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Measurement of small and large distances - class-VI

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From the following, the biggest unit of energy is?

  1. Joule

  2. Kilowatt hour

  3. Erg

  4. Electron volt


Correct Option: B
Explanation:

Since, we know that SI unit of energy is represented by Joule. Hence, let us convert all the different units in the form of Joules.


$1\,Joule=1\,Joule$

$1\,Kwh=3600000\,Joule$

$1\,erg=10^{-7}\,Joule$

$1\,eV=1.6\times 10^{-19}\,Joule$

Hence, Kilowatt hour is the largest unit of energy and hence it is practically used in everyday life of electric consumption.

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

  1. Echo method

  2. Parallax method

  3. Triangulation method

  4. None of these


Correct Option: B
Explanation:

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

'The parallax angle in radians is: $\theta = \left( 1 + \frac { 54 } { 60 } \right) \times \frac { \pi } { 180 } = 0.03316 \mathrm { rad }$
  Hence, the distance between moon and earth: 

  1. $\dfrac { \text { Diameter of Earth } } { \theta }$

  2. $\dfrac { 1.276 \times 10 ^ { 7 } m } { 0.03316 }$

  3. $3.84 \times 10 ^ { 8 } m$

  4. nnone of these


Correct Option: C
Explanation:

Given,

$\theta =0.03316\,\,rad$

Also $b = \mathrm { AB } =$ diameter of earth $= 1.276 \times 10 ^ { 7 } \mathrm { m }$

Now d $=\dfrac{b}{\theta }=\dfrac{1.276\times {{10}^{7}}}{0.03316\,}=3.84\times {{10}^{8}}\text{m}$

Hence, distance between earth and moon is $3.84\times {{10}^{8}}\text{m}$

Which of the following instrument can be used to determine the radius of curvature of a spherical surface?

  1. Spherometer

  2. Vernier callipers

  3. Screw gauge

  4. Simple pendulum


Correct Option: A
Explanation:

Spherometer is an instrument which is used to measure the curvature of spherical surfaces.

Parallax method is suitable for the measurement of:

  1. ball

  2. planet

  3. vehicle

  4. distance between cities


Correct Option: B
Explanation:

Parallax method is used for the measurement of planet distance and also used to measure distance of starts and planets

option B is correct.

Which of the following is the best method to measure microscopic distances?

  1. Optical microscope

  2. Meter scale

  3. Screw gauge

  4. Diffraction pattern


Correct Option: A
Explanation:

Optical microscopes are enabled with objective fitted with lenses with different magnifications. This helps to magnify a small objects and so necessary measurements.

What is the approximation made in the parallax method?

  1. All distances measured between two points on earth is zero.

  2. All distances measured between two points on earth is constant.

  3. Distance between a point on the earth and the planet is very large as compared to the distance between two points on earth's surface.

  4. No approximation is made.


Correct Option: C
Explanation:

In parallax method, an approximation is made that distance between a point on the earth and the planet is very large as compared to the distance between two points on the earth's surface.

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?

  1. The parallax of $S _1$ and $S _2$ are same.

  2. The parallax of $S _1$ is twice as that of $S _2$.

  3. The parallax of $S _1$ is greater than parallax of $S _2$

  4. The parallax of $S _2$ is greater than parallax of $S _1$


Correct Option: D
Explanation:

$Answer:-$ 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.

Parallax is the apparent displacement of an object because of:

  1. change in observer's point of view

  2. change in object's position

  3. changes both in observers point of view and object's position

  4. consistency in observer's point of view


Correct Option: A
Explanation:

Astronomers use an effect called parallax to measure distances to nearby stars. Parallax is the apparent displacement of an object because of a change in the observer's point of view.

As the Earth orbits the sun, a nearby star will appear to move against the more distant background stars. Astronomers can measure a star's position once, and then again 6 months later and calculate the apparent change in position.

Parallax method is useful 

  1. for measuring speed of the light.

  2. for measuring distances of star.

  3. for finding the intensity of the light.

  4. None of the above.


Correct Option: B
Explanation:

$Answer:-$ B

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. Astronomers use the principle of parallax to measure distances to the closer stars.

To minimise parallax error, the observer should place the object :

  1. as near to the scale of the ruler as possible and the eye must be directly above the scale

  2. as far to the scale of the ruler as possible and the eye must be directly above the scale

  3. as near to the scale of the ruler as possible and the eye must be to the right of the scale

  4. as near to the scale of the ruler as possible and the eye must be to the left of the scale


Correct Option: A
Explanation:

$Answer:-$ A

There are three simple ways of reducing parallax error in that case: 
1. Attach a straight object (like a ruler or straightened paperclip) to the thing you're measuring the displacement of, it should stick out of the object perpendicularly to the scale you're measuring the displacement on. 
2. Make sure that object is as close to your scale as possible, but not touching. 
3. Put your eyes level with the object, and as close to it as you safely can.

The effect of parallax is used to measure:

  1. distances to nearby objects

  2. distances to nearby stars

  3. nearness of atoms in substances

  4. the object and image distance in optical experiments


Correct Option: B
Explanation:

Astronomers use an effect called parallax to measure distances to nearby stars. Parallax is the apparent displacement of an object because of a change in the observer's point of view.

As the Earth orbits the sun, a nearby star will appear to move against the more distant background stars. Astronomers can measure a star's position once, and then again 6 months later and calculate the apparent change in position.

