A screw gauge with a pitch of 1 mm has 100 divisions on its circular scale. When it is used to measure the diameter of a thin wire, the main scale reading is 3 mm and the circular scale reading is 45. If D is the diameter of the wire in m m, then find the value of D.

Question

A screw gauge with a pitch of 1 mm has 100 divisions on its circular scale. When it is used to measure the diameter of a thin wire, the main scale reading is 3 mm and the circular scale reading is 45. If D is the diameter of the wire in m m, then find the value of D. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Least count =(P i t c h/N o . o f d i v i s i o n)=(1 m m/100)=0.01 mm Diameter of wire = main scale + circular scale reading =3 mm+(L . C)× No. of division =3 mm+0.01× 45 D=3.45 mm

Q. A screw gauge with a pitch of $1 mm$ has $100$ divisions on its circular scale. When it is used to measure the diameter of a thin wire, the main scale reading is $3 mm$ and the circular scale reading is $45.$ If $D$ is the diameter of the wire in m $m,$ then find the value of $D .$

$L e a s t co u n t = \frac{P i t c h}{N o . o f d i v i s i o n} = \frac{1 mm}{100} = 0.01 mm$
$D iam e t er o f w i re = main sc a l e + c i rc u l a r sc a l e re a d in g$
$= 3 mm + (L . C) \times N o . o f d i v i s i o n$
$= 3 mm + 0.01 \times 45$
$D = 3.45 mm$

Q. A screw gauge with a pitch of 1mm has 100 divisions on its circular scale. When it is used to measure the diameter of a thin wire, the main scale reading is 3mm and the circular scale reading is 45. If D is the diameter of the wire in m m, then find the value of D.

Leastcount=No.ofdivisionPitch​=1001mm​=0.01mm Diameterofwire=mainscale+circularscalereading =3mm+(L.C)×No.ofdivision =3mm+0.01×45 D=3.45mm

Q. A screw gauge with a pitch of $1 mm$ has $100$ divisions on its circular scale. When it is used to measure the diameter of a thin wire, the main scale reading is $3 mm$ and the circular scale reading is $45.$ If $D$ is the diameter of the wire in m $m,$ then find the value of $D .$

$L e a s t co u n t = \frac{P i t c h}{N o . o f d i v i s i o n} = \frac{1 mm}{100} = 0.01 mm$
$D iam e t er o f w i re = main sc a l e + c i rc u l a r sc a l e re a d in g$
$= 3 mm + (L . C) \times N o . o f d i v i s i o n$
$= 3 mm + 0.01 \times 45$
$D = 3.45 mm$