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Units and Measurements question

2022 · 29 Jun · Shift 2 · Q62
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Units and Measurements question

2022 · 29 Jun · Shift 2 · Q62

JEE MainPhysicsUnits and MeasurementsNumerical+4 / −1
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 ‾×\underline{\hspace{2cm}}\times​× 10 −-− 2 cm.
Numerical answer
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Correct answer: 180

  1. Given data
  • Vernier constant (least count) =0.1 mm= 0.1\text{ mm}=0.1 mm
  • Zero error =−0.05 cm= -0.05\text{ cm}=−0.05 cm
  • Main scale reading =1.7 cm= 1.7\text{ cm}=1.7 cm
  • Coinciding vernier division =5= 5=5
  1. Convert least count to cm

0.1 mm=0.01 cm0.1\text{ mm} = 0.01\text{ cm}0.1 mm=0.01 cm

  1. Observed reading

Vernier contribution:

5×0.01=0.05 cm5 \times 0.01 = 0.05\text{ cm}5×0.01=0.05 cm

So observed diameter is

1.7+0.05=1.75 cm1.7 + 0.05 = 1.75\text{ cm}1.7+0.05=1.75 cm

  1. Apply zero correction

Zero error is −0.05 cm-0.05\text{ cm}−0.05 cm.

Hence zero correction is

−(zero error)=−(−0.05)=+0.05 cm-(\text{zero error}) = -(-0.05) = +0.05\text{ cm}−(zero error)=−(−0.05)=+0.05 cm

Therefore corrected diameter:

1.75+0.05=1.80 cm1.75 + 0.05 = 1.80\text{ cm}1.75+0.05=1.80 cm

  1. Match with required form

We need

corrected diameter=‾×10−2 cm\text{corrected diameter} = \underline{\hspace{1cm}} \times 10^{-2}\text{ cm}corrected diameter=​×10−2 cm

Now,

1.80 cm=180×10−2 cm1.80\text{ cm} = 180 \times 10^{-2}\text{ cm}1.80 cm=180×10−2 cm

So the required integer is

180\boxed{180}180​

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