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Mechanical Properties of Solids

140958 A copper wire of length $2.4 \mathrm{~m}$ and an aluminum wire of length $0.7 \mathrm{~m}$. Both having diameter $2 \mathrm{~mm}$ are connected end to end. When stretched by a load, the obtained elongation is found to be $0.6 \mathrm{~mm}$. The applied load is
(Young's modulus of copper $=1.2 \times 10^{11} \mathrm{Nm}^{-2}$ and Young's modulus of aluminum $=0.7 \times 10^{11}$ $\mathrm{Nm}^{-2}$ )

1 $12 \pi \mathrm{N}$

2 $24 \pi \mathrm{N}$

3 $20 \pi \mathrm{N}$

4 $80 \pi \mathrm{N}$

Explanation:

C Given,
$\mathrm{L}_{\mathrm{C}}=2.4 \mathrm{~m} \quad, \quad \mathrm{Y}_{\mathrm{C}}=1.2 \times 10^{11} \mathrm{Nm}^{-2}$
$\mathrm{L}_{\mathrm{A}}=0.7 \mathrm{~m} \quad, \quad \mathrm{Y}_{\mathrm{A}}=0.7 \times 10^{11} \mathrm{Nm}^{-2}$
Load $=\mathrm{W}$
Diameter $=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$
So radius $=1 \times 10^{-3} \mathrm{~m}$
Total elongation $=0.6 \mathrm{~mm}=0.6 \times 10^{-3} \mathrm{~m}$ As we know,
Stress $=$ Young's modulus $\times$ strain
$\frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y} \times \frac{\Delta \mathrm{L}}{\mathrm{L}}$
$\therefore \quad \frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y}_{\mathrm{C}} \times \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{C}}}=\mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}$
or, $\quad \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\Delta \mathrm{L}_{\mathrm{A}}}=\frac{\mathrm{Y}_{\mathrm{A}}}{\mathrm{Y}_{\mathrm{C}}} \times \frac{\mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{A}}}$
$=\frac{0.7 \times 10^{11}}{1.2 \times 10^{11}} \times \frac{2.4}{0.7}$
$\frac{\Delta \mathrm{L}_{\mathrm{C}}}{\Delta \mathrm{L}_{\mathrm{A}}}=\frac{2}{1}$
Net elongation $=0.6 \times 10^{-3} \mathrm{~m}$
$\Delta \mathrm{L}_{\mathrm{C}}+\Delta \mathrm{L}_{\mathrm{A}}=0.6 \times 10^{-3} \mathrm{~m}$
From equation (ii) and (iii),
$\Delta \mathrm{L}_{\mathrm{C}} =0.4 \times 10^{-3} \mathrm{~m}$
$\Delta \mathrm{L}_{\mathrm{A}} =0.2 \times 10^{-3} \mathrm{~m}$
From equation (i), we get
$\frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}=\mathrm{Y}_{\mathrm{C}} \times \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{C}}}$
$\mathrm{W}=\mathrm{A} \times \mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}$
Put the value of $\triangle \mathrm{L}_{\mathrm{A}}$ and $\triangle \mathrm{L}_{\mathrm{C}}$, we get
$\mathrm{W}=\pi \mathrm{r}^{2} \times 0.7 \times 10^{11} \times \frac{0.2 \times 10^{-3}}{0.7}$
$\mathrm{~W}=\pi \times\left(1 \times 10^{-3}\right)^{2} \times 0.7 \times 10^{11} \times \frac{0.2 \times 10^{-3}}{0.7}$
$\mathrm{~W}=20 \pi$

Shift-II]

Mechanical Properties of Solids

140959 A metallic rod breaks when strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$. The area of cross section of support a load of $10^{4} \mathrm{~N}$ is

