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PHXI02:UNITS AND MEASUREMENTS

367499 The period of oscillation of a simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ where $l$ is about 100 $c m$ and is known to have 1 $m m$ accuracy. The period is about 2 $s$ . The time of 100 oscillations is measured by a stop watch of least count 0.1 sec. The percentage error in $g$ is

1

0.8 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

$T = 2 π \sqrt{l / g} \Rightarrow T^{2} = 4 π^{2} l / g \Rightarrow g = \frac{4 π^{2} l}{T^{2}}$
Here % error in
$l = \frac{1 m m}{100 c m} \times 100 = \frac{0.1}{100} \times 100 = 0.1 %$
and % error in $T = \frac{0.1}{2 \times 100} \times 100 = 0.05 %$
$∴ %$ error in $g = %$ error in $l + 2$ (% error in $T$ )
$= 0.2 %$

PHXI02:UNITS AND MEASUREMENTS

367500 The length and breadth of a rectangular sheet are 16.2 $c m$ and 10.1 $c m$ , respectively. The area of the sheet in appropriate significant figures and error is

1

163.62 \pm 2.6 c m^{2}

2

164 \pm 3 c m^{2}

3

163.62 \pm 3 c m^{2}

4

163.6 \pm 2.6 c m^{2}

Explanation:

Given that
$l = (16.2 \pm 0.1) c m$
Breadth $b = (10.1 \pm 0.1) c m$
$Area A = l \times b = 16.2 \times 10.1 = 163.62 c m^{2}$
Rounding off to three significant digits, area $A = 164 c m^{2}$
$\frac{Δ A}{A} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{16.2} + \frac{0.1}{10.1}$
$= \frac{1.01 + 1.62}{16.2 \times 10.1} = \frac{2.63}{163.62}$
$\Rightarrow Δ A = A \times \frac{2.63}{163.62}$
$= 163.62 \times \frac{2.63}{163.62} = 2.63 c m^{2}$
$Δ A = 3 c m^{2}$ (By rounding off to one significant figure)
Area, $A = A \pm Δ A = (164 \pm 3) c m^{2}$

PHXI02:UNITS AND MEASUREMENTS

367501 If $X = A \times B, Δ X, Δ A$ and $Δ B$ are maximum absolute errors in $X$ , $A$ and $B$ respectively, then the maximum fractional error in $X$ is given by

1

Δ X = Δ A - Δ B

2

\frac{Δ X}{X} = \frac{Δ A}{A} + \frac{Δ B}{B}

3

Δ X = Δ A + Δ B

4

\frac{Δ X}{X} = \frac{Δ A}{A} - \frac{Δ B}{B}

Explanation:

When two quantities are multiplied, their maximum relative errors are added up
hence: $\frac{Δ X}{X} = \frac{Δ A}{A} + \frac{Δ B}{B}$

PHXI02:UNITS AND MEASUREMENTS

367502 To estimate ‘ $g$ ’ $(from g = 4 π^{2} \frac{L}{T^{2}}),$ error in measurement of $L$ is $\pm 2 %$ and error in measurement of $T$ is $\pm 3 %$ . The error in estimated ‘ $g$ ’ will be

1

\pm 3 %

2

\pm 5 %

3

\pm 8 %

4

\pm 6 %

Explanation:

$g = 4 π^{2} \frac{ℓ}{T^{2}}$
$\frac{Δ ℓ}{ℓ} = 2 % = \pm 2 \times 10^{- 2}$
$\frac{Δ T}{T} = \pm 2 % = \pm 3 \times 10^{- 2}$
$\Rightarrow \frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + \frac{2 Δ T}{T} = 2 \times 10^{- 2} + 2 \times 3 \times 10^{- 2}$
$= 8 \times 10^{- 2} = \pm 8 %$

PHXI02:UNITS AND MEASUREMENTS

367499 The period of oscillation of a simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ where $l$ is about 100 $c m$ and is known to have 1 $m m$ accuracy. The period is about 2 $s$ . The time of 100 oscillations is measured by a stop watch of least count 0.1 sec. The percentage error in $g$ is

1

0.8 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

$T = 2 π \sqrt{l / g} \Rightarrow T^{2} = 4 π^{2} l / g \Rightarrow g = \frac{4 π^{2} l}{T^{2}}$
Here % error in
$l = \frac{1 m m}{100 c m} \times 100 = \frac{0.1}{100} \times 100 = 0.1 %$
and % error in $T = \frac{0.1}{2 \times 100} \times 100 = 0.05 %$
$∴ %$ error in $g = %$ error in $l + 2$ (% error in $T$ )
$= 0.2 %$

PHXI02:UNITS AND MEASUREMENTS

367500 The length and breadth of a rectangular sheet are 16.2 $c m$ and 10.1 $c m$ , respectively. The area of the sheet in appropriate significant figures and error is

1

163.62 \pm 2.6 c m^{2}

2

164 \pm 3 c m^{2}

3

163.62 \pm 3 c m^{2}

4

163.6 \pm 2.6 c m^{2}

Explanation:

