A student performs an experiment of measuring thickness of a slab with a vernier calliper whose 50 d

English

Gujarati

Hindi

1.Units, Dimensions and Measurement

medium

A student performs an experiment of measuring the thickness of a slab with a vernier calliper whose $50$ divisions of the vernier scale are equal to $49$ divisions of the main scale. He noted that zero of the vernier scale is between $7.00\; cm$ and $7.05 \;cm$ mark of the main scale and $23^{rd}$ division of the vernier scale exactly coincides with the main scale. The measured value of the thickness of the given slab using the calliper will be

A$7.23$

B$7.023$

C$7.073$

D$7.73$

(NEET-2017)

Solution

$a=$ Main Scale division
$b=$ vernier scale division
$L . C .=\left(\frac{ b – a }{ b }\right) M$
$=\frac{50-49}{50} M$ $M=$ Main scale reading
$L C =\frac{ M }{50}$ $M =7.05-7.00$
$L C \cdot=\frac{0.05}{50} \quad M =0.05$
L.C. $=0.001$
Reading $=7+0.001 \times 23=7.023 cm$

Standard 11

Physics

Share

Similar Questions

In an experiment the angles are required to be measured using an instrument, $29$ divisions of the main scale exactly coincide with the $30$ divisions of the vernier scale. If the smallest division of the main scale is half- a degree $(= 0.5^o )$, then the least count of the instrument is

medium

(AIEEE-2009)

A student measured the diameter of a small steel ball using a screw gauge of least count $0.001\, cm.$ The main scale reading is $5\, mm$ and zero of circular scale division coincides with $25$ divisions above the reference level. If screw gauge has a zero error of $-0.004 \,cm,$ the correct diameter of the ball is

medium

(NEET-2018)

A student measured the length of a rod and wrote it as $3.50\;cm$. Which instrument did he use to measure it?

medium

(JEE MAIN-2014)

Diameter of a steel ball is measured using a Vernier callipers which has divisions of $0. 1\,cm$ on its main scale $(MS)$ and $10$ divisions of its vernier scale $(VS)$ match $9$ divisions on the main scale. Three such measurements for a ball are given as

S.No. $MS\;(cm)$ $VS$ divisions

$(1)$ $0.5$ $8$

$(2)$ $0.5$ $4$

$(3)$ $0.5$ $6$

If the zero error is $- 0.03\,cm,$ then mean corrected diameter is ……….. $cm$

medium

(JEE MAIN-2015)

When the gap is closed without placing any object in the screw gauge whose least count is $0.005\ mm$, the $5^{th}$ division on its circular scale with the reference line on main scale, and when a small sphere is placed reading on main scale advances by $4$ divisions, whereas circular scale reading advances by five times to the corresponding reading when no object was placed. There are $200$ divisions on the circular scale. The radius of the sphere is ………. $mm$

hard