1 $7.1 \times 10^{-4} \mathrm{~m}^{2}$

2 $7.1 \times 10^{-2} \mathrm{~m}^{2}$

3 $7.1 \times 10^{-8} \mathrm{~m}^{2}$

4 $7.1 \times 10^{-6} \mathrm{~m}^{2}$

Karnataka CET-2022

Mechanical Properties of Solids

140960 A metal cube has an edge length $0.1 \mathrm{~m}$. The bulk modulus of the metal is $1.4 \times 10^{11} \mathrm{Nm}^{-2}$. When subjected to hydraulic pressure, the volume contraction of the cube is $5 \times 10^{-2} \mathrm{~cm}^{3}$. The applied pressure is

1 $2 \times 10^{6} \mathrm{~Pa}$

2 $1.4 \times 10^{7} \mathrm{~Pa}$

3 $7 \times 10^{6} \mathrm{~Pa}$

4 $5 \times 10^{5} \mathrm{~Pa}$

Shift-I]

Mechanical Properties of Solids

140961 A horizontal metal rod of diameter $4 \mathrm{~cm}$ projects $6 \mathrm{~cm}$ from a wall. The shear modulus of the metal is $3 \times 10^{10} \mathrm{~N} \mathrm{~m}^{-2}$. An object of mass $1100 \mathrm{~kg}$ is suspended from the free end of the rod. The free end of the rod moves down by a distance of (Acceleration due to gravity $=10$ $\mathrm{ms}^{-2}$, neglect the mass of the rod)

1 $2.5 \times 10^{-6} \mathrm{~m}$

2 $6.75 \times 10^{-5} \mathrm{~m}$

3 $2.25 \times 10^{-6} \mathrm{~m}$

4 $1.75 \times 10^{-5} \mathrm{~m}$

Shift-II]

Mechanical Properties of Solids

140958 A copper wire of length $2.4 \mathrm{~m}$ and an aluminum wire of length $0.7 \mathrm{~m}$. Both having diameter $2 \mathrm{~mm}$ are connected end to end. When stretched by a load, the obtained elongation is found to be $0.6 \mathrm{~mm}$. The applied load is
(Young's modulus of copper $=1.2 \times 10^{11} \mathrm{Nm}^{-2}$ and Young's modulus of aluminum $=0.7 \times 10^{11}$ $\mathrm{Nm}^{-2}$ )

1 $12 \pi \mathrm{N}$

2 $24 \pi \mathrm{N}$

3 $20 \pi \mathrm{N}$

4 $80 \pi \mathrm{N}$

Explanation:

C Given,
$\mathrm{L}_{\mathrm{C}}=2.4 \mathrm{~m} \quad, \quad \mathrm{Y}_{\mathrm{C}}=1.2 \times 10^{11} \mathrm{Nm}^{-2}$
$\mathrm{L}_{\mathrm{A}}=0.7 \mathrm{~m} \quad, \quad \mathrm{Y}_{\mathrm{A}}=0.7 \times 10^{11} \mathrm{Nm}^{-2}$
Load $=\mathrm{W}$
Diameter $=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$
So radius $=1 \times 10^{-3} \mathrm{~m}$
Total elongation $=0.6 \mathrm{~mm}=0.6 \times 10^{-3} \mathrm{~m}$ As we know,
Stress $=$ Young's modulus $\times$ strain
$\frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y} \times \frac{\Delta \mathrm{L}}{\mathrm{L}}$
$\therefore \quad \frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y}_{\mathrm{C}} \times \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{C}}}=\mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}$
or, $\quad \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\Delta \mathrm{L}_{\mathrm{A}}}=\frac{\mathrm{Y}_{\mathrm{A}}}{\mathrm{Y}_{\mathrm{C}}} \times \frac{\mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{A}}}$
$=\frac{0.7 \times 10^{11}}{1.2 \times 10^{11}} \times \frac{2.4}{0.7}$
$\frac{\Delta \mathrm{L}_{\mathrm{C}}}{\Delta \mathrm{L}_{\mathrm{A}}}=\frac{2}{1}$
Net elongation $=0.6 \times 10^{-3} \mathrm{~m}$
$\Delta \mathrm{L}_{\mathrm{C}}+\Delta \mathrm{L}_{\mathrm{A}}=0.6 \times 10^{-3} \mathrm{~m}$
From equation (ii) and (iii),
$\Delta \mathrm{L}_{\mathrm{C}} =0.4 \times 10^{-3} \mathrm{~m}$
$\Delta \mathrm{L}_{\mathrm{A}} =0.2 \times 10^{-3} \mathrm{~m}$
From equation (i), we get
$\frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}=\mathrm{Y}_{\mathrm{C}} \times \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{C}}}$
$\mathrm{W}=\mathrm{A} \times \mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}$
Put the value of $\triangle \mathrm{L}_{\mathrm{A}}$ and $\triangle \mathrm{L}_{\mathrm{C}}$, we get
$\mathrm{W}=\pi \mathrm{r}^{2} \times 0.7 \times 10^{11} \times \frac{0.2 \times 10^{-3}}{0.7}$
$\mathrm{~W}=\pi \times\left(1 \times 10^{-3}\right)^{2} \times 0.7 \times 10^{11} \times \frac{0.2 \times 10^{-3}}{0.7}$
$\mathrm{~W}=20 \pi$

