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PHXI02:UNITS AND MEASUREMENTS

367550 One centimeter on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers ?

1

0.05 c m

2

0.04 c m

3

0.06 c m

4

0.02 c m

Explanation:

$L . C = (\frac{b - a}{b}) M = (\frac{20 - 8}{20}) (\frac{1}{10}) c m$
$= \frac{3}{50} c m = 0.06 c m$

PHXI02:UNITS AND MEASUREMENTS

367551 One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of the vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is

1

0.01 c m

2

0.02 c m

3

0.05 c m

4

0.005 c m

Explanation:

Least count $= 1 M . S . D - 1 V . S . D$
From the given information $8 M . S . D = 10 V . S . D$
$\Rightarrow 1 V . S . D = \frac{8}{10} M . S . D$
$\Rightarrow L C = M . S . D (1 - \frac{8}{10})$
$= (1 - \frac{8}{10}) = \frac{2}{10} m m = 0.02 c m$

PHXI02:UNITS AND MEASUREMENTS

367552 The main scale of vernier calliper is divided into $0.5 m m$ and its least count is $0.005 c m$ . The number of vernier divisions are

1 10

2 20

3 30

4 40

Explanation:

Given, L.C. $= \frac{S}{n}$
$∴$ No. of divisions on main scale $(n)$
$= \frac{S}{L.C.} = \frac{0.5 m m}{0.05 m m} = 10$

PHXI02:UNITS AND MEASUREMENTS

367550 One centimeter on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers ?

1

0.05 c m

2

0.04 c m

3

0.06 c m

4

0.02 c m

Explanation:

$L . C = (\frac{b - a}{b}) M = (\frac{20 - 8}{20}) (\frac{1}{10}) c m$
$= \frac{3}{50} c m = 0.06 c m$

PHXI02:UNITS AND MEASUREMENTS

367551 One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of the vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is

1

0.01 c m

2

0.02 c m

3

0.05 c m

4

0.005 c m

Explanation:

Least count $= 1 M . S . D - 1 V . S . D$
From the given information $8 M . S . D = 10 V . S . D$
$\Rightarrow 1 V . S . D = \frac{8}{10} M . S . D$
$\Rightarrow L C = M . S . D (1 - \frac{8}{10})$
$= (1 - \frac{8}{10}) = \frac{2}{10} m m = 0.02 c m$

PHXI02:UNITS AND MEASUREMENTS

367552 The main scale of vernier calliper is divided into $0.5 m m$ and its least count is $0.005 c m$ . The number of vernier divisions are

1 10

2 20

3 30

4 40

Explanation:

Given, L.C. $= \frac{S}{n}$
$∴$ No. of divisions on main scale $(n)$
$= \frac{S}{L.C.} = \frac{0.5 m m}{0.05 m m} = 10$

PHXI02:UNITS AND MEASUREMENTS

367550 One centimeter on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers ?

1

0.05 c m

2

0.04 c m

3

0.06 c m

4

0.02 c m

Explanation:

$L . C = (\frac{b - a}{b}) M = (\frac{20 - 8}{20}) (\frac{1}{10}) c m$
$= \frac{3}{50} c m = 0.06 c m$

PHXI02:UNITS AND MEASUREMENTS

367551 One centimetre on the main scale of vernier callipers is divided into ten equal parts. If 10 divisions of the vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is

1

0.01 c m

2

0.02 c m

3

0.05 c m

4

0.005 c m

Explanation:

Least count $= 1 M . S . D - 1 V . S . D$
From the given information $8 M . S . D = 10 V . S . D$
$\Rightarrow 1 V . S . D = \frac{8}{10} M . S . D$
$\Rightarrow L C = M . S . D (1 - \frac{8}{10})$
$= (1 - \frac{8}{10}) = \frac{2}{10} m m = 0.02 c m$

PHXI02:UNITS AND MEASUREMENTS

367552 The main scale of vernier calliper is divided into $0.5 m m$ and its least count is $0.005 c m$ . The number of vernier divisions are

1 10

2 20

3 30

4 40

Explanation:

Given, L.C. $= \frac{S}{n}$
$∴$ No. of divisions on main scale $(n)$
$= \frac{S}{L.C.} = \frac{0.5 m m}{0.05 m m} = 10$