Question

3. On vernier calipers, one centimeter if the main scale is divided into 10 equal parts. If the
number of divisions on the vernier scale are 10. Find the least count of the vernier caliper.
In a measure of the correct reading of an object. The main scale reading lies between 5.6
cm and 5.7 cm. vernier scale reading is 4. The zero error is - 0.02 cm. Find the correct
reading of the object. vry important pls solve the answer fast​

Answer 1

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3 mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

$\underline{\blacksquare\:\:\:\footnotesize{\red{SolutioN}}}$

$\footnotesize{\therefore\:\text{10 vernier scale division = 9 main scale division}}$

$\footnotesize{\implies \text{10 v.d.= 9 s.d.}}$

$\footnotesize{\implies \:1 v.d.=\dfrac{9}{10} \:s.d.}$

$\footnotesize{\therefore\:\text{1 vernier constant =1 s.d. - 1 v.d.}}$

$\footnotesize{\implies\:\text{1 V.C.=1 s.d. -}\:\dfrac{9}{10}\:s.d.}$

$\footnotesize{\implies\:1 V.C.\: =\:\dfrac{10-9}{10}\:s.d.}$

$\footnotesize{\implies\:1 V.C.\: =\:\dfrac{1}{10}\:s.d.}$

$\footnotesize{\implies\:1 V.C.\: =\:0.1\:s.d.}$

$\footnotesize{\implies\:1 V.C.\: =\:0.1\times 1s.d.}$

$\footnotesize{\implies\:1 V.C.\: =\:0.1\:\times1\:mm\:\:\: (\because\:1\:s.d.\:=\:1\:mm)}$

$\footnotesize{\implies\:\boxed{\red{1 V.C.\: =\:0.1\:mm=0.01\:cm}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5 mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

$\footnotesize{\therefore\:\text{Main Reading}\:=[5.6+(v.r.\times 1 v.c.)]\pm error}$

$\footnotesize{\text{Main Reading}\:=[5.6+(4\times0.01)]\:\pm(-0.02\:cm)}$

$\footnotesize{\implies\:\text{Main Reading}\:=[(5.6+0.04)-0.02]\: cm}$

$\footnotesize{\implies\:\text{Main Reading}\:=[(5.6+0.04)-0.02]\:cm}$

$\footnotesize{\implies\:\text{Main Reading}\:=[5.6+0.02]\:cm}$

$\footnotesize{\implies\:\red{\boxed{\text{Main Reading}\:=5.62\:cm}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3 mm}\put(1,1){\line(1,0){6.8}}\end{picture}$