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PHXI02:UNITS AND MEASUREMENTS

367494 The length and breadth of a rectangular sheet have been measured by a measuring scale to an accuracy of $0.1 c m$ . If the measured values are $35.2 c m a n d 21.7 c m$ , respectively, then find the relative error in the area of the sheet.

1

\pm 8 \times 10^{- 2}

2

\pm 4 \times 10^{- 3}

3

\pm 7.45 \times 10^{- 3}

4

\pm 9 \times 10^{- 6}

Explanation:

Here, length of sheet
$l = (35.2 \pm 0.1) c m a n d b r e a d t h o f s h e e t$
$b = (21.7 \pm 0.1) c m$
$∴ R e l a t i v e e r r o r i n s u r f a c e a r e a o f t h e s h e e t,$
${(\frac{Δ s}{s})}_{max} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{35.2} + \frac{0.1}{21.7}$
$= 2.841 \times 10^{- 3} + 4.608 \times 10^{- 3}$
$= 7.449 \times 10^{- 3} = \pm 7.45 \times 10^{- 3}$

PHXI02:UNITS AND MEASUREMENTS

367495 Estimate the percentage error in the determination of density of a material taken in the form of a cylinder with 2 % error in mass , 1 % error in radius and 3 % error in height .

1

3 %

2

4 %

3

7 %

4

5 %

Explanation:

Density of material of a cylinder,
$ρ = \frac{M a s s}{V o l u m e} = \frac{M}{π r^{2} h}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ M}{M} + 2 \frac{Δ r}{r} + \frac{Δ h}{h}) \times 100$
$= 2 % + 2 \times 1 % + 3 % = 7 %$

PHXI02:UNITS AND MEASUREMENTS

367496 The percentage errors in the measurement of mass and speed are $1 %$ and $2 %$ respectively. What is the percentage error in the kinetic energy

1

5 %

2

2.5 %

3

3 %

4

1.5 %

Explanation:

$K = \frac{1}{2} m v^{2}$
$∴ \frac{Δ K}{K} = \frac{Δ m}{m} + \frac{2 Δ v}{v}$
$%$ error in $K = (%$ error in $m) + 2 \times (%$ error in $v)$
$= 1 % + 2 \times 2 % = 5 %$ .
So correct option is (1)

PHXI02:UNITS AND MEASUREMENTS

367497 A person measures two quantities as $A = 1.0 m \pm 0.2 m,$
$B = 2.0 m \pm 0.2 m .$ We should report correct value for $\sqrt{A B}$ as

1

1.4 m + 0.3 m

2

1.4 m \pm 0.2 m

3

1.4 m \pm 0.4 m

4

1.41 m \pm 0.15 m

Explanation:

Given, $A = 1.0 m \pm 0.2 m, B = 2.0 \pm 0.2 m$
Let, $Y = \sqrt{A B} = \sqrt{(1.0) (2.0)} = 1.414 m$
Rounding off to two significant digit $Y = 1.4 m$
$\frac{Δ Y}{Y} = \frac{1}{2} [\frac{Δ A}{A} + \frac{Δ B}{B}]$
$= \frac{1}{2} [\frac{0.2}{1.0} + \frac{0.2}{2.0}] = \frac{0.6}{2 \times 2.0}$
$\Rightarrow Δ Y = \frac{0.6 Y}{2 \times 2.0} = \frac{0.6 \times 1.4}{2 \times 20} = 0.212$
Rounding off to one significant digit
$Δ Y = 0.2 m$
Thus, correct value for
$\sqrt{A B} = Y + Δ Y = 1.4 \pm 0.2 m$

PHXI02:UNITS AND MEASUREMENTS

367498 The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) V$ and $I = (20 \pm 0.2)$ $A$ , the percentage error in the measurement of $R$ is

1

5.5 %

2

7 %

3

3 %

4

3.5 %

Explanation:

As, $R = \frac{V}{I}$
To find $%$ error in $R$ ;
$\frac{Δ R}{R} \times 100 % = \frac{Δ V}{V} \times 100 % + \frac{Δ I}{I} \times 100 %$
$= \frac{5}{200} \times 100 + \frac{0.2}{20} \times 100 = 3.5 %$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367494 The length and breadth of a rectangular sheet have been measured by a measuring scale to an accuracy of $0.1 c m$ . If the measured values are $35.2 c m a n d 21.7 c m$ , respectively, then find the relative error in the area of the sheet.