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

  1. 1.38 parsecs

  2. 2.38 parsecs

  3. 3.38 parsecs

  4. 4.38 parsecs


Correct Option: A
Explanation:

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, $d=\dfrac{1}{0.723}=1.38 parsecs$

Error due to eye vision is termed as :

  1. climax error

  2. sight error

  3. parallax error

  4. visional error


Correct Option: C
Explanation:

$Answer:-$ C

Parallax also affects optical instruments such as rifle scopes, binoculars,microscope and twin lens reflex cameras that view objects from slightly different angles. Many animals, including humans, have two eyes with overlapping visual fields that use parallax to gain depth perception; this process is known as stereopsis.

A star's distance ($d$) and its parallax angle ($p$) are related to each other as:

  1. $d=\dfrac{1}{p}$

  2. $d=\dfrac{1}{p^2}$

  3. $p=\dfrac{1}{d^2}$

  4. none of these


Correct Option: A
Explanation:

Astronomers use an effect called parallax to measure distances to nearby stars. Parallax is the apparent displacement of an object because of a change in the observer's point of view.

The 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 arc seconds.

Parallax method is based on which of the following principle?

  1. Disparity

  2. Lutz Kelker bias

  3. Trilateration

  4. Triangulation


Correct Option: D
Explanation:

Distance measurement by parallax is a special case of the principle of triangulation, which states that one can solve for all the sides and angles in a network of triangles if, in addition to all the angles in the network, the length of at least one side has been measured. Thus, the careful measurement of the length of one baseline can fix the scale of an entire triangulation network. In parallax, the triangle is extremely long and narrow, and by measuring both its shortest side (the motion of the observer) and the small top angle (always less than 1 arcsecond leaving the other two close to 90 degrees), the length of the long sides can be determined.

Parallax angles _______ $0.01/ arcsec$ are very difficult to measure from Earth.

  1. more than

  2. less than

  3. equal to

  4. greater than or equal to


Correct Option: B
Explanation:

Parallax effect depends upon the path of light that travels from object to the observer. For distant objects observed from Earth, the light that reaches Earth has to go through a number of layers in Earth's atmosphere, during which it refracts and disperses and hence decreases the accuracy of the method. Resultantly parallax angles less than $0.01/arcsec$ are very difficult to measure from Earth.

A student measure the height h of a convex mirror using spherometer. The legs of the spherometer are 4 cm apart and there are 10 divisions per cm on its linear scale and circular scale has 50 divisions. The student takes 2 as linear scale division and 40 as circular scale division. What is the radius of curvature of the convex mirror ?   

  1. $9.06 cm$

  2. $20.66 cm$

  3. $5.66 cm$

  4. $9.66 cm$


Correct Option: D
Explanation:

Least count of the spherometer is $LC=$ pitch/number of circular divisions $=\dfrac{1/10}{50}=0.002 cm$
The total reading for height is $h=MSR+(CSR\times LC)=(2\times 0.1)+(40\times 0.002)=0.28 cm$
The formula for radius of curvature is $R=\dfrac{l^2}{6h}+\dfrac{h}{2}=\dfrac{4^2}{6(0.28)}+\dfrac{0.28}{2}=9.66 cm$

A spherometer has 10 threads per cm and its circular scale has 50 divisions. The least count of the instrument is  

  1. $0.01 cm$

  2. $0.02 cm$

  3. $0.002 cm$

  4. $0.2 cm$


Correct Option: C
Explanation:

The least count of the spherometer, $LC=$pitch/ total number of circular divisions $=\dfrac{p}{n}$
Here, pitch $p=1/20=0.1\ cm$ and $n=50$
Thus, $LC=\dfrac{0.1}{50}=0.002\ cm$

If a star is $5.2\times 10^{16}\ m$ away. What is the parallax angle in degrees?

  1. $1.67 \times 10^{-4}$ degrees

  2. $1.67 \times 10^{-5}$ degrees

  3. $0.67 \times 10^{-4}$ degrees

  4. $2.3 \times 10^{-4}$ degrees


Correct Option: A
Explanation:

Given :    $1$ AU $ = 1.5\times 10^11$ m                $d = 5.2\times 10^{15}$ m

Parallax angle:     $\alpha = \dfrac{1 AU}{d} =\dfrac{1.5\times 10^{11}}{5.2\times 10^{16}} = 0.288\times 10^{-5}$  radians
$\implies$   $\alpha = \dfrac{180}{\pi} \times 0.288\times 10^{-5} = 1.67\times 10^{-4}$  degrees

A star is $1.45\ parsec$ light years away. How much parallax would this star show when viewed from two locations of the earth six months apart in its orbit around the sun?

  1. $2\ Parsec$

  2. $0.725\ Parsec$

  3. $1.45\ Parsec$

  4. $2.9\ Parsec$


Correct Option: A
Explanation:

One light year $=$ speed of light $\times$ one year

or $1 ly=3\times 10^8\times (24\times 3600)=94608\times 10^{11} m$
So, $4.29 ly=4.29\times (94608\times 10^{11})=4.058\times 10^{16} m$
As $1 $ parsec $=3.08\times 10^{16} m$
(Parsec is a unit of length used to measure large distances to objects outside our Solar System)
Thus, $4.29 ly=\dfrac{4.058\times 10^{16}}{3.08\times 10^{16}}=1.32$ parsec
Now angular displacement , $\theta=\dfrac{d}{D}$
where $d=$ diameter of earth's orbit $= 3\times 10^{11} m$ and 
$D=$ distance of star from the earth $=4.058\times 10^{16} m$
So, $\theta=\dfrac{3\times 10^{11}}{4.058\times 10^{16}}=7.39\times 10^{-6}$ rad
As $1 sec=4.85\times 10^{-6} rad$ so, $7.39\times 10^{-6} rad=\dfrac{7.39\times 10^{-6}}{4.85\times 10^{-6}}=1.52 sec$

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