Shift-II]

Mechanical Properties of Solids

140959 A metallic rod breaks when strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$. The area of cross section of support a load of $10^{4} \mathrm{~N}$ is

1 $7.1 \times 10^{-4} \mathrm{~m}^{2}$

2 $7.1 \times 10^{-2} \mathrm{~m}^{2}$

3 $7.1 \times 10^{-8} \mathrm{~m}^{2}$

4 $7.1 \times 10^{-6} \mathrm{~m}^{2}$

Karnataka CET-2022

Mechanical Properties of Solids

140960 A metal cube has an edge length $0.1 \mathrm{~m}$. The bulk modulus of the metal is $1.4 \times 10^{11} \mathrm{Nm}^{-2}$. When subjected to hydraulic pressure, the volume contraction of the cube is $5 \times 10^{-2} \mathrm{~cm}^{3}$. The applied pressure is

1 $2 \times 10^{6} \mathrm{~Pa}$

2 $1.4 \times 10^{7} \mathrm{~Pa}$

3 $7 \times 10^{6} \mathrm{~Pa}$

4 $5 \times 10^{5} \mathrm{~Pa}$

Shift-I]

Mechanical Properties of Solids

140961 A horizontal metal rod of diameter $4 \mathrm{~cm}$ projects $6 \mathrm{~cm}$ from a wall. The shear modulus of the metal is $3 \times 10^{10} \mathrm{~N} \mathrm{~m}^{-2}$. An object of mass $1100 \mathrm{~kg}$ is suspended from the free end of the rod. The free end of the rod moves down by a distance of (Acceleration due to gravity $=10$ $\mathrm{ms}^{-2}$, neglect the mass of the rod)

1 $2.5 \times 10^{-6} \mathrm{~m}$

2 $6.75 \times 10^{-5} \mathrm{~m}$

3 $2.25 \times 10^{-6} \mathrm{~m}$

4 $1.75 \times 10^{-5} \mathrm{~m}$

Shift-II]

Mechanical Properties of Solids

140958 A copper wire of length $2.4 \mathrm{~m}$ and an aluminum wire of length $0.7 \mathrm{~m}$. Both having diameter $2 \mathrm{~mm}$ are connected end to end. When stretched by a load, the obtained elongation is found to be $0.6 \mathrm{~mm}$. The applied load is
(Young's modulus of copper $=1.2 \times 10^{11} \mathrm{Nm}^{-2}$ and Young's modulus of aluminum $=0.7 \times 10^{11}$ $\mathrm{Nm}^{-2}$ )

1 $12 \pi \mathrm{N}$

2 $24 \pi \mathrm{N}$

3 $20 \pi \mathrm{N}$

4 $80 \pi \mathrm{N}$

Explanation:

