Units and Measurements Test

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Units and Measurements Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Calculate area enclosed by a circle of diameter $$1.06\ m$$ to correct number of significant figures.

A
$$ 0.883 m^2$$

B
$$ 0.0883 m^2$$

C
$$ 0.88333 m^2$$

D
$$ 0.8830 m^2$$

Solution

Area = $$ \pi (\dfrac{d}{2})^2 $$
$$ \Rightarrow Area = \dfrac{22}{7}\times 0.53^2 = 0.882828 $$
But, no of significant in diameter should be equal to the no. of significant in Area, so Ans is $$ 0.883 m^2$$
Question 2
1 / -0

The dimension of permittivity ($${ \varepsilon }_{ 0 }$$) are ______. Take $$Q$$ as the dimension of charge.

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

B
$${ M }^{ -1 }{ L }^{ -3 }{ T }^{ 2 }{ Q }^{ 2 }$$

C
$${ M }^{ -1 }{ L }^{ 2 }{ T }^{ -3 }{ Q }^{ -1 }\quad $$

D
$${ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }{ Q }^{ -2 }\quad $$

Solution

We know that electrostatic force is given as:
$$F=\cfrac { K{ Q }^{ 2 } }{ { r }^{ 2 } }$$

We know that the dimension of force is $$ [F]=ML{ T }^{ -2 }$$

$$ \therefore ML{ T }^{ -2 }=\cfrac { K{ Q }^{ 2 } }{ { r }^{ 2 } } \\ [K]=M{ L }^{ 3 }{ T }^{ -2 }{ Q }^{ -2 }\\ K=\cfrac { 1 }{ 4\pi { \varepsilon }_{ o } } \\ [{ \varepsilon }_{ o }]={ M }^{ -1 }{ L }^{ -3 }{ T }^{ 2 }{ Q }^{ 2 }$$
Question 3
1 / -0

The order of magnitude of $$379$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Order of magnitude of $$379$$ is $$\dfrac { 379 }{ { 10 }^{ 2 } } =3.79<{ 5 }^{ -1 }$$ i.e, $$379\approx { 10 }^{ 2 }$$
order of magnitude $$=2$$.
Question 4
1 / -0

Which of the following is dimensionless?

A
Force/acceleration

B
Velocity/acceleration

C
Volume/area

D
Energy/work

Solution

$$\textbf{Correct option: Option (D)}$$
$$\textbf{Hint:}$$ If two quantity have same dimension then their ratio will be dimensionless.

$$\textbf{Explanation:}$$ A dimensionless quantity means the quantity which has no physical dimension or has a dimension one, they are also referred to as scalar quantities. They are often obtained as the ratio of two or more quantities.

Since the dimension of work and energy is the same, i.e., $$M{{L}^{2}}{{T}^{-2}}$$, when we take their ratio both the dimensions will cancel out and the result will be a dimensionless quantity.

$$\textbf{Thus, option (D) is correct.}$$
Question 5
1 / -0

If $$y$$ represents the pressure and $$x$$ represents the position then dimensional expression of $$\int { \dfrac { dx }{ \sqrt { p^{ 2 }- y^2} } } $$

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

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

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

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

Solution

Dimension of pressure = $$ ML^{-1}T^{-2}$$
Dimensions of p and y has to be equal
Dimensions of $$ dx = L^{1}$$
$$ \therefore $$ Dimensions of $$\int { \dfrac { dx }{ \sqrt { p^2 - y^2 } } }$$
Dimension of $$ \cfrac{dx}{y} = \cfrac{L^{1}}{ML^{-1}T^{-2}}$$
$$ = M^{-1}L^2T^2$$
Question 6
1 / -0

If the speed of light (c), acceleration due to gravity (g) and pressure (p) are taken as fundamental units, the dimensions of gravitational constant (G) are:

