Measurements and Experimentation Numericals Class 9
Measurements and Experimentation
Numericals
1. The wavelength of light of a particular colour is 5800 Ã… . Express it in (a) nanometer and (b) metre.
Ans : 1 Ã… = 10^-10 m
5800 Ã… = 5800 * 10^-10 m
(a) Now 10^-9 = 1nm
( 5800 * 10^-10 m) = 58 * 10^ -10+2 / 10^-9 nm
= 58 + 10^-8 + 9
= 58 * 10 nm = 580 nm
(b) 1 nm = 10^-9 = 1/ 10^9 m
580 nm = 580 * 1/ 10^9 m
= 580 /100 * 10^-7m
= 5,8 * 10^-7 m
2. The size of bacteria is 1 μ. Find the number of bacteria in 1 m length.
Ans : 1 μ = 10^-6 m = 1/10^6 m
In 1/10^6 m has 1 bacteria
1 m has 1 * 10^6 = 10^6 bacteria
3. The distance of a galaxy from the earth is 5.6 *10 ^25 m. Assuming the speed of light to be 3*10^8 ms^-1. Find the time taken by light to travel this distance.
Ans : The distance travelled by light = 5.6 * 10^25 m
The speed of light = 3 * 10^8 ms^-1
We know, Time taken = distance travelled / speed
= 5.6 * 10^25/ 3*10^8 seconds
= 1.866 * 10^17 s
= 1.87 / 10 ^17 s
4 . The wavelength of light is 589 nm. What is its wavelength in Ã… ?
Ans : The wavelength of light = 589 nm
we know, 1 nm = 10 Ã…
589 nm = 10 * 589 Ã… = 5890 Ã…
5. The distance of the nearest star, Proxima Centauri, from the Earth is 4.0 * 10^13 km. Express it in light year.
Ans ; Distance of Proxima Centauri from Earth
= 4.0 * 10^13 km 1 light year
= 9.46 * 10^12 km
Distance of Proxima Centauri in light year
= 4 * 10 ^ 13 / 9.46 * 10^12
= 4.228
= 4.2 light year
6. It takes time 8 min for light to reach from the sun to the earth surface. If speed of light is taken to be 3 * 10 ^8 ms^1, find the distance from the sun to the earth in km.
Ans : Distance in ( 8 * 60) sec = ( 3 * 10^8 * 8 * 60)
= 3 * 48 * 10^9 / 100 km
= 144 * 10^9 / 100 * 10
= 1.44 * 10^8 km
= 1.44 * 10 ^ 8 km
7. The distance of star from the earth is 8.33 light minutes. What do you mean by this statement ? Express the distance in metre.
Ans : The distance of stars from earth is generally expressed in light years.
1 light minute = 1.8 * 10^9 m
The distance of a star from the earth is 8.33 light minutes.
8.33 light minutes = 8.33 * 1.8 * 10^9 m
= 14.99 * 10^9 m
= 1.5 * 10^10 m
Numericals on Vernier Calliper and Screw guage
1. A stop watch has 10 divisions graduated between the 0 and 5 s marks. What is its least count ?
Ans : Least count of stop watch = 5/10 = 0.5 s
2. A vernier has 10 divisions and they are equal to 9 divisions of main scale in length. If the main scale is calibrated in mm, what is its least count ?
Ans. Value of 1 M.S.D. = 1 mm
Number of divisions on V.S. = 10
L.C. of vernier = Value of one M.S.D / Number of division on V.S
= 1/10 = 0.1 mm = 0.01 cm
3. A microscope is provided with a main scale graduated with 20 divisions in 1 cm and a vernier scale with 50 divisions on it of length same as of 49 divisions of main scale. Find the least count of the microscope.
Ans : Least count of microscope = Value of 1 main scale division / Number of divisions on vernier scale
= (1/20cm) / 50
= 1/1000 cm
= 0.001 cm
4. A boy uses a vernier callipers to measure thickness of his pencil. He measures it to be 1.4 mm. If the zero error of vernier callipers is + 0. 02 cm, what is correct thickness of pencil ?
Ans : Observed thickness of pencil = 1.4 mm
Zero error = + 0.02 cm = + 0.02 * 10 mm = 0,2 mm
Correct thickness of pencil = observed thickness - zero error
= 1.4 mm - 0.2 mm = 1.2 mm
5. A vernier callipers has its main scale graduated in mm and 10 divisions on its vernier scale are equal in length to 9 mm. When the two jaws are in contact, the zero of vernier scale is ahead of zero of main scale and 3rd division of vernier scale coincides with the main scale division. Find the least count and the zero error of the vernier callipers.
Ans . Value of 1 MSD = 1 mm = u
number of division on VS = 10
Least count of vernier = Value of 1 MSD / number of division on VS
= 1/10 = 0.1 mm = 0.01 cm
Zero error of vernier callipers
Now 3rd division of VS coincides with main scale division
Zero error is +ve and is +3 * LC = 3 * 0.01 = + 0.03 cm
6. The main scale of Vernier calipers is calibrated in mm and 19 division of main scale are equal in length to 20 divisions of Vernier scale. In measuring the diameter of a cylinder by this instrument, the main scale reads 35 divisions and 4th division of Vernier scale coincides with the main scale division. Find least count and radius of cylinder
Ans : Value of 1 MSD = 1 mm = x
number of division on VS = 20
Least count of Vernier = Value of 1 MSD / number of division on VS
= 1/20 = 0.05 mm = 0.005 cm
diameter of cylinder = main scale reading + VST coinciding * LC
= 35 + 4 * 0.05 mm = ( 35 + 0.20) mm
= 35.2 mm
diameter = 35.2 / 10 = 3.52 cm
Radius of cylinder = 3.52 /2 = 1.76 cm
7 . In a Vernier callipers there are 10 divisions on the Vernier scale and 1 cm on main scale is divided in 10 parts. While measuring a length, the zero of the Vernier lies just ahead of 1.8 centimeters mark and 4th division of Vernier coincides with main scale division. Find the length and If zero error of Vernier callipers is - 0.02 cm, what is the correct length ?
