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Mechanical Properties of Solids

141124 A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of $2 \times 10^{- 3}$ , then the percentage change in volume is

1 0.6

2 0.4

3 0.2

4 zero

Explanation:

D Given that, $v = 0.50, \frac{Δ l}{l} = 2 \times 10^{- 3}$
We know that,
$v = \frac{\frac{- Δ r}{r}}{\frac{Δ l}{l}} \Rightarrow \frac{Δ r}{r} = - v \times \frac{Δ l}{l}$
$\frac{Δ r}{r} = - 0.5 \times 2 \times 10^{- 3}$
$\frac{Δ r}{r} = - 10^{- 3}$
Volume of Rod
$V = π r^{2} l$
Taking the logarithm and partial differentiation we get,
$\frac{Δ V}{V} \times 100 = (\frac{Δ l}{l} + \frac{2 Δ r}{r}) \times 100$
$= [2 \times 10^{- 3} + 2 \times (- 10^{- 3})] \times 100$
$= [2 \times 10^{- 3} - 2 \times 10^{- 3}] \times 100$
$\frac{Δ V}{V} = 0 %$

WB JEE 2011

Mechanical Properties of Solids

141125 The Poisson's ratio of a material is 0.5 . If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by $4 %$ . The percentage increase in the length is

1

1 %

2

2 %

3

2.5 %

4

4 %

Explanation:

D Poisson's ratio $= 0.5$
Decrease in the cross-section area $\frac{Δ A}{A} = 4 %$
Since, density is constant. Therefore change in volume is zero, we have
$V = A \times l$
Taking partial differentiation with logarithm we get,
$\frac{Δ A}{A} + \frac{Δ l}{l} = 0$
$\frac{Δ l}{l} = \frac{- Δ A}{A}$
Percentage increases in length $= 4 %$

WB JEE 2009

Mechanical Properties of Solids

141126 If for a material Young's modulus $= 6.6 \times 10^{10}$
${Nm}^{- 2}$ and Bulk modulus $= 11 \times 10^{10} {Nm}^{- 2}$ , its Poisson's ratio is

1 0.1

2 0.2

3 0.3

4 0.4

Explanation:

D Given,
$Y = 6.6 \times 10^{10} {Nm}^{- 2}$
$B = 11 \times 10^{10} {Nm}^{- 2}$
The relation is between young modulus and Bulk modulus,
$Y = 3 B (1 - 2 v)$
$1 - 2 v = \frac{Y}{3 B}$
$2 v = 1 - \frac{Y}{3 B}$
$v = \frac{1}{2} - \frac{Y}{6 B} = \frac{1}{2} - \frac{6.6 \times 10^{10}}{6 \times 11 \times 10^{10}} = 0.5 - 0.1$
$v = 0.4$

EAMCET-1993

Mechanical Properties of Solids

141128 When a wire of length $10 m$ is subjected to a force of $100 N$ along its length, the lateral strain produced is $0.01 \times 10^{- 3}$ . The Poisson's ratio was found to be 0.4 . If the area of cross-section of wire is $0.025 m^{2}$ , its Young's modulus is

1

1.6 \times 10^{8} {Nm}^{- 2}

2

2.5 \times 10^{10} {Nm}^{- 2}

3

1.26 \times 10^{11} {Nm}^{- 2}

4

16 \times 10^{9} {Nm}^{- 2}

Explanation:

A Given, Wire Length $(l) = 10 m$ , Force $(F) =$ $100 N$ , Lateral Strain $(ε_{L}) = 0.01 \times 10^{- 3}$ , Possion's ratio $(σ) = 0.4$ , Area of cross-section $(A) = 0.025 m^{2}$
$∵ σ = \frac{lateral strain (ε_{L})}{longitudinal strain (ε)}$
Longitudinal strain $(ε) = \frac{0.01 \times 10^{- 3}}{0.4}$
$ε = 0.25 \times 10^{- 4}$
Young's modulus $(Y) = \frac{F}{A ε}$
$= \frac{100}{0.025 \times 0.25 \times 10^{- 4}}$
$Y = 1.6 \times 10^{8} {Nm}^{- 2}$

EAMCET-2007

Mechanical Properties of Solids

141129 When a wire is subjected to a force along its length, its length increases by $0.4 %$ and its radius decreases by $0.2 %$ . Then the Poisson's ratio of the material of the wire is.

