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PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369775 Assertion :
When a spring is loaded, then shearing stress is produced in it.
Reason :
Shape of spring remains unchanged under the application of tangential stress.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369776 A river $10 m$ deep is flowing at $5 m / s$ . The shearing strees between horizontal layers of the river ( $η = 10^{- 3}$ SI units)

1

10^{- 3} N / m^{2}

2

0.8 \times 10^{- 3} N / m^{2}

3

0.5 \times 10^{- 3} N / m^{2}

4

1 N / m^{2}

Explanation:

shearing stress $= \frac{viscous force}{area}$
$= \frac{η A \frac{d v}{d x}}{A} = 10^{- 3} \times \frac{5}{10} = 0.5 \times 10^{- 3} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369777 Calculate the force $F$ needed to punch a 1.46 $c m$ diameter hole in a steel plate $1.27 c m$ thick. The ultimate shear strength of steel is
$345 M N / m^{2}$ .
supporting img

1

300 K N

2

200 K N

3

350 K N

4

225 K N

Explanation:

$F >$ (shear stress) $\times$ area
$\begin{matrix} ∴ F_{min} = (3.45 \times 10^{8}) (2 π r l) \\ = (3.45 \times 10^{8}) (2 \times 3.14 \times 0.73 \times 10^{- 2} \times 1.27 \times 10^{- 2}) \\ = 200 K N \end{matrix}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369778 The $Y_{e q}$ of the composite rod is (assume A $=$ Area of cross section of each rod), then $x = l_{1} = l_{2} = 1 m, F = 2 N / m^{2}, A = 1 s q \cdot m^{2}$
$Y_{1} = 2 \times 10^{11} N / m^{2}, Y_{2} = 3 \times 10^{11} N / m^{2}$
supporting img

1

2.4 \times 10^{11} N / m^{2}

2

2.86 \times 10^{11} N / m^{2}

3

2.88 \times 10^{11} N / m^{2}

4

2.4 \times 10^{8} N / m^{2}

Explanation:

The system can be considered as two springs connected in series.
$K = \frac{K_{1} K_{2}}{K_{1} + K_{2}}$
$K_{1} = \frac{Y_{1} A}{l_{1}}, K_{2} = \frac{Y_{2} A}{l_{2}}$
$K = \frac{Y_{1} Y_{2} A}{Y_{1} l_{2} + Y_{2} l_{1}} = \frac{Y A}{(l_{1} + l_{2})}$
$\Rightarrow Y = 2 \cdot 4 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369779 A steel rod of length $3 m$ and cross-sectional area $1 m^{2}$ ia acted upon by forces shown in figure.Determine the elongation of the length BC of the bar. Take $Y = 2.0 \times 10^{11} N / m^{2} .$
supporting img

1

4.5 \times 10^{- 7} m

2

13 \times 10^{- 7} m

3

5 \times 10^{- 7} m

4

3.5 \times 10^{- 7} m

Explanation:

The bar is at equilibrium. The net force from right or left of a section of BC is $70 K N$ . We know that the extension due to external forces F is given
$Δ l = \frac{F l}{A Y}; Δ l_{B C} = \frac{(70 \times 10^{3}) \times 1}{1 \times 2 \times 10^{11}} = 3.5 \times 10^{- 7} m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369775 Assertion :
When a spring is loaded, then shearing stress is produced in it.
Reason :
Shape of spring remains unchanged under the application of tangential stress.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369776 A river $10 m$ deep is flowing at $5 m / s$ . The shearing strees between horizontal layers of the river ( $η = 10^{- 3}$ SI units)

1

10^{- 3} N / m^{2}

2

0.8 \times 10^{- 3} N / m^{2}

3

0.5 \times 10^{- 3} N / m^{2}

4

1 N / m^{2}

Explanation:

shearing stress $= \frac{viscous force}{area}$
$= \frac{η A \frac{d v}{d x}}{A} = 10^{- 3} \times \frac{5}{10} = 0.5 \times 10^{- 3} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369777 Calculate the force $F$ needed to punch a 1.46 $c m$ diameter hole in a steel plate $1.27 c m$ thick. The ultimate shear strength of steel is
$345 M N / m^{2}$ .
supporting img

1

300 K N

2

200 K N

3

350 K N

4

225 K N

Explanation:

$F >$ (shear stress) $\times$ area
$\begin{matrix} ∴ F_{min} = (3.45 \times 10^{8}) (2 π r l) \\ = (3.45 \times 10^{8}) (2 \times 3.14 \times 0.73 \times 10^{- 2} \times 1.27 \times 10^{- 2}) \\ = 200 K N \end{matrix}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369778 The $Y_{e q}$ of the composite rod is (assume A $=$ Area of cross section of each rod), then $x = l_{1} = l_{2} = 1 m, F = 2 N / m^{2}, A = 1 s q \cdot m^{2}$
$Y_{1} = 2 \times 10^{11} N / m^{2}, Y_{2} = 3 \times 10^{11} N / m^{2}$
supporting img

1

2.4 \times 10^{11} N / m^{2}

2

2.86 \times 10^{11} N / m^{2}

3

2.88 \times 10^{11} N / m^{2}

4

2.4 \times 10^{8} N / m^{2}

Explanation:

The system can be considered as two springs connected in series.
$K = \frac{K_{1} K_{2}}{K_{1} + K_{2}}$
$K_{1} = \frac{Y_{1} A}{l_{1}}, K_{2} = \frac{Y_{2} A}{l_{2}}$
$K = \frac{Y_{1} Y_{2} A}{Y_{1} l_{2} + Y_{2} l_{1}} = \frac{Y A}{(l_{1} + l_{2})}$
$\Rightarrow Y = 2 \cdot 4 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369779 A steel rod of length $3 m$ and cross-sectional area $1 m^{2}$ ia acted upon by forces shown in figure.Determine the elongation of the length BC of the bar. Take $Y = 2.0 \times 10^{11} N / m^{2} .$
supporting img

1

4.5 \times 10^{- 7} m

2

13 \times 10^{- 7} m

3

5 \times 10^{- 7} m

4

3.5 \times 10^{- 7} m

Explanation:

The bar is at equilibrium. The net force from right or left of a section of BC is $70 K N$ . We know that the extension due to external forces F is given
$Δ l = \frac{F l}{A Y}; Δ l_{B C} = \frac{(70 \times 10^{3}) \times 1}{1 \times 2 \times 10^{11}} = 3.5 \times 10^{- 7} m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369775 Assertion :
When a spring is loaded, then shearing stress is produced in it.
Reason :
Shape of spring remains unchanged under the application of tangential stress.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369776 A river $10 m$ deep is flowing at $5 m / s$ . The shearing strees between horizontal layers of the river ( $η = 10^{- 3}$ SI units)

1

10^{- 3} N / m^{2}

2

0.8 \times 10^{- 3} N / m^{2}

3

0.5 \times 10^{- 3} N / m^{2}

4

1 N / m^{2}

Explanation:

shearing stress $= \frac{viscous force}{area}$
$= \frac{η A \frac{d v}{d x}}{A} = 10^{- 3} \times \frac{5}{10} = 0.5 \times 10^{- 3} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369777 Calculate the force $F$ needed to punch a 1.46 $c m$ diameter hole in a steel plate $1.27 c m$ thick. The ultimate shear strength of steel is
$345 M N / m^{2}$ .
supporting img

1

300 K N

2

200 K N

3

350 K N

4

225 K N

Explanation:

$F >$ (shear stress) $\times$ area
$\begin{matrix} ∴ F_{min} = (3.45 \times 10^{8}) (2 π r l) \\ = (3.45 \times 10^{8}) (2 \times 3.14 \times 0.73 \times 10^{- 2} \times 1.27 \times 10^{- 2}) \\ = 200 K N \end{matrix}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369778 The $Y_{e q}$ of the composite rod is (assume A $=$ Area of cross section of each rod), then $x = l_{1} = l_{2} = 1 m, F = 2 N / m^{2}, A = 1 s q \cdot m^{2}$
$Y_{1} = 2 \times 10^{11} N / m^{2}, Y_{2} = 3 \times 10^{11} N / m^{2}$
supporting img

