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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369911 If the work done in stretching a wire by $1 m m$ is $2 J$ , the work necessary for stretching another wire of same material but with double the radius of cross-section and half the length by $1 m m$ is

1

4 J

2

\frac{1}{4} J

3

16 J

4

8 J

Explanation:

Work done in stretching the wire is
$\begin{aligned} W = \frac{1}{2} K x^{2} (K = \frac{Y A}{L}) \\ W = \frac{1}{2} \frac{Y A x^{2}}{L} \\ \frac{W_{1}}{W_{2}} = \frac{r_{1}^{2}}{L_{1}} \times \frac{L_{2}}{r_{2}^{2}} = \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2} \\ \frac{W_{1}}{W_{2}} = \frac{1}{2} \times {(\frac{1}{2})}^{2} = \frac{1}{8} \\ W_{2} = 8 W_{1} = 16 J \end{aligned}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369912 The increase in energy of a metal bar of length $L$ and cross - sectional area $A$ when compressed with a load $F$ along its length is (where, $Y =$ Young's modulus of the material of metal bar)

1

\frac{F L}{2 A Y}

2

\frac{F^{2} L}{2 A Y}

3

\frac{F L}{A Y}

4

\frac{F^{2} L^{2}}{2 A Y}

Explanation:

Increase in elastic potential energy $U$
$= \frac{1}{2} \times ($ stress $) \times ($ strain $) \times ($ volume $)$
$= \frac{1}{2} \times \frac{(stress)^{2}}{Y} \times ($ volume $)$
$= \frac{1}{2} \times \frac{{(\frac{F}{A})}^{2}}{Y} \times (L . A) = \frac{F^{2} L}{2 A Y}$
$[$ as stress $= \frac{F}{A}$ , volume $= L \times A]$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369913 Statement A :
Identical springs of steel and copper are equally stretched. More work will be done on the steel spring
Statement B :
Steel is more elastic than copper.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

$= \frac{1}{2} \times Stress \times Strain = \frac{1}{2} \times Y \times (Strain)^{2}$
Since, elasticity of steel is more than copper, hence more work has to be done in order to stretch the steel. So option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369914 A catapult has rubber cord of length $10 c m$ and cross-sectional area $20 m m^{2} .$ It is stretched by $4 c m$ to project a pebble of $10 g .$ Find velocity (in $m / s$ ) of the pebble just after its launch is
( $Y_{r u b b e r} = 10^{8} N / m^{2}$ and $\sqrt{2} = 1.41$ )

1

75.2 m / s

2

56.4 m / s

3

33.6 m / s

4

22.6 m / s

Explanation:

Elastic potential energy $=$ kinetic energy
$∴ \frac{1}{2} \frac{Y A l^{2}}{L} = \frac{1}{2} m v^{2}$
$∴ v = \sqrt{\frac{Y A l^{2}}{m L}}$
$= \sqrt{\frac{10^{8} \times 20 \times 10^{- 6} \times {(4 \times 10^{- 2})}^{2}}{10 \times 10^{- 3} \times 10 \times 10^{- 2}}}$
$∴ v = 40 \sqrt{2} = 56.4 m / s$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369911 If the work done in stretching a wire by $1 m m$ is $2 J$ , the work necessary for stretching another wire of same material but with double the radius of cross-section and half the length by $1 m m$ is

1

4 J

2

\frac{1}{4} J

3

16 J

4

8 J

Explanation:

Work done in stretching the wire is
$\begin{aligned} W = \frac{1}{2} K x^{2} (K = \frac{Y A}{L}) \\ W = \frac{1}{2} \frac{Y A x^{2}}{L} \\ \frac{W_{1}}{W_{2}} = \frac{r_{1}^{2}}{L_{1}} \times \frac{L_{2}}{r_{2}^{2}} = \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2} \\ \frac{W_{1}}{W_{2}} = \frac{1}{2} \times {(\frac{1}{2})}^{2} = \frac{1}{8} \\ W_{2} = 8 W_{1} = 16 J \end{aligned}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369912 The increase in energy of a metal bar of length $L$ and cross - sectional area $A$ when compressed with a load $F$ along its length is (where, $Y =$ Young's modulus of the material of metal bar)

