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Units and Measurements

139946 Assertion: The number of significant figures depends on the least count of measuring instrument.
Reason: Significant figures define the accuracy of measuring instrument.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2016

Units and Measurements

139953 A screw gauge of pitch $0.5 mm$ is used to measure the diameter of uniform wire of length $6.8 cm$ , the main scale reading is $1.5 mm$ and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is:
[Screw gauge has 50 divisions on the circular scale]

1

6.8 {cm}^{2}

2

3.4 {cm}^{2}

3

3.9 {cm}^{2}

4

2.4 {cm}^{2}

Explanation:

B Given that, screw gauge pitch $= 0.5 mm$
Length of wire $l = 6.8 cm$
Circular scale reading $= 7$
Least count $= \frac{0.5}{50} mm = 0.01 mm$
Diameter $(d) =$ main scale reading + circular scale reading $\times$ Least count
$d = 1.5 mm + 7 \times 0.01$
$d = 1.57 mm = 0.157 cm$
Curved surface area of wire $= π d l {cm}^{2}$
$= 3.14 \times 0.157 \times 6.8 {cm}^{2}$
$= 3.35 {cm}^{2}$
$≅ 3.4 {cm}^{2}$

JEE Main-26.07.2022

Units and Measurements

139954 One main scale division of a vernier callipers is a $cm$ and $n^{th}$ division of the vernier scale coincide with $(n - 1)^{th}$ division of the main scale. The least count of the callipers (in $m m$ ) is

1

\frac{10 na}{(n - 1)}

2

\frac{10 a}{(n - 1)}

3

(\frac{n - 1}{10 n}) a

4

\frac{10 a}{n}

Explanation:

D 1 MSD of vernier callipers $= a cm = 10 a mm$
$n^{th}$ division of vernier scale coincide with $(n - 1)^{th}$ division of the main scale
So, $n \times$ V.S.D. $= (n - 1) \times$ M.S.D.
$1 \times V.S.D = \frac{(n - 1)}{n} M.S.D.$
Least count $(LC) = 1$ M.S.D. -1 V.S.D.
$= 1 MSD - (\frac{n - 1}{n}) M.S.D.$
$= [1 - (\frac{n - 1}{n})] M.S.D.$
$= \frac{1}{n} MSD$
Least count $= \frac{1}{n} \times 10 a mm (∵ 1 MSD = 10 a mm)$
$LC = (\frac{10 a}{n}) mm$

JEE Main-16.03.2021

Units and Measurements

139927 Number of significant figures in expression $\frac{4.327 g}{2.51 {cm}^{3}}$ is

1 2

2 4

3 3

4 5

Explanation:

C Number of significant figures in $\frac{4.327}{2.51}$
$= 1.7239 = 1.72$
$(1, 7, 2)$ there are three significant figure

BITSAT-2019

Units and Measurements

139946 Assertion: The number of significant figures depends on the least count of measuring instrument.
Reason: Significant figures define the accuracy of measuring instrument.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2016

Units and Measurements

139953 A screw gauge of pitch $0.5 mm$ is used to measure the diameter of uniform wire of length $6.8 cm$ , the main scale reading is $1.5 mm$ and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is:
[Screw gauge has 50 divisions on the circular scale]

1

6.8 {cm}^{2}

2

3.4 {cm}^{2}

3

3.9 {cm}^{2}

4

2.4 {cm}^{2}

Explanation:

B Given that, screw gauge pitch $= 0.5 mm$
Length of wire $l = 6.8 cm$
Circular scale reading $= 7$
Least count $= \frac{0.5}{50} mm = 0.01 mm$
Diameter $(d) =$ main scale reading + circular scale reading $\times$ Least count
$d = 1.5 mm + 7 \times 0.01$
$d = 1.57 mm = 0.157 cm$
Curved surface area of wire $= π d l {cm}^{2}$
$= 3.14 \times 0.157 \times 6.8 {cm}^{2}$
$= 3.35 {cm}^{2}$
$≅ 3.4 {cm}^{2}$

JEE Main-26.07.2022

Units and Measurements

139954 One main scale division of a vernier callipers is a $cm$ and $n^{th}$ division of the vernier scale coincide with $(n - 1)^{th}$ division of the main scale. The least count of the callipers (in $m m$ ) is

1

\frac{10 na}{(n - 1)}

2

\frac{10 a}{(n - 1)}

3

(\frac{n - 1}{10 n}) a

4

\frac{10 a}{n}

Explanation:

D 1 MSD of vernier callipers $= a cm = 10 a mm$
$n^{th}$ division of vernier scale coincide with $(n - 1)^{th}$ division of the main scale
So, $n \times$ V.S.D. $= (n - 1) \times$ M.S.D.
$1 \times V.S.D = \frac{(n - 1)}{n} M.S.D.$
Least count $(LC) = 1$ M.S.D. -1 V.S.D.
$= 1 MSD - (\frac{n - 1}{n}) M.S.D.$
$= [1 - (\frac{n - 1}{n})] M.S.D.$
$= \frac{1}{n} MSD$
Least count $= \frac{1}{n} \times 10 a mm (∵ 1 MSD = 10 a mm)$
$LC = (\frac{10 a}{n}) mm$

JEE Main-16.03.2021

Units and Measurements

139927 Number of significant figures in expression $\frac{4.327 g}{2.51 {cm}^{3}}$ is

1 2

2 4

3 3

4 5

Explanation:

C Number of significant figures in $\frac{4.327}{2.51}$
$= 1.7239 = 1.72$
$(1, 7, 2)$ there are three significant figure

