Q}$$
So dimension of potential is $$[V]=\dfrac{[ML^2T^{-2}]}{Q}$$
$$[V]=[ML^2T^{-2}Q^{-1}]$$

$$\textbf{Step 3-Dimensions of capacitance}$$
$$[C]=\dfrac{[Q]}{[ML^2T^{-2}Q^{-1}]}$$
$$[C]=[M^{-1}L^{-2}T^2Q^{2}]$$

Hence the correct answer is (D).

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Units and Measurements Test - 58

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Units and Measurements Test - 58

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Question 1

$$ML^8T^{-2}$$ and $$L^{3/2}$$

$$ML^5T^{-2}$$ and $$L^{3}$$

$$ML^5T^{-2}$$ and $$L$$

$$ML^6T^{-2}$$ and $$L^{3/2}$$

Solution

Question 2

force

coefficient of friction

modulus of elasticity

energy

Solution

Question 3

$$X+Y$$

$$X-Y$$

$$\dfrac{X}{Y}$$

None of these

Solution

Question 4

I

II

III

IV

Solution

Question 5

Planck's constant

Gravitational constant

Dielectric constant

None of these

Solution

Question 6

$$x=1, y=-\dfrac{3}{2}, z=\dfrac{1}{2}$$

$$x=\dfrac{1}{2}, y=\dfrac{1}{2}, z=\dfrac{3}{2}$$

$$x=\dfrac{1}{2}, y=-\dfrac{3}{2}, z=\dfrac{1}{2}$$

$$x=-\dfrac{3}{2}, y=\dfrac{1}{2}, z=\dfrac{1}{2}$$

Solution

Question 7

$$\displaystyle [LT^{-3}]$$

$$\displaystyle [ML^3 T^{-1}]$$

$$\displaystyle [M^2 L T^{-3}]$$

$$\displaystyle [MT^{-2}]$$

Solution

Question 8

$$e^{2}/h$$

$$h/e^{2}$$

$$h^{2}/e$$

$$e/h^{2}$$

Solution

Question 9

$$6.01 \times 10^{-3}$$

$$6.0006 \times 10^{-3}$$

$$6.00 \times 10^{-3}$$

$$6.0 \times 10^{-3}$$

Solution

Question 10

$$\displaystyle \left[ { ML }^{ -2 }{ Q }^{ -2 }{ T }^{ 2 } \right] $$

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

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

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

Solution

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