Ans : Value of 1 MSD = x = 1 mm
Number of division on VS = n = 10
LC of Vernier = x/n = 1/10 = 0.1 mm = 0.01 cm
Now, correct length = Main scale division + LC * ( VSD coinciding with MS) - error
= 1.8 + 0.01 (4) - (-0.02) cm
= 1.8 + 0.04 + 0.02
= 1.86 cm
Length of rod = 1.84 cm
Correct length = 1.86 cm
8. While measuring the length of a rod with the Vernier callipers, Fig. given below shows the position of its scales. What is the length of the rod ?
Ans:
As zero of the Vernier scale lies between 3.3 and 3.4 of main scale.
Reading on the main scale = 3.3 cm
VSD coinciding with the MSD = 6th
Correct length of rod is 3.3 + 6 * LC
Value of 1 MSD = 1 mm
x = 1 mm = [3.3 + 6 * 0.01] cm
and n = number of Vernier division = 10
= 3.36 cm
L.C = 1/10 = 0.1 mm = 0.01 cm
9. The pitch of a screw gauge is 0.5 mm and the head scale is divided in 100 parts. What is the least count of screw gauge ?
Ans .
Pitch of a screw = 0.5 mm
Number of division on his scale =100
LC = Pitch of screw / Total number of divisions on head scale
= 0.5 /100 = 0.005 = 0.0005 cm
10. The thimble of a screw gauge has 50 divisions. The spindle advance 1 mm when the screw is turned through to revolution.
i) What is the pitch of screw gauge ?
ii) What is the least count of screw gauge ?
Ans : Pitch of screw gauge is the distance moved by spindle in one revolution = 1/2 = 0.5 mm
Pitch = 0.5 mm
Number of divisions on thimble = 50
LC of screw = Pitch / 50 = 1/2 /50 = 1/ 2 * 50 = 1/ 100 = 0.01 mm
11. The pitch of screw gauge is 1 mm and the circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on a circular scale coincides with the base line. Find the least count and the diameter of the wire.
Ans :
Pitch of the screw gauge = 1 mm
Number of divisions on circular scale n =100
LC of screw gauge = Pitch/n = 1 / 100 = 0.01 mm
LC = 0.01 mm / 10 = 0.001 cm
Reading on the main scale = 2 mm
division on circular scale coinciding with the baseline = 45th
correct diameter = 2 mm + 45 * LC
= 2 + 45 * 0.01
= 2 + 0.45
= 2.45 mm = 0.245 cm
correct diameter = 0.245 cm
12. When is screw watch of least count of 0.01 mm is used to measure the diameter of a wire, the reading on the sleeve is found to be 1 mm and the reading on the thimble is found to be 27 divisions
i) What is the diameter of the wire in cm
ii) if the zero error is + 0.005 cm, what is the current diameter ?
Ans : Least count of instruments = 0.01 mm
Reading on the main scale = 1 mm
Reading on thimble = 27th
i) Diameter of the wire = 1 mm + 27 * LC
= 1 mm + 27 * 0.01mm
= 1 + 0.27 = 1.27 = 1.27 / 10 = 0.127 cm
ii) Zero error = + 0.005 cm
Correct diameter = 0.127 cm - 0.005 cm = 0.122 cm
13 . A screw gauge has 50 divisions on its circular scale and its screw moves by 1 mm on turning it by two revolutions. When the flat end of the school is in contact with the stud, the zero of circular scale lies below the base line and 4th division of circular scale is in line with the base line. Find the pitch, the least count and the zero error of the screw gauge.
Ans : Number of divisions on circular scale = 50
Distance moved in 2 revolutions = 1 mm
Distance moved in 1 revolution = 1/2 mm
Pitch = 0.5 mm = 0.05 cm
Least count = Pitch / Number of division on circular scale
= 0.5 / 50 = 0.01 mm
Zero error : When the flat end of screw is in contact with stud and zero of circular scale lies 4th divisions = P of circular scale below the base line
Zero error = p * LC = +4 *0.01 = + 0.04 mm
14 . Figure given below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by one division on main scale when circular head is rotated once .
Find pitch of the screw gauge, least count of screw gauge and the diameter of the wire.
Ans :
Pitch of screw gauge = 1 mm
Number of divisions on circular scale = 50
LC of screw gauge = Pitch / 50 = 1/50 = 2/100 = 0.02 mm = 0.002 cm
Diameter of wire = Reading on MS + Division on circular scale coinciding * LC
= 4 + 47 * 0.02
= ( 4 + 0.94 ) mm
= 4.94 mm
15 . A screw has a pitch equal to 0.5 mm. What should be the number of divisions on its head so as to read correct up to 0.001 mm with its help ?
Ans :
Pitch of screw = 0.5 mm
LC = 0.001 mm
LC = Pitch / number of division on head
Number of divisions = Pitch / LC = 0.5 mm / 0.001 mm
= (5/10 ) / (1/ 1000)
= 5/10 * 1000/1
= 500
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