QBManager

Units and Measurements

139885 The velocity of a transverse wave in a string is directly proportional to $\sqrt{T}$ and inversely proportional to $\sqrt{μ}$ . In a measurement, the mass applied at the end of string is $3.0 gm$ , length of string is $1 m$ and mass of string is 5 gm. If possible error in measuring mass is 0.1 $g m$ and that of length is $1 mm$ , the parentage error in measurement of velocity is

1

4.5 %

2

2.7 %

3

2.1 %

4

3.7 %

Explanation:

B Given,
$m = 3 gm$
$l = 1 m$
according to question
$v \propto \sqrt{\frac{T}{μ}} = k \sqrt{\frac{T}{μ}}$
$μ = \frac{M}{L} and T = m^{'} g$
$v = K \sqrt{\frac{T L}{M}} = K \sqrt{\frac{m^{' g L}}{M}}$
$\frac{Δ v}{v} = \frac{1}{2} \times \frac{Δ m^{'}}{m^{'}} + \frac{1}{2} \times \frac{Δ L}{L} + \frac{1}{2} \times \frac{Δ M}{M}$
$= \frac{1}{2} \times \frac{0.1}{5} + \frac{1}{2} \times \frac{1 \times 10^{- 3}}{1} + \frac{1}{2} \times \frac{0.1}{3}$
$= 0.01 + 0.0005 + 0.016$
$= 0.0271$
$\frac{Δ v}{v} \times 100 = 2.7 %$

AMU-2016

Units and Measurements

139886 A physical quantity $Q$ is related to four observables $x, y, z$ ad $t$ by the relation
$Q = \frac{x^{2 / 5} z^{3}}{y \sqrt{t}}$
The percentage errors of measurement in $x, y$ , $z$ , and $t$ are $2.5 %, 2 %, 0.5 %$ and $1 %$ respectively. The percentage error in $Q$ will be

1

5 %

2

4.5 %

3

8 %

4

7.75 %

Explanation:

A Given,
$\frac{Δ x}{x} = 2.5 %, \frac{Δ y}{y} = 2 %, \frac{Δ z}{z} = 0.5 % and$
$\frac{Δ t}{t} = 1 %$
Given, $Q = \frac{x^{\frac{2}{5}} z^{3}}{y \sqrt{t}}$
The percentage error in $Q$ in given by,
$\frac{Δ Q}{Q} \times 100 = \frac{2}{5} (\frac{Δ x}{x} \times 100) + 3 (\frac{Δ Z}{Z} \times 100)$
$+ (\frac{Δ y}{y} \times 100) + \frac{1}{2} (\frac{Δ t}{t} \times 100)$
$\frac{Δ Q}{Q} \times 100 = (\frac{2}{5} \times 2.5 % + 3 \times 0.5 % + 2 % + \frac{1}{2} \times 1 %)$
$= 5 %$

AMU-2006

Units and Measurements

139888 What is the approximate percentage error in the measurement of time period of a simple pendulum if maximum errors in the measurement of length $l$ and gravitational acceleration $g$ are $3 %$ and $7 %$ respectively?

1

5 %

2

10 %

3

2 %

4

3 %

Explanation:

A We know that time period of a simple pendulum
$T = 2 π \sqrt{\frac{l}{g}}$
Where, $l =$ length of pendulum
$g =$ acceleration due to gravity
$T = 2 π \sqrt{\frac{l}{g}}$
Take $\log$ and differentiate
$\frac{Δ T}{T} = \frac{1}{2} [\frac{Δ l}{l} - \frac{Δ g}{g}]$
For maximize error consider $+ ve$ sign
$\frac{Δ T}{T} \times 100 = \frac{1}{2} [\frac{Δ l}{l} + \frac{Δ g}{g}] \times 100$
$= (\frac{3}{2} + \frac{7}{2}) % = \frac{10}{2} % = 5 %$
$\frac{Δ T}{T} \times 100 = 5 %$

UPSEE - 2016

Units and Measurements

139889 The relation gives the value of $x$ as $x = \frac{(a^{3} b^{3})}{(c \sqrt{d})}$
Find the percentage error in $x$ , if the percentage error in a, b, c, d, are $2 %, 1 %, 2 %$ and $4 %$ respectively :

1

\pm 13 %

2

\pm 5 %

3

\pm 9 %

4

\pm 8 %

Explanation:

A Given,
Relation gives value of, $x = \frac{a^{3} b^{3}}{c \sqrt{d}}$
Relative error is-
$\frac{Δ x}{x} = \pm 3 \frac{Δ a}{a} \pm 3 \frac{Δ b}{b} \pm \frac{Δ c}{c} \pm \frac{1}{2} \frac{Δ d}{d}$
Percentage error is-
$\frac{Δ x}{x} \times 100 = \pm 3 (\frac{Δ a}{a} \times 100) \pm 3 (\frac{Δ b}{b} \times 100)$
$\pm (\frac{Δ c}{c} \times 100) \pm \frac{1}{2} (\frac{Δ d}{d} \times 100)$
$= \pm 3 \times 2 \pm 3 \times 1 \pm 1 \times 2 \pm \frac{1}{2} \times 4$
$= \pm 6 \pm 3 \pm 2 \pm 2$
$%$ error in $x = \pm 13 %$

UPSEE - 2005

Units and Measurements

139890 When the voltage and current in a conductor are measured as $(100 \pm 4) V$ and $(5 \pm 0.2) A$ , then the percentage of error in the calculation of resistance is

1

8 %

2

4 %

3

20 %

4

10 %

5

6 %

Explanation:

A Given,
$(V \pm Δ V) = (100 \pm 4) V$
$(I \pm Δ I) = (5 \pm 0.2) A$
We know that,
$R = \frac{V}{I}$
$\frac{Δ R}{R} \times 100 = \frac{Δ V}{V} \times 100 + \frac{Δ I}{I} \times 100$
$= \frac{4}{100} \times 100 + \frac{0.2}{5} \times 100$
$%$ error in resistance $= 4 % + 4 % = 8 %$

Kerala CEE - 2016

Units and Measurements

139885 The velocity of a transverse wave in a string is directly proportional to $\sqrt{T}$ and inversely proportional to $\sqrt{μ}$ . In a measurement, the mass applied at the end of string is $3.0 gm$ , length of string is $1 m$ and mass of string is 5 gm. If possible error in measuring mass is 0.1 $g m$ and that of length is $1 mm$ , the parentage error in measurement of velocity is

1

4.5 %

2

2.7 %

3

2.1 %

4

3.7 %

Explanation:

B Given,
$m = 3 gm$
$l = 1 m$
according to question
$v \propto \sqrt{\frac{T}{μ}} = k \sqrt{\frac{T}{μ}}$
$μ = \frac{M}{L} and T = m^{'} g$
$v = K \sqrt{\frac{T L}{M}} = K \sqrt{\frac{m^{' g L}}{M}}$
$\frac{Δ v}{v} = \frac{1}{2} \times \frac{Δ m^{'}}{m^{'}} + \frac{1}{2} \times \frac{Δ L}{L} + \frac{1}{2} \times \frac{Δ M}{M}$
$= \frac{1}{2} \times \frac{0.1}{5} + \frac{1}{2} \times \frac{1 \times 10^{- 3}}{1} + \frac{1}{2} \times \frac{0.1}{3}$
$= 0.01 + 0.0005 + 0.016$
$= 0.0271$
$\frac{Δ v}{v} \times 100 = 2.7 %$

AMU-2016

Units and Measurements

139886 A physical quantity $Q$ is related to four observables $x, y, z$ ad $t$ by the relation
$Q = \frac{x^{2 / 5} z^{3}}{y \sqrt{t}}$
The percentage errors of measurement in $x, y$ , $z$ , and $t$ are $2.5 %, 2 %, 0.5 %$ and $1 %$ respectively. The percentage error in $Q$ will be

1

5 %

2

4.5 %

3

8 %

4

7.75 %

Explanation:

A Given,
$\frac{Δ x}{x} = 2.5 %, \frac{Δ y}{y} = 2 %, \frac{Δ z}{z} = 0.5 % and$
$\frac{Δ t}{t} = 1 %$
Given, $Q = \frac{x^{\frac{2}{5}} z^{3}}{y \sqrt{t}}$
The percentage error in $Q$ in given by,
$\frac{Δ Q}{Q} \times 100 = \frac{2}{5} (\frac{Δ x}{x} \times 100) + 3 (\frac{Δ Z}{Z} \times 100)$
$+ (\frac{Δ y}{y} \times 100) + \frac{1}{2} (\frac{Δ t}{t} \times 100)$
$\frac{Δ Q}{Q} \times 100 = (\frac{2}{5} \times 2.5 % + 3 \times 0.5 % + 2 % + \frac{1}{2} \times 1 %)$
$= 5 %$

AMU-2006

Units and Measurements

139888 What is the approximate percentage error in the measurement of time period of a simple pendulum if maximum errors in the measurement of length $l$ and gravitational acceleration $g$ are $3 %$ and $7 %$ respectively?

