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1.Units, Dimensions and Measurement
hard
The main scale of vernier callipers reads in millimetre and its vernier is divided into $8$ divisions, which coincides with $5$ divisions of main scale. When two jaws of instrument touch each other, the zero of the vernier coincides with the zero of main scale. A rod is tightly placed along its length between both jaws. It is observed that the zero of vernier scale lies just left to $36^{th}$ division of main scale and fourth division of vernier scale coincides with the main scale Then the measured value is .......... $cm$
A$3.66$
B$3.55$
C$3.65$
D$3.56$
Solution
M.S. reading 35 mm
$X+4 V=n M$
$X=n M-4 \frac{5}{8} M$
$X<1 m \text { for } n=3 \text { only }$
So, $X=3-\frac{5}{2}=\frac{1}{2} mm$
$\therefore$ reading $y=35+0.5 mm$
$=3.55 cm$
$X+4 V=n M$
$X=n M-4 \frac{5}{8} M$
$X<1 m \text { for } n=3 \text { only }$
So, $X=3-\frac{5}{2}=\frac{1}{2} mm$
$\therefore$ reading $y=35+0.5 mm$
$=3.55 cm$
Standard 11
Physics
Similar Questions
hard
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 ……….. $cm$