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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369883 A $100 m$ long wire having cross-section area $6.25 \times 10^{- 4} m^{2}$ and Young's modulus is $10^{10} N m^{- 2}$ is subjected to a load of $250 N$ , then the elongation in the wire will be

1

6.25 \times 10^{- 3} m

2

4 \times 10^{- 3} m

3

4 \times 10^{- 4} m

4

6.25 \times 10^{- 6} m

Explanation:

Given, $A = 6.25 \times 10^{- 4} m^{2},$
$Y = 10^{10} N m^{- 2}$
$F = 250 N$
Young's modulus of the wire is given by
$Y = \frac{(F / A)}{(\frac{Δ L}{L})} = \frac{250}{6.25 \times 10^{- 4}} \times \frac{L}{Δ L}$
$\Rightarrow 10^{10} = \frac{250}{6.25 \times 10^{- 4}} \times \frac{100}{Δ L}$
$\Rightarrow Δ L = \frac{25}{6.25} \times 10^{- 3}$
$\Rightarrow Δ L = 4 \times 10^{- 3} m$
So option (2) is correct.

JEE - 2023

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369884 Four identical hollow cylindrical columns of steel support a big structure of mass $50, 000 k g$ The inner and outer radii of each column are $30 c m$ and $40 c m$ respectively, Assuming the load distribution to be uniform calculate the compressional strain of each column, the Young's modulus of steel is $2 \times 10^{11} P a$ .

1

2.78 \times 10^{- 6}

2

3.78 \times 10^{- 6}

3

2.78 \times 10^{- 4}

4

3.78 \times 10^{- 4}

Explanation:

$\frac{Δ ℓ}{ℓ} = \frac{M g}{4 A Y}$
$\begin{aligned} = \frac{50000 \times 10}{4 \times π (1600 - 900) \times 10^{- 4} \times 2 \times 10^{11}} \\ = 2.78 \times 10^{- 6} . \end{aligned}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369885 Graph between the applied force $&$ the change in length of a wire with in elastic limit is a

1 Straight line with negative slope

2 Straight line with zero slope

3 Straight line with positive slope

4 None of the above

Explanation:

$F = \frac{Y A Δ l}{l}$
The graph between $F & Δ l$ is a straight line.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369886 The Young's modulus of air is

1 Infinity

2 More than 1 but not infinity

3 Less than 1 but not zero

4 Zero

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369883 A $100 m$ long wire having cross-section area $6.25 \times 10^{- 4} m^{2}$ and Young's modulus is $10^{10} N m^{- 2}$ is subjected to a load of $250 N$ , then the elongation in the wire will be

1

6.25 \times 10^{- 3} m

2

4 \times 10^{- 3} m

3

4 \times 10^{- 4} m

4

6.25 \times 10^{- 6} m

Explanation:

Given, $A = 6.25 \times 10^{- 4} m^{2},$
$Y = 10^{10} N m^{- 2}$
$F = 250 N$
Young's modulus of the wire is given by
$Y = \frac{(F / A)}{(\frac{Δ L}{L})} = \frac{250}{6.25 \times 10^{- 4}} \times \frac{L}{Δ L}$
$\Rightarrow 10^{10} = \frac{250}{6.25 \times 10^{- 4}} \times \frac{100}{Δ L}$
$\Rightarrow Δ L = \frac{25}{6.25} \times 10^{- 3}$
$\Rightarrow Δ L = 4 \times 10^{- 3} m$
So option (2) is correct.

JEE - 2023

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369884 Four identical hollow cylindrical columns of steel support a big structure of mass $50, 000 k g$ The inner and outer radii of each column are $30 c m$ and $40 c m$ respectively, Assuming the load distribution to be uniform calculate the compressional strain of each column, the Young's modulus of steel is $2 \times 10^{11} P a$ .

1

2.78 \times 10^{- 6}

2

3.78 \times 10^{- 6}

3

2.78 \times 10^{- 4}

4

3.78 \times 10^{- 4}

Explanation:

$\frac{Δ ℓ}{ℓ} = \frac{M g}{4 A Y}$
$\begin{aligned} = \frac{50000 \times 10}{4 \times π (1600 - 900) \times 10^{- 4} \times 2 \times 10^{11}} \\ = 2.78 \times 10^{- 6} . \end{aligned}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369885 Graph between the applied force $&$ the change in length of a wire with in elastic limit is a

1 Straight line with negative slope

2 Straight line with zero slope

3 Straight line with positive slope

4 None of the above

Explanation:

$F = \frac{Y A Δ l}{l}$
The graph between $F & Δ l$ is a straight line.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369886 The Young's modulus of air is