1

\frac{F L}{2 A Y}

2

\frac{F^{2} L}{2 A Y}

3

\frac{F L}{A Y}

4

\frac{F^{2} L^{2}}{2 A Y}

Explanation:

Increase in elastic potential energy $U$
$= \frac{1}{2} \times ($ stress $) \times ($ strain $) \times ($ volume $)$
$= \frac{1}{2} \times \frac{(stress)^{2}}{Y} \times ($ volume $)$
$= \frac{1}{2} \times \frac{{(\frac{F}{A})}^{2}}{Y} \times (L . A) = \frac{F^{2} L}{2 A Y}$
$[$ as stress $= \frac{F}{A}$ , volume $= L \times A]$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369913 Statement A :
Identical springs of steel and copper are equally stretched. More work will be done on the steel spring
Statement B :
Steel is more elastic than copper.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

$= \frac{1}{2} \times Stress \times Strain = \frac{1}{2} \times Y \times (Strain)^{2}$
Since, elasticity of steel is more than copper, hence more work has to be done in order to stretch the steel. So option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369914 A catapult has rubber cord of length $10 c m$ and cross-sectional area $20 m m^{2} .$ It is stretched by $4 c m$ to project a pebble of $10 g .$ Find velocity (in $m / s$ ) of the pebble just after its launch is
( $Y_{r u b b e r} = 10^{8} N / m^{2}$ and $\sqrt{2} = 1.41$ )

1

75.2 m / s

2

56.4 m / s

3

33.6 m / s

4

22.6 m / s

Explanation:

Elastic potential energy $=$ kinetic energy
$∴ \frac{1}{2} \frac{Y A l^{2}}{L} = \frac{1}{2} m v^{2}$
$∴ v = \sqrt{\frac{Y A l^{2}}{m L}}$
$= \sqrt{\frac{10^{8} \times 20 \times 10^{- 6} \times {(4 \times 10^{- 2})}^{2}}{10 \times 10^{- 3} \times 10 \times 10^{- 2}}}$
$∴ v = 40 \sqrt{2} = 56.4 m / s$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369911 If the work done in stretching a wire by $1 m m$ is $2 J$ , the work necessary for stretching another wire of same material but with double the radius of cross-section and half the length by $1 m m$ is

1

4 J

2

\frac{1}{4} J

3

16 J

4

8 J

Explanation:

Work done in stretching the wire is
$\begin{aligned} W = \frac{1}{2} K x^{2} (K = \frac{Y A}{L}) \\ W = \frac{1}{2} \frac{Y A x^{2}}{L} \\ \frac{W_{1}}{W_{2}} = \frac{r_{1}^{2}}{L_{1}} \times \frac{L_{2}}{r_{2}^{2}} = \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2} \\ \frac{W_{1}}{W_{2}} = \frac{1}{2} \times {(\frac{1}{2})}^{2} = \frac{1}{8} \\ W_{2} = 8 W_{1} = 16 J \end{aligned}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369912 The increase in energy of a metal bar of length $L$ and cross - sectional area $A$ when compressed with a load $F$ along its length is (where, $Y =$ Young's modulus of the material of metal bar)

1

\frac{F L}{2 A Y}

2

\frac{F^{2} L}{2 A Y}

3

\frac{F L}{A Y}

4

\frac{F^{2} L^{2}}{2 A Y}

Explanation:

Increase in elastic potential energy $U$
$= \frac{1}{2} \times ($ stress $) \times ($ strain $) \times ($ volume $)$
$= \frac{1}{2} \times \frac{(stress)^{2}}{Y} \times ($ volume $)$
$= \frac{1}{2} \times \frac{{(\frac{F}{A})}^{2}}{Y} \times (L . A) = \frac{F^{2} L}{2 A Y}$
$[$ as stress $= \frac{F}{A}$ , volume $= L \times A]$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369913 Statement A :
Identical springs of steel and copper are equally stretched. More work will be done on the steel spring
Statement B :
Steel is more elastic than copper.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

