Question

If the zero of the vernier lies on the right hand side and fourth division coincide with the main scale division when the jaws are in contacts so the correction will be–

• A.
  • 0.04 cm
• B.
  • 0.06 cm
• C. –0.04 cm
• D. –0.06 cm

Answer 1

Answer: (C)

–0.04 cm

Answer 2

By using $y = \frac{qELD}{mv^{2}}$; where y = deflection on screen due to electric field only

⇒ $\frac{q}{m} = \frac{yv^{2}}{ELD} = \frac{2 \times 10^{- 2} \times (10^{6})^{2}}{20 \times 10^{3} \times 5 \times 10^{- 2} \times 0.2} = 10^{8}C/kg.$