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PHXI02:UNITS AND MEASUREMENTS

367654 The length, breadth and thickness of a block are given by $l = 12 c m$ , $b = 6 c m$ and $t = 2.45 c m$ . The volume of the block according to the idea of significant figures should be

1

2 \times 10^{2} c m^{3}

2

1.763 \times 10^{2} c m^{3}

3

1 \times 10^{2} c m^{3}

4

N o n e o f t h e s e

Explanation:

Volume $V = l \times b \times t$
$= 12 \times 6 \times 2.45 = 176.4 c m^{3}$
$V = 1.764 \times 10^{2} c m^{3}$
Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.

PHXI02:UNITS AND MEASUREMENTS

367655 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:

$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529235}{1.592}$
$= 0.9605747 = 0.961$
(Rounding off to three significant digits)

PHXI02:UNITS AND MEASUREMENTS

367656 More the number of significant figures shows more the

1 Accuracy

2 Error

3 Number of figures

4 Value

PHXI02:UNITS AND MEASUREMENTS

367657 If $x = 0.72 + 0.8 + 3.87 - 1.089$ , then number of significant digits in the value of $x$ is

1 1

2 4

3 2

4 6

Explanation:

$x = 0.72 + 0.8 + 3.87 - 1.089$
$= 0.7 + 0.8 + 3.9 - 1.1 = 4.3$
Thus, the number of significant digit is 2.

PHXI02:UNITS AND MEASUREMENTS

367654 The length, breadth and thickness of a block are given by $l = 12 c m$ , $b = 6 c m$ and $t = 2.45 c m$ . The volume of the block according to the idea of significant figures should be

1

2 \times 10^{2} c m^{3}

2

1.763 \times 10^{2} c m^{3}

3

1 \times 10^{2} c m^{3}

4

N o n e o f t h e s e

Explanation:

Volume $V = l \times b \times t$
$= 12 \times 6 \times 2.45 = 176.4 c m^{3}$
$V = 1.764 \times 10^{2} c m^{3}$
Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.

PHXI02:UNITS AND MEASUREMENTS

367655 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:

$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529235}{1.592}$
$= 0.9605747 = 0.961$
(Rounding off to three significant digits)

PHXI02:UNITS AND MEASUREMENTS

367656 More the number of significant figures shows more the

1 Accuracy

2 Error

3 Number of figures

4 Value

PHXI02:UNITS AND MEASUREMENTS

367657 If $x = 0.72 + 0.8 + 3.87 - 1.089$ , then number of significant digits in the value of $x$ is

1 1

2 4

3 2

4 6

Explanation:

$x = 0.72 + 0.8 + 3.87 - 1.089$
$= 0.7 + 0.8 + 3.9 - 1.1 = 4.3$
Thus, the number of significant digit is 2.

PHXI02:UNITS AND MEASUREMENTS

367654 The length, breadth and thickness of a block are given by $l = 12 c m$ , $b = 6 c m$ and $t = 2.45 c m$ . The volume of the block according to the idea of significant figures should be

1

2 \times 10^{2} c m^{3}

2

1.763 \times 10^{2} c m^{3}

3

1 \times 10^{2} c m^{3}

4

N o n e o f t h e s e

Explanation:

Volume $V = l \times b \times t$
$= 12 \times 6 \times 2.45 = 176.4 c m^{3}$
$V = 1.764 \times 10^{2} c m^{3}$
Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.

PHXI02:UNITS AND MEASUREMENTS

367655 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:

$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529235}{1.592}$
$= 0.9605747 = 0.961$
(Rounding off to three significant digits)

PHXI02:UNITS AND MEASUREMENTS

367656 More the number of significant figures shows more the

1 Accuracy

2 Error

3 Number of figures

4 Value

PHXI02:UNITS AND MEASUREMENTS

367657 If $x = 0.72 + 0.8 + 3.87 - 1.089$ , then number of significant digits in the value of $x$ is

1 1

2 4

3 2

4 6

Explanation:

$x = 0.72 + 0.8 + 3.87 - 1.089$
$= 0.7 + 0.8 + 3.9 - 1.1 = 4.3$
Thus, the number of significant digit is 2.

PHXI02:UNITS AND MEASUREMENTS

367654 The length, breadth and thickness of a block are given by $l = 12 c m$ , $b = 6 c m$ and $t = 2.45 c m$ . The volume of the block according to the idea of significant figures should be

1

2 \times 10^{2} c m^{3}

2

1.763 \times 10^{2} c m^{3}

3

1 \times 10^{2} c m^{3}

4

N o n e o f t h e s e

Explanation:

Volume $V = l \times b \times t$
$= 12 \times 6 \times 2.45 = 176.4 c m^{3}$
$V = 1.764 \times 10^{2} c m^{3}$
Since, the minimum number of significant figure is one in breadth, hence volume will also contain only one significant figure.

PHXI02:UNITS AND MEASUREMENTS

367655 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:

$\frac{1.53 \times 0.9995}{1.592} = \frac{1.529235}{1.592}$
$= 0.9605747 = 0.961$
(Rounding off to three significant digits)

PHXI02:UNITS AND MEASUREMENTS

367656 More the number of significant figures shows more the

1 Accuracy

2 Error

3 Number of figures

4 Value

PHXI02:UNITS AND MEASUREMENTS

367657 If $x = 0.72 + 0.8 + 3.87 - 1.089$ , then number of significant digits in the value of $x$ is

1 1

2 4

3 2

4 6

Explanation:

$x = 0.72 + 0.8 + 3.87 - 1.089$
$= 0.7 + 0.8 + 3.9 - 1.1 = 4.3$
Thus, the number of significant digit is 2.