1

\pm 8 \times 10^{- 2}

2

\pm 4 \times 10^{- 3}

3

\pm 7.45 \times 10^{- 3}

4

\pm 9 \times 10^{- 6}

Explanation:

Here, length of sheet
$l = (35.2 \pm 0.1) c m a n d b r e a d t h o f s h e e t$
$b = (21.7 \pm 0.1) c m$
$∴ R e l a t i v e e r r o r i n s u r f a c e a r e a o f t h e s h e e t,$
${(\frac{Δ s}{s})}_{max} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{35.2} + \frac{0.1}{21.7}$
$= 2.841 \times 10^{- 3} + 4.608 \times 10^{- 3}$
$= 7.449 \times 10^{- 3} = \pm 7.45 \times 10^{- 3}$

PHXI02:UNITS AND MEASUREMENTS

367495 Estimate the percentage error in the determination of density of a material taken in the form of a cylinder with 2 % error in mass , 1 % error in radius and 3 % error in height .

1

3 %

2

4 %

3

7 %

4

5 %

Explanation:

Density of material of a cylinder,
$ρ = \frac{M a s s}{V o l u m e} = \frac{M}{π r^{2} h}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ M}{M} + 2 \frac{Δ r}{r} + \frac{Δ h}{h}) \times 100$
$= 2 % + 2 \times 1 % + 3 % = 7 %$

PHXI02:UNITS AND MEASUREMENTS

367496 The percentage errors in the measurement of mass and speed are $1 %$ and $2 %$ respectively. What is the percentage error in the kinetic energy

1

5 %

2

2.5 %

3

3 %

4

1.5 %

Explanation:

$K = \frac{1}{2} m v^{2}$
$∴ \frac{Δ K}{K} = \frac{Δ m}{m} + \frac{2 Δ v}{v}$
$%$ error in $K = (%$ error in $m) + 2 \times (%$ error in $v)$
$= 1 % + 2 \times 2 % = 5 %$ .
So correct option is (1)

PHXI02:UNITS AND MEASUREMENTS

367497 A person measures two quantities as $A = 1.0 m \pm 0.2 m,$
$B = 2.0 m \pm 0.2 m .$ We should report correct value for $\sqrt{A B}$ as

1

1.4 m + 0.3 m

2

1.4 m \pm 0.2 m

3

1.4 m \pm 0.4 m

4

1.41 m \pm 0.15 m

Explanation:

Given, $A = 1.0 m \pm 0.2 m, B = 2.0 \pm 0.2 m$
Let, $Y = \sqrt{A B} = \sqrt{(1.0) (2.0)} = 1.414 m$
Rounding off to two significant digit $Y = 1.4 m$
$\frac{Δ Y}{Y} = \frac{1}{2} [\frac{Δ A}{A} + \frac{Δ B}{B}]$
$= \frac{1}{2} [\frac{0.2}{1.0} + \frac{0.2}{2.0}] = \frac{0.6}{2 \times 2.0}$
$\Rightarrow Δ Y = \frac{0.6 Y}{2 \times 2.0} = \frac{0.6 \times 1.4}{2 \times 20} = 0.212$
Rounding off to one significant digit
$Δ Y = 0.2 m$
Thus, correct value for
$\sqrt{A B} = Y + Δ Y = 1.4 \pm 0.2 m$

PHXI02:UNITS AND MEASUREMENTS

367498 The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) V$ and $I = (20 \pm 0.2)$ $A$ , the percentage error in the measurement of $R$ is

1

5.5 %

2

7 %

3

3 %

4

3.5 %

Explanation:

As, $R = \frac{V}{I}$
To find $%$ error in $R$ ;
$\frac{Δ R}{R} \times 100 % = \frac{Δ V}{V} \times 100 % + \frac{Δ I}{I} \times 100 %$
$= \frac{5}{200} \times 100 + \frac{0.2}{20} \times 100 = 3.5 %$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367494 The length and breadth of a rectangular sheet have been measured by a measuring scale to an accuracy of $0.1 c m$ . If the measured values are $35.2 c m a n d 21.7 c m$ , respectively, then find the relative error in the area of the sheet.

