Measurements and Experimentation Exe-1B Short Answer for ICSE Physics Class-9 Selina Publishers

Measurements and Experimentation Exe-1B Measurement of Length Short Answer for ICSE Physics Class-9 Selina Publishers Solutions Chapter-1. There is the solutions of short Answer type Questions of your latest textbook which is applicable in 2023-24 academic sessionVisit official Website CISCE for detail information about ICSE Board Class-9.

Measurements and Experimentation Exe-1B Measurement of Length

Short Answer Type for ICSE Class-9 Physics Selina Concise Solutions

Board ICSE
Class 9
Subject Physics
Writer / Publication Concise selina Publishers
Chapter-1 Measurements and Experimentation 
Exe – 1 B Measurement of Length
Topics Solution of Exe-1(B) Short Answer Type
Academic Session 2023-2024

Measurements and Experimentation Exe-1B Short Answer for ICSE Physics Class-9 Selina Publishers

Length is the measurement that describes how long an object is. It is the distance from end to end. Length is defined as the straight line distance between two points. Its SI unit is metre

Short Answer Type of Exe-1(B) Measurement of Length

Measurements and Experimentation Class-9 Physics Concise selina Solutions  

Page 22

Question 1. Explain the meaning of the term ‘least count of an instrument’ by taking a suitable example.

Answer : The least count of an instrument is the smallest measurement that can be taken accurately with it. For example, if an ammeter has 5 divisions between the marks 0 and 1A, then its least count is 1/5 = 0.2 A or it can measure current up to the value 0.2 accurately.

Question 2. A boy makes a rular with graduations in cm on it, i.e. 100 divisions in 1 m. To what accuracy can this scale measure? How can this accuracy be increased?

Answer : Total length of the scale = 1 m = 100 cm

No. of divisions = 100

Length of each division = Total length/total no. of divisions

= 100 cm/100

= 1 cm

Thus, this scale can measure with an accuracy of 1 cm.

To increase the accuracy, the total number of divisions on the scale must be increased.

Question 3. A boy measures the length of a pencil and expresses it to be 2.6 cm. What is the accuracy of his measurement ? Can he write it as 2.60 cm?

Answer : Given, the length of the pencil is 2.6 cm.

As the length is measured in cm till one place of decimal, we can assume that it was measured using a metre rule.

So, the measurement is accurate

No, it cannot be written as 2.60, as it would mean that length is measured precisely up to second decimal place using vernier callipers or a screw gauge.

Question 4. Define the least count of vernier callipers. How do you determine it?

Answer : The least count of vernier callipers is equal to the difference between the values of one main scale division and one vernier scale division.

Let n divisions on vernier callipers be of length equal to that of (n – 1) divisions on the main scale and the value of 1 main scale division be x. Then Value of n divisions on vernier = (n – 1) x

Alternatively, value of 1 division on vernier =(n – 1) x/n

Hence, Least count =

L.C. = (Value of one main scale division)/(Total no. of divisions on vernier callipers)

Value of one main scale division = 1 mm

Total no. of divisions on vernier = 10

Therefore, L.C. =1/10 cm=.1 cm=1mm

Question 5. A vernier callipers has a zero error of + 0.06 cm. Draw a neat labelled diagram to represent it.

Answer : 

 Measurements and Experimentation Class-9 Selina Physics Class-9 Chapter-1 Ans 8 exe 1B

The least count of the vernier callipers shown in the diagram is 0.01 cm and the 6th division of the vernier scale, coincides with a main scale division.

∴ Zero Error=6×L.C.

Zero Error=6×0.01cm

⇒Zero Error=0.06cm

Hence, we get the zero error = 0.06 cm.

Question 6. State three uses of the vernier callipers.

Answer : Three uses of vernier callipers are

(a) Measuring the internal diameter of a tube or a cylinder.

(b) Measuring the length of an object.

(c) Measuring the depth of a beaker or a bottle.

Question 7. Name the two scales of a vernier callipers and explain how it is used to measure length correct up to 0.01 cm.

Answer : Two scales of vernier calipers are

(a) Main scale

(b) Vernier scale

The main scale is graduated to read up to 1 mm and on vernier scale, the length of 10 divisions is equal to the length of 9 divisions on the main scale.

Value of 1 division on the main scale= 1 mm

Total no. of divisions on the vernier scale = 10

Thus, L.C. = 1 mm /10 = 0.1 mm = 0.01 cm.

Hence, a vernier callipers can measure length correct up to 0.01 cm.

Question 8. Explain the terms (i) Pitch and (ii) Least count of a screw gauge. How are they determined?

Answer : (i) Pitch: The pitch of a screw gauge is the distance moved by the screw along its axis in one complete rotation.

(ii) Least count (L.C.) of a screw gauge: The L.C. of a screw gauge is the distance moved by it in rotating the circular scale by one division.

Thus, L.C. = Pitch of the screw gauge/total no. of divisions on its circular scale.

If a screw moves by 1 mm in one rotation and it has 100 divisions on its circular scale, then pitch of screw = 1 mm.

Thus, L.C. = 1 mm / 100 = 0.01 mm = 0.001 cm

Question 9. How can the least count of a screw gauge be decreased.

Answer : The least count of a screw gauge be decreased by

(i) Decreasing pitch

(ii) Increasing total number of division on circular scale.

Question 10. What do you mean by zero error of a screw gauge? How is it accounted for?

Answer : Due to mechanical errors, sometimes when the anvil and spindle end are brought in contact, the zero mark of the circular scale does not coincide with the base line of main scale. so It is either above or below the base line of the main scale, in which case the screw gauge is said to have a zero error. It can be both positive and negative.

hence It is accounted by subtracting the zero error (with sign) from the observed reading in order to get the correct reading.

Correct reading = Observed reading – zero error (with sign)

Question 11. A screw gauge has a least count 0.001 cm and zero error + 0.007 cm. Draw a neat diagram to represent it.

Answer : Diagram of a screw gauge with L.C. 0.001 cm and zero error +0.007 cm.

Question 12. What is backlash error? Why is it caused? How is it avoided?

Answer : If by reversing the direction of rotation of the thimble, the tip of the screw does not start moving in the opposite direction immediately but remains stationary for a part of rotation; it is called backlash error.

Reason: It happens due to wear and tear of the screw threads.

To avoid the backlash error, while taking the measurements the screw should be rotated in one direction only. If the direction of rotation of the screw needs to be changed, then it should be stopped for a while and then rotated in the reverse direction.

—  : End of Measurements and Experimentation Exe-1B Short Answer Class-9 ICSE Physics Solutions :–

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