## The vernier constant of a vernier callipers is 0.1 mm and it has zero error of 0.04 cm . While measuring diameter of a rod , the main scale reading is 1.2 cm and 5th vernier division is coinciding with any scale division . The corrected diameter of the rod is

Q: The vernier constant of a vernier callipers is 0.1 mm and it has zero error of 0.04 cm . While measuring diameter of a rod , the main scale reading is 1.2 cm and 5th vernier division is coinciding with any scale division . The corrected diameter of the rod is

(a) 1.21 cm

(b) 1.21 mm

(c) 1.29 mm

(d) 1.29 cm

Sol: V.C = 0.1 mm = 0.01 cm ; zero error = 0.04 cm

Zero correction = -0.04 cm

Observed diameter = (1.2 + 5 × 0.01) cm = 1.25 cm

Corrected diameter = (1.25 – 0.04) cm = 1.21 cm

## The centimeter scale on a vernier calliper is divided into eight equal parts . Ten vernier divisions are contained in one cm . Calculate vernier constant .

Q: The centimeter scale on a vernier calliper is divided into eight equal parts . Ten vernier divisions are contained in one cm . Calculate vernier constant .

(a) 0.1 cm

(b) 0.01 cm

(c) 0.025 cm

(d) 0.001 cm

Sol: 1 S.D = 1/8 cm .  Also , 10 V.D = 8 S.D

$\large 1 V.D = \frac{8}{10} S.D$

Vernier constant = 1 S.D – 1 V.D

$\large V.C = 1 S.D – \frac{8}{10} S.D$

$\large V.C = \frac{1}{5} S.D = \frac{1}{5} \times \frac{1}{8} cm = \frac{1}{40} cm$

V.C = 0.025 cm

## Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale . The total number of divisions on the circular scale is 50 . Further , it is found that the screw gauge has a zero error of -0.03 mm . While measuring the diameter of a thin wire , a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35 . The diameter of the wire is

Q: Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale . The total number of divisions on the circular scale is 50 . Further , it is found that the screw gauge has a zero error of -0.03 mm . While measuring the diameter of a thin wire , a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35 . The diameter of the wire is

(a) 3.32 mm

(b) 3.73 mm

(c) 3.67 mm

(d) 3.38 mm

Sol: Pitch of screw gauge = 1/2 mm = 0.5 mm

Least count of screw gauge = 0.5/50 mm = 0.01 mm

Zero error = – 0.03 mm

Zero correction = + 0.03 mm

Observed diameter of wire = 3 mm + 35 x 0.01 mm = 3.35 mm

Corrected diameter of wire = (3.35 + 0.03 ) mm = 3.38 mm

## Add 17.35 g , 25.6 g and 8.498 g and write the result with the correct number of significant figures .

Q: Add 17.35 g , 25.6 g and 8.498 g and write the result with the correct number of significant figures .

Sol: Out of three given values 25.6 g is least accurate . Other two values have to be rounded off to one place of decimal i.e. 17.35 ≈ 17.4 g and 8.498 ≈ 8.5 g

Hence , 17.4 + 25.6 + 8.5 = 51.5 g

## Which of the following length measurement is most accurate and why ? (i) 4.00 cm (ii) 0.004 mm (iii) 40.00 cm

Q: Which of the following length measurement is most accurate and why ?

(i) 4.00 cm (ii) 0.004 mm (iii) 40.00 cm

Sol: $\frac{\Delta x}{x} = \frac{0.01}{4.00} = 0.0025$

(ii) $\frac{\Delta x}{x} = \frac{0.001}{0.004} = 0.25$

(iii) $\frac{\Delta x}{x} = \frac{0.01}{0.004} = 0.00025$

The last observation has least fractional error and hence it is most accurate .

## What is the difference between nm , mN , Nm ?

Q: What is the difference between nm , mN , Nm ?

Sol: nm stands for nanometre , 1 nm = 10-9 m

mN stands for milli newton , 1 m N = 10-3 N

Nm stands for newton metre .

## What is the difference between 4.0 and 4.000 ?

Q: What is the difference between 4.0 and 4.000 ?

Sol : As per rules , 4.0 has two significant figures and 4.000 has four significant figures . Therefore , 4.000 is more accurate than 4.0 .