1

2.4 \times 10^{11} N / m^{2}

2

2.86 \times 10^{11} N / m^{2}

3

2.88 \times 10^{11} N / m^{2}

4

2.4 \times 10^{8} N / m^{2}

Explanation:

The system can be considered as two springs connected in series.
$K = \frac{K_{1} K_{2}}{K_{1} + K_{2}}$
$K_{1} = \frac{Y_{1} A}{l_{1}}, K_{2} = \frac{Y_{2} A}{l_{2}}$
$K = \frac{Y_{1} Y_{2} A}{Y_{1} l_{2} + Y_{2} l_{1}} = \frac{Y A}{(l_{1} + l_{2})}$
$\Rightarrow Y = 2 \cdot 4 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369779 A steel rod of length $3 m$ and cross-sectional area $1 m^{2}$ ia acted upon by forces shown in figure.Determine the elongation of the length BC of the bar. Take $Y = 2.0 \times 10^{11} N / m^{2} .$
supporting img

1

4.5 \times 10^{- 7} m

2

13 \times 10^{- 7} m

3

5 \times 10^{- 7} m

4

3.5 \times 10^{- 7} m

Explanation:

The bar is at equilibrium. The net force from right or left of a section of BC is $70 K N$ . We know that the extension due to external forces F is given
$Δ l = \frac{F l}{A Y}; Δ l_{B C} = \frac{(70 \times 10^{3}) \times 1}{1 \times 2 \times 10^{11}} = 3.5 \times 10^{- 7} m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369775 Assertion :
When a spring is loaded, then shearing stress is produced in it.
Reason :
Shape of spring remains unchanged under the application of tangential stress.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369776 A river $10 m$ deep is flowing at $5 m / s$ . The shearing strees between horizontal layers of the river ( $η = 10^{- 3}$ SI units)

1

10^{- 3} N / m^{2}

2

0.8 \times 10^{- 3} N / m^{2}

3

0.5 \times 10^{- 3} N / m^{2}

4

1 N / m^{2}

Explanation:

shearing stress $= \frac{viscous force}{area}$
$= \frac{η A \frac{d v}{d x}}{A} = 10^{- 3} \times \frac{5}{10} = 0.5 \times 10^{- 3} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369777 Calculate the force $F$ needed to punch a 1.46 $c m$ diameter hole in a steel plate $1.27 c m$ thick. The ultimate shear strength of steel is
$345 M N / m^{2}$ .
supporting img

1

300 K N

2

200 K N

3

350 K N

4

225 K N

Explanation:

$F >$ (shear stress) $\times$ area
$\begin{matrix} ∴ F_{min} = (3.45 \times 10^{8}) (2 π r l) \\ = (3.45 \times 10^{8}) (2 \times 3.14 \times 0.73 \times 10^{- 2} \times 1.27 \times 10^{- 2}) \\ = 200 K N \end{matrix}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369778 The $Y_{e q}$ of the composite rod is (assume A $=$ Area of cross section of each rod), then $x = l_{1} = l_{2} = 1 m, F = 2 N / m^{2}, A = 1 s q \cdot m^{2}$
$Y_{1} = 2 \times 10^{11} N / m^{2}, Y_{2} = 3 \times 10^{11} N / m^{2}$
supporting img

1

2.4 \times 10^{11} N / m^{2}

2

2.86 \times 10^{11} N / m^{2}

3

2.88 \times 10^{11} N / m^{2}

4

2.4 \times 10^{8} N / m^{2}

Explanation:

The system can be considered as two springs connected in series.
$K = \frac{K_{1} K_{2}}{K_{1} + K_{2}}$
$K_{1} = \frac{Y_{1} A}{l_{1}}, K_{2} = \frac{Y_{2} A}{l_{2}}$
$K = \frac{Y_{1} Y_{2} A}{Y_{1} l_{2} + Y_{2} l_{1}} = \frac{Y A}{(l_{1} + l_{2})}$
$\Rightarrow Y = 2 \cdot 4 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369779 A steel rod of length $3 m$ and cross-sectional area $1 m^{2}$ ia acted upon by forces shown in figure.Determine the elongation of the length BC of the bar. Take $Y = 2.0 \times 10^{11} N / m^{2} .$
supporting img

