In this article, we shall study the parallax and other methods to find the distance between two heavenly bodies.

Parallax:

Hold a pencil vertically in front of you at a certain distance against a point on a background like a wall. Now view the pencil through your left eye closing your right eye and then view the pencil through your right eye closing your left eye. You will notice that with respect to the fixed point on the wall the position of the pencil is changing. This phenomenon is known as parallax. The distance between the two points of observation is called the basis. In this example, the basis is the distance between the eyes.

Parallax is defined as the apparent shift of a body with respect to another, with the shift of eye. The distances between planets and stars from the Earth cannot be measured directly with a metre scale. Hence indirect method called the parallax method is used.

Example – 01:

Calculate the angle of a) 1^O (degree) b) 1’ (minute of arc or arcmin) and c) 1’’ (second of arc or arc second) in radians.

Solution:

Determination of the Distance Between a Faraway Planet or a Near Star and the Earth (Parallax Method – I):

To measure the distance d of a faraway planet or near star S₁ by the parallax method, we observe it from two different positions (observatories) A and B on the Earth, separated by distance b at the same time as shown in Fig. Some distant star S is taken as reference. We measure the angle between the two directions along which the planet is viewed at these two points. The∠ASB called parallax angle = θ = θ₁ + θ₂, Now the distance d << b. 

sinθ = θ = b/d
∴  d = b/θ

θ should be measured in radians.

Determination of the Distance Between a Faraway Planet or a Near Star and the Earth (Parallax Method – II):

To measure the distance d of a faraway planet, the moon or the near star (the sun) S₁ by the parallax method, we observe it from two diametrically opposite positions on the earth. A and B on the Earth, separated by distance b at the same time as shown in Fig. Some distant star S is taken as reference. We measure the angle between the two directions along which the planet is viewed at these two points. The∠ASB called parallax angle = θ = θ₁ + θ₂, Now the distance d << b. 

b = AB = 2R = 2 x radius of earth

sinθ = θ = b/d
∴  d = b/θ

θ should be measured in radians.

Example – 02:

The moon is observed from two diametrically opposite points A and B on Earth. The angle θ subtended at the moon by the two directions of observation is 1°54’. Given the diameter of the Earth to be about 1.276 x 10⁷ m, compute the distance of the moon from the Earth.

We have θ = 1^O54’ = 1 x 60’ + 54 = 114’
θ = 114’ x 2.91 x 10^-4 = 3.32 x 10^-2   rad

Also b = AB = diameter of earth = 1.276 x 10⁷ m

Ans: The distance of the moon from the earth is 3.84 x 10⁵ km

Example – 03:

Two parallax of a heavenly body measured from two points diametrically opposite on equator of Earth is 2.0′. Calculate the distance of the heavenly body if the radius of the earth is 6400 km.

We have θ = 2’ = 2 x 2.91 x 10^-4 = 5.82 x 10^-4   rad

Also b = AB = diameter of earth = 2 x 6400 km = 12800 km

Now, d = b/θ = 12800 / 5.82 x 10^-4

d = 2.2 x 10⁷ km = 2.2 x 10¹⁰ m

Ans: The distance of the heavenly body from the earth is 2.2 x 10¹⁰ m

Determination of The Distance of a Faraway Planet or a Near Star from the Earth (Parallax Method – III):

To measure the distance d of a faraway planet or the near star S₁ by the parallax method, we observe it from two diametrically opposite positions on the earth’s orbit around the sun at a time gap of six months. A and B on the Earth’s orbit, separated by distance b as shown in Fig. Some distant star S is taken as reference. We measure the angle between the two directions along which the planet is viewed at these two points. The∠ASB called parallax angle = θ = θ₁ + θ₂, Now the distance d << b.

b = AB = 2 AU = 2 x 1.496 x 10¹¹ m

sinθ = θ = b/d
∴  d = b/θ

θ should be measured in radians.

Example – 04:

When the observations are taken at an interval of 6 months, the angle of parallax for star is 0.4’’. Find the distance of star in light year and parsec.

θ = 0^.44’’ = 0.4 x 4.847 x 10^-6 = 1.939  x 10^-6  rad

Ans: The distance of the star from the earth is 16.3 lightyears or 5 parsec.

Determination of Diameter of the Moon or the Sun:

The angular diameter of the moon (or the sun) is the angle subtended by two diametrically opposite ends of the moon (or the sun) at a point on the Earth.

Let D be the diameter of the moon (or the sun) and d be the distance between the moon (or the sun) from the earth. D << d

arc length AB = D = d θ

θ should be measured in radians.

Example – 05:

The Sun’s angular diameter is measured to be 1920”. The distance D of the Sun from the Earth is 1.496 × 10¹¹ m. What is the diameter of the Sun?

Sun’s angular diameter θ = 1920″ = 1920 x 4.85 x 10^-6 = 9.31 x 10^-3 rad
Sun’s diameter = D = d θ

d = (1.496 × 10¹¹) x (9.31 x 10^-3 ) = 1.39 x 10⁹  m

Ans: The diameter of the sun is 1.39 x 10⁹ m

Example – 06:

The angle subtended by the Moon at a point on the Earth is 0^o31’. If the distance of the Moon from the Erath is 3.84 x 10⁵ km, find the diameter of the moon.

θ = 0^o31’ = 31 x 2.908 x 10^-4 = 9.01 x 10^-3 rad

Now D = d θ = 3.84 x 10⁵ x 9.01 x 10^-3 = 3.46 x 10³  km

Ans: Diameter of the moon is 3.46 x 10³ km

Determination of Angular Distance Between Two Stars:

Let θ₁  be the elevation of a distant star S₁ and θ₂  be the elevation of another distant star S₂ such that (θ₂ > θ₁). Then the quantity (θ₂ – θ₁) is called the angular distance between the stars.

It is to be noted that the smaller angular distance between two distant stars does not imply that the two stars are close to each other. There is a possibility that distance between them is very very large (hundreds of light-years).

Determination of the Distance of Inferior Planets from the Earth (Copernicus Method):

The planets which are closer to the Sun than the Earth are called inferior planets. The Mercury and Venus are two inferior planets. Copernicus assumed the orbits of inferior planets as perfectly circular.

The angle formed at the Earth between the Earth planet direction and the Earth-Sun Direction is called the planet elongation and is denoted by symbol ε. The planets elongation changes continuously.  When elongation acquires maximum value it appears to be farthest from the sun, At this time the Sun and the Earth subtends an angle of 90° at the planet.  This maximum elongation is noted.

SE = 1 AU = 1.496 x 10¹¹ m

Δ EPS is a right-angled triangle

sin ε = SP/ SE

∴  SP = SE sin ε

Example – 07:

Copernicus assumed the orbit of Mercury to be circular and estimated its orbital radius as 0.38 AU. Using this radius derive angle of maximum elongation for mercury and its distance from the Earth when the elongation is maximum.

We have SP = SE sin ε

Where SP = orbital radius of Planet = 0.38

SE = orbital radius of the Earth = 1 AU

0.38 = 1 x sin ε

ε = sin^-1(0.38) = 22°20′

Applying Pythagoras theorem to ΔEPS

SE² = EP² + SP²

∴ EP² = SE² – SP²

∴ EP² = (1)² – (0.38)² = (1+0.38)(1-0.38)

∴ EP² = 1.38 x 0.62 = 0.8556

∴ EP = 0.925 AU

∴ EP = 0.925 x 1.496 × 10¹¹  m = 1.384 ×  10¹¹  m

Ans: The angle of maximum elongation for Mercury is 22°20′ and the distance between the Earth and Mercury at maximum elongation is 1.384 × 10¹¹  m

Example – 08:

For Venus the angle of maximum elongation is found to be approximately equal to 47°. Determine the distance between the Sun and Venus and that between the Erath and the Venus. Also, find orbital period of Venus.

We have SP = SE sin ε

Where SP = orbital radius of Planet

SE = orbital radius of the Earth = 1 AU

SP = 1 x sin 47° = 1 x 0.7314 = 0.7314 AU

∴ SP = 0.7314 x 1.496 × 10¹¹  m = 1.094 ×  10¹¹  m

Applying Pythagoras theorem to ΔEPS

SE² = EP² + SP²

∴ EP² = SE² – SP²

∴ EP² = (1)² – (0.73)² = (1+0.73(1-0.73)

∴ EP² = 1.73 x 0.27 = 0.4671

∴ EP = 0.6834 AU

∴ EP = 0.6834 x 1.496 × 10¹¹  m = 1.022 ×  10¹¹  m

Ans: The distance between the sun and venus is 1.094 × 10¹¹  m and the distance between the Earth and Venus at maximum elongation is 1.022 × 10¹¹  m, Period of Venus is 226 days

Determination of the Distance of Superior Planets from the Sun:

The planets which are farther to the Sun than the Earth are called inferior planets. The Mars, The Jupiter, The Saturn, The Neptune and The Uranus are superior planets.

By Keppler’s period law of orbiting satellite “The square of the time period of the satellite is directly proportional to the cube of the semimajor axis of the orbit.

Example – 09:

A Period of a planet orbiting around the sun is half that of the earth. What is the orbital radius of the planet?

∴  r_P = 0.63 x 1.496 ×  10¹¹  m = 9.42 ×  10¹⁰  m

Thus r_P/ r_E = 0.63/1 = 0.63

Ans: The orbital size of the planet is smaller than that of the Earth by a factor of 0.63

Measurement of Distances of Star by Intensity Method (Optical or Spectroscopic Method):

This method uses the inverse square law of photometry, it states that “The intensity of illumination at a point varies inversely as the square of the distance from the source of light

Thus I ∝ 1/r²

This method is used to find the distance of a star from the earth. The assumption of this method is the intrinsic brightness of all the stars is the same. This method gives the approximate result.

In the above formula, I₁ is the apparent brightness of a star at a distance r₁ from the earth and I₂ is the apparent brightness of a star at a distance r₂ from the earth. If we know the intensity ratio and the distance of one star from the earth, the distance of another star from the earth can be calculated.

Measurement of Distances of Star by Doppler Shift Method (Optical or Spectroscopic Method):

When a body that is emitting radiation has a non-zero velocity relative to an observer, the wavelength of the emission will be shortened or lengthened, depending upon whether the body is moving towards or away from an observer. This change in observed wavelength or frequency is known as the Doppler shift.

If the object is moving towards an observer, then the emission will be blueshifted – i.e. the wavelength of the emission will be shortened, moving it towards the blue end of the spectrum. If the object is moving away from an observer, then the emission will be redshifted i.e. the wavelength of the emission will increase, moving it towards the red end of the spectrum.

Generally, by determining the redshift and using doppler’s formulae the distance of the star and its speed at which it is moving away from the earth can be found.

For More Topics in Measurement of Length, Area, and Volume Click Here

For More Topics in Physics Click Here

The post Distances of Heavenly Bodies from the Earth appeared first on The Fact Factor.

Physics is a science of measurement. In science and engineering, we perform experiments. During experiments, we have to take readings. Thus all these experiments require some measurements to be made. During the production of mechanical products, we have to measure the parts so as to find whether the part is made as per the specifications. Thus measurements are necessary for production and quality control. A measurement is a quantitative description of one or more fundamental properties compared to a standard. To measure length is a very important step during the performance of experiments. Measurement can be done directly or indirectly. In this article, we shall study the use of a micrometer screw gauge to measure length, diameter, etc.

Principle:

A screw gauge works on the principle of a screw in a nut. i.e. in one rotation screw moves forward through a distance equal to the pitch of the screw. The distance between two successive threads of a screw is called its pitch. Or the pitch of a screw may be defined as the distance that the tip of the screw advances when its head is given one complete rotation. The measurements are done actually through a precisely made integrated screw with a pitch of usually 2 threads per millimetre, which means that on completion of one revolution the displacement achieved is 0.5 millimetre. Screw gauges are more precise than Vernier calipers

Construction:

A screw gauge is a device incorporating a calibrated screw that is widely used for the precise measurement of components in the manufacturing of mechanical parts. The body used to hold the anvil and barrel firmly in its place is called a frame, in micrometer screw gauges, thick C-shaped frames are used. It is the fixed part mounted at one end of the frame exactly parallel to the moving spindle which moves towards it. The object whose dimension is to be measured is held between the anvil and the spindle.

The cylindrical part which displaces by rotation of thimble decreasing the distance between itself and anvil until the object being measured becomes stable between the two of them is called the spindle. The stationary part with having a linear scale onto it is called the main scale. It covers the screw mechanism of the screw gauge.

Thimble is the part through which the measuring screw is rotated, this screwing results in the displacement of the spindle and thimble itself. A ratchet is a small device which is used to provide a limited applied force.

A screw gauge has two scales, one rotating scale which can be found on its rotating cylindrical part it is also called a circular scale and the other one can be found on its stationary sleeve which is called the main scale or sleeve scale.

Generally, the least count on the main scale is 0.5 mm. The circular scale is divided into 50 or 100 equal parts.

Least Count:

The minimum length that can be measured using the Vernier calipers is called its least count.

Zero Errors of Micrometer Screw Gauge:

When the anvil and spindle of the screw gauge are made to touch each other, then the zero on the main scale should match with a zero on the circular scale. However due to wear and tear or manufacturing defect the two zeros usually do not coincide with each other, then the vernier is said to have zero error. There are two types of zero errors.

Positive zero error:

If on bringing the anvil and spindle of micrometer screw gauge together, the zero mark of the circular scale is below the main scale line, then the zero error is said to be positive.

To find positive zero error, note the division on the circular scale (C.S.R.) coinciding with the line of the main scale line. Then the positive error is equal to the product of the number of divisions on a circular scale matching with the main scale line and the least count of the micrometer screw gauge. To get the correct reading this error is to be subtracted from the overall reading.

Negative zero error:

If on bringing the anvil and spindle of micrometer screw gauge together, the zero mark of the circular scale is above the main scale line, then the zero error is said to be negative.

To find negative zero error, note the division on the circular scale (C.S.R.) coinciding with the line of the main scale line. Then the negative error is equal to the product of the number of divisions on a circular scale matching with the main scale line and the least count of the micrometer screw gauge. To get the correct reading this error is to be added to the overall reading.

Back-lash Error:

Sometimes the tip of the screw does not move backward for a part of a rotation of the head in the reverse direction, because of the wear and tear of threads. This is called a back-lash error. The back-lash error can be avoided by not rotating the head in the reverse direction once the object is held between the anvil and spindle.

Use of Screw Gauge:

• Hold the object whose dimensions is to be measured between the anvil and spindle of the screw gauge with gentle pressure.
• Note down the main scale reading just before zero of a circular scale. This is called the main scale reading (M.S.R.)
• Note down the number of circular scale division (n) which coincides with the main scale line.  Then circular scale reading, (C.S.R.) = n X Least Count.
• Add the M.S.R. and the C.S.R. to get reading.
• Subtract the zero error with a proper sign from the above reading to get the correct reading.

Example: Let us consider a screw gauge with the least count of 0.01 mm.

The main scale reading is 2.5 mm and circular scale reading is 38. Hence the total reading  = MSR + CSR x LC = 2.5 +38 x 0.01 = 2.5 + 0.38 = 2.88 mm

To Measure Inside Dimensions:

To Measure Depth:

Micron Micrometer Screw Gauge:

In this case, the least count of the main scale is 1mm, the circular scale is divided into 100 parts and each division on the circular scale is divided into 10 parts.

Least count of the main scale 1mm

Least count of circular scale = 1mm/100 = 0.01 mm

Least count of micron micrometer = 0.01/10 = 0.001 mm = 1 x 10^-6 m = 1 micrometre

Micron micrometre screw gauge can be used for measuring the thickness up to 1 micrometre.

Problems on Use of Micrometer Screw Gauge:

Example 01:

The screw of the micrometer screw gauge moves through a distance of 2 mm when it is turned through 4 rotations. Find the pitch of the screw. If the circular scale is divided into 100 equal parts. Find the least count of the micrometer screw gauge.

Solution:

Distance traveled by screw = 2 mm

No. of rotations given = 4

Pitch of screw = Distance traveled by screw / No. of rotations given = 2mm /4 = 0.5 mm

No. of divisions on circular scale = 100

Least count of the micrometer screw gauge = Pitch of screw / No. of circular scale division = 0.5 mm / 100 = 0.005 mm.

Hence the pitch of the screw is 0.5 mm and the least count of the micrometer screw gauge is 0,005 mm.

Example 02:

If the pitch of the micrometer screw gauge screw is 0.1 mm and its circular scale is divided into 100 equal parts. Find the least count of the micrometer screw gauge.

Solution:

Pitch of screw = 0.1 mm

No. of divisions on circular scale = 100

Least count of the micrometer screw gauge = Pitch of screw / No. of circular scale division = 0.1 mm / 100 = 0.001 mm.

Hence the least count of the micrometer screw gauge is 0,001 mm.

Example 03:

If the pitch of the micrometer screw gauge screw is 0.1 mm and its circular scale is divided into 100 equal parts. Find the correct reading for reading as shown. The no-zero error.

Solution:

Pitch of screw = 0.1 cm = 1 mm

No. of divisions on circular scale = 100

Least count of the micrometer screw gauge = Pitch of screw / No. of circular scale division = 1 mm / 100 = 0.01 mm.

Main scale reading = 4.5 mm

Circular scale reading = 28

Shown reading = M.S.R. + C.S.R. x Least Count = 4.5 + 28 x 0.01 = 4.5 + 0.28 = 4.78 mm

Corrected reading = shown reading – zero error with proper sign = 4.78 – 0 = 4.78 mm

Hence the corrected reading is 4.78 mm.

Example 04:

When a screw gauge with a least count 0.01 mm is used to measure the diameter of a wire, the reading on the sleeve is found to be 0.5 mm and the reading on the thimble is found to be 27 divisions. What is the correct diameter of the wire if the zero error for the gauge is +0.005 cm?

Solution:

Least count of the micrometer screw gauge = 0.01 mm.

Main scale reading = 0.5 mm

Circular scale reading = 27

Shown reading = M.S.R. + C.S.R. x Least Count = 0.5 + 27 x 0.01 = 0.5 + 0.27 = 0.77 mm

Zero error = +0.005 cm = 0.05 mm

Corrected reading = shown reading – zero error with proper sign = 0.77 – 0.o5 = 0.728 mm

Hence correct diameter of wire is 0.72 mm.

Example 05:

When a screw gauge with a least count 0.01 mm is used to measure the diameter of a rod, the reading on the sleeve is found to be 1.6 cm and the reading on the thimble is found to be 48 divisions. What is the correct diameter of the rod if the zero error for the gauge is – 0.003 cm?

Solution:

Least count of the micrometer screw gauge = 0.01 mm.

Main scale reading = 1.6 cm = 16 mm

Circular scale reading = 48

Shown reading = M.S.R. + C.S.R. x Least Count = 16 + 48 x 0.01 = 16 + 0.48 = 16.48 mm = 1.648 cm

Zero error = – 0.003 cm

Corrected reading = shown reading – zero error with proper sign = 1.648 + 0.o03 = 1.651 cm

Hence correct diameter of wire is 1.651 cm.

For More Topics in Measurement of Length, Area, and Volume Click Here

For More Topics in Physics Click Here

The post Use of Micrometer Screw Gauge appeared first on The Fact Factor.

Physics is a science of measurement. In science and engineering, we perform experiments. During experiments, we have to take readings. Thus all these experiments require some measurements to be made. During the production of mechanical products, we have to measure the parts so as to find whether the part is made as per the specifications. Thus measurements are necessary for production and quality control. A measurement is a quantitative description of one or more fundamental properties compared to a standard. To measure length is a very important step during the performance of experiments. Measurement can be done directly or indirectly. For direct methods metre scale, vernier calipers, micrometer screw gauges are used. In this article, we shall study the use of Vernier Callipers to measure length, diameter, etc.

Construction:

Vernier calipers is a device is used to measure the internal and external diameter of a tube, the external diameter of a sphere, depth of a vessel, the diameter of thick wires and cylinders.

Typical Vernier calipers consist of a steel strip which is generally marked in centimetre and millimetre along its lower edge this is a fixed scale called the main scale. The end of the main scale is provided with a fixed jaw forming external jaw on the lower side and internal jaw on the upper side. A sliding frame with graduation on the lower side slides over the main scale. This sliding scale is called the Vernier scale. The inside end of a Vernier scale is provided with a fixed jaw forming external jaw on the lower side and internal jaw on the upper side.

Generally, the Vernier scale is provided with N divisions matching with (N – 1) divisions of the main scale.

Least Count of Vernier Scale:

The minimum length that can be measured using the Vernier calipers is called its least count.

Zero Errors of Vernier Calipers:

When the two jaws of Vernier calipers are made to touch each other, then the zero on the main scale should match with a zero on the Vernier scale. However due to wear and tear or manufacturing defect the two zeros usually do not coincide with each other, then the Vernier is said to have zero error. There are two types of zero errors.

Positive zero error:

If on bringing both the jaws together, the zero mark of the Vernier scale is on the right side of a zero mark of the main scale, then the zero error is said to be positive.

To find positive zero error, note the division on the Vernier scale (V.S.R.) coinciding with the division on the main scale. Then the positive error is equal to the product of the number of division on Vernier scale matching with main scale division and the least count of the Vernier. To get correct reading this error is to be subtracted from the overall reading.

Negative zero error:

If on bringing both the jaws together, the zero mark of the Vernier scale is on the left side of a zero mark of the main scale, then the zero error is said to be negative.

To find a negative zero error, note the division on the Vernier scale (V.S.R.) coinciding with the division on the main scale. Subtract this number from a total number of divisions on the Vernier scale to obtain V.S.R. Then the negative error is equal to the product of V.S.R. and the least count of the Vernier. To get correct reading this error is to be added to the overall reading.

Use of Vernier Calipers:

• Hold the object whose dimensions is to be measured in the jaws of the calipers with gentle pressure.
• Note down the main scale reading just before zero of vernier calipers. This is called the main scale reading (M.S.R.)
• Note down the number of vernier scale division (n) which coincides with any division of the main scale.  Then vernier scale reading, (V.S.R.) = n X Least Count.
• Add the M.S.R. and the V.S.R. to get reading.
• Subtract the zero error with a proper sign from above reading to get the correct reading.

Uses of Different Parts of Vernier Calipers:

• The outside jaw is used to measure outside dimensions like the length of a rod, the external diameter of the sphere or cylinder.
• The inside jaw is used to measure inside dimensions like the internal diameter of ring or hollow cylinder or pipe
• The strip is used to measure the depth of a beaker or bottle.
• The main scale is used to measure the length of an object correct up to 1 mm.
• Vernier scale is used to measure the length of an object correct up to 0.1 mm

Dial Vernier Calipers (Direct Reading)

Digital Vernier Calipers (Direct Reading)

Problems on Use of Vernier Calipers:

Example 01:

The main scale of a Vernier scale has the least count of 0.5 mm. If 20 divisions of this scale are divided into 25 equal parts of the Vernier scale, what is the least count of the Vernier caliper?

Solution:

1 M.S.D. = 0.5 mm

25 V.S.D. = 20 M.S.D.

Thus,    1 V.S.D. = 20/25 M.S.D. = 0.8 M.S.D. = 0.8 x 0.5 mm = 0.4 mm

Now, Least Count of Vernier Caliper = 1 M.S.D. – 1.V.S.D. = 0.5 mm – 0.4 mm = 0.1 mm

Hence least count of Vernier caliper is 0.1 mm

Example 02:

On Vernier calipers, one centimetre of the main scale is divided into 10 equal parts. If 10 divisions of the Vernier scale coincide with 9 divisions of the main scale, then what will be the least count of the Vernier caliper?

Solution:

1 M.S.D.    =   cm = 0.1 cm

10 V.S.D.   = 9 M.S.D.

Thus,    1 V.S.D = 9/10 M.S.D.   =   0.9 M.S.D. = 0.9 x 0.1 cm = 0.09 cm

Now, Least Count of Vernier Caliper = 1 M.S.D. – 1.V.S.D.    =    0.1 cm – 0.09 cm = 0.01 cm

Hence least count of Vernier caliper is 0.01 cm

Example 03:

On Vernier calipers, one centimetre of the main scale is divided into 20 equal parts. If the number of divisions on the Vernier scale is 25. Find the least count of the Vernier caliper. In a measurement of the length of an object, the main scale reading lies between 2.35 cm and 2.40 cm. Vernier scale reading is 6.  Find the length of the object.

Solution:

On main scale 1 cm is divided into 20 equal parts. Hence 1 M.S.D.    =  1/20 cm = 0.05 cm.

Now least count = (1/No. of divisions of Vernier scale) x M.S.D. = (1/25) x 0.05 cm = 0.002 cm.

Reading = Main scale reading + Vernier scale reading x Least Count Reading = 2.35 + 6 x 0.002 =    2.35 + 0.012 = 2.362 cm

Hence the length of the object is 2.362 cm

Example 04:

On Vernier calipers, one centimetre of the main scale is divided into 20 equal parts. If the number of divisions on the Vernier scale is 25. Find the least count of the Vernier caliper. In a measurement of the length of an object, the main scale reading lies between 2.75 cm and 2.80 cm. Vernier scale reading is 9.  The zero error is + 0.024 cm. Find the length of the object.

Solution:

On main scale 1 cm is divided into 20 equal parts. Hence 1 M.S.D.    =  1/20 cm = 0.05 cm.

Now least count = (1/No. of divisions of vernier scale) x M.S.D. = (1/25) x 0.05 cm = 0.002 cm.

Reading = Main scale reading + Vernier scale reading x Least Count – Zero error Reading

= 2.75 + 9 x 0.002  – (+ 0.024)   =    2.75 + 0.018 – 0.024 = 2.744 cm

Hence the length of the object is 2.744 cm

Example 05:

On Vernier calipers, one centimetre of the main scale is divided into 10 equal parts. If the number of divisions on the Vernier scale are 10. Find the least count of the Vernier caliper.  In a measurement of the length of an object, the main scale reading lies between 5.6 cm and 5.7 cm. Vernier scale reading is 4.  The zero error is – 0.02 cm. Find the length of the object.

Solution:

On main scale 1 cm is divided into 10 equal parts. Hence 1 M.S.D.    =  1/10 cm = 0.1 cm.

Now least count = (1/No. of divisions of vernier scale) x M.S.D. = (1/10) x 0.1 cm = 0.01 cm.

Reading = Main scale reading + Vernier scale reading Least Count – Zero error Reading

= 5.6 + 4 x 0.01  – (- 0.02)  =   5.6 +  0.04  +0.02  = 5.66 cm

Hence the length of the object is 5.66 cm

For More Topics in Measurement of Length, Area, and Volume Click Here

For More Topics in Physics Click Here

The post Use of Vernier Calipers appeared first on The Fact Factor.