1 Infinity

2 More than 1 but not infinity

3 Less than 1 but not zero

4 Zero

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369883 A $100 m$ long wire having cross-section area $6.25 \times 10^{- 4} m^{2}$ and Young's modulus is $10^{10} N m^{- 2}$ is subjected to a load of $250 N$ , then the elongation in the wire will be

1

6.25 \times 10^{- 3} m

2

4 \times 10^{- 3} m

3

4 \times 10^{- 4} m

4

6.25 \times 10^{- 6} m

Explanation:

Given, $A = 6.25 \times 10^{- 4} m^{2},$
$Y = 10^{10} N m^{- 2}$
$F = 250 N$
Young's modulus of the wire is given by
$Y = \frac{(F / A)}{(\frac{Δ L}{L})} = \frac{250}{6.25 \times 10^{- 4}} \times \frac{L}{Δ L}$
$\Rightarrow 10^{10} = \frac{250}{6.25 \times 10^{- 4}} \times \frac{100}{Δ L}$
$\Rightarrow Δ L = \frac{25}{6.25} \times 10^{- 3}$
$\Rightarrow Δ L = 4 \times 10^{- 3} m$
So option (2) is correct.

JEE - 2023

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369884 Four identical hollow cylindrical columns of steel support a big structure of mass $50, 000 k g$ The inner and outer radii of each column are $30 c m$ and $40 c m$ respectively, Assuming the load distribution to be uniform calculate the compressional strain of each column, the Young's modulus of steel is $2 \times 10^{11} P a$ .

1

2.78 \times 10^{- 6}

2

3.78 \times 10^{- 6}

3

2.78 \times 10^{- 4}

4

3.78 \times 10^{- 4}

Explanation:

$\frac{Δ ℓ}{ℓ} = \frac{M g}{4 A Y}$
$\begin{aligned} = \frac{50000 \times 10}{4 \times π (1600 - 900) \times 10^{- 4} \times 2 \times 10^{11}} \\ = 2.78 \times 10^{- 6} . \end{aligned}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369885 Graph between the applied force $&$ the change in length of a wire with in elastic limit is a

1 Straight line with negative slope

2 Straight line with zero slope

3 Straight line with positive slope

4 None of the above

Explanation:

$F = \frac{Y A Δ l}{l}$
The graph between $F & Δ l$ is a straight line.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369886 The Young's modulus of air is

1 Infinity

2 More than 1 but not infinity

3 Less than 1 but not zero

4 Zero

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369883 A $100 m$ long wire having cross-section area $6.25 \times 10^{- 4} m^{2}$ and Young's modulus is $10^{10} N m^{- 2}$ is subjected to a load of $250 N$ , then the elongation in the wire will be

1

6.25 \times 10^{- 3} m

2

4 \times 10^{- 3} m

3

4 \times 10^{- 4} m

4

6.25 \times 10^{- 6} m

Explanation:

Given, $A = 6.25 \times 10^{- 4} m^{2},$
$Y = 10^{10} N m^{- 2}$
$F = 250 N$
Young's modulus of the wire is given by
$Y = \frac{(F / A)}{(\frac{Δ L}{L})} = \frac{250}{6.25 \times 10^{- 4}} \times \frac{L}{Δ L}$
$\Rightarrow 10^{10} = \frac{250}{6.25 \times 10^{- 4}} \times \frac{100}{Δ L}$
$\Rightarrow Δ L = \frac{25}{6.25} \times 10^{- 3}$
$\Rightarrow Δ L = 4 \times 10^{- 3} m$
So option (2) is correct.

JEE - 2023

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369884 Four identical hollow cylindrical columns of steel support a big structure of mass $50, 000 k g$ The inner and outer radii of each column are $30 c m$ and $40 c m$ respectively, Assuming the load distribution to be uniform calculate the compressional strain of each column, the Young's modulus of steel is $2 \times 10^{11} P a$ .

1

2.78 \times 10^{- 6}

2

3.78 \times 10^{- 6}

3

2.78 \times 10^{- 4}

4

3.78 \times 10^{- 4}

Explanation:

$\frac{Δ ℓ}{ℓ} = \frac{M g}{4 A Y}$
$\begin{aligned} = \frac{50000 \times 10}{4 \times π (1600 - 900) \times 10^{- 4} \times 2 \times 10^{11}} \\ = 2.78 \times 10^{- 6} . \end{aligned}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369885 Graph between the applied force $&$ the change in length of a wire with in elastic limit is a

1 Straight line with negative slope

2 Straight line with zero slope

3 Straight line with positive slope

4 None of the above

Explanation:

$F = \frac{Y A Δ l}{l}$
The graph between $F & Δ l$ is a straight line.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369886 The Young's modulus of air is

1 Infinity

2 More than 1 but not infinity

3 Less than 1 but not zero

4 Zero