1

4.5 \times 10^{- 7} m

2

13 \times 10^{- 7} m

3

5 \times 10^{- 7} m

4

3.5 \times 10^{- 7} m

Explanation:

The bar is at equilibrium. The net force from right or left of a section of BC is $70 K N$ . We know that the extension due to external forces F is given
$Δ l = \frac{F l}{A Y}; Δ l_{B C} = \frac{(70 \times 10^{3}) \times 1}{1 \times 2 \times 10^{11}} = 3.5 \times 10^{- 7} m$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369775 Assertion :
When a spring is loaded, then shearing stress is produced in it.
Reason :
Shape of spring remains unchanged under the application of tangential stress.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369776 A river $10 m$ deep is flowing at $5 m / s$ . The shearing strees between horizontal layers of the river ( $η = 10^{- 3}$ SI units)

1

10^{- 3} N / m^{2}

2

0.8 \times 10^{- 3} N / m^{2}

3

0.5 \times 10^{- 3} N / m^{2}

4

1 N / m^{2}

Explanation:

shearing stress $= \frac{viscous force}{area}$
$= \frac{η A \frac{d v}{d x}}{A} = 10^{- 3} \times \frac{5}{10} = 0.5 \times 10^{- 3} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369777 Calculate the force $F$ needed to punch a 1.46 $c m$ diameter hole in a steel plate $1.27 c m$ thick. The ultimate shear strength of steel is
$345 M N / m^{2}$ .
supporting img

1

300 K N

2

200 K N

3

350 K N

4

225 K N

Explanation:

$F >$ (shear stress) $\times$ area
$\begin{matrix} ∴ F_{min} = (3.45 \times 10^{8}) (2 π r l) \\ = (3.45 \times 10^{8}) (2 \times 3.14 \times 0.73 \times 10^{- 2} \times 1.27 \times 10^{- 2}) \\ = 200 K N \end{matrix}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369778 The $Y_{e q}$ of the composite rod is (assume A $=$ Area of cross section of each rod), then $x = l_{1} = l_{2} = 1 m, F = 2 N / m^{2}, A = 1 s q \cdot m^{2}$
$Y_{1} = 2 \times 10^{11} N / m^{2}, Y_{2} = 3 \times 10^{11} N / m^{2}$
supporting img

1

2.4 \times 10^{11} N / m^{2}

2

2.86 \times 10^{11} N / m^{2}

3

2.88 \times 10^{11} N / m^{2}

4

2.4 \times 10^{8} N / m^{2}

Explanation:

The system can be considered as two springs connected in series.
$K = \frac{K_{1} K_{2}}{K_{1} + K_{2}}$
$K_{1} = \frac{Y_{1} A}{l_{1}}, K_{2} = \frac{Y_{2} A}{l_{2}}$
$K = \frac{Y_{1} Y_{2} A}{Y_{1} l_{2} + Y_{2} l_{1}} = \frac{Y A}{(l_{1} + l_{2})}$
$\Rightarrow Y = 2 \cdot 4 \times 10^{11} N / m^{2}$

PHXI09:MECHANICAL PROPERTIES OF SOLIDS

369779 A steel rod of length $3 m$ and cross-sectional area $1 m^{2}$ ia acted upon by forces shown in figure.Determine the elongation of the length BC of the bar. Take $Y = 2.0 \times 10^{11} N / m^{2} .$
supporting img

1

4.5 \times 10^{- 7} m

2

13 \times 10^{- 7} m

3

5 \times 10^{- 7} m

4

3.5 \times 10^{- 7} m

Explanation:

The bar is at equilibrium. The net force from right or left of a section of BC is $70 K N$ . We know that the extension due to external forces F is given
$Δ l = \frac{F l}{A Y}; Δ l_{B C} = \frac{(70 \times 10^{3}) \times 1}{1 \times 2 \times 10^{11}} = 3.5 \times 10^{- 7} m$

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