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PHXI02:UNITS AND MEASUREMENTS

367524 A circular park has a radius of $(12.1 m \pm 0.1 m)$ . Find the area of the park with percentage error.

1

460 m^{2} \pm 1.65 %

2

450 m^{2} \pm 1.50 %

3

460 m^{2} \pm 1.25 %

4

460 m^{2} \pm 1.45 %

Explanation:

$r = 12.1 m \pm 0.1 m$
$Area = π r^{2} = π (12.1)^{2} = 460 m^{2}$
Percentage error,
$\frac{Δ A}{A} = 2 \frac{Δ r}{r} = 2 \times \frac{0.1}{12.1} \times 100 = 1.65 %$
$∴$ Area $= 460 m^{2} \pm 1.65 %$

PHXI02:UNITS AND MEASUREMENTS

367525 A student measures the distance travelled in free fall of a body, initially at rest in a given time. He uses this data to estimate $g$ , the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are $e_{1}$ and $e_{2}$ respectively, the percentage error in the estimation of $g$ is $(s = \frac{1}{2} g t^{2})$ :

1

e_{2} - e_{1}

2

e_{1} + 2 e_{2}

3

e_{1} + e_{2}

4

e_{1} - 2 e_{2}

Explanation:

$g = \frac{s}{t^{2}} \Rightarrow {(\frac{Δ g}{g})}_{max} = \frac{Δ s}{s} + 2 \frac{Δ t}{t}$
${(\frac{Δ g}{g} \times 100)}_{max} = \frac{Δ s}{s} \times 100 + 2 \frac{Δ t}{t} \times 100$
$= e_{1} + 2 e_{2}$

PHXI02:UNITS AND MEASUREMENTS

367526 An object covers $(16.0 \pm 0.4) m$ distance in $(4.0 \pm 0.2) s$ . Find out its speed.

1

(4.0 \pm 0.2) m / s

2

(4.0 \pm 0.3) m / s

3

(4.0 \pm 0.5) m / s

4

(4.0 \pm 0.1) m / s

Explanation:

Speed $v =$ $\frac{d i s t a n c e}{t i m e}$ $= \frac{16.0}{4.0} = 4.0 m / s;$
Error in speed
$Δ v = \pm (\frac{Δ s}{s} + \frac{Δ t}{t}) v = (\frac{0.4}{16.0} + \frac{0.2}{4.0}) (4.0)$
$v = (4.0 \pm 0.3) m / s$

PHXI02:UNITS AND MEASUREMENTS

367527 A metal wire has mass $(0.4 \pm 0.002) g$ , radius
$(0.3 \pm 0.001) m m$ and length $(5 \pm 0.02) c m$ .
The maximum possible percentage error in the measurement of density will nearly be:

1

1.3 %

2

1.6 %

3

1.4 %

4

1.2 %

Explanation:

$ρ = \frac{M}{V} = \frac{M}{π r^{2} l}$
$\Rightarrow \frac{Δ ρ}{ρ} = \frac{Δ M}{M} + \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\Rightarrow \frac{Δ ρ}{ρ} % = [\frac{0.002}{0.4} + \frac{2 (0.001)}{(0.3)} + \frac{0.02}{5}] \times 100 %$
$= \frac{1}{2} % + \frac{2}{3} % + \frac{2}{5} % = 1.6 %$
Correct option is (2).

NEET - 2023

PHXI02:UNITS AND MEASUREMENTS

367528 Assertion :
If errors in measurement of distance and time 3% and 2% respectively, error in calculation of velocity is 5%
Reason :
$V e l o c i t y = \frac{D i s t a n c e}{T i m e}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

In fact $[v] = \frac{[L]}{[T]}$
$∴ \frac{Δ v}{v} \times 100 = \pm (\frac{Δ L}{L} + \frac{Δ T}{T}) \times 100$
$= \pm (3 % + 2 %) = \pm 5 %$
So, option (1) is correct.

PHXI02:UNITS AND MEASUREMENTS

367524 A circular park has a radius of $(12.1 m \pm 0.1 m)$ . Find the area of the park with percentage error.

1

460 m^{2} \pm 1.65 %

2

450 m^{2} \pm 1.50 %

3

460 m^{2} \pm 1.25 %

4

460 m^{2} \pm 1.45 %

Explanation:

$r = 12.1 m \pm 0.1 m$
$Area = π r^{2} = π (12.1)^{2} = 460 m^{2}$
Percentage error,
$\frac{Δ A}{A} = 2 \frac{Δ r}{r} = 2 \times \frac{0.1}{12.1} \times 100 = 1.65 %$
$∴$ Area $= 460 m^{2} \pm 1.65 %$

PHXI02:UNITS AND MEASUREMENTS

367525 A student measures the distance travelled in free fall of a body, initially at rest in a given time. He uses this data to estimate $g$ , the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are $e_{1}$ and $e_{2}$ respectively, the percentage error in the estimation of $g$ is $(s = \frac{1}{2} g t^{2})$ :

1

e_{2} - e_{1}

2

e_{1} + 2 e_{2}

3

e_{1} + e_{2}

4

e_{1} - 2 e_{2}

Explanation:

$g = \frac{s}{t^{2}} \Rightarrow {(\frac{Δ g}{g})}_{max} = \frac{Δ s}{s} + 2 \frac{Δ t}{t}$
${(\frac{Δ g}{g} \times 100)}_{max} = \frac{Δ s}{s} \times 100 + 2 \frac{Δ t}{t} \times 100$
$= e_{1} + 2 e_{2}$

PHXI02:UNITS AND MEASUREMENTS

367526 An object covers $(16.0 \pm 0.4) m$ distance in $(4.0 \pm 0.2) s$ . Find out its speed.

1

(4.0 \pm 0.2) m / s

2

(4.0 \pm 0.3) m / s

3

(4.0 \pm 0.5) m / s

4

(4.0 \pm 0.1) m / s

Explanation:

Speed $v =$ $\frac{d i s t a n c e}{t i m e}$ $= \frac{16.0}{4.0} = 4.0 m / s;$
Error in speed
$Δ v = \pm (\frac{Δ s}{s} + \frac{Δ t}{t}) v = (\frac{0.4}{16.0} + \frac{0.2}{4.0}) (4.0)$
$v = (4.0 \pm 0.3) m / s$

PHXI02:UNITS AND MEASUREMENTS

367527 A metal wire has mass $(0.4 \pm 0.002) g$ , radius
$(0.3 \pm 0.001) m m$ and length $(5 \pm 0.02) c m$ .
The maximum possible percentage error in the measurement of density will nearly be:

1

1.3 %

2

1.6 %

3

1.4 %

4

1.2 %

Explanation:

$ρ = \frac{M}{V} = \frac{M}{π r^{2} l}$
$\Rightarrow \frac{Δ ρ}{ρ} = \frac{Δ M}{M} + \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\Rightarrow \frac{Δ ρ}{ρ} % = [\frac{0.002}{0.4} + \frac{2 (0.001)}{(0.3)} + \frac{0.02}{5}] \times 100 %$
$= \frac{1}{2} % + \frac{2}{3} % + \frac{2}{5} % = 1.6 %$
Correct option is (2).

NEET - 2023

PHXI02:UNITS AND MEASUREMENTS

367528 Assertion :
If errors in measurement of distance and time 3% and 2% respectively, error in calculation of velocity is 5%
Reason :
$V e l o c i t y = \frac{D i s t a n c e}{T i m e}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

In fact $[v] = \frac{[L]}{[T]}$
$∴ \frac{Δ v}{v} \times 100 = \pm (\frac{Δ L}{L} + \frac{Δ T}{T}) \times 100$
$= \pm (3 % + 2 %) = \pm 5 %$
So, option (1) is correct.

