You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Screw Gauge

The least cou...

Question

The least count of the main scale of a screw gauge is $1 mm$ . The minimum number of divisions on its circular scale required to measure $5 μ m$ diameter of a wire is:

100

No worries! We‘ve got your back. Try BYJU‘S free classes today!

50

No worries! We‘ve got your back. Try BYJU‘S free classes today!

500

No worries! We‘ve got your back. Try BYJU‘S free classes today!

200

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $200$
Least count of main scale of screw gauge $= 1 mm$

Least count of screw gauge
$= \frac{Pitch}{Number of division on circular scale}$

I.e. $5 \times 10^{- 6} = \frac{10^{- 3}}{N}$

$\Rightarrow N = 200$

Suggest Corrections

Similar questions

Q. The least count of the main scale of a screw gauge is

1 m m

.The minimum number of division on its circular scale required to measure

5 μ m

diameter of wire is:

Q. The pitch of a screw gauge is

1

mm and its circular scale has

100

divisions. In measurements of the diameter of a wire, the main scale reads

2

mm and

45

th mark on circular scale coincides with the baseline. Find the least count ?

Q. A screw gauge gives the following reading for measuring the diameter of a wire.
Main scale reading :

0 mm

Circular scale reading :

60

divisions
Given, that

1 mm

on main scale corresponds to

100

divisions on circular scale. The diameter of wire is,

Q. The least count of a screw gauge having 1 mm pitch and 100 circular scale divisions is ____

μ m

The pitch of a screw gauge is 1 mm and its circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on circular scale coincides with the base line. Find :

(i) the least count, and

(ii) the diameter of the wire.

Join BYJU'S Learning Program

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a wire is:

The correct option is D 200Least count of main scale of screw gauge =1 mm Least count of screw gauge =PitchNumber of division on circular scale I.e. 5×10−6=10−3N ⇒N=200

The least count of the main scale of a screw gauge is $1 mm$ . The minimum number of divisions on its circular scale required to measure $5 μ m$ diameter of a wire is:

The correct option is D $200$
Least count of main scale of screw gauge $= 1 mm$

Least count of screw gauge
$= \frac{Pitch}{Number of division on circular scale}$

I.e. $5 \times 10^{- 6} = \frac{10^{- 3}}{N}$

$\Rightarrow N = 200$