1 0.8

2 0.5

3 0.2

4 0.1

Explanation:

B Given that, $\frac{δ d}{d} = 0.4 %$
$\frac{δ d}{d} = \frac{0.4}{100}$
$\frac{δ r}{r} = 0.2 %$
$\frac{δ r}{r} = \frac{0.2}{100}$
Then poission's ratio $(v) = \frac{δ r}{r} / \frac{δ d}{d}$
$= \frac{0.2 / 100}{0.4 / 100}$
$= \frac{0.2}{0.4} = \frac{1}{2}$
$v = \frac{1}{2} = 0.5$

AP EMCET(Medical)-2008

Mechanical Properties of Solids

141124 A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of $2 \times 10^{- 3}$ , then the percentage change in volume is

1 0.6

2 0.4

3 0.2

4 zero

Explanation:

D Given that, $v = 0.50, \frac{Δ l}{l} = 2 \times 10^{- 3}$
We know that,
$v = \frac{\frac{- Δ r}{r}}{\frac{Δ l}{l}} \Rightarrow \frac{Δ r}{r} = - v \times \frac{Δ l}{l}$
$\frac{Δ r}{r} = - 0.5 \times 2 \times 10^{- 3}$
$\frac{Δ r}{r} = - 10^{- 3}$
Volume of Rod
$V = π r^{2} l$
Taking the logarithm and partial differentiation we get,
$\frac{Δ V}{V} \times 100 = (\frac{Δ l}{l} + \frac{2 Δ r}{r}) \times 100$
$= [2 \times 10^{- 3} + 2 \times (- 10^{- 3})] \times 100$
$= [2 \times 10^{- 3} - 2 \times 10^{- 3}] \times 100$
$\frac{Δ V}{V} = 0 %$

WB JEE 2011

Mechanical Properties of Solids

141125 The Poisson's ratio of a material is 0.5 . If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by $4 %$ . The percentage increase in the length is

1

1 %

2

2 %

3

2.5 %

4

4 %

Explanation:

D Poisson's ratio $= 0.5$
Decrease in the cross-section area $\frac{Δ A}{A} = 4 %$
Since, density is constant. Therefore change in volume is zero, we have
$V = A \times l$
Taking partial differentiation with logarithm we get,
$\frac{Δ A}{A} + \frac{Δ l}{l} = 0$
$\frac{Δ l}{l} = \frac{- Δ A}{A}$
Percentage increases in length $= 4 %$

WB JEE 2009

Mechanical Properties of Solids

141126 If for a material Young's modulus $= 6.6 \times 10^{10}$
${Nm}^{- 2}$ and Bulk modulus $= 11 \times 10^{10} {Nm}^{- 2}$ , its Poisson's ratio is

1 0.1

2 0.2

3 0.3

4 0.4

Explanation:

D Given,
$Y = 6.6 \times 10^{10} {Nm}^{- 2}$
$B = 11 \times 10^{10} {Nm}^{- 2}$
The relation is between young modulus and Bulk modulus,
$Y = 3 B (1 - 2 v)$
$1 - 2 v = \frac{Y}{3 B}$
$2 v = 1 - \frac{Y}{3 B}$
$v = \frac{1}{2} - \frac{Y}{6 B} = \frac{1}{2} - \frac{6.6 \times 10^{10}}{6 \times 11 \times 10^{10}} = 0.5 - 0.1$
$v = 0.4$

EAMCET-1993

Mechanical Properties of Solids

141128 When a wire of length $10 m$ is subjected to a force of $100 N$ along its length, the lateral strain produced is $0.01 \times 10^{- 3}$ . The Poisson's ratio was found to be 0.4 . If the area of cross-section of wire is $0.025 m^{2}$ , its Young's modulus is

1

1.6 \times 10^{8} {Nm}^{- 2}

2

2.5 \times 10^{10} {Nm}^{- 2}

3

1.26 \times 10^{11} {Nm}^{- 2}

4

16 \times 10^{9} {Nm}^{- 2}

Explanation:

A Given, Wire Length $(l) = 10 m$ , Force $(F) =$ $100 N$ , Lateral Strain $(ε_{L}) = 0.01 \times 10^{- 3}$ , Possion's ratio $(σ) = 0.4$ , Area of cross-section $(A) = 0.025 m^{2}$
$∵ σ = \frac{lateral strain (ε_{L})}{longitudinal strain (ε)}$
Longitudinal strain $(ε) = \frac{0.01 \times 10^{- 3}}{0.4}$
$ε = 0.25 \times 10^{- 4}$
Young's modulus $(Y) = \frac{F}{A ε}$
$= \frac{100}{0.025 \times 0.25 \times 10^{- 4}}$
$Y = 1.6 \times 10^{8} {Nm}^{- 2}$

EAMCET-2007

Mechanical Properties of Solids

141129 When a wire is subjected to a force along its length, its length increases by $0.4 %$ and its radius decreases by $0.2 %$ . Then the Poisson's ratio of the material of the wire is.

1 0.8

2 0.5

3 0.2

4 0.1

Explanation:

B Given that, $\frac{δ d}{d} = 0.4 %$
$\frac{δ d}{d} = \frac{0.4}{100}$
$\frac{δ r}{r} = 0.2 %$
$\frac{δ r}{r} = \frac{0.2}{100}$
Then poission's ratio $(v) = \frac{δ r}{r} / \frac{δ d}{d}$
$= \frac{0.2 / 100}{0.4 / 100}$
$= \frac{0.2}{0.4} = \frac{1}{2}$
$v = \frac{1}{2} = 0.5$

AP EMCET(Medical)-2008

Mechanical Properties of Solids

141124 A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of $2 \times 10^{- 3}$ , then the percentage change in volume is

1 0.6

2 0.4

3 0.2

4 zero

Explanation:

D Given that, $v = 0.50, \frac{Δ l}{l} = 2 \times 10^{- 3}$
We know that,
$v = \frac{\frac{- Δ r}{r}}{\frac{Δ l}{l}} \Rightarrow \frac{Δ r}{r} = - v \times \frac{Δ l}{l}$
$\frac{Δ r}{r} = - 0.5 \times 2 \times 10^{- 3}$
$\frac{Δ r}{r} = - 10^{- 3}$
Volume of Rod
$V = π r^{2} l$
Taking the logarithm and partial differentiation we get,
$\frac{Δ V}{V} \times 100 = (\frac{Δ l}{l} + \frac{2 Δ r}{r}) \times 100$
$= [2 \times 10^{- 3} + 2 \times (- 10^{- 3})] \times 100$
$= [2 \times 10^{- 3} - 2 \times 10^{- 3}] \times 100$
$\frac{Δ V}{V} = 0 %$

WB JEE 2011

Mechanical Properties of Solids

141125 The Poisson's ratio of a material is 0.5 . If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by $4 %$ . The percentage increase in the length is

1

1 %

2

2 %

3

2.5 %

4

4 %

Explanation:

D Poisson's ratio $= 0.5$
Decrease in the cross-section area $\frac{Δ A}{A} = 4 %$
Since, density is constant. Therefore change in volume is zero, we have
$V = A \times l$
Taking partial differentiation with logarithm we get,
$\frac{Δ A}{A} + \frac{Δ l}{l} = 0$
$\frac{Δ l}{l} = \frac{- Δ A}{A}$
Percentage increases in length $= 4 %$

WB JEE 2009

Mechanical Properties of Solids

141126 If for a material Young's modulus $= 6.6 \times 10^{10}$
${Nm}^{- 2}$ and Bulk modulus $= 11 \times 10^{10} {Nm}^{- 2}$ , its Poisson's ratio is

1 0.1

2 0.2

3 0.3

4 0.4

Explanation:

D Given,
$Y = 6.6 \times 10^{10} {Nm}^{- 2}$
$B = 11 \times 10^{10} {Nm}^{- 2}$
The relation is between young modulus and Bulk modulus,
$Y = 3 B (1 - 2 v)$
$1 - 2 v = \frac{Y}{3 B}$
$2 v = 1 - \frac{Y}{3 B}$
$v = \frac{1}{2} - \frac{Y}{6 B} = \frac{1}{2} - \frac{6.6 \times 10^{10}}{6 \times 11 \times 10^{10}} = 0.5 - 0.1$
$v = 0.4$

EAMCET-1993

Mechanical Properties of Solids

141128 When a wire of length $10 m$ is subjected to a force of $100 N$ along its length, the lateral strain produced is $0.01 \times 10^{- 3}$ . The Poisson's ratio was found to be 0.4 . If the area of cross-section of wire is $0.025 m^{2}$ , its Young's modulus is

1

1.6 \times 10^{8} {Nm}^{- 2}

2

2.5 \times 10^{10} {Nm}^{- 2}

3

1.26 \times 10^{11} {Nm}^{- 2}

4

16 \times 10^{9} {Nm}^{- 2}

Explanation:

A Given, Wire Length $(l) = 10 m$ , Force $(F) =$ $100 N$ , Lateral Strain $(ε_{L}) = 0.01 \times 10^{- 3}$ , Possion's ratio $(σ) = 0.4$ , Area of cross-section $(A) = 0.025 m^{2}$
$∵ σ = \frac{lateral strain (ε_{L})}{longitudinal strain (ε)}$
Longitudinal strain $(ε) = \frac{0.01 \times 10^{- 3}}{0.4}$
$ε = 0.25 \times 10^{- 4}$
Young's modulus $(Y) = \frac{F}{A ε}$
$= \frac{100}{0.025 \times 0.25 \times 10^{- 4}}$
$Y = 1.6 \times 10^{8} {Nm}^{- 2}$

EAMCET-2007

Mechanical Properties of Solids

141129 When a wire is subjected to a force along its length, its length increases by $0.4 %$ and its radius decreases by $0.2 %$ . Then the Poisson's ratio of the material of the wire is.

1 0.8

2 0.5

3 0.2

4 0.1

Explanation:

B Given that, $\frac{δ d}{d} = 0.4 %$
$\frac{δ d}{d} = \frac{0.4}{100}$
$\frac{δ r}{r} = 0.2 %$
$\frac{δ r}{r} = \frac{0.2}{100}$
Then poission's ratio $(v) = \frac{δ r}{r} / \frac{δ d}{d}$
$= \frac{0.2 / 100}{0.4 / 100}$
$= \frac{0.2}{0.4} = \frac{1}{2}$
$v = \frac{1}{2} = 0.5$

AP EMCET(Medical)-2008

Mechanical Properties of Solids

141124 A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of $2 \times 10^{- 3}$ , then the percentage change in volume is

1 0.6

2 0.4

3 0.2

4 zero

Explanation:

D Given that, $v = 0.50, \frac{Δ l}{l} = 2 \times 10^{- 3}$
We know that,
$v = \frac{\frac{- Δ r}{r}}{\frac{Δ l}{l}} \Rightarrow \frac{Δ r}{r} = - v \times \frac{Δ l}{l}$
$\frac{Δ r}{r} = - 0.5 \times 2 \times 10^{- 3}$
$\frac{Δ r}{r} = - 10^{- 3}$
Volume of Rod
$V = π r^{2} l$
Taking the logarithm and partial differentiation we get,
$\frac{Δ V}{V} \times 100 = (\frac{Δ l}{l} + \frac{2 Δ r}{r}) \times 100$
$= [2 \times 10^{- 3} + 2 \times (- 10^{- 3})] \times 100$
$= [2 \times 10^{- 3} - 2 \times 10^{- 3}] \times 100$
$\frac{Δ V}{V} = 0 %$

WB JEE 2011

Mechanical Properties of Solids

141125 The Poisson's ratio of a material is 0.5 . If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by $4 %$ . The percentage increase in the length is

1

1 %

2

2 %

3

2.5 %

4

4 %

Explanation:

D Poisson's ratio $= 0.5$
Decrease in the cross-section area $\frac{Δ A}{A} = 4 %$
Since, density is constant. Therefore change in volume is zero, we have
$V = A \times l$
Taking partial differentiation with logarithm we get,
$\frac{Δ A}{A} + \frac{Δ l}{l} = 0$
$\frac{Δ l}{l} = \frac{- Δ A}{A}$
Percentage increases in length $= 4 %$

WB JEE 2009

Mechanical Properties of Solids

141126 If for a material Young's modulus $= 6.6 \times 10^{10}$
${Nm}^{- 2}$ and Bulk modulus $= 11 \times 10^{10} {Nm}^{- 2}$ , its Poisson's ratio is

1 0.1

2 0.2

3 0.3

4 0.4

Explanation:

D Given,
$Y = 6.6 \times 10^{10} {Nm}^{- 2}$
$B = 11 \times 10^{10} {Nm}^{- 2}$
The relation is between young modulus and Bulk modulus,
$Y = 3 B (1 - 2 v)$
$1 - 2 v = \frac{Y}{3 B}$
$2 v = 1 - \frac{Y}{3 B}$
$v = \frac{1}{2} - \frac{Y}{6 B} = \frac{1}{2} - \frac{6.6 \times 10^{10}}{6 \times 11 \times 10^{10}} = 0.5 - 0.1$
$v = 0.4$

EAMCET-1993

Mechanical Properties of Solids

141128 When a wire of length $10 m$ is subjected to a force of $100 N$ along its length, the lateral strain produced is $0.01 \times 10^{- 3}$ . The Poisson's ratio was found to be 0.4 . If the area of cross-section of wire is $0.025 m^{2}$ , its Young's modulus is

1

1.6 \times 10^{8} {Nm}^{- 2}

2

2.5 \times 10^{10} {Nm}^{- 2}

3

1.26 \times 10^{11} {Nm}^{- 2}

4

16 \times 10^{9} {Nm}^{- 2}

Explanation:

A Given, Wire Length $(l) = 10 m$ , Force $(F) =$ $100 N$ , Lateral Strain $(ε_{L}) = 0.01 \times 10^{- 3}$ , Possion's ratio $(σ) = 0.4$ , Area of cross-section $(A) = 0.025 m^{2}$
$∵ σ = \frac{lateral strain (ε_{L})}{longitudinal strain (ε)}$
Longitudinal strain $(ε) = \frac{0.01 \times 10^{- 3}}{0.4}$
$ε = 0.25 \times 10^{- 4}$
Young's modulus $(Y) = \frac{F}{A ε}$
$= \frac{100}{0.025 \times 0.25 \times 10^{- 4}}$
$Y = 1.6 \times 10^{8} {Nm}^{- 2}$

EAMCET-2007

Mechanical Properties of Solids

141129 When a wire is subjected to a force along its length, its length increases by $0.4 %$ and its radius decreases by $0.2 %$ . Then the Poisson's ratio of the material of the wire is.

1 0.8

2 0.5

3 0.2

4 0.1

Explanation:

B Given that, $\frac{δ d}{d} = 0.4 %$
$\frac{δ d}{d} = \frac{0.4}{100}$
$\frac{δ r}{r} = 0.2 %$
$\frac{δ r}{r} = \frac{0.2}{100}$
Then poission's ratio $(v) = \frac{δ r}{r} / \frac{δ d}{d}$
$= \frac{0.2 / 100}{0.4 / 100}$
$= \frac{0.2}{0.4} = \frac{1}{2}$
$v = \frac{1}{2} = 0.5$

AP EMCET(Medical)-2008

Mechanical Properties of Solids

141124 A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of $2 \times 10^{- 3}$ , then the percentage change in volume is

1 0.6

2 0.4

3 0.2

4 zero

Explanation:

D Given that, $v = 0.50, \frac{Δ l}{l} = 2 \times 10^{- 3}$
We know that,
$v = \frac{\frac{- Δ r}{r}}{\frac{Δ l}{l}} \Rightarrow \frac{Δ r}{r} = - v \times \frac{Δ l}{l}$
$\frac{Δ r}{r} = - 0.5 \times 2 \times 10^{- 3}$
$\frac{Δ r}{r} = - 10^{- 3}$
Volume of Rod
$V = π r^{2} l$
Taking the logarithm and partial differentiation we get,
$\frac{Δ V}{V} \times 100 = (\frac{Δ l}{l} + \frac{2 Δ r}{r}) \times 100$
$= [2 \times 10^{- 3} + 2 \times (- 10^{- 3})] \times 100$
$= [2 \times 10^{- 3} - 2 \times 10^{- 3}] \times 100$
$\frac{Δ V}{V} = 0 %$

WB JEE 2011

Mechanical Properties of Solids

141125 The Poisson's ratio of a material is 0.5 . If a force is applied to a wire of this material, there is a decrease in the cross-sectional area by $4 %$ . The percentage increase in the length is

1

1 %

2

2 %

3

2.5 %

4

4 %

Explanation:

D Poisson's ratio $= 0.5$
Decrease in the cross-section area $\frac{Δ A}{A} = 4 %$
Since, density is constant. Therefore change in volume is zero, we have
$V = A \times l$
Taking partial differentiation with logarithm we get,
$\frac{Δ A}{A} + \frac{Δ l}{l} = 0$
$\frac{Δ l}{l} = \frac{- Δ A}{A}$
Percentage increases in length $= 4 %$

WB JEE 2009

Mechanical Properties of Solids

141126 If for a material Young's modulus $= 6.6 \times 10^{10}$
${Nm}^{- 2}$ and Bulk modulus $= 11 \times 10^{10} {Nm}^{- 2}$ , its Poisson's ratio is

1 0.1

2 0.2

3 0.3

4 0.4

Explanation:

D Given,
$Y = 6.6 \times 10^{10} {Nm}^{- 2}$
$B = 11 \times 10^{10} {Nm}^{- 2}$
The relation is between young modulus and Bulk modulus,
$Y = 3 B (1 - 2 v)$
$1 - 2 v = \frac{Y}{3 B}$
$2 v = 1 - \frac{Y}{3 B}$
$v = \frac{1}{2} - \frac{Y}{6 B} = \frac{1}{2} - \frac{6.6 \times 10^{10}}{6 \times 11 \times 10^{10}} = 0.5 - 0.1$
$v = 0.4$

EAMCET-1993

Mechanical Properties of Solids

141128 When a wire of length $10 m$ is subjected to a force of $100 N$ along its length, the lateral strain produced is $0.01 \times 10^{- 3}$ . The Poisson's ratio was found to be 0.4 . If the area of cross-section of wire is $0.025 m^{2}$ , its Young's modulus is

1

1.6 \times 10^{8} {Nm}^{- 2}

2

2.5 \times 10^{10} {Nm}^{- 2}

3

1.26 \times 10^{11} {Nm}^{- 2}

4

16 \times 10^{9} {Nm}^{- 2}

Explanation:

A Given, Wire Length $(l) = 10 m$ , Force $(F) =$ $100 N$ , Lateral Strain $(ε_{L}) = 0.01 \times 10^{- 3}$ , Possion's ratio $(σ) = 0.4$ , Area of cross-section $(A) = 0.025 m^{2}$
$∵ σ = \frac{lateral strain (ε_{L})}{longitudinal strain (ε)}$
Longitudinal strain $(ε) = \frac{0.01 \times 10^{- 3}}{0.4}$
$ε = 0.25 \times 10^{- 4}$
Young's modulus $(Y) = \frac{F}{A ε}$
$= \frac{100}{0.025 \times 0.25 \times 10^{- 4}}$
$Y = 1.6 \times 10^{8} {Nm}^{- 2}$

EAMCET-2007

Mechanical Properties of Solids

141129 When a wire is subjected to a force along its length, its length increases by $0.4 %$ and its radius decreases by $0.2 %$ . Then the Poisson's ratio of the material of the wire is.

1 0.8

2 0.5

3 0.2

4 0.1

Explanation:

B Given that, $\frac{δ d}{d} = 0.4 %$
$\frac{δ d}{d} = \frac{0.4}{100}$
$\frac{δ r}{r} = 0.2 %$
$\frac{δ r}{r} = \frac{0.2}{100}$
Then poission's ratio $(v) = \frac{δ r}{r} / \frac{δ d}{d}$
$= \frac{0.2 / 100}{0.4 / 100}$
$= \frac{0.2}{0.4} = \frac{1}{2}$
$v = \frac{1}{2} = 0.5$

AP EMCET(Medical)-2008