1

\pm 8 \times 10^{- 2}

2

\pm 4 \times 10^{- 3}

3

\pm 7.45 \times 10^{- 3}

4

\pm 9 \times 10^{- 6}

Explanation:

Here, length of sheet
$l = (35.2 \pm 0.1) c m a n d b r e a d t h o f s h e e t$
$b = (21.7 \pm 0.1) c m$
$∴ R e l a t i v e e r r o r i n s u r f a c e a r e a o f t h e s h e e t,$
${(\frac{Δ s}{s})}_{max} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{35.2} + \frac{0.1}{21.7}$
$= 2.841 \times 10^{- 3} + 4.608 \times 10^{- 3}$
$= 7.449 \times 10^{- 3} = \pm 7.45 \times 10^{- 3}$

PHXI02:UNITS AND MEASUREMENTS

367495 Estimate the percentage error in the determination of density of a material taken in the form of a cylinder with 2 % error in mass , 1 % error in radius and 3 % error in height .

1

3 %

2

4 %

3

7 %

4

5 %

Explanation:

Density of material of a cylinder,
$ρ = \frac{M a s s}{V o l u m e} = \frac{M}{π r^{2} h}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ M}{M} + 2 \frac{Δ r}{r} + \frac{Δ h}{h}) \times 100$
$= 2 % + 2 \times 1 % + 3 % = 7 %$

PHXI02:UNITS AND MEASUREMENTS

367496 The percentage errors in the measurement of mass and speed are $1 %$ and $2 %$ respectively. What is the percentage error in the kinetic energy

1

5 %

2

2.5 %

3

3 %

4

1.5 %

Explanation:

$K = \frac{1}{2} m v^{2}$
$∴ \frac{Δ K}{K} = \frac{Δ m}{m} + \frac{2 Δ v}{v}$
$%$ error in $K = (%$ error in $m) + 2 \times (%$ error in $v)$
$= 1 % + 2 \times 2 % = 5 %$ .
So correct option is (1)

PHXI02:UNITS AND MEASUREMENTS

367497 A person measures two quantities as $A = 1.0 m \pm 0.2 m,$
$B = 2.0 m \pm 0.2 m .$ We should report correct value for $\sqrt{A B}$ as

1

1.4 m + 0.3 m

2

1.4 m \pm 0.2 m

3

1.4 m \pm 0.4 m

4

1.41 m \pm 0.15 m

Explanation:

Given, $A = 1.0 m \pm 0.2 m, B = 2.0 \pm 0.2 m$
Let, $Y = \sqrt{A B} = \sqrt{(1.0) (2.0)} = 1.414 m$
Rounding off to two significant digit $Y = 1.4 m$
$\frac{Δ Y}{Y} = \frac{1}{2} [\frac{Δ A}{A} + \frac{Δ B}{B}]$
$= \frac{1}{2} [\frac{0.2}{1.0} + \frac{0.2}{2.0}] = \frac{0.6}{2 \times 2.0}$
$\Rightarrow Δ Y = \frac{0.6 Y}{2 \times 2.0} = \frac{0.6 \times 1.4}{2 \times 20} = 0.212$
Rounding off to one significant digit
$Δ Y = 0.2 m$
Thus, correct value for
$\sqrt{A B} = Y + Δ Y = 1.4 \pm 0.2 m$

PHXI02:UNITS AND MEASUREMENTS

367498 The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) V$ and $I = (20 \pm 0.2)$ $A$ , the percentage error in the measurement of $R$ is

1

5.5 %

2

7 %

3

3 %

4

3.5 %

Explanation:

As, $R = \frac{V}{I}$
To find $%$ error in $R$ ;
$\frac{Δ R}{R} \times 100 % = \frac{Δ V}{V} \times 100 % + \frac{Δ I}{I} \times 100 %$
$= \frac{5}{200} \times 100 + \frac{0.2}{20} \times 100 = 3.5 %$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367494 The length and breadth of a rectangular sheet have been measured by a measuring scale to an accuracy of $0.1 c m$ . If the measured values are $35.2 c m a n d 21.7 c m$ , respectively, then find the relative error in the area of the sheet.

