A student measured the thickness of a glass slip using a spherometer with least count 0.001 cm . The correct listing is

Q: A student measured the thickness of a glass slip using a spherometer with least count 0.001 cm . The correct listing is

(a) 0.23 cm

(b) 0.234 mm

(c) 2.34 mm

(d) 0.234 cm

Ans: (d)

Sol: As least count of spherometer is 0.001 cm , the measurement in centimetre must be correct upto three places of decimal .
The correct listing is 0.234 cm

Two spherometers A ans B have the same pitch , A has 100 divisions on periphery of its circular disc and B has 200 divisions on periphery of  its circular disc . Then

Q: Two spherometers A ans B have the same pitch , A has 100 divisions on periphery of its circular disc and B has 200 divisions on periphery of  its circular disc . Then

(a) Both A and B have same least count

(b) L.C of A is twice the L.C of B

(c) L.C of A is half the L.C of B

(d) Nothing can be said

Sol: $L.C = \frac{Pitch}{Number \; of \;divisions \; on \;Circular \; Scale} $

As B has twice the number of divisions of A .

Hence , L.C of A is twice the L.C of B .

When screw gauge is completely closed , zero of circular scale is 4 divisions below the reference line of graduation . If least count of screw gauge is 0.001 cm , the zero correction is

Q: When screw gauge is completely closed , zero of circular scale is 4 divisions below the reference line of graduation . If least count of screw gauge is 0.001 cm , the zero correction is

(a) -0.004 cm

(b) +0.004 cm

(c) -0.004 mm

(d) +0.004 mm

Sol: As zero of circular scale is below the reference line of graduation , zero correction is negative .

Its magnitude is 4 × 0.001 cm = 0.004 cm

Ans : (a)

A spherometer has 200 equal divisions marked on the periphery of its disc . On giving five complete rotations , the central screw advances by 5 mm , Least count of spherometer is

Q: A spherometer has 200 equal divisions marked on the periphery of its disc . On giving five complete rotations , the central screw advances by 5 mm , Least count of spherometer is

(a) 0.001 cm

(b) 0.001 mm

(c) 0.0005 cm

(d) 0.0005 mm

Ans: (c)

Sol: $ Pitch = \frac{5}{5}mm $

Pitch = 1 mm

$ Least \; count = \frac{Pitch}{200} = \frac{1}{200} mm $

L.C = 0.005 mm = 0.0005 cm

While measuring diameter of a wire using a screw gauge , the main scale reading is 7 mm and zero of circular scale is 35 division above the reference line . If screw gauge has a zero error of -0.003 cm , the correct diameter of wire is

Q: While measuring diameter of a wire using a screw gauge , the main scale reading is 7 mm and zero of circular scale is 35 division above the reference line . If screw gauge has a zero error of -0.003 cm , the correct diameter of wire is (given Least count = 0.001 cm)

(a) 0.735 cm

(b) 0.732 cm

(c) 0.738 cm

(d) 7.38 cm

Ans: (c)

Sol: Observed diameter = 0.7 cm + 35 x 0.001 cm

= 0.735 cm

Corrected diameter = (0.735 + 0.003) cm = 0.738 cm

The circular scale of a screw gauge has 200 divisions . When it is given 4 complete rotations , it moves through 2 mm . The least count of screw gauge is

Q: The circular scale of a screw gauge has 200 divisions . When it is given 4 complete rotations , it moves through 2 mm . The least count of screw gauge is

(a) 0.25 x 10-2 cm

(b) 0.25 x 10-3 cm

(c) 0.001 cm

(d) 0.001 mm

Ans: (b)

Sol: $Pitch = \frac{2}{4} mm = 0.5 mm $

$\large L.C = \frac{Pitch}{200} = \frac{0.5}{200} mm $

Pitch = 0.25 x 10-2 mm

= 0.25 x 10-3 cm

When the two jaws of a vernier calliper are in touch , zero of vernier scale lies to the right of zero of main scale and coinciding vernier division is 3 . If vernier constant is 0.1 mm , the zero correction is 

Q: When the two jaws of a vernier calliper are in touch , zero of vernier scale lies to the right of zero of main scale and coinciding vernier division is 3 . If vernier constant is 0.1 mm , the zero correction is

(a) -0.03 cm

(b) +0.03 cm

(c) -0.03 mm

(d) + 0.03 mm

Sol: As Zero of vernier scale is to the right of zero of main scale , therefore , zero correction is negative .

Its magnitude = 3 × 0.1 mm = 0.3 mm = 0.03 cm

Ans: (a)