PHXI02:UNITS AND MEASUREMENTS

367524 A circular park has a radius of $(12.1 m \pm 0.1 m)$ . Find the area of the park with percentage error.

1

460 m^{2} \pm 1.65 %

2

450 m^{2} \pm 1.50 %

3

460 m^{2} \pm 1.25 %

4

460 m^{2} \pm 1.45 %

Explanation:

$r = 12.1 m \pm 0.1 m$
$Area = π r^{2} = π (12.1)^{2} = 460 m^{2}$
Percentage error,
$\frac{Δ A}{A} = 2 \frac{Δ r}{r} = 2 \times \frac{0.1}{12.1} \times 100 = 1.65 %$
$∴$ Area $= 460 m^{2} \pm 1.65 %$

PHXI02:UNITS AND MEASUREMENTS

367525 A student measures the distance travelled in free fall of a body, initially at rest in a given time. He uses this data to estimate $g$ , the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are $e_{1}$ and $e_{2}$ respectively, the percentage error in the estimation of $g$ is $(s = \frac{1}{2} g t^{2})$ :

1

e_{2} - e_{1}

2

e_{1} + 2 e_{2}

3

e_{1} + e_{2}

4

e_{1} - 2 e_{2}

Explanation:

$g = \frac{s}{t^{2}} \Rightarrow {(\frac{Δ g}{g})}_{max} = \frac{Δ s}{s} + 2 \frac{Δ t}{t}$
${(\frac{Δ g}{g} \times 100)}_{max} = \frac{Δ s}{s} \times 100 + 2 \frac{Δ t}{t} \times 100$
$= e_{1} + 2 e_{2}$

PHXI02:UNITS AND MEASUREMENTS

367526 An object covers $(16.0 \pm 0.4) m$ distance in $(4.0 \pm 0.2) s$ . Find out its speed.

1

(4.0 \pm 0.2) m / s

2

(4.0 \pm 0.3) m / s

3

(4.0 \pm 0.5) m / s

4

(4.0 \pm 0.1) m / s

Explanation:

Speed $v =$ $\frac{d i s t a n c e}{t i m e}$ $= \frac{16.0}{4.0} = 4.0 m / s;$
Error in speed
$Δ v = \pm (\frac{Δ s}{s} + \frac{Δ t}{t}) v = (\frac{0.4}{16.0} + \frac{0.2}{4.0}) (4.0)$
$v = (4.0 \pm 0.3) m / s$

PHXI02:UNITS AND MEASUREMENTS

367527 A metal wire has mass $(0.4 \pm 0.002) g$ , radius
$(0.3 \pm 0.001) m m$ and length $(5 \pm 0.02) c m$ .
The maximum possible percentage error in the measurement of density will nearly be:

1

1.3 %

2

1.6 %

3

1.4 %

4

1.2 %

Explanation:

$ρ = \frac{M}{V} = \frac{M}{π r^{2} l}$
$\Rightarrow \frac{Δ ρ}{ρ} = \frac{Δ M}{M} + \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\Rightarrow \frac{Δ ρ}{ρ} % = [\frac{0.002}{0.4} + \frac{2 (0.001)}{(0.3)} + \frac{0.02}{5}] \times 100 %$
$= \frac{1}{2} % + \frac{2}{3} % + \frac{2}{5} % = 1.6 %$
Correct option is (2).

NEET - 2023

PHXI02:UNITS AND MEASUREMENTS

367528 Assertion :
If errors in measurement of distance and time 3% and 2% respectively, error in calculation of velocity is 5%
Reason :
$V e l o c i t y = \frac{D i s t a n c e}{T i m e}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

In fact $[v] = \frac{[L]}{[T]}$
$∴ \frac{Δ v}{v} \times 100 = \pm (\frac{Δ L}{L} + \frac{Δ T}{T}) \times 100$
$= \pm (3 % + 2 %) = \pm 5 %$
So, option (1) is correct.

PHXI02:UNITS AND MEASUREMENTS

367524 A circular park has a radius of $(12.1 m \pm 0.1 m)$ . Find the area of the park with percentage error.

1

460 m^{2} \pm 1.65 %

2

450 m^{2} \pm 1.50 %

3

460 m^{2} \pm 1.25 %

4

460 m^{2} \pm 1.45 %

Explanation:

$r = 12.1 m \pm 0.1 m$
$Area = π r^{2} = π (12.1)^{2} = 460 m^{2}$
Percentage error,
$\frac{Δ A}{A} = 2 \frac{Δ r}{r} = 2 \times \frac{0.1}{12.1} \times 100 = 1.65 %$
$∴$ Area $= 460 m^{2} \pm 1.65 %$

PHXI02:UNITS AND MEASUREMENTS

367525 A student measures the distance travelled in free fall of a body, initially at rest in a given time. He uses this data to estimate $g$ , the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are $e_{1}$ and $e_{2}$ respectively, the percentage error in the estimation of $g$ is $(s = \frac{1}{2} g t^{2})$ :

1

e_{2} - e_{1}

2

e_{1} + 2 e_{2}

3

e_{1} + e_{2}

4

e_{1} - 2 e_{2}

Explanation:

$g = \frac{s}{t^{2}} \Rightarrow {(\frac{Δ g}{g})}_{max} = \frac{Δ s}{s} + 2 \frac{Δ t}{t}$
${(\frac{Δ g}{g} \times 100)}_{max} = \frac{Δ s}{s} \times 100 + 2 \frac{Δ t}{t} \times 100$
$= e_{1} + 2 e_{2}$

PHXI02:UNITS AND MEASUREMENTS

367526 An object covers $(16.0 \pm 0.4) m$ distance in $(4.0 \pm 0.2) s$ . Find out its speed.

1

(4.0 \pm 0.2) m / s

2

(4.0 \pm 0.3) m / s

3

(4.0 \pm 0.5) m / s

4

(4.0 \pm 0.1) m / s

Explanation:

Speed $v =$ $\frac{d i s t a n c e}{t i m e}$ $= \frac{16.0}{4.0} = 4.0 m / s;$
Error in speed
$Δ v = \pm (\frac{Δ s}{s} + \frac{Δ t}{t}) v = (\frac{0.4}{16.0} + \frac{0.2}{4.0}) (4.0)$
$v = (4.0 \pm 0.3) m / s$

PHXI02:UNITS AND MEASUREMENTS

367527 A metal wire has mass $(0.4 \pm 0.002) g$ , radius
$(0.3 \pm 0.001) m m$ and length $(5 \pm 0.02) c m$ .
The maximum possible percentage error in the measurement of density will nearly be:

1

1.3 %

2

1.6 %

3

1.4 %

4

1.2 %

Explanation:

$ρ = \frac{M}{V} = \frac{M}{π r^{2} l}$
$\Rightarrow \frac{Δ ρ}{ρ} = \frac{Δ M}{M} + \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\Rightarrow \frac{Δ ρ}{ρ} % = [\frac{0.002}{0.4} + \frac{2 (0.001)}{(0.3)} + \frac{0.02}{5}] \times 100 %$
$= \frac{1}{2} % + \frac{2}{3} % + \frac{2}{5} % = 1.6 %$
Correct option is (2).

NEET - 2023

PHXI02:UNITS AND MEASUREMENTS

367528 Assertion :
If errors in measurement of distance and time 3% and 2% respectively, error in calculation of velocity is 5%
Reason :
$V e l o c i t y = \frac{D i s t a n c e}{T i m e}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

In fact $[v] = \frac{[L]}{[T]}$
$∴ \frac{Δ v}{v} \times 100 = \pm (\frac{Δ L}{L} + \frac{Δ T}{T}) \times 100$
$= \pm (3 % + 2 %) = \pm 5 %$
So, option (1) is correct.