1

5 %

2

10 %

3

2 %

4

3 %

Explanation:

A We know that time period of a simple pendulum
$T = 2 π \sqrt{\frac{l}{g}}$
Where, $l =$ length of pendulum
$g =$ acceleration due to gravity
$T = 2 π \sqrt{\frac{l}{g}}$
Take $\log$ and differentiate
$\frac{Δ T}{T} = \frac{1}{2} [\frac{Δ l}{l} - \frac{Δ g}{g}]$
For maximize error consider $+ ve$ sign
$\frac{Δ T}{T} \times 100 = \frac{1}{2} [\frac{Δ l}{l} + \frac{Δ g}{g}] \times 100$
$= (\frac{3}{2} + \frac{7}{2}) % = \frac{10}{2} % = 5 %$
$\frac{Δ T}{T} \times 100 = 5 %$

UPSEE - 2016

Units and Measurements

139889 The relation gives the value of $x$ as $x = \frac{(a^{3} b^{3})}{(c \sqrt{d})}$
Find the percentage error in $x$ , if the percentage error in a, b, c, d, are $2 %, 1 %, 2 %$ and $4 %$ respectively :

1

\pm 13 %

2

\pm 5 %

3

\pm 9 %

4

\pm 8 %

Explanation:

A Given,
Relation gives value of, $x = \frac{a^{3} b^{3}}{c \sqrt{d}}$
Relative error is-
$\frac{Δ x}{x} = \pm 3 \frac{Δ a}{a} \pm 3 \frac{Δ b}{b} \pm \frac{Δ c}{c} \pm \frac{1}{2} \frac{Δ d}{d}$
Percentage error is-
$\frac{Δ x}{x} \times 100 = \pm 3 (\frac{Δ a}{a} \times 100) \pm 3 (\frac{Δ b}{b} \times 100)$
$\pm (\frac{Δ c}{c} \times 100) \pm \frac{1}{2} (\frac{Δ d}{d} \times 100)$
$= \pm 3 \times 2 \pm 3 \times 1 \pm 1 \times 2 \pm \frac{1}{2} \times 4$
$= \pm 6 \pm 3 \pm 2 \pm 2$
$%$ error in $x = \pm 13 %$

UPSEE - 2005

Units and Measurements

139890 When the voltage and current in a conductor are measured as $(100 \pm 4) V$ and $(5 \pm 0.2) A$ , then the percentage of error in the calculation of resistance is

1

8 %

2

4 %

3

20 %

4

10 %

5

6 %

Explanation:

A Given,
$(V \pm Δ V) = (100 \pm 4) V$
$(I \pm Δ I) = (5 \pm 0.2) A$
We know that,
$R = \frac{V}{I}$
$\frac{Δ R}{R} \times 100 = \frac{Δ V}{V} \times 100 + \frac{Δ I}{I} \times 100$
$= \frac{4}{100} \times 100 + \frac{0.2}{5} \times 100$
$%$ error in resistance $= 4 % + 4 % = 8 %$

Kerala CEE - 2016

Units and Measurements

139885 The velocity of a transverse wave in a string is directly proportional to $\sqrt{T}$ and inversely proportional to $\sqrt{μ}$ . In a measurement, the mass applied at the end of string is $3.0 gm$ , length of string is $1 m$ and mass of string is 5 gm. If possible error in measuring mass is 0.1 $g m$ and that of length is $1 mm$ , the parentage error in measurement of velocity is

1

4.5 %

2

2.7 %

3

2.1 %

4

3.7 %

Explanation:

B Given,
$m = 3 gm$
$l = 1 m$
according to question
$v \propto \sqrt{\frac{T}{μ}} = k \sqrt{\frac{T}{μ}}$
$μ = \frac{M}{L} and T = m^{'} g$
$v = K \sqrt{\frac{T L}{M}} = K \sqrt{\frac{m^{' g L}}{M}}$
$\frac{Δ v}{v} = \frac{1}{2} \times \frac{Δ m^{'}}{m^{'}} + \frac{1}{2} \times \frac{Δ L}{L} + \frac{1}{2} \times \frac{Δ M}{M}$
$= \frac{1}{2} \times \frac{0.1}{5} + \frac{1}{2} \times \frac{1 \times 10^{- 3}}{1} + \frac{1}{2} \times \frac{0.1}{3}$
$= 0.01 + 0.0005 + 0.016$
$= 0.0271$
$\frac{Δ v}{v} \times 100 = 2.7 %$

AMU-2016

Units and Measurements

139886 A physical quantity $Q$ is related to four observables $x, y, z$ ad $t$ by the relation
$Q = \frac{x^{2 / 5} z^{3}}{y \sqrt{t}}$
The percentage errors of measurement in $x, y$ , $z$ , and $t$ are $2.5 %, 2 %, 0.5 %$ and $1 %$ respectively. The percentage error in $Q$ will be

1

5 %

2

4.5 %

3

8 %

4

7.75 %

Explanation:

A Given,
$\frac{Δ x}{x} = 2.5 %, \frac{Δ y}{y} = 2 %, \frac{Δ z}{z} = 0.5 % and$
$\frac{Δ t}{t} = 1 %$
Given, $Q = \frac{x^{\frac{2}{5}} z^{3}}{y \sqrt{t}}$
The percentage error in $Q$ in given by,
$\frac{Δ Q}{Q} \times 100 = \frac{2}{5} (\frac{Δ x}{x} \times 100) + 3 (\frac{Δ Z}{Z} \times 100)$
$+ (\frac{Δ y}{y} \times 100) + \frac{1}{2} (\frac{Δ t}{t} \times 100)$
$\frac{Δ Q}{Q} \times 100 = (\frac{2}{5} \times 2.5 % + 3 \times 0.5 % + 2 % + \frac{1}{2} \times 1 %)$
$= 5 %$

AMU-2006

Units and Measurements

139888 What is the approximate percentage error in the measurement of time period of a simple pendulum if maximum errors in the measurement of length $l$ and gravitational acceleration $g$ are $3 %$ and $7 %$ respectively?

1

5 %

2

10 %

3

2 %

4

3 %

Explanation:

A We know that time period of a simple pendulum
$T = 2 π \sqrt{\frac{l}{g}}$
Where, $l =$ length of pendulum
$g =$ acceleration due to gravity
$T = 2 π \sqrt{\frac{l}{g}}$
Take $\log$ and differentiate
$\frac{Δ T}{T} = \frac{1}{2} [\frac{Δ l}{l} - \frac{Δ g}{g}]$
For maximize error consider $+ ve$ sign
$\frac{Δ T}{T} \times 100 = \frac{1}{2} [\frac{Δ l}{l} + \frac{Δ g}{g}] \times 100$
$= (\frac{3}{2} + \frac{7}{2}) % = \frac{10}{2} % = 5 %$
$\frac{Δ T}{T} \times 100 = 5 %$

UPSEE - 2016

Units and Measurements

139889 The relation gives the value of $x$ as $x = \frac{(a^{3} b^{3})}{(c \sqrt{d})}$
Find the percentage error in $x$ , if the percentage error in a, b, c, d, are $2 %, 1 %, 2 %$ and $4 %$ respectively :

1

\pm 13 %

2

\pm 5 %

3

\pm 9 %

4

\pm 8 %

Explanation:

A Given,
Relation gives value of, $x = \frac{a^{3} b^{3}}{c \sqrt{d}}$
Relative error is-
$\frac{Δ x}{x} = \pm 3 \frac{Δ a}{a} \pm 3 \frac{Δ b}{b} \pm \frac{Δ c}{c} \pm \frac{1}{2} \frac{Δ d}{d}$
Percentage error is-
$\frac{Δ x}{x} \times 100 = \pm 3 (\frac{Δ a}{a} \times 100) \pm 3 (\frac{Δ b}{b} \times 100)$
$\pm (\frac{Δ c}{c} \times 100) \pm \frac{1}{2} (\frac{Δ d}{d} \times 100)$
$= \pm 3 \times 2 \pm 3 \times 1 \pm 1 \times 2 \pm \frac{1}{2} \times 4$
$= \pm 6 \pm 3 \pm 2 \pm 2$
$%$ error in $x = \pm 13 %$

UPSEE - 2005

Units and Measurements

139890 When the voltage and current in a conductor are measured as $(100 \pm 4) V$ and $(5 \pm 0.2) A$ , then the percentage of error in the calculation of resistance is

1

8 %

2

4 %

3

20 %

4

10 %

5

6 %

Explanation:

A Given,
$(V \pm Δ V) = (100 \pm 4) V$
$(I \pm Δ I) = (5 \pm 0.2) A$
We know that,
$R = \frac{V}{I}$
$\frac{Δ R}{R} \times 100 = \frac{Δ V}{V} \times 100 + \frac{Δ I}{I} \times 100$
$= \frac{4}{100} \times 100 + \frac{0.2}{5} \times 100$
$%$ error in resistance $= 4 % + 4 % = 8 %$

Kerala CEE - 2016

Units and Measurements

139885 The velocity of a transverse wave in a string is directly proportional to $\sqrt{T}$ and inversely proportional to $\sqrt{μ}$ . In a measurement, the mass applied at the end of string is $3.0 gm$ , length of string is $1 m$ and mass of string is 5 gm. If possible error in measuring mass is 0.1 $g m$ and that of length is $1 mm$ , the parentage error in measurement of velocity is

1

4.5 %

2

2.7 %

3

2.1 %

4

3.7 %

Explanation:

B Given,
$m = 3 gm$
$l = 1 m$
according to question
$v \propto \sqrt{\frac{T}{μ}} = k \sqrt{\frac{T}{μ}}$
$μ = \frac{M}{L} and T = m^{'} g$
$v = K \sqrt{\frac{T L}{M}} = K \sqrt{\frac{m^{' g L}}{M}}$
$\frac{Δ v}{v} = \frac{1}{2} \times \frac{Δ m^{'}}{m^{'}} + \frac{1}{2} \times \frac{Δ L}{L} + \frac{1}{2} \times \frac{Δ M}{M}$
$= \frac{1}{2} \times \frac{0.1}{5} + \frac{1}{2} \times \frac{1 \times 10^{- 3}}{1} + \frac{1}{2} \times \frac{0.1}{3}$
$= 0.01 + 0.0005 + 0.016$
$= 0.0271$
$\frac{Δ v}{v} \times 100 = 2.7 %$

AMU-2016

Units and Measurements

139886 A physical quantity $Q$ is related to four observables $x, y, z$ ad $t$ by the relation
$Q = \frac{x^{2 / 5} z^{3}}{y \sqrt{t}}$
The percentage errors of measurement in $x, y$ , $z$ , and $t$ are $2.5 %, 2 %, 0.5 %$ and $1 %$ respectively. The percentage error in $Q$ will be

1

5 %

2

4.5 %

3

8 %

4

7.75 %

Explanation:

A Given,
$\frac{Δ x}{x} = 2.5 %, \frac{Δ y}{y} = 2 %, \frac{Δ z}{z} = 0.5 % and$
$\frac{Δ t}{t} = 1 %$
Given, $Q = \frac{x^{\frac{2}{5}} z^{3}}{y \sqrt{t}}$
The percentage error in $Q$ in given by,
$\frac{Δ Q}{Q} \times 100 = \frac{2}{5} (\frac{Δ x}{x} \times 100) + 3 (\frac{Δ Z}{Z} \times 100)$
$+ (\frac{Δ y}{y} \times 100) + \frac{1}{2} (\frac{Δ t}{t} \times 100)$
$\frac{Δ Q}{Q} \times 100 = (\frac{2}{5} \times 2.5 % + 3 \times 0.5 % + 2 % + \frac{1}{2} \times 1 %)$
$= 5 %$

AMU-2006

Units and Measurements

139888 What is the approximate percentage error in the measurement of time period of a simple pendulum if maximum errors in the measurement of length $l$ and gravitational acceleration $g$ are $3 %$ and $7 %$ respectively?

1

5 %

2

10 %

3

2 %

4

3 %

Explanation:

A We know that time period of a simple pendulum
$T = 2 π \sqrt{\frac{l}{g}}$
Where, $l =$ length of pendulum
$g =$ acceleration due to gravity
$T = 2 π \sqrt{\frac{l}{g}}$
Take $\log$ and differentiate
$\frac{Δ T}{T} = \frac{1}{2} [\frac{Δ l}{l} - \frac{Δ g}{g}]$
For maximize error consider $+ ve$ sign
$\frac{Δ T}{T} \times 100 = \frac{1}{2} [\frac{Δ l}{l} + \frac{Δ g}{g}] \times 100$
$= (\frac{3}{2} + \frac{7}{2}) % = \frac{10}{2} % = 5 %$
$\frac{Δ T}{T} \times 100 = 5 %$

UPSEE - 2016

Units and Measurements

139889 The relation gives the value of $x$ as $x = \frac{(a^{3} b^{3})}{(c \sqrt{d})}$
Find the percentage error in $x$ , if the percentage error in a, b, c, d, are $2 %, 1 %, 2 %$ and $4 %$ respectively :

1

\pm 13 %

2

\pm 5 %

3

\pm 9 %

4

\pm 8 %

Explanation:

A Given,
Relation gives value of, $x = \frac{a^{3} b^{3}}{c \sqrt{d}}$
Relative error is-
$\frac{Δ x}{x} = \pm 3 \frac{Δ a}{a} \pm 3 \frac{Δ b}{b} \pm \frac{Δ c}{c} \pm \frac{1}{2} \frac{Δ d}{d}$
Percentage error is-
$\frac{Δ x}{x} \times 100 = \pm 3 (\frac{Δ a}{a} \times 100) \pm 3 (\frac{Δ b}{b} \times 100)$
$\pm (\frac{Δ c}{c} \times 100) \pm \frac{1}{2} (\frac{Δ d}{d} \times 100)$
$= \pm 3 \times 2 \pm 3 \times 1 \pm 1 \times 2 \pm \frac{1}{2} \times 4$
$= \pm 6 \pm 3 \pm 2 \pm 2$
$%$ error in $x = \pm 13 %$

UPSEE - 2005

Units and Measurements

139890 When the voltage and current in a conductor are measured as $(100 \pm 4) V$ and $(5 \pm 0.2) A$ , then the percentage of error in the calculation of resistance is

1

8 %

2

4 %

3

20 %

4

10 %

5

6 %

Explanation:

A Given,
$(V \pm Δ V) = (100 \pm 4) V$
$(I \pm Δ I) = (5 \pm 0.2) A$
We know that,
$R = \frac{V}{I}$
$\frac{Δ R}{R} \times 100 = \frac{Δ V}{V} \times 100 + \frac{Δ I}{I} \times 100$
$= \frac{4}{100} \times 100 + \frac{0.2}{5} \times 100$
$%$ error in resistance $= 4 % + 4 % = 8 %$

Kerala CEE - 2016

Units and Measurements

139885 The velocity of a transverse wave in a string is directly proportional to $\sqrt{T}$ and inversely proportional to $\sqrt{μ}$ . In a measurement, the mass applied at the end of string is $3.0 gm$ , length of string is $1 m$ and mass of string is 5 gm. If possible error in measuring mass is 0.1 $g m$ and that of length is $1 mm$ , the parentage error in measurement of velocity is

1

4.5 %

2

2.7 %

3

2.1 %

4

3.7 %

Explanation:

B Given,
$m = 3 gm$
$l = 1 m$
according to question
$v \propto \sqrt{\frac{T}{μ}} = k \sqrt{\frac{T}{μ}}$
$μ = \frac{M}{L} and T = m^{'} g$
$v = K \sqrt{\frac{T L}{M}} = K \sqrt{\frac{m^{' g L}}{M}}$
$\frac{Δ v}{v} = \frac{1}{2} \times \frac{Δ m^{'}}{m^{'}} + \frac{1}{2} \times \frac{Δ L}{L} + \frac{1}{2} \times \frac{Δ M}{M}$
$= \frac{1}{2} \times \frac{0.1}{5} + \frac{1}{2} \times \frac{1 \times 10^{- 3}}{1} + \frac{1}{2} \times \frac{0.1}{3}$
$= 0.01 + 0.0005 + 0.016$
$= 0.0271$
$\frac{Δ v}{v} \times 100 = 2.7 %$

AMU-2016

Units and Measurements

139886 A physical quantity $Q$ is related to four observables $x, y, z$ ad $t$ by the relation
$Q = \frac{x^{2 / 5} z^{3}}{y \sqrt{t}}$
The percentage errors of measurement in $x, y$ , $z$ , and $t$ are $2.5 %, 2 %, 0.5 %$ and $1 %$ respectively. The percentage error in $Q$ will be

