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PHXI02:UNITS AND MEASUREMENTS

367537 The pitch of a screw gauge is 1 $m m$ and there are 100 divisions on the circular scale. In measuring the diameter of a sphere there are seven divisions on the linear scale and forty five divisions on circular scale coincides with the reference line. Find the diameter of the sphere.

1

7.45 m m

2

8.23 m m

3

6.56 m m

4

5.20 m m

Explanation:

$L . C = \frac{1}{100} = 0.01 m m$
Linear scale reading $= 7 \times pitch = 7 m m$
Circular scale reading $= 45 \times 0.01 = 0.45 m m$
$∴$ Total reading $= (7 + 0.45) = 7.45 m m$

PHXI02:UNITS AND MEASUREMENTS

367538 A screw gauge having 100 equal divisions and a pitch of length 1 $m m$ is used to measure the diameter of a wire of length 5.6 $c m$ . The main scale reading is 1 $m m$ and $47^{t h}$ circular division coincides with the main scale. Find the curved surface area of the wire in $c m^{2}$ to approriate significant figures $(Use π = \frac{22}{7})$

1

2.6 c m^{2}

2

2.2 c m^{2}

3

3.0 c m^{2}

4

2.0 c m^{2}

Explanation:

Least count $= \frac{1 m m}{100} = 0.01 m m$
Diameter $= M S R + C S R = 1 m m + 47 (0.01) m m = 1.47 m m$
Surface area $= 2 π r l = π D l = \frac{22}{7} \times 0.147 c m \times 5.6 c m$
$= 2.58724 c m^{2} = 2.6 c m^{2}$

PHXI02:UNITS AND MEASUREMENTS

367539 For the pitch of a screw 0.1 cm and 200 divisions on circular scale, the least count will be

1

0.5 m m

2

0.05 m m

3

0.005 m m

4

0.0005 m m

Explanation:

$L . C . = \frac{0.1 c m}{200} = \frac{1}{2000}$
$= 0.0005 c m = 0.005 m m$

PHXI02:UNITS AND MEASUREMENTS

367540 A student measured the diameter of a wire using a screw gauge with least count 0.001 $c m$ and listed the measurements. The correct measurement is

1

5.3 c m

2

5.32 c m

3

5.320 c m

4

5.3200 c m

Explanation:

The measurement is correct upto third decimal. Therefore, it must be 5.320 $c m$

PHXI02:UNITS AND MEASUREMENTS

367537 The pitch of a screw gauge is 1 $m m$ and there are 100 divisions on the circular scale. In measuring the diameter of a sphere there are seven divisions on the linear scale and forty five divisions on circular scale coincides with the reference line. Find the diameter of the sphere.

1

7.45 m m

2

8.23 m m

3

6.56 m m

4

5.20 m m

Explanation:

$L . C = \frac{1}{100} = 0.01 m m$
Linear scale reading $= 7 \times pitch = 7 m m$
Circular scale reading $= 45 \times 0.01 = 0.45 m m$
$∴$ Total reading $= (7 + 0.45) = 7.45 m m$

PHXI02:UNITS AND MEASUREMENTS

367538 A screw gauge having 100 equal divisions and a pitch of length 1 $m m$ is used to measure the diameter of a wire of length 5.6 $c m$ . The main scale reading is 1 $m m$ and $47^{t h}$ circular division coincides with the main scale. Find the curved surface area of the wire in $c m^{2}$ to approriate significant figures $(Use π = \frac{22}{7})$

1

2.6 c m^{2}

2

2.2 c m^{2}

3

3.0 c m^{2}

4

2.0 c m^{2}

Explanation:

Least count $= \frac{1 m m}{100} = 0.01 m m$
Diameter $= M S R + C S R = 1 m m + 47 (0.01) m m = 1.47 m m$
Surface area $= 2 π r l = π D l = \frac{22}{7} \times 0.147 c m \times 5.6 c m$
$= 2.58724 c m^{2} = 2.6 c m^{2}$

PHXI02:UNITS AND MEASUREMENTS

367539 For the pitch of a screw 0.1 cm and 200 divisions on circular scale, the least count will be

1

0.5 m m

2

0.05 m m

3

0.005 m m

4

0.0005 m m

Explanation:

$L . C . = \frac{0.1 c m}{200} = \frac{1}{2000}$
$= 0.0005 c m = 0.005 m m$

PHXI02:UNITS AND MEASUREMENTS

367540 A student measured the diameter of a wire using a screw gauge with least count 0.001 $c m$ and listed the measurements. The correct measurement is

1

5.3 c m

2

5.32 c m

3

5.320 c m

4

5.3200 c m

Explanation:

The measurement is correct upto third decimal. Therefore, it must be 5.320 $c m$

PHXI02:UNITS AND MEASUREMENTS

367537 The pitch of a screw gauge is 1 $m m$ and there are 100 divisions on the circular scale. In measuring the diameter of a sphere there are seven divisions on the linear scale and forty five divisions on circular scale coincides with the reference line. Find the diameter of the sphere.