1

\pm 8 \times 10^{- 2}

2

\pm 4 \times 10^{- 3}

3

\pm 7.45 \times 10^{- 3}

4

\pm 9 \times 10^{- 6}

Explanation:

Here, length of sheet
$l = (35.2 \pm 0.1) c m a n d b r e a d t h o f s h e e t$
$b = (21.7 \pm 0.1) c m$
$∴ R e l a t i v e e r r o r i n s u r f a c e a r e a o f t h e s h e e t,$
${(\frac{Δ s}{s})}_{max} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{35.2} + \frac{0.1}{21.7}$
$= 2.841 \times 10^{- 3} + 4.608 \times 10^{- 3}$
$= 7.449 \times 10^{- 3} = \pm 7.45 \times 10^{- 3}$

PHXI02:UNITS AND MEASUREMENTS

367495 Estimate the percentage error in the determination of density of a material taken in the form of a cylinder with 2 % error in mass , 1 % error in radius and 3 % error in height .

1

3 %

2

4 %

3

7 %

4

5 %

Explanation:

Density of material of a cylinder,
$ρ = \frac{M a s s}{V o l u m e} = \frac{M}{π r^{2} h}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ M}{M} + 2 \frac{Δ r}{r} + \frac{Δ h}{h}) \times 100$
$= 2 % + 2 \times 1 % + 3 % = 7 %$

PHXI02:UNITS AND MEASUREMENTS

367496 The percentage errors in the measurement of mass and speed are $1 %$ and $2 %$ respectively. What is the percentage error in the kinetic energy

1

5 %

2

2.5 %

3

3 %

4

1.5 %

Explanation:

$K = \frac{1}{2} m v^{2}$
$∴ \frac{Δ K}{K} = \frac{Δ m}{m} + \frac{2 Δ v}{v}$
$%$ error in $K = (%$ error in $m) + 2 \times (%$ error in $v)$
$= 1 % + 2 \times 2 % = 5 %$ .
So correct option is (1)

PHXI02:UNITS AND MEASUREMENTS

367497 A person measures two quantities as $A = 1.0 m \pm 0.2 m,$
$B = 2.0 m \pm 0.2 m .$ We should report correct value for $\sqrt{A B}$ as

1

1.4 m + 0.3 m

2

1.4 m \pm 0.2 m

3

1.4 m \pm 0.4 m

4

1.41 m \pm 0.15 m

Explanation:

Given, $A = 1.0 m \pm 0.2 m, B = 2.0 \pm 0.2 m$
Let, $Y = \sqrt{A B} = \sqrt{(1.0) (2.0)} = 1.414 m$
Rounding off to two significant digit $Y = 1.4 m$
$\frac{Δ Y}{Y} = \frac{1}{2} [\frac{Δ A}{A} + \frac{Δ B}{B}]$
$= \frac{1}{2} [\frac{0.2}{1.0} + \frac{0.2}{2.0}] = \frac{0.6}{2 \times 2.0}$
$\Rightarrow Δ Y = \frac{0.6 Y}{2 \times 2.0} = \frac{0.6 \times 1.4}{2 \times 20} = 0.212$
Rounding off to one significant digit
$Δ Y = 0.2 m$
Thus, correct value for
$\sqrt{A B} = Y + Δ Y = 1.4 \pm 0.2 m$

PHXI02:UNITS AND MEASUREMENTS

367498 The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) V$ and $I = (20 \pm 0.2)$ $A$ , the percentage error in the measurement of $R$ is

1

5.5 %

2

7 %

3

3 %

4

3.5 %

Explanation:

As, $R = \frac{V}{I}$
To find $%$ error in $R$ ;
$\frac{Δ R}{R} \times 100 % = \frac{Δ V}{V} \times 100 % + \frac{Δ I}{I} \times 100 %$
$= \frac{5}{200} \times 100 + \frac{0.2}{20} \times 100 = 3.5 %$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367494 The length and breadth of a rectangular sheet have been measured by a measuring scale to an accuracy of $0.1 c m$ . If the measured values are $35.2 c m a n d 21.7 c m$ , respectively, then find the relative error in the area of the sheet.