$= \frac{1}{2} \times Stress \times Strain = \frac{1}{2} \times Y \times (Strain)^{2}$
Since, elasticity of steel is more than copper, hence more work has to be done in order to stretch the steel. So option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369914 A catapult has rubber cord of length $10 c m$ and cross-sectional area $20 m m^{2} .$ It is stretched by $4 c m$ to project a pebble of $10 g .$ Find velocity (in $m / s$ ) of the pebble just after its launch is
( $Y_{r u b b e r} = 10^{8} N / m^{2}$ and $\sqrt{2} = 1.41$ )

1

75.2 m / s

2

56.4 m / s

3

33.6 m / s

4

22.6 m / s

Explanation:

Elastic potential energy $=$ kinetic energy
$∴ \frac{1}{2} \frac{Y A l^{2}}{L} = \frac{1}{2} m v^{2}$
$∴ v = \sqrt{\frac{Y A l^{2}}{m L}}$
$= \sqrt{\frac{10^{8} \times 20 \times 10^{- 6} \times {(4 \times 10^{- 2})}^{2}}{10 \times 10^{- 3} \times 10 \times 10^{- 2}}}$
$∴ v = 40 \sqrt{2} = 56.4 m / s$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369911 If the work done in stretching a wire by $1 m m$ is $2 J$ , the work necessary for stretching another wire of same material but with double the radius of cross-section and half the length by $1 m m$ is

1

4 J

2

\frac{1}{4} J

3

16 J

4

8 J

Explanation:

Work done in stretching the wire is
$\begin{aligned} W = \frac{1}{2} K x^{2} (K = \frac{Y A}{L}) \\ W = \frac{1}{2} \frac{Y A x^{2}}{L} \\ \frac{W_{1}}{W_{2}} = \frac{r_{1}^{2}}{L_{1}} \times \frac{L_{2}}{r_{2}^{2}} = \frac{L_{2}}{L_{1}} \times {(\frac{r_{1}}{r_{2}})}^{2} \\ \frac{W_{1}}{W_{2}} = \frac{1}{2} \times {(\frac{1}{2})}^{2} = \frac{1}{8} \\ W_{2} = 8 W_{1} = 16 J \end{aligned}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369912 The increase in energy of a metal bar of length $L$ and cross - sectional area $A$ when compressed with a load $F$ along its length is (where, $Y =$ Young's modulus of the material of metal bar)

1

\frac{F L}{2 A Y}

2

\frac{F^{2} L}{2 A Y}

3

\frac{F L}{A Y}

4

\frac{F^{2} L^{2}}{2 A Y}

Explanation:

Increase in elastic potential energy $U$
$= \frac{1}{2} \times ($ stress $) \times ($ strain $) \times ($ volume $)$
$= \frac{1}{2} \times \frac{(stress)^{2}}{Y} \times ($ volume $)$
$= \frac{1}{2} \times \frac{{(\frac{F}{A})}^{2}}{Y} \times (L . A) = \frac{F^{2} L}{2 A Y}$
$[$ as stress $= \frac{F}{A}$ , volume $= L \times A]$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369913 Statement A :
Identical springs of steel and copper are equally stretched. More work will be done on the steel spring
Statement B :
Steel is more elastic than copper.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

$= \frac{1}{2} \times Stress \times Strain = \frac{1}{2} \times Y \times (Strain)^{2}$
Since, elasticity of steel is more than copper, hence more work has to be done in order to stretch the steel. So option (3) is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369914 A catapult has rubber cord of length $10 c m$ and cross-sectional area $20 m m^{2} .$ It is stretched by $4 c m$ to project a pebble of $10 g .$ Find velocity (in $m / s$ ) of the pebble just after its launch is
( $Y_{r u b b e r} = 10^{8} N / m^{2}$ and $\sqrt{2} = 1.41$ )

1

75.2 m / s

2

56.4 m / s

3

33.6 m / s

4

22.6 m / s

Explanation:

Elastic potential energy $=$ kinetic energy
$∴ \frac{1}{2} \frac{Y A l^{2}}{L} = \frac{1}{2} m v^{2}$
$∴ v = \sqrt{\frac{Y A l^{2}}{m L}}$
$= \sqrt{\frac{10^{8} \times 20 \times 10^{- 6} \times {(4 \times 10^{- 2})}^{2}}{10 \times 10^{- 3} \times 10 \times 10^{- 2}}}$
$∴ v = 40 \sqrt{2} = 56.4 m / s$

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