1

7.45 m m

2

8.23 m m

3

6.56 m m

4

5.20 m m

Explanation:

$L . C = \frac{1}{100} = 0.01 m m$
Linear scale reading $= 7 \times pitch = 7 m m$
Circular scale reading $= 45 \times 0.01 = 0.45 m m$
$∴$ Total reading $= (7 + 0.45) = 7.45 m m$

PHXI02:UNITS AND MEASUREMENTS

367538 A screw gauge having 100 equal divisions and a pitch of length 1 $m m$ is used to measure the diameter of a wire of length 5.6 $c m$ . The main scale reading is 1 $m m$ and $47^{t h}$ circular division coincides with the main scale. Find the curved surface area of the wire in $c m^{2}$ to approriate significant figures $(Use π = \frac{22}{7})$

1

2.6 c m^{2}

2

2.2 c m^{2}

3

3.0 c m^{2}

4

2.0 c m^{2}

Explanation:

Least count $= \frac{1 m m}{100} = 0.01 m m$
Diameter $= M S R + C S R = 1 m m + 47 (0.01) m m = 1.47 m m$
Surface area $= 2 π r l = π D l = \frac{22}{7} \times 0.147 c m \times 5.6 c m$
$= 2.58724 c m^{2} = 2.6 c m^{2}$

PHXI02:UNITS AND MEASUREMENTS

367539 For the pitch of a screw 0.1 cm and 200 divisions on circular scale, the least count will be

1

0.5 m m

2

0.05 m m

3

0.005 m m

4

0.0005 m m

Explanation:

$L . C . = \frac{0.1 c m}{200} = \frac{1}{2000}$
$= 0.0005 c m = 0.005 m m$

PHXI02:UNITS AND MEASUREMENTS

367540 A student measured the diameter of a wire using a screw gauge with least count 0.001 $c m$ and listed the measurements. The correct measurement is

1

5.3 c m

2

5.32 c m

3

5.320 c m

4

5.3200 c m

Explanation:

The measurement is correct upto third decimal. Therefore, it must be 5.320 $c m$

PHXI02:UNITS AND MEASUREMENTS

367537 The pitch of a screw gauge is 1 $m m$ and there are 100 divisions on the circular scale. In measuring the diameter of a sphere there are seven divisions on the linear scale and forty five divisions on circular scale coincides with the reference line. Find the diameter of the sphere.

1

7.45 m m

2

8.23 m m

3

6.56 m m

4

5.20 m m

Explanation:

$L . C = \frac{1}{100} = 0.01 m m$
Linear scale reading $= 7 \times pitch = 7 m m$
Circular scale reading $= 45 \times 0.01 = 0.45 m m$
$∴$ Total reading $= (7 + 0.45) = 7.45 m m$

PHXI02:UNITS AND MEASUREMENTS

367538 A screw gauge having 100 equal divisions and a pitch of length 1 $m m$ is used to measure the diameter of a wire of length 5.6 $c m$ . The main scale reading is 1 $m m$ and $47^{t h}$ circular division coincides with the main scale. Find the curved surface area of the wire in $c m^{2}$ to approriate significant figures $(Use π = \frac{22}{7})$

1

2.6 c m^{2}

2

2.2 c m^{2}

3

3.0 c m^{2}

4

2.0 c m^{2}

Explanation:

Least count $= \frac{1 m m}{100} = 0.01 m m$
Diameter $= M S R + C S R = 1 m m + 47 (0.01) m m = 1.47 m m$
Surface area $= 2 π r l = π D l = \frac{22}{7} \times 0.147 c m \times 5.6 c m$
$= 2.58724 c m^{2} = 2.6 c m^{2}$

PHXI02:UNITS AND MEASUREMENTS

367539 For the pitch of a screw 0.1 cm and 200 divisions on circular scale, the least count will be

1

0.5 m m

2

0.05 m m

3

0.005 m m

4

0.0005 m m

Explanation:

$L . C . = \frac{0.1 c m}{200} = \frac{1}{2000}$
$= 0.0005 c m = 0.005 m m$

PHXI02:UNITS AND MEASUREMENTS

367540 A student measured the diameter of a wire using a screw gauge with least count 0.001 $c m$ and listed the measurements. The correct measurement is

1

5.3 c m

2

5.32 c m

3

5.320 c m

4

5.3200 c m

Explanation:

The measurement is correct upto third decimal. Therefore, it must be 5.320 $c m$