Units and Measurements Test

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Units and Measurements Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The moon is observed from two diametrically opposite points on earth. The angle subtended at the moon by the two directions of observations is $$1^o 54'$$. Given the diameter of the earth to be about $$2.276 \times 10^7 m$$, compute the distance of moon from the earth.

A
$$1 \times 10^8 m$$

B
$$6.84 \times 10^8 m$$

C
$$6.84 \times 10^{10} m$$

D
$$38.4 \times 10^8 m$$

Solution

Here D be the distance between the moon and the earth, d be the diameter of the earth.
Here, $$\theta=1^o54'=1.9^o=\dfrac{\pi}{180}\times 1.9 rad=0.033$$ radian

From the figure, $$\tan\theta=d/D$$

As $$\theta$$ is very small, $$\theta=d/D$$

$$D=\dfrac{d}{\theta}=\dfrac{2.276\times 10^7}{0.033}=6.84\times 10^8$$ m
Question 2
1 / -0

The mass of the electron is given $$9.1\times 10^{-31} kg$$. Find the order of magnitude of the number of electrons of $$1 kg$$.

A
$$10^{31}$$

B
$$10^{-30}$$

C
$$10^{30}$$

D
$$10^{-31}$$

Solution

Let n be the required number of electrons.
Thus, $$n\times (9.1\times 10^{-31})=1$$
or $$n=1.1\times 10^{30}$$
According to rule of order of magnitude, as $$1.1<5$$ so it will be taken as 1.
Thus, $$n=10^{30}$$
Question 3
1 / -0

What is the order of magnitude of the distance of a quasar from us if light takes 2.9 billion years to reach us ?

A
$$2.7\times 10^{25} m$$

B
$$ 10^{25} m$$

C
$$10^24 m$$

D
$$3\times 10^{25} m$$

Solution

Here, time taken $$t=2.9$$ billion years $$=2.9\times 10^9$$ years $$ =2.9\times 10^9\times (365\times 24\times 3600) s=9.14\times 10^{16} s$$
Distance $$=$$ velocity $$\times $$ time
or $$d=(3\times 10^8)\times (9.14\times 10^{16})=2.74\times 10^{25} m$$
As $$2.74 <5$$ so it will be taken as 1.
Thus, the order of magnitude of distance $$d=1\times 10^{25} m=10^{25} m$$
Question 4
1 / -0

The radius of proton is $$0.6\times 10^{-15} m$$. What is the order of magnitude of it volume?

A
$$10^{-47} m^3$$

B
$$10^{-44} m^3$$

C
$$10^{-45} m^3$$

D
$$10^{-46} m^3$$

Solution

The volume of the proton is $$V=\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\times 3.14\times (0.6\times 10^{-15})^3=9.04\times 10^{-46} m^3$$
According to rule of order of magnitude , as $$9.04>5$$ so it will be taken 10.
Thus, order of magnitude of volume $$=10\times 10^{-46}=10^{-45} m^3$$
Question 5
1 / -0

What is the order of magnitude of one light year?

A
$$10^{15} m$$

B
$$10^{10} m$$

C
$$9.2 \times 10^{15} m$$

D
$$10^{16} m$$

Solution

One light year is the distance of light at a speed $$3\times 10^8 m/s$$ in 1 year.
Thus, $$1 $$ light year $$= 3\times 10^8 m/s\times 1$$ year $$=3\times 10^8 m/s\times (365\times 24\times 3600 s)=9.5\times 10^{15} m$$
According to rule of order of magnitude, as $$9.5>5$$ so 9.5 will be taken as 10.
Thus, order of magnitude of one light year $$=10\times 10^{15}=10^{16} m$$
Question 6
1 / -0

Match the following:

Column - I

Column - II
A
Deca
p
$$10^3$$
B
Hecto
q
$$10^6$$
C
Kilo
r
$$10^1$$
D
Mega
s
$$10^2$$
E
Giga
t
$$10^8$$

u
$$10^9$$

A
A-r, B-s, C-p, D-q, E-u

B
A-s, B-r, C-p, D-q, E-u

C
A-r, B-s, C-u, D-q, E-p

D
A-q, B-s, C-p, D-r, E-u

Solution

Here column-I represents the some prefixes and column-II represents the power of ten corresponding prefixes.
Deca $$=10^1$$, hecto =$$10^2$$, kilo =$$10^3$$, mega $$=10^6$$, giga $$=10^9$$
Thus, A-r, B-s, C-p, D-q, E-u
Question 7
1 / -0

The radius of the sun is $$7\times 10^8 m$$ and its mass is $$2\times 10^{30} kg$$. What is the order of magnitude of density of the sun?

