QBManager

Units and Measurements

139941 Find the value of $\frac{1.53 \times 0.9995}{1.592}$ with due regard for significant figures

1 0.961

2 0.123

3 0.921

4 0.913

Explanation:

A Significant figure rule
$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529}{1.592} = \frac{1.53}{1.592} = 0.961$
$(9, 6, 1)$ There are three significant figures.

BITSAT-2007

Units and Measurements

139942 How many significant figures are there in 0.30100?

1 1

2 3

3 5

4 None of these

Explanation:

C No. of significant figure in 0.30100
$= (3, 0, 1, 0, 0)$
$= (5 significant figure)$

BITSAT-2010

Units and Measurements

139945 The number of significant figures in 0.002305 is

1 6

2 4

3 7

4 2

5 3

Explanation:

B The number of significant figures in 0.002305
Digits $= (2, 3, 0, 5) = 4$ significant figures
$0 . \underset{―}{002305}$
Non-significant figure (Left zeros)

Kerala CEE - 2010

Units and Measurements

139947 The mass of a metal cube is $5.74 g$ and its volume is $1.2 {cm}^{3}$ . Then its density expressed up to appropriate significant figures will be

1

4.8 g {cm}^{- 3}

2

4.78 g {cm}^{- 3}

3

4.783 g {cm}^{- 3}

4

5.0 g {cm}^{- 3}

Explanation:

A As we know that, density $ρ = \frac{m}{V}$
Given, $m = 5.74 g, V = 1.2 {cm}^{3}$
$\begin{array}{r} ∴ ρ = \frac{5.74}{1.2} = 4.78333 g / {cm}^{3} \\ \approx 4.8 g / {cm}^{3} \end{array}$

SCRA-2013

Units and Measurements

139948 The length and breadth of a metal sheet are $3.124 m$ and $3.002 m$ respectively. The area of this sheet upto four correct significant figure is:

1

9.378 m^{2}

2

9.37 m^{2}

3

9.378248 m^{2}

4

9.3782 m^{2}

Explanation:

A Given, Length $= 3.124 m$
$Breadth = 3.002 m$
$= L \times B Area$
$= 3.124 \times 3.002 = 9.378 m^{2}$
F. measuring Instrument

AIIMS-2001

Units and Measurements

139941 Find the value of $\frac{1.53 \times 0.9995}{1.592}$ with due regard for significant figures

1 0.961

2 0.123

3 0.921

4 0.913

Explanation:

A Significant figure rule
$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529}{1.592} = \frac{1.53}{1.592} = 0.961$
$(9, 6, 1)$ There are three significant figures.

BITSAT-2007

Units and Measurements

139942 How many significant figures are there in 0.30100?

1 1

2 3

3 5

4 None of these

Explanation:

C No. of significant figure in 0.30100
$= (3, 0, 1, 0, 0)$
$= (5 significant figure)$

BITSAT-2010

Units and Measurements

139945 The number of significant figures in 0.002305 is

1 6

2 4

3 7

4 2

5 3

Explanation:

B The number of significant figures in 0.002305
Digits $= (2, 3, 0, 5) = 4$ significant figures
$0 . \underset{―}{002305}$
Non-significant figure (Left zeros)

Kerala CEE - 2010

Units and Measurements

139947 The mass of a metal cube is $5.74 g$ and its volume is $1.2 {cm}^{3}$ . Then its density expressed up to appropriate significant figures will be

1

4.8 g {cm}^{- 3}

2

4.78 g {cm}^{- 3}

3

4.783 g {cm}^{- 3}

4

5.0 g {cm}^{- 3}

Explanation:

A As we know that, density $ρ = \frac{m}{V}$
Given, $m = 5.74 g, V = 1.2 {cm}^{3}$
$\begin{array}{r} ∴ ρ = \frac{5.74}{1.2} = 4.78333 g / {cm}^{3} \\ \approx 4.8 g / {cm}^{3} \end{array}$

SCRA-2013

Units and Measurements

139948 The length and breadth of a metal sheet are $3.124 m$ and $3.002 m$ respectively. The area of this sheet upto four correct significant figure is:

1

9.378 m^{2}

2

9.37 m^{2}

3

9.378248 m^{2}

4

9.3782 m^{2}

Explanation:

A Given, Length $= 3.124 m$
$Breadth = 3.002 m$
$= L \times B Area$
$= 3.124 \times 3.002 = 9.378 m^{2}$
F. measuring Instrument

AIIMS-2001

Units and Measurements

139941 Find the value of $\frac{1.53 \times 0.9995}{1.592}$ with due regard for significant figures

1 0.961

2 0.123

3 0.921

4 0.913

Explanation:

A Significant figure rule
$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529}{1.592} = \frac{1.53}{1.592} = 0.961$
$(9, 6, 1)$ There are three significant figures.

