Consider the Vernier caliper as shown, the instrument has no zero error. The length of the rod (in mm ) shown is α , if 1MSD=1mm . Use 7MSD=8VSD . Write the value of 2α .

Question

Consider the Vernier caliper as shown, the instrument has no zero error. The length of the rod (in mm ) shown is α , if 1MSD=1mm . Use 7MSD=8VSD . Write the value of 2α . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-hcabrsfbl3c9nokx.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given, 7MSD=8VSD ⇒ 1VSD=(7/8)MSD ⇒ LC=(1 - (7/8))mm=(1/8)mm Also, MSR=4mm V.S.R=4 value =(4+(4)((1/8))) mm =4.5mm Hence, 2α =9

Q. Consider the Vernier caliper as shown, the instrument has no zero error. The length of the rod (in $mm$ ) shown is $α$ , if $1 MS D = 1 mm$ . Use $7 MS D = 8 V S D$ . Write the value of $2 α$ .

Given, $7 MS D = 8 V S D$
$\Rightarrow 1 V S D = \frac{7}{8} MS D$
$\Rightarrow L C = (1 - \frac{7}{8}) mm = \frac{1}{8} mm$
Also, $MSR = 4 mm$
$V . S . R = 4$
value $= (4 + (4) (\frac{1}{8})) mm$
$= 4.5 mm$
Hence, $2 α = 9$

Q. Consider the Vernier caliper as shown, the instrument has no zero error. The length of the rod (in mm ) shown is α , if 1MSD=1mm . Use 7MSD=8VSD . Write the value of 2α .

Given, 7MSD=8VSD ⇒1VSD=87​MSD ⇒LC=(1−87​)mm=81​mm Also, MSR=4mm V.S.R=4 value =(4+(4)(81​))mm =4.5mm Hence, 2α=9

Q. Consider the Vernier caliper as shown, the instrument has no zero error. The length of the rod (in $mm$ ) shown is $α$ , if $1 MS D = 1 mm$ . Use $7 MS D = 8 V S D$ . Write the value of $2 α$ .

Given, $7 MS D = 8 V S D$
$\Rightarrow 1 V S D = \frac{7}{8} MS D$
$\Rightarrow L C = (1 - \frac{7}{8}) mm = \frac{1}{8} mm$
Also, $MSR = 4 mm$
$V . S . R = 4$
value $= (4 + (4) (\frac{1}{8})) mm$
$= 4.5 mm$
Hence, $2 α = 9$