BITSAT-2019

Units and Measurements

139946 Assertion: The number of significant figures depends on the least count of measuring instrument.
Reason: Significant figures define the accuracy of measuring instrument.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2016

Units and Measurements

139953 A screw gauge of pitch $0.5 mm$ is used to measure the diameter of uniform wire of length $6.8 cm$ , the main scale reading is $1.5 mm$ and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is:
[Screw gauge has 50 divisions on the circular scale]

1

6.8 {cm}^{2}

2

3.4 {cm}^{2}

3

3.9 {cm}^{2}

4

2.4 {cm}^{2}

Explanation:

B Given that, screw gauge pitch $= 0.5 mm$
Length of wire $l = 6.8 cm$
Circular scale reading $= 7$
Least count $= \frac{0.5}{50} mm = 0.01 mm$
Diameter $(d) =$ main scale reading + circular scale reading $\times$ Least count
$d = 1.5 mm + 7 \times 0.01$
$d = 1.57 mm = 0.157 cm$
Curved surface area of wire $= π d l {cm}^{2}$
$= 3.14 \times 0.157 \times 6.8 {cm}^{2}$
$= 3.35 {cm}^{2}$
$≅ 3.4 {cm}^{2}$

JEE Main-26.07.2022

Units and Measurements

139954 One main scale division of a vernier callipers is a $cm$ and $n^{th}$ division of the vernier scale coincide with $(n - 1)^{th}$ division of the main scale. The least count of the callipers (in $m m$ ) is

1

\frac{10 na}{(n - 1)}

2

\frac{10 a}{(n - 1)}

3

(\frac{n - 1}{10 n}) a

4

\frac{10 a}{n}

Explanation:

D 1 MSD of vernier callipers $= a cm = 10 a mm$
$n^{th}$ division of vernier scale coincide with $(n - 1)^{th}$ division of the main scale
So, $n \times$ V.S.D. $= (n - 1) \times$ M.S.D.
$1 \times V.S.D = \frac{(n - 1)}{n} M.S.D.$
Least count $(LC) = 1$ M.S.D. -1 V.S.D.
$= 1 MSD - (\frac{n - 1}{n}) M.S.D.$
$= [1 - (\frac{n - 1}{n})] M.S.D.$
$= \frac{1}{n} MSD$
Least count $= \frac{1}{n} \times 10 a mm (∵ 1 MSD = 10 a mm)$
$LC = (\frac{10 a}{n}) mm$

JEE Main-16.03.2021

Units and Measurements

139927 Number of significant figures in expression $\frac{4.327 g}{2.51 {cm}^{3}}$ is

1 2

2 4

3 3

4 5

Explanation:

C Number of significant figures in $\frac{4.327}{2.51}$
$= 1.7239 = 1.72$
$(1, 7, 2)$ there are three significant figure

BITSAT-2019

Units and Measurements

139946 Assertion: The number of significant figures depends on the least count of measuring instrument.
Reason: Significant figures define the accuracy of measuring instrument.

1 If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

3 If Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

AIIMS-2016

Units and Measurements

139953 A screw gauge of pitch $0.5 mm$ is used to measure the diameter of uniform wire of length $6.8 cm$ , the main scale reading is $1.5 mm$ and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is:
[Screw gauge has 50 divisions on the circular scale]

1

6.8 {cm}^{2}

2

3.4 {cm}^{2}

3

3.9 {cm}^{2}

4

2.4 {cm}^{2}

Explanation:

B Given that, screw gauge pitch $= 0.5 mm$
Length of wire $l = 6.8 cm$
Circular scale reading $= 7$
Least count $= \frac{0.5}{50} mm = 0.01 mm$
Diameter $(d) =$ main scale reading + circular scale reading $\times$ Least count
$d = 1.5 mm + 7 \times 0.01$
$d = 1.57 mm = 0.157 cm$
Curved surface area of wire $= π d l {cm}^{2}$
$= 3.14 \times 0.157 \times 6.8 {cm}^{2}$
$= 3.35 {cm}^{2}$
$≅ 3.4 {cm}^{2}$

JEE Main-26.07.2022

Units and Measurements

139954 One main scale division of a vernier callipers is a $cm$ and $n^{th}$ division of the vernier scale coincide with $(n - 1)^{th}$ division of the main scale. The least count of the callipers (in $m m$ ) is

1

\frac{10 na}{(n - 1)}

2

\frac{10 a}{(n - 1)}

3

(\frac{n - 1}{10 n}) a

4

\frac{10 a}{n}

Explanation:

D 1 MSD of vernier callipers $= a cm = 10 a mm$
$n^{th}$ division of vernier scale coincide with $(n - 1)^{th}$ division of the main scale
So, $n \times$ V.S.D. $= (n - 1) \times$ M.S.D.
$1 \times V.S.D = \frac{(n - 1)}{n} M.S.D.$
Least count $(LC) = 1$ M.S.D. -1 V.S.D.
$= 1 MSD - (\frac{n - 1}{n}) M.S.D.$
$= [1 - (\frac{n - 1}{n})] M.S.D.$
$= \frac{1}{n} MSD$
Least count $= \frac{1}{n} \times 10 a mm (∵ 1 MSD = 10 a mm)$
$LC = (\frac{10 a}{n}) mm$

JEE Main-16.03.2021

Units and Measurements

139927 Number of significant figures in expression $\frac{4.327 g}{2.51 {cm}^{3}}$ is

1 2

2 4

3 3

4 5

Explanation:

C Number of significant figures in $\frac{4.327}{2.51}$
$= 1.7239 = 1.72$
$(1, 7, 2)$ there are three significant figure

BITSAT-2019