1

\pm 8 \times 10^{- 2}

2

\pm 4 \times 10^{- 3}

3

\pm 7.45 \times 10^{- 3}

4

\pm 9 \times 10^{- 6}

Explanation:

Here, length of sheet
$l = (35.2 \pm 0.1) c m a n d b r e a d t h o f s h e e t$
$b = (21.7 \pm 0.1) c m$
$∴ R e l a t i v e e r r o r i n s u r f a c e a r e a o f t h e s h e e t,$
${(\frac{Δ s}{s})}_{max} = \frac{Δ l}{l} + \frac{Δ b}{b} = \frac{0.1}{35.2} + \frac{0.1}{21.7}$
$= 2.841 \times 10^{- 3} + 4.608 \times 10^{- 3}$
$= 7.449 \times 10^{- 3} = \pm 7.45 \times 10^{- 3}$

PHXI02:UNITS AND MEASUREMENTS

367495 Estimate the percentage error in the determination of density of a material taken in the form of a cylinder with 2 % error in mass , 1 % error in radius and 3 % error in height .

1

3 %

2

4 %

3

7 %

4

5 %

Explanation:

Density of material of a cylinder,
$ρ = \frac{M a s s}{V o l u m e} = \frac{M}{π r^{2} h}$
$\frac{Δ ρ}{ρ} \times 100 = (\frac{Δ M}{M} + 2 \frac{Δ r}{r} + \frac{Δ h}{h}) \times 100$
$= 2 % + 2 \times 1 % + 3 % = 7 %$

PHXI02:UNITS AND MEASUREMENTS

367496 The percentage errors in the measurement of mass and speed are $1 %$ and $2 %$ respectively. What is the percentage error in the kinetic energy

1

5 %

2

2.5 %

3

3 %

4

1.5 %

Explanation:

$K = \frac{1}{2} m v^{2}$
$∴ \frac{Δ K}{K} = \frac{Δ m}{m} + \frac{2 Δ v}{v}$
$%$ error in $K = (%$ error in $m) + 2 \times (%$ error in $v)$
$= 1 % + 2 \times 2 % = 5 %$ .
So correct option is (1)

PHXI02:UNITS AND MEASUREMENTS

367497 A person measures two quantities as $A = 1.0 m \pm 0.2 m,$
$B = 2.0 m \pm 0.2 m .$ We should report correct value for $\sqrt{A B}$ as

1

1.4 m + 0.3 m

2

1.4 m \pm 0.2 m

3

1.4 m \pm 0.4 m

4

1.41 m \pm 0.15 m

Explanation:

Given, $A = 1.0 m \pm 0.2 m, B = 2.0 \pm 0.2 m$
Let, $Y = \sqrt{A B} = \sqrt{(1.0) (2.0)} = 1.414 m$
Rounding off to two significant digit $Y = 1.4 m$
$\frac{Δ Y}{Y} = \frac{1}{2} [\frac{Δ A}{A} + \frac{Δ B}{B}]$
$= \frac{1}{2} [\frac{0.2}{1.0} + \frac{0.2}{2.0}] = \frac{0.6}{2 \times 2.0}$
$\Rightarrow Δ Y = \frac{0.6 Y}{2 \times 2.0} = \frac{0.6 \times 1.4}{2 \times 20} = 0.212$
Rounding off to one significant digit
$Δ Y = 0.2 m$
Thus, correct value for
$\sqrt{A B} = Y + Δ Y = 1.4 \pm 0.2 m$

PHXI02:UNITS AND MEASUREMENTS

367498 The resistance $R = \frac{V}{I}$ where $V = (200 \pm 5) V$ and $I = (20 \pm 0.2)$ $A$ , the percentage error in the measurement of $R$ is

1

5.5 %

2

7 %

3

3 %

4

3.5 %

Explanation:

As, $R = \frac{V}{I}$
To find $%$ error in $R$ ;
$\frac{Δ R}{R} \times 100 % = \frac{Δ V}{V} \times 100 % + \frac{Δ I}{I} \times 100 %$
$= \frac{5}{200} \times 100 + \frac{0.2}{20} \times 100 = 3.5 %$

JEE - 2024