PHXI02:UNITS AND MEASUREMENTS

367524 A circular park has a radius of $(12.1 m \pm 0.1 m)$ . Find the area of the park with percentage error.

1

460 m^{2} \pm 1.65 %

2

450 m^{2} \pm 1.50 %

3

460 m^{2} \pm 1.25 %

4

460 m^{2} \pm 1.45 %

Explanation:

$r = 12.1 m \pm 0.1 m$
$Area = π r^{2} = π (12.1)^{2} = 460 m^{2}$
Percentage error,
$\frac{Δ A}{A} = 2 \frac{Δ r}{r} = 2 \times \frac{0.1}{12.1} \times 100 = 1.65 %$
$∴$ Area $= 460 m^{2} \pm 1.65 %$

PHXI02:UNITS AND MEASUREMENTS

367525 A student measures the distance travelled in free fall of a body, initially at rest in a given time. He uses this data to estimate $g$ , the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are $e_{1}$ and $e_{2}$ respectively, the percentage error in the estimation of $g$ is $(s = \frac{1}{2} g t^{2})$ :

1

e_{2} - e_{1}

2

e_{1} + 2 e_{2}

3

e_{1} + e_{2}

4

e_{1} - 2 e_{2}

Explanation:

$g = \frac{s}{t^{2}} \Rightarrow {(\frac{Δ g}{g})}_{max} = \frac{Δ s}{s} + 2 \frac{Δ t}{t}$
${(\frac{Δ g}{g} \times 100)}_{max} = \frac{Δ s}{s} \times 100 + 2 \frac{Δ t}{t} \times 100$
$= e_{1} + 2 e_{2}$

PHXI02:UNITS AND MEASUREMENTS

367526 An object covers $(16.0 \pm 0.4) m$ distance in $(4.0 \pm 0.2) s$ . Find out its speed.

1

(4.0 \pm 0.2) m / s

2

(4.0 \pm 0.3) m / s

3

(4.0 \pm 0.5) m / s

4

(4.0 \pm 0.1) m / s

Explanation:

Speed $v =$ $\frac{d i s t a n c e}{t i m e}$ $= \frac{16.0}{4.0} = 4.0 m / s;$
Error in speed
$Δ v = \pm (\frac{Δ s}{s} + \frac{Δ t}{t}) v = (\frac{0.4}{16.0} + \frac{0.2}{4.0}) (4.0)$
$v = (4.0 \pm 0.3) m / s$

PHXI02:UNITS AND MEASUREMENTS

367527 A metal wire has mass $(0.4 \pm 0.002) g$ , radius
$(0.3 \pm 0.001) m m$ and length $(5 \pm 0.02) c m$ .
The maximum possible percentage error in the measurement of density will nearly be:

1

1.3 %

2

1.6 %

3

1.4 %

4

1.2 %

Explanation:

$ρ = \frac{M}{V} = \frac{M}{π r^{2} l}$
$\Rightarrow \frac{Δ ρ}{ρ} = \frac{Δ M}{M} + \frac{2 Δ r}{r} + \frac{Δ l}{l}$
$\Rightarrow \frac{Δ ρ}{ρ} % = [\frac{0.002}{0.4} + \frac{2 (0.001)}{(0.3)} + \frac{0.02}{5}] \times 100 %$
$= \frac{1}{2} % + \frac{2}{3} % + \frac{2}{5} % = 1.6 %$
Correct option is (2).

NEET - 2023

PHXI02:UNITS AND MEASUREMENTS

367528 Assertion :
If errors in measurement of distance and time 3% and 2% respectively, error in calculation of velocity is 5%
Reason :
$V e l o c i t y = \frac{D i s t a n c e}{T i m e}$

1 Both assertion and reason are correct and reason is the correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

In fact $[v] = \frac{[L]}{[T]}$
$∴ \frac{Δ v}{v} \times 100 = \pm (\frac{Δ L}{L} + \frac{Δ T}{T}) \times 100$
$= \pm (3 % + 2 %) = \pm 5 %$
So, option (1) is correct.

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