Question

measurements:
(i) 0.0025
(ii) 0.0250
(iv) 5.98 x 1024
(iii) 0.002500
(Ans.: 2, 3, 4, 3)
2.2
The length of an object measured by vernier calliper
is 4.78 cm. If the L.C. of vernier caliper is 0.01 cm.
Calculate percentage error.
(Ans. : 0.209 %)
The diameter of a wire measures by means of
micrometer screw gauge of least count 0.001 cm is
102 mm. Calculate the percentage error in
measurement.
(Ans. : 0.98 %)
please guys right answer beta na please ​

Answer 1

Answer:

L.C. = 0.01 cm

Error = + 0.02 cm ⇒ Positive zero

error = + 0.02 cm

main scale reading (MSR) = 3.60 cm 8

th

vernier scale coincides with main scale

⇒ vernier coincidence (VC) = 8

We know that

Correct diameter = [main scale reading+ (L.C × V.C)] - (error)

= [3.60 + (0.01 × 8)] - [+ 0.002]

= (3.60 + 0.08)- (+0.02)

= 3.68 - 0.02

= 3.66 cm

∴correctradius=

2

diameter

=

2

3.66

=1.83cm