1

5 %

2

4.5 %

3

8 %

4

7.75 %

Explanation:

A Given,
$\frac{Δ x}{x} = 2.5 %, \frac{Δ y}{y} = 2 %, \frac{Δ z}{z} = 0.5 % and$
$\frac{Δ t}{t} = 1 %$
Given, $Q = \frac{x^{\frac{2}{5}} z^{3}}{y \sqrt{t}}$
The percentage error in $Q$ in given by,
$\frac{Δ Q}{Q} \times 100 = \frac{2}{5} (\frac{Δ x}{x} \times 100) + 3 (\frac{Δ Z}{Z} \times 100)$
$+ (\frac{Δ y}{y} \times 100) + \frac{1}{2} (\frac{Δ t}{t} \times 100)$
$\frac{Δ Q}{Q} \times 100 = (\frac{2}{5} \times 2.5 % + 3 \times 0.5 % + 2 % + \frac{1}{2} \times 1 %)$
$= 5 %$

AMU-2006

Units and Measurements

139888 What is the approximate percentage error in the measurement of time period of a simple pendulum if maximum errors in the measurement of length $l$ and gravitational acceleration $g$ are $3 %$ and $7 %$ respectively?

1

5 %

2

10 %

3

2 %

4

3 %

Explanation:

A We know that time period of a simple pendulum
$T = 2 π \sqrt{\frac{l}{g}}$
Where, $l =$ length of pendulum
$g =$ acceleration due to gravity
$T = 2 π \sqrt{\frac{l}{g}}$
Take $\log$ and differentiate
$\frac{Δ T}{T} = \frac{1}{2} [\frac{Δ l}{l} - \frac{Δ g}{g}]$
For maximize error consider $+ ve$ sign
$\frac{Δ T}{T} \times 100 = \frac{1}{2} [\frac{Δ l}{l} + \frac{Δ g}{g}] \times 100$
$= (\frac{3}{2} + \frac{7}{2}) % = \frac{10}{2} % = 5 %$
$\frac{Δ T}{T} \times 100 = 5 %$

UPSEE - 2016

Units and Measurements

139889 The relation gives the value of $x$ as $x = \frac{(a^{3} b^{3})}{(c \sqrt{d})}$
Find the percentage error in $x$ , if the percentage error in a, b, c, d, are $2 %, 1 %, 2 %$ and $4 %$ respectively :

1

\pm 13 %

2

\pm 5 %

3

\pm 9 %

4

\pm 8 %

Explanation:

A Given,
Relation gives value of, $x = \frac{a^{3} b^{3}}{c \sqrt{d}}$
Relative error is-
$\frac{Δ x}{x} = \pm 3 \frac{Δ a}{a} \pm 3 \frac{Δ b}{b} \pm \frac{Δ c}{c} \pm \frac{1}{2} \frac{Δ d}{d}$
Percentage error is-
$\frac{Δ x}{x} \times 100 = \pm 3 (\frac{Δ a}{a} \times 100) \pm 3 (\frac{Δ b}{b} \times 100)$
$\pm (\frac{Δ c}{c} \times 100) \pm \frac{1}{2} (\frac{Δ d}{d} \times 100)$
$= \pm 3 \times 2 \pm 3 \times 1 \pm 1 \times 2 \pm \frac{1}{2} \times 4$
$= \pm 6 \pm 3 \pm 2 \pm 2$
$%$ error in $x = \pm 13 %$

UPSEE - 2005

Units and Measurements

139890 When the voltage and current in a conductor are measured as $(100 \pm 4) V$ and $(5 \pm 0.2) A$ , then the percentage of error in the calculation of resistance is

1

8 %

2

4 %

3

20 %

4

10 %

5

6 %

Explanation:

A Given,
$(V \pm Δ V) = (100 \pm 4) V$
$(I \pm Δ I) = (5 \pm 0.2) A$
We know that,
$R = \frac{V}{I}$
$\frac{Δ R}{R} \times 100 = \frac{Δ V}{V} \times 100 + \frac{Δ I}{I} \times 100$
$= \frac{4}{100} \times 100 + \frac{0.2}{5} \times 100$
$%$ error in resistance $= 4 % + 4 % = 8 %$

Kerala CEE - 2016