A
$$c^0gp^{-3}$$

B
$$c^2g^3p^{-2}$$

C
$$c^0g^2p^{-1}$$

D
$$c^2g^2p^{-2}$$

Solution

Dimension of $$ G= M^{-1}L^3T^{-2}$$
$$[c] = [LT^{-1}]$$
$$[g] = [LT^{-2}]$$
$$[p] = [ML^{-1}T^{-2}]$$
Equate dimensions
$$[M^{-1}L^3T^{-2}] = [LT^{-1}]^\alpha [LT^{-2}]^\beta [ML^{-1}T^{-2}]^\gamma$$
$$ -1 = \gamma$$
$$ 3 = \alpha + \beta - \gamma$$
$$ \alpha + \beta =2$$ (1)
And, $$ -2 = -\alpha -2\beta -2\gamma $$
$$ \alpha + 2\beta =4$$ (2)
Solving (1) and (2) we get,
$$ \alpha =0, \beta =2$$
$$ \therefore Dimension = c^0g^2p^{-1}$$
Question 7
1 / -0

The number of significant figures in $$0.00210$$ is

A
five

B
six

C
four

D
three

Solution

In the given number $$0.00210$$, only last digits $$210$$ are considered as significant figures. Thus the given number has three significant figures.
Question 8
1 / -0

Subtract $$2.5 \times {10}^{4}$$ from $$3.9 \times {10}^{5}$$ and give the answer to correct number of significant figures.

A
$$ 3.6\times 10^6$$

B
$$ 3.6\times 10^5$$

C
$$ 36\times 10^5$$

D
$$ 36\times 10^6$$

Solution

We have $$ 3.9 \times 10^5- 2.5 \times 10^4$$
$$ = 3.9 \times 10^5- 0.25 \times 10^5$$
$$ =3.65 \times 10^5$$
But our answer should be rounded off upto two significant digits.
So the ans is $$ 3.6\times 10^5$$
Question 9
1 / -0

The order of magnitude of 0.00701 is:

A
-3

B
-1

C
2

D
1

Solution

the order of magnitude of $$0.00701$$ is $$-3$$
we know that,
$$ N=a\times {{10}^{b}} $$

$$ 0.00701=7.01\times {{10}^{-3}} $$

Where b is the order of magnitude

So, the order of magnitude is -3
Note: For taking the order of magnitude, the number should always be written in the scientific notation first.
Question 10
1 / -0

The amplitude of a damped oscillator of mass $$'m'$$ varies with the time 't' as $$A=A_0e^{-at/m}$$. The dimension of $$'a'$$ are:

A
$$ML^0T^{-1}$$

B
$$M^0LT^{-1}$$

C
$$MLT^{-1}$$

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

Solution

Given, $$Mass=m,time=t.A=A_0e^{\dfrac{-at}{m}}$$

Since the term $$\dfrac{-at}{m}$$ is the exponential term so it is a dimension less

We know that,

The dimension of $$T=[M^0L^0T^1]$$ The dimension of $$m=[M^1L^0T^0]$$

So,

$$\Rightarrow \dfrac{[a][M^0L^0T^1]}{[M^1L^0T^0]}=1\Rightarrow a=\dfrac{[M^1L^0T^0]}{[M^0L^0T^1]}=[M^1L^0T^{-1}]$$

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Units and Measurements Test - 39

Result

Units and Measurements Test - 39

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Question 1

$$ 0.883 m^2$$

$$ 0.0883 m^2$$

$$ 0.88333 m^2$$

$$ 0.8830 m^2$$

Solution

Question 2

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

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

$${ M }^{ -1 }{ L }^{ 2 }{ T }^{ -3 }{ Q }^{ -1 }\quad $$

$${ M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 }{ Q }^{ -2 }\quad $$

Solution

Question 3

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 4

Force/acceleration

Velocity/acceleration

Volume/area

Energy/work

Solution

Question 5

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

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

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

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

Solution

Question 6

$$c^0gp^{-3}$$

$$c^2g^3p^{-2}$$

$$c^0g^2p^{-1}$$

$$c^2g^2p^{-2}$$

Solution

Question 7

five

six

four

three

Solution

Question 8

$$ 3.6\times 10^6$$

$$ 3.6\times 10^5$$

$$ 36\times 10^5$$

$$ 36\times 10^6$$

Solution

Question 9

-3

-1

2

1

Solution

Question 10

$$ML^0T^{-1}$$

$$M^0LT^{-1}$$

$$MLT^{-1}$$

$$ML^{-1}T$$

Solution

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