Units and Measurements Test

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Units and Measurements Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

If $$L,C$$ and $$R$$ denote the inductance, capacitance and resistance respectively, the dimensional formula for $${C}^{2}LR$$ is :

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

B
$$\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 3 }{ I }^{ 0 } \right] $$

C
$$\left[ { M }^{ -1 }{ L }^{ -2 }{ T }^{ 6 }{ I }^{ 2 } \right] $$

D
$$\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 2 }{ I }^{ 0 } \right] $$

Solution

Given $$\left[ { C }^{ 2 }LR \right] =\left[ { C }^{ 2 }{ L }^{ 2 }\cfrac { R }{ L } \right] =\left[ { (LC) }^{ 2 }\cfrac { R }{ L } \right] $$
ad we know that frequency of $$LC$$ circuits
$$f=\cfrac { 1 }{ 2\pi } \cfrac { 1 }{ \sqrt { LC } } $$ Here the dimension of $$LC$$ is equal to $$\left[ { T }^{ 2 } \right] $$
$$\left[ \cfrac { L }{ R } \right] $$ gives the time constant of $$L-R$$ circuit, so that the dimension of $$\left[ \cfrac { L }{ R } \right] $$ is equal to $$\left[ T \right] $$
Hence the required dimensions
$$\left[ \cfrac { L }{ R } \right] \quad \left[ { (LC) }^{ 2 }\cfrac { R }{ L } \right] ={ \left[ { T }^{ 2 } \right] }^{ 2 }\left[ { T }^{ -1 } \right] =\left[ { T }^{ 3 } \right] $$
Question 2
1 / -0

The dimensions of angular momentum/ magnetic moment are

A
$$[MA^{-1}T^{-1}]$$

B
$$[M^{-1}AT^{-1}]$$

C
$$[MAT^{-1}]$$

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

Solution

Angular momentum
$$= l\omega = [M^{1}L^{2}T^{-1}]$$
Magnetic moment
$$= [AL^2]$$
$$\therefore = \dfrac {Angular\ momentum}{Magnetic\ moment}$$
$$= \dfrac {[M^{1}L^{2}T^{-1}]}{[AL^{2}]} = [MA^{-1}T^{-1}]$$.
Question 3
1 / -0

The dimensions of $$\dfrac {\alpha}{\beta}$$ in the equation $$F = \dfrac {\alpha - t^{2}}{\beta v^{2}}$$, where $$F$$ is force, $$v$$ is velocity and $$t$$ is time, is :

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

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

C
$$[ML^{3}T^{-4}]$$

D
$$[ML^{2}T^{-4}]$$

Solution

The given equation is   $$F = \dfrac{\alpha - t^2}{\beta v^2}$$
Dimension of $$\alpha$$ is same as that of square of time i.e. $$\alpha = [T^2]$$
Dimensoin of $$\beta$$ is   $$\beta = \dfrac{\alpha - t^2}{Fv^2} = \dfrac{[T^2]}{[MLT^{-2}][L^2T^{-2}]} = [M^{-1}L^{-3}T^6]$$
Thus   $$\dfrac{\alpha}{\beta} = \dfrac{[T^2]}{[M^{-1}L^{-3}T^6]} = [ML^3T^{-4}]$$
Question 4
1 / -0

While measuring the length of a wooden box, the reading at one end is 1.5 cm and the other end is 4.7 cm. What is the length of the wooden box?

A
4.7 cm

B
1.5 cm

C
3.2 cm

D
6.2 cm

Solution

The length of the wooden box = Difference between the reading at both the ends = 4.7 - 1.5 = 3.2 cm
Question 5
1 / -0

The position x of a particle at time t is given by $$x = \dfrac{v_{0}}{a}(1 - e^{-at})$$ where $$v_{0}$$ is a constant and a > 0. The dimensional formula of $$v_{0}$$ and a are

A
$$[M^{0}L^{1}T^{-1}] and [M^{0}L^{0}T^{-1}]$$

B
$$[M^{0}L^{1}T^{0}] and [M^{0}L^{0}T^{-1}]$$

C
$$[M^{0}LT^{-1}] and [M^{0}L^{1}T^{-2}]$$

D
$$[M^{0}L^{1}T^{-1}] and [M^{0}L^{0}T^{1}]$$

Solution

For finding $$v_0=\cfrac{ax}{1-e^{a/t}}\\ \quad=\cfrac{m/s\times m/s^2}{1-e^{-m/s^2\times s}}\\ \quad=\cfrac{m^2}{s^2}\times\cfrac{s}{m}\\ \quad=m/s[M^0LT^{-1}]$$
For finding,
$$v_0=\cfrac{xa}{1-e^{-at}}\\ \cfrac{v_0}{x}=\cfrac{a}{1-e^{-at}}\\ \cfrac{x}{v_0}=\cfrac{1-e^{-at}}{a}\\a=\cfrac{v_0}{x}\\=\quad \cfrac{m/s}{m}\\a=\cfrac{1}{s}\\a=s^{-1}\\ \quad=[M^0L^0T^{-1}]$$
Question 6
1 / -0