Given that
$l = (16.2 \pm 0.1) c m$
Breadth $b = (10.1 \pm 0.1) c m$
$Area A = l \times b = 16.2 \times 10.1 = 163.62 c m^{2}$
Rounding off to three significant digits, area $A = 164 c m^{2}$
$\frac{Δ A}{A} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{16.2} + \frac{0.1}{10.1}$
$= \frac{1.01 + 1.62}{16.2 \times 10.1} = \frac{2.63}{163.62}$
$\Rightarrow Δ A = A \times \frac{2.63}{163.62}$
$= 163.62 \times \frac{2.63}{163.62} = 2.63 c m^{2}$
$Δ A = 3 c m^{2}$ (By rounding off to one significant figure)
Area, $A = A \pm Δ A = (164 \pm 3) c m^{2}$

PHXI02:UNITS AND MEASUREMENTS

367501 If $X = A \times B, Δ X, Δ A$ and $Δ B$ are maximum absolute errors in $X$ , $A$ and $B$ respectively, then the maximum fractional error in $X$ is given by

1

Δ X = Δ A - Δ B

2

\frac{Δ X}{X} = \frac{Δ A}{A} + \frac{Δ B}{B}

3

Δ X = Δ A + Δ B

4

\frac{Δ X}{X} = \frac{Δ A}{A} - \frac{Δ B}{B}

Explanation:

When two quantities are multiplied, their maximum relative errors are added up
hence: $\frac{Δ X}{X} = \frac{Δ A}{A} + \frac{Δ B}{B}$

PHXI02:UNITS AND MEASUREMENTS

367502 To estimate ‘ $g$ ’ $(from g = 4 π^{2} \frac{L}{T^{2}}),$ error in measurement of $L$ is $\pm 2 %$ and error in measurement of $T$ is $\pm 3 %$ . The error in estimated ‘ $g$ ’ will be

1

\pm 3 %

2

\pm 5 %

3

\pm 8 %

4

\pm 6 %

Explanation:

$g = 4 π^{2} \frac{ℓ}{T^{2}}$
$\frac{Δ ℓ}{ℓ} = 2 % = \pm 2 \times 10^{- 2}$
$\frac{Δ T}{T} = \pm 2 % = \pm 3 \times 10^{- 2}$
$\Rightarrow \frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + \frac{2 Δ T}{T} = 2 \times 10^{- 2} + 2 \times 3 \times 10^{- 2}$
$= 8 \times 10^{- 2} = \pm 8 %$

PHXI02:UNITS AND MEASUREMENTS

367499 The period of oscillation of a simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ where $l$ is about 100 $c m$ and is known to have 1 $m m$ accuracy. The period is about 2 $s$ . The time of 100 oscillations is measured by a stop watch of least count 0.1 sec. The percentage error in $g$ is

1

0.8 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

$T = 2 π \sqrt{l / g} \Rightarrow T^{2} = 4 π^{2} l / g \Rightarrow g = \frac{4 π^{2} l}{T^{2}}$
Here % error in
$l = \frac{1 m m}{100 c m} \times 100 = \frac{0.1}{100} \times 100 = 0.1 %$
and % error in $T = \frac{0.1}{2 \times 100} \times 100 = 0.05 %$
$∴ %$ error in $g = %$ error in $l + 2$ (% error in $T$ )
$= 0.2 %$

PHXI02:UNITS AND MEASUREMENTS

367500 The length and breadth of a rectangular sheet are 16.2 $c m$ and 10.1 $c m$ , respectively. The area of the sheet in appropriate significant figures and error is

1

163.62 \pm 2.6 c m^{2}

2

164 \pm 3 c m^{2}

3

163.62 \pm 3 c m^{2}

4

163.6 \pm 2.6 c m^{2}

Explanation:

Given that
$l = (16.2 \pm 0.1) c m$
Breadth $b = (10.1 \pm 0.1) c m$
$Area A = l \times b = 16.2 \times 10.1 = 163.62 c m^{2}$
Rounding off to three significant digits, area $A = 164 c m^{2}$
$\frac{Δ A}{A} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{16.2} + \frac{0.1}{10.1}$
$= \frac{1.01 + 1.62}{16.2 \times 10.1} = \frac{2.63}{163.62}$
$\Rightarrow Δ A = A \times \frac{2.63}{163.62}$
$= 163.62 \times \frac{2.63}{163.62} = 2.63 c m^{2}$
$Δ A = 3 c m^{2}$ (By rounding off to one significant figure)
Area, $A = A \pm Δ A = (164 \pm 3) c m^{2}$

PHXI02:UNITS AND MEASUREMENTS

367501 If $X = A \times B, Δ X, Δ A$ and $Δ B$ are maximum absolute errors in $X$ , $A$ and $B$ respectively, then the maximum fractional error in $X$ is given by

1

Δ X = Δ A - Δ B

2

\frac{Δ X}{X} = \frac{Δ A}{A} + \frac{Δ B}{B}

3

Δ X = Δ A + Δ B

4

\frac{Δ X}{X} = \frac{Δ A}{A} - \frac{Δ B}{B}

Explanation:

When two quantities are multiplied, their maximum relative errors are added up
hence: $\frac{Δ X}{X} = \frac{Δ A}{A} + \frac{Δ B}{B}$

PHXI02:UNITS AND MEASUREMENTS

367502 To estimate ‘ $g$ ’ $(from g = 4 π^{2} \frac{L}{T^{2}}),$ error in measurement of $L$ is $\pm 2 %$ and error in measurement of $T$ is $\pm 3 %$ . The error in estimated ‘ $g$ ’ will be

1

\pm 3 %

2

\pm 5 %

3

\pm 8 %

4

\pm 6 %

Explanation:

$g = 4 π^{2} \frac{ℓ}{T^{2}}$
$\frac{Δ ℓ}{ℓ} = 2 % = \pm 2 \times 10^{- 2}$
$\frac{Δ T}{T} = \pm 2 % = \pm 3 \times 10^{- 2}$
$\Rightarrow \frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + \frac{2 Δ T}{T} = 2 \times 10^{- 2} + 2 \times 3 \times 10^{- 2}$
$= 8 \times 10^{- 2} = \pm 8 %$

PHXI02:UNITS AND MEASUREMENTS

367499 The period of oscillation of a simple pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$ where $l$ is about 100 $c m$ and is known to have 1 $m m$ accuracy. The period is about 2 $s$ . The time of 100 oscillations is measured by a stop watch of least count 0.1 sec. The percentage error in $g$ is

1

0.8 %

2

0.2 %

3

0.1 %

4

1 %

Explanation:

$T = 2 π \sqrt{l / g} \Rightarrow T^{2} = 4 π^{2} l / g \Rightarrow g = \frac{4 π^{2} l}{T^{2}}$
Here % error in
$l = \frac{1 m m}{100 c m} \times 100 = \frac{0.1}{100} \times 100 = 0.1 %$
and % error in $T = \frac{0.1}{2 \times 100} \times 100 = 0.05 %$
$∴ %$ error in $g = %$ error in $l + 2$ (% error in $T$ )
$= 0.2 %$

PHXI02:UNITS AND MEASUREMENTS

367500 The length and breadth of a rectangular sheet are 16.2 $c m$ and 10.1 $c m$ , respectively. The area of the sheet in appropriate significant figures and error is

1

163.62 \pm 2.6 c m^{2}

2

164 \pm 3 c m^{2}

3

163.62 \pm 3 c m^{2}

4

163.6 \pm 2.6 c m^{2}

Explanation:

Given that
$l = (16.2 \pm 0.1) c m$
Breadth $b = (10.1 \pm 0.1) c m$
$Area A = l \times b = 16.2 \times 10.1 = 163.62 c m^{2}$
Rounding off to three significant digits, area $A = 164 c m^{2}$
$\frac{Δ A}{A} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{16.2} + \frac{0.1}{10.1}$
$= \frac{1.01 + 1.62}{16.2 \times 10.1} = \frac{2.63}{163.62}$
$\Rightarrow Δ A = A \times \frac{2.63}{163.62}$
$= 163.62 \times \frac{2.63}{163.62} = 2.63 c m^{2}$
$Δ A = 3 c m^{2}$ (By rounding off to one significant figure)
Area, $A = A \pm Δ A = (164 \pm 3) c m^{2}$

PHXI02:UNITS AND MEASUREMENTS

367501 If $X = A \times B, Δ X, Δ A$ and $Δ B$ are maximum absolute errors in $X$ , $A$ and $B$ respectively, then the maximum fractional error in $X$ is given by

1

Δ X = Δ A - Δ B

2

\frac{Δ X}{X} = \frac{Δ A}{A} + \frac{Δ B}{B}

3

Δ X = Δ A + Δ B

4

\frac{Δ X}{X} = \frac{Δ A}{A} - \frac{Δ B}{B}

Explanation:

When two quantities are multiplied, their maximum relative errors are added up
hence: $\frac{Δ X}{X} = \frac{Δ A}{A} + \frac{Δ B}{B}$

PHXI02:UNITS AND MEASUREMENTS

367502 To estimate ‘ $g$ ’ $(from g = 4 π^{2} \frac{L}{T^{2}}),$ error in measurement of $L$ is $\pm 2 %$ and error in measurement of $T$ is $\pm 3 %$ . The error in estimated ‘ $g$ ’ will be

1

\pm 3 %

2

\pm 5 %

3

\pm 8 %

4

\pm 6 %

Explanation:

$g = 4 π^{2} \frac{ℓ}{T^{2}}$
$\frac{Δ ℓ}{ℓ} = 2 % = \pm 2 \times 10^{- 2}$
$\frac{Δ T}{T} = \pm 2 % = \pm 3 \times 10^{- 2}$
$\Rightarrow \frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + \frac{2 Δ T}{T} = 2 \times 10^{- 2} + 2 \times 3 \times 10^{- 2}$
$= 8 \times 10^{- 2} = \pm 8 %$

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