BITSAT-2007

Units and Measurements

139942 How many significant figures are there in 0.30100?

1 1

2 3

3 5

4 None of these

Explanation:

C No. of significant figure in 0.30100
$= (3, 0, 1, 0, 0)$
$= (5 significant figure)$

BITSAT-2010

Units and Measurements

139945 The number of significant figures in 0.002305 is

1 6

2 4

3 7

4 2

5 3

Explanation:

B The number of significant figures in 0.002305
Digits $= (2, 3, 0, 5) = 4$ significant figures
$0 . \underset{―}{002305}$
Non-significant figure (Left zeros)

Kerala CEE - 2010

Units and Measurements

139947 The mass of a metal cube is $5.74 g$ and its volume is $1.2 {cm}^{3}$ . Then its density expressed up to appropriate significant figures will be

1

4.8 g {cm}^{- 3}

2

4.78 g {cm}^{- 3}

3

4.783 g {cm}^{- 3}

4

5.0 g {cm}^{- 3}

Explanation:

A As we know that, density $ρ = \frac{m}{V}$
Given, $m = 5.74 g, V = 1.2 {cm}^{3}$
$\begin{array}{r} ∴ ρ = \frac{5.74}{1.2} = 4.78333 g / {cm}^{3} \\ \approx 4.8 g / {cm}^{3} \end{array}$

SCRA-2013

Units and Measurements

139948 The length and breadth of a metal sheet are $3.124 m$ and $3.002 m$ respectively. The area of this sheet upto four correct significant figure is:

1

9.378 m^{2}

2

9.37 m^{2}

3

9.378248 m^{2}

4

9.3782 m^{2}

Explanation:

A Given, Length $= 3.124 m$
$Breadth = 3.002 m$
$= L \times B Area$
$= 3.124 \times 3.002 = 9.378 m^{2}$
F. measuring Instrument

AIIMS-2001

Units and Measurements

139941 Find the value of $\frac{1.53 \times 0.9995}{1.592}$ with due regard for significant figures

1 0.961

2 0.123

3 0.921

4 0.913

Explanation:

A Significant figure rule
$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529}{1.592} = \frac{1.53}{1.592} = 0.961$
$(9, 6, 1)$ There are three significant figures.

BITSAT-2007

Units and Measurements

139942 How many significant figures are there in 0.30100?

1 1

2 3

3 5

4 None of these

Explanation:

C No. of significant figure in 0.30100
$= (3, 0, 1, 0, 0)$
$= (5 significant figure)$

BITSAT-2010

Units and Measurements

139945 The number of significant figures in 0.002305 is

1 6

2 4

3 7

4 2

5 3

Explanation:

B The number of significant figures in 0.002305
Digits $= (2, 3, 0, 5) = 4$ significant figures
$0 . \underset{―}{002305}$
Non-significant figure (Left zeros)

Kerala CEE - 2010

Units and Measurements

139947 The mass of a metal cube is $5.74 g$ and its volume is $1.2 {cm}^{3}$ . Then its density expressed up to appropriate significant figures will be

1

4.8 g {cm}^{- 3}

2

4.78 g {cm}^{- 3}

3

4.783 g {cm}^{- 3}

4

5.0 g {cm}^{- 3}

Explanation:

A As we know that, density $ρ = \frac{m}{V}$
Given, $m = 5.74 g, V = 1.2 {cm}^{3}$
$\begin{array}{r} ∴ ρ = \frac{5.74}{1.2} = 4.78333 g / {cm}^{3} \\ \approx 4.8 g / {cm}^{3} \end{array}$

SCRA-2013

Units and Measurements

139948 The length and breadth of a metal sheet are $3.124 m$ and $3.002 m$ respectively. The area of this sheet upto four correct significant figure is:

1

9.378 m^{2}

2

9.37 m^{2}

3

9.378248 m^{2}

4

9.3782 m^{2}

Explanation:

A Given, Length $= 3.124 m$
$Breadth = 3.002 m$
$= L \times B Area$
$= 3.124 \times 3.002 = 9.378 m^{2}$
F. measuring Instrument

AIIMS-2001

Units and Measurements

139941 Find the value of $\frac{1.53 \times 0.9995}{1.592}$ with due regard for significant figures

1 0.961

2 0.123

3 0.921

4 0.913

Explanation:

A Significant figure rule
$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529}{1.592} = \frac{1.53}{1.592} = 0.961$
$(9, 6, 1)$ There are three significant figures.

BITSAT-2007

Units and Measurements

139942 How many significant figures are there in 0.30100?

1 1

2 3

3 5

4 None of these

Explanation:

C No. of significant figure in 0.30100
$= (3, 0, 1, 0, 0)$
$= (5 significant figure)$

BITSAT-2010

Units and Measurements

139945 The number of significant figures in 0.002305 is

1 6

2 4

3 7

4 2

5 3

Explanation:

B The number of significant figures in 0.002305
Digits $= (2, 3, 0, 5) = 4$ significant figures
$0 . \underset{―}{002305}$
Non-significant figure (Left zeros)

Kerala CEE - 2010

Units and Measurements

139947 The mass of a metal cube is $5.74 g$ and its volume is $1.2 {cm}^{3}$ . Then its density expressed up to appropriate significant figures will be

1

4.8 g {cm}^{- 3}

2

4.78 g {cm}^{- 3}

3

4.783 g {cm}^{- 3}

4

5.0 g {cm}^{- 3}

Explanation:

A As we know that, density $ρ = \frac{m}{V}$
Given, $m = 5.74 g, V = 1.2 {cm}^{3}$
$\begin{array}{r} ∴ ρ = \frac{5.74}{1.2} = 4.78333 g / {cm}^{3} \\ \approx 4.8 g / {cm}^{3} \end{array}$

SCRA-2013

Units and Measurements

139948 The length and breadth of a metal sheet are $3.124 m$ and $3.002 m$ respectively. The area of this sheet upto four correct significant figure is:

1

9.378 m^{2}

2

9.37 m^{2}

3

9.378248 m^{2}

4

9.3782 m^{2}

Explanation:

A Given, Length $= 3.124 m$
$Breadth = 3.002 m$
$= L \times B Area$
$= 3.124 \times 3.002 = 9.378 m^{2}$
F. measuring Instrument

AIIMS-2001

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