C Given,
$\mathrm{L}_{\mathrm{C}}=2.4 \mathrm{~m} \quad, \quad \mathrm{Y}_{\mathrm{C}}=1.2 \times 10^{11} \mathrm{Nm}^{-2}$
$\mathrm{L}_{\mathrm{A}}=0.7 \mathrm{~m} \quad, \quad \mathrm{Y}_{\mathrm{A}}=0.7 \times 10^{11} \mathrm{Nm}^{-2}$
Load $=\mathrm{W}$
Diameter $=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$
So radius $=1 \times 10^{-3} \mathrm{~m}$
Total elongation $=0.6 \mathrm{~mm}=0.6 \times 10^{-3} \mathrm{~m}$ As we know,
Stress $=$ Young's modulus $\times$ strain
$\frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y} \times \frac{\Delta \mathrm{L}}{\mathrm{L}}$
$\therefore \quad \frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y}_{\mathrm{C}} \times \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{C}}}=\mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}$
or, $\quad \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\Delta \mathrm{L}_{\mathrm{A}}}=\frac{\mathrm{Y}_{\mathrm{A}}}{\mathrm{Y}_{\mathrm{C}}} \times \frac{\mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{A}}}$
$=\frac{0.7 \times 10^{11}}{1.2 \times 10^{11}} \times \frac{2.4}{0.7}$
$\frac{\Delta \mathrm{L}_{\mathrm{C}}}{\Delta \mathrm{L}_{\mathrm{A}}}=\frac{2}{1}$
Net elongation $=0.6 \times 10^{-3} \mathrm{~m}$
$\Delta \mathrm{L}_{\mathrm{C}}+\Delta \mathrm{L}_{\mathrm{A}}=0.6 \times 10^{-3} \mathrm{~m}$
From equation (ii) and (iii),
$\Delta \mathrm{L}_{\mathrm{C}} =0.4 \times 10^{-3} \mathrm{~m}$
$\Delta \mathrm{L}_{\mathrm{A}} =0.2 \times 10^{-3} \mathrm{~m}$
From equation (i), we get
$\frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}=\mathrm{Y}_{\mathrm{C}} \times \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{C}}}$
$\mathrm{W}=\mathrm{A} \times \mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}$
Put the value of $\triangle \mathrm{L}_{\mathrm{A}}$ and $\triangle \mathrm{L}_{\mathrm{C}}$, we get
$\mathrm{W}=\pi \mathrm{r}^{2} \times 0.7 \times 10^{11} \times \frac{0.2 \times 10^{-3}}{0.7}$
$\mathrm{~W}=\pi \times\left(1 \times 10^{-3}\right)^{2} \times 0.7 \times 10^{11} \times \frac{0.2 \times 10^{-3}}{0.7}$
$\mathrm{~W}=20 \pi$

Shift-II]

Mechanical Properties of Solids

140959 A metallic rod breaks when strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$. The area of cross section of support a load of $10^{4} \mathrm{~N}$ is

1 $7.1 \times 10^{-4} \mathrm{~m}^{2}$

2 $7.1 \times 10^{-2} \mathrm{~m}^{2}$

3 $7.1 \times 10^{-8} \mathrm{~m}^{2}$

4 $7.1 \times 10^{-6} \mathrm{~m}^{2}$

Karnataka CET-2022

Mechanical Properties of Solids

140960 A metal cube has an edge length $0.1 \mathrm{~m}$. The bulk modulus of the metal is $1.4 \times 10^{11} \mathrm{Nm}^{-2}$. When subjected to hydraulic pressure, the volume contraction of the cube is $5 \times 10^{-2} \mathrm{~cm}^{3}$. The applied pressure is

1 $2 \times 10^{6} \mathrm{~Pa}$

2 $1.4 \times 10^{7} \mathrm{~Pa}$

3 $7 \times 10^{6} \mathrm{~Pa}$

4 $5 \times 10^{5} \mathