The vernier constant of Vernier callipers is 0.1 mm and it has zero error of (-0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be × 10-2 cm.

Question

The vernier constant of Vernier callipers is 0.1 mm and it has zero error of (-0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be × 10-2 cm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Measured diameter = MSR + VSR × VC =1.7+0.01 × 5 =1.75 Corrected = Measured - Error =1.75-(-0.05) =1.80 cm =180 × 10-2 cm 180

Q. The vernier constant of Vernier callipers is $0.1 mm$ and it has zero error of $(- 0.05) c m$ . While measuring diameter of a sphere, the main scale reading is $1.7 c m$ and coinciding vernier division is $5$ . The corrected diameter will be_____ $\times 1 0^{- 2} c m$ .

Measured diameter $= MSR + V SR \times V C$
$= 1.7 + 0.01 \times 5$
$= 1.75$
Corrected $=$ Measured $-$ Error
$= 1.75 - (- 0.05)$
$= 1.80 c m$
$= 180 \times 1 0^{- 2} c m$
$180$

Q. The vernier constant of Vernier callipers is 0.1mm and it has zero error of (−0.05)cm. While measuring diameter of a sphere, the main scale reading is 1.7cm and coinciding vernier division is 5. The corrected diameter will be_____ ×10−2cm.

Measured diameter =MSR+VSR×VC =1.7+0.01×5 =1.75 Corrected = Measured − Error =1.75−(−0.05) =1.80cm =180×10−2cm 180

Q. The vernier constant of Vernier callipers is $0.1 mm$ and it has zero error of $(- 0.05) c m$ . While measuring diameter of a sphere, the main scale reading is $1.7 c m$ and coinciding vernier division is $5$ . The corrected diameter will be_____ $\times 1 0^{- 2} c m$ .

Measured diameter $= MSR + V SR \times V C$
$= 1.7 + 0.01 \times 5$
$= 1.75$
Corrected $=$ Measured $-$ Error
$= 1.75 - (- 0.05)$
$= 1.80 c m$
$= 180 \times 1 0^{- 2} c m$
$180$