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1.Units, Dimensions and Measurement
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A vernier callipers used by student has $20$ divisions in $1\;cm$ on main scale. $10$ vernier divisions coincide with $9$ main scale divisions. When jaws are closed, zero of main scale is on left of zero of vernier scale and $6\ th$ division of vernier scale coincides with any of main scale divisions. He places a wooden cylinder in between the jaws and measures the length. The zero of vernier scale is on right of $3.20\;cm$ and $8\ th$ vernier division coincides with any main scale division. When he measures thickness of cylinder he finds that zero of vernier scale lies on right of $1.50\;cm$ mark of main scale and $6\ th$ division of vernier scale coincides with any main scale division. The correct values of measured length and diameter are respectively
A$3.21\ cm,$ $1.50$ $cm$
B$3.210$ $cm,$ $1.500$ $cm$
C$3.27$ $cm,$ $1.93$ $cm$
D$3.270$ $cm,$ $1.560$ $cm$
Solution
$M \cdot S \cdot \quad 20 d i v=1 \mathrm{cm}$
$1 \mathrm{MsD}=\frac{1}{20} \mathrm{cm}=0.05 \mathrm{cm}$
$10 \mathrm{VsD}=9 \mathrm{MsD}$
$\mid V S D=\frac{9}{10} \times M S D=\frac{9}{10} \times 0.05 \mathrm{cm}$
$L \cdot c=1 \mathrm{MsD} -1 \mathrm{VsD}$
$=0.05-\frac{9}{10} \times 0.05=\frac{1}{10} \times 0.05$
$L c=0.005 \mathrm{cm}$
zero error: $=6 \times L C=6 \times 0.005$
Zero error $=0.03 \mathrm{cm}$
Reading $=\mathrm{M} \cdot \mathrm{sD}+\mathrm{VsD}$
$=3 \cdot 2+8 \times L \cdot c$
$=3 \cdot 2+8 \times 0.005$
$=3 \cdot 2+0.04=3.24 \mathrm{cm}$
Actual reading = Reading – zero emor
$=3 \cdot 2 4-0.03=3 \cdot 21 \mathrm{cm}$
Reading $=\operatorname{MsR}+V S R$
$=1.5+6 \times 10$
$=1.5+6 \times 0.005=1.53$
$=1.5+0.03=1.53$
A ctual reading = Reading – zero error
$=1.530-0.03=1.5 \mathrm{cm}$
$1 \mathrm{MsD}=\frac{1}{20} \mathrm{cm}=0.05 \mathrm{cm}$
$10 \mathrm{VsD}=9 \mathrm{MsD}$
$\mid V S D=\frac{9}{10} \times M S D=\frac{9}{10} \times 0.05 \mathrm{cm}$
$L \cdot c=1 \mathrm{MsD} -1 \mathrm{VsD}$
$=0.05-\frac{9}{10} \times 0.05=\frac{1}{10} \times 0.05$
$L c=0.005 \mathrm{cm}$
zero error: $=6 \times L C=6 \times 0.005$
Zero error $=0.03 \mathrm{cm}$
Reading $=\mathrm{M} \cdot \mathrm{sD}+\mathrm{VsD}$
$=3 \cdot 2+8 \times L \cdot c$
$=3 \cdot 2+8 \times 0.005$
$=3 \cdot 2+0.04=3.24 \mathrm{cm}$
Actual reading = Reading – zero emor
$=3 \cdot 2 4-0.03=3 \cdot 21 \mathrm{cm}$
Reading $=\operatorname{MsR}+V S R$
$=1.5+6 \times 10$
$=1.5+6 \times 0.005=1.53$
$=1.5+0.03=1.53$
A ctual reading = Reading – zero error
$=1.530-0.03=1.5 \mathrm{cm}$
Standard 11
Physics
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