The dimensional formula for permittivity of free space $$(\epsilon_{0})$$ in the equation $$F = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q_{1}q_{2}}{r^{2}}$$ where, symbols have their usual meaning is

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

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

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

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

Solution

Permittivity of free space $$\epsilon_o = \dfrac{q_1q_2}{4\pi Fr^2}$$
Dimensions of $$[F] = [MLT^{-2}]$$
Dimensions of $$[q] = [AT]$$
Dimensions of $$[r] = [L]$$
Thus dimensional formula of $$[\epsilon_o] = \dfrac{[AT][AT]}{[MLT^{-2}][L^2]}$$
$$\implies$$ $$[\epsilon_o] = [M^{-1}L^{-3} T^4A^2]$$
Question 7
1 / -0

The sun's angular diameter is measured to be 1920". The distance of the sun from the earth is $$1.496 \times 10^{11} m$$. What is the diameter of the sun then?

A
$$1.39 \times 10^9 m$$.

B
$$1.39 \times 10^{10} m$$.

C
$$1.39 \times 10^{11} m$$.

D
$$1.39 \times 10^{12} m$$.

Solution

The Sun’s diameter (d) = $${1.494\times {{10}^{11}}m}$$

And Sun’s angular diameter ($$\alpha $$) = $$1920''$$

To convert the angular diameter into radians, converting $$1920''$$into radians.

$$1''=4.85\times {{10}^{-6}}rad$$

⸫ $$1920''$$= $$1920\times 4.85\times {{10}^{-6}}=9.31\times {{10}^{-3}}rad$$= $$\alpha $$

Now, Sun’s diameter (D) = $$\alpha \times d$$

D = ($$9.31\times {{10}^{-3}}\times 1.494\times {{10}^{11}}$$)m

⸫ D = $$1.39\times {{10}^{9}}m$$

$$\textbf{Answer:}$$

$$\textbf{Hence, the correct option is (a)}$$ $$1.39\times {{10}^{9}}m$$
Question 8
1 / -0

Four charges are placed at the circumference of the dial of a clock as shown in figure. If the clock has only hour hand, then the resultant force on a positive charge $$q_{0}$$ placed at the centre, points in the direction which shows the time as :-

A
1:30

B
7:30

C
4:30

D
10:30

Solution

Where $$F=\cfrac{kq^2}{r^2}\\F_{net}=\sqrt{(2F)^2+(2F)^2}$$
Between $$6:00$$ and $$9:00$$
$$Time =7:30$$
Question 9
1 / -0

Four pieces of wooden sticks P, Q, R and S are placed along the length of 15 cm long scale as shown in figure. What is the average length of these sticks?

A
2.0 cm

B
2.5 cm

C
2.6 cm

D
2.9 cm

Solution

The length of the different sticks is as follows:
$$P = 2.1 cm;\\ Q = 3.0 cm;\\ R = 3.1 cm;\\ S = 2.5 cm$$

Average length of these sticks
$$=\dfrac{P+Q+R+S}{4}$$

$$= 2.6 cm$$
Question 10
1 / -0

Which one of the following methods is used to measure distance of a planet or a star from the earth?

A
Echo method

B
Parallax method

C
Triangulation method

D
None of these

Solution

Astronomers estimate the distance of nearby objects in space by using a method called stellar parallax, or trigonometric parallax. Simply put, they measure a star's apparent movement against the background of more distant stars as Earth revolves around the sun

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Units and Measurements Test - 60

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Units and Measurements Test - 60

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Question 1

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

$$\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 3 }{ I }^{ 0 } \right] $$

$$\left[ { M }^{ -1 }{ L }^{ -2 }{ T }^{ 6 }{ I }^{ 2 } \right] $$

$$\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 2 }{ I }^{ 0 } \right] $$

Solution

Question 2

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

$$[M^{-1}AT^{-1}]$$

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

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

Solution

Question 3

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

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

$$[ML^{3}T^{-4}]$$

$$[ML^{2}T^{-4}]$$

Solution

Question 4

4.7 cm

1.5 cm

3.2 cm

6.2 cm

Solution

Question 5

$$[M^{0}L^{1}T^{-1}] and [M^{0}L^{0}T^{-1}]$$

$$[M^{0}L^{1}T^{0}] and [M^{0}L^{0}T^{-1}]$$

$$[M^{0}LT^{-1}] and [M^{0}L^{1}T^{-2}]$$

$$[M^{0}L^{1}T^{-1}] and [M^{0}L^{0}T^{1}]$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$1.39 \times 10^9 m$$.

$$1.39 \times 10^{10} m$$.

$$1.39 \times 10^{11} m$$.

$$1.39 \times 10^{12} m$$.

Solution

Question 8

1:30

7:30

4:30

10:30

Solution

Question 9

2.0 cm

2.5 cm

2.6 cm

2.9 cm

Solution

Question 10

Echo method

Parallax method

Triangulation method

None of these

Solution

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