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1.Units, Dimensions and Measurement
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In an experiment to find out the diameter of wire using screw gauge, the following observation were noted.
$(a)$ Screw moves $0.5\,mm$ on main scale in one complete rotation
$(b)$ Total divisions on circular scale $=50$
$(c)$ Main scale reading is $2.5\,mm$
$(d)$ $45^{\text {th }}$ division of circular scale is in the pitch line
$(e)$ Instrument has $0.03 \;mm$ negative error
Then the diameter of wire is $...........\,mm$
A$2.92$
B$2.54$
C$2.98$
D$3.45$
(JEE MAIN-2022)
Solution
$MSR =2.5\,mm$
$CSR =45 \times \frac{0.5}{50}\,mm$
$=0.45\,mm$
Diameter reading $= MSR + CSR -$ zero error
$=2.5+0.45-(-0.03)$
$=2.98\,mm$
$CSR =45 \times \frac{0.5}{50}\,mm$
$=0.45\,mm$
Diameter reading $= MSR + CSR -$ zero error
$=2.5+0.45-(-0.03)$
$=2.98\,mm$
Standard 11
Physics
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