The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale division is 1 mm. The main scale reading is 10 mm and 8th division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of 0.04 cm, then the radius of the bob is \underline\hspace2cm\times 10 - 2 cm.

Units and Measurements PYQ — JEE Main 2021 · 31 Aug · Shift 2 · Q67 with Solution

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Correct answer: 52

Given:

So, $9\text{ MSD} = 10\text{ VSD} = 9\text{ mm}$ Hence, $1\text{ VSD} = \frac{9}{10}\text{ mm} = 0.9\text{ mm}$

Least count: $LC = 1\text{ MSD} - 1\text{ VSD} = 1 - 0.9 = 0.1\text{ mm}$

Main scale reading: $MSR = 10\text{ mm}$

8th vernier division coincides, so vernier reading: $VR = 8 \times LC = 8 \times 0.1 = 0.8\text{ mm}$

Therefore observed diameter: $D_{obs} = 10 + 0.8 = 10.8\text{ mm}$

Positive zero error $= 0.04\text{ cm}$

Convert to mm: $0.04\text{ cm} = 0.4\text{ mm}$

For positive zero error, correction is subtracted: $D_{true} = 10.8 - 0.4 = 10.4\text{ mm}$

$r = \frac{D_{true}}{2} = \frac{10.4}{2} = 5.2\text{ mm}$

Convert to cm: $5.2\text{ mm} = 0.52\text{ cm}$

Now write in the form: $r = \underline{\hspace{1cm}} \times 10^{-2}\text{ cm}$

Since, $0.52\text{ cm} = 52 \times 10^{-2}\text{ cm}$

So the required integer is: $\boxed{52}$

Stored correct answer = $52$

Our derived answer also = $52$ .