A
$$1.4\times 10^3 kg/m^3$$

B
$$ 10^7 kg/m^3$$

C
$$1.5\times 10^3 kg/m^3$$

D
$$10^3 kg/m^3$$

Solution

Given : $$R = 7\times 10^8 \ m$$ and $$M = 2\times 10^{30} \ kg$$
Volume of sun $$V = \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi\times (7\times 10^8)^3 = 1.4\times 10^{27} \ m^3$$
Density of sun $$\rho = \dfrac{M}{V} = \dfrac{2\times 10^{30}}{1.4\times 10^{27}} = 1.43\times 10^3 \ kg/m^3$$
Thus order of magnitude of density of sun is $$10^3 \ kg/m^3$$.
Question 8
1 / -0

The radius of the earth is given $$6.4\times 10^6 m$$. What is the order of magnitude of the size of the earth?

A
$$10^6 m$$

B
$$6 \times 10^6 m$$

C
$$10^7 m$$

D
$$5 \times 10^6 m$$

Solution

When a number rounded to the nearest power of 10, it is called order of magnitude. Here a number which is less than 5 is taken as 1 and a number greater than 5 is taken as 10.
As $$6.4>10, $$ so it would be takes as 10.
The order of magnitude of earth's size $$=10\times 10^6=10^7 m$$
Question 9
1 / -0

A star has a parallax angle p of 0.723 arcseconds. What is the distance of the star?

A
1.38 parsecs

B
2.38 parsecs

C
3.38 parsecs

D
4.38 parsecs

Solution

Relationship between a star's distance and its parallax angle:
$$d=\dfrac{1}{p}$$
The distance $$d$$ is measured in parsecs and the parallax angle $$p$$ is measured in arcseconds.
Hence, $$d=\dfrac{1}{0.723}=1.38 parsecs$$
Question 10
1 / -0

Two stars $$S_1$$ and $$S_2$$ are located at distances $$d_1$$ and $$d_2$$ respectively. Also if $$d_1>d_2$$ then which of the following statements is true?

A
The parallax of $$S_1$$ and $$S_2$$ are same.

B
The parallax of $$S_1$$ is twice as that of $$S_2$$.

C
The parallax of $$S_1$$ is greater than parallax of $$S_2$$

D
The parallax of $$S_2$$ is greater than parallax of $$S_1$$

Solution

$$Answer:-$$ D
Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines.Due to foreshortning, nearby objects have a larger parallax than more distant objects when observed from different positions, so parallax can be used to determine distances.
Hence parallax of $$S_2$$ is greater than that of $$S_1$$
Astronomersuse the principle of parallax to measure distances to the closer stars. Here, the term "parallax" is the semi-angle of inclination between two sight-lines to the star, as observed when the Earth is on opposite sides of the Sun in its orbit.

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Re-Attempt Weekly Quiz Competition

	Column - I		Column - II
A	Deca	p	$$10^3$$
B	Hecto	q	$$10^6$$
C	Kilo	r	$$10^1$$
D	Mega	s	$$10^2$$
E	Giga	t	$$10^8$$
		u	$$10^9$$

Units and Measurements Test - 56

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Units and Measurements Test - 56

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SHARING IS CARING

Question 1

$$1 \times 10^8 m$$

$$6.84 \times 10^8 m$$

$$6.84 \times 10^{10} m$$

$$38.4 \times 10^8 m$$

Solution

Question 2

$$10^{31}$$

$$10^{-30}$$

$$10^{30}$$

$$10^{-31}$$

Solution

Question 3

$$2.7\times 10^{25} m$$

$$ 10^{25} m$$

$$10^24 m$$

$$3\times 10^{25} m$$

Solution

Question 4

$$10^{-47} m^3$$

$$10^{-44} m^3$$

$$10^{-45} m^3$$

$$10^{-46} m^3$$

Solution

Question 5

$$10^{15} m$$

$$10^{10} m$$

$$9.2 \times 10^{15} m$$

$$10^{16} m$$

Solution

Question 6

A-r, B-s, C-p, D-q, E-u

A-s, B-r, C-p, D-q, E-u

A-r, B-s, C-u, D-q, E-p

A-q, B-s, C-p, D-r, E-u

Solution

Question 7

$$1.4\times 10^3 kg/m^3$$

$$ 10^7 kg/m^3$$

$$1.5\times 10^3 kg/m^3$$

$$10^3 kg/m^3$$

Solution

Question 8

$$10^6 m$$

$$6 \times 10^6 m$$

$$10^7 m$$

$$5 \times 10^6 m$$

Solution

Question 9

1.38 parsecs

2.38 parsecs

3.38 parsecs

4.38 parsecs

Solution

Question 10

The parallax of $$S_1$$ and $$S_2$$ are same.

The parallax of $$S_1$$ is twice as that of $$S_2$$.

The parallax of $$S_1$$ is greater than parallax of $$S_2$$

The parallax of $$S_2$$ is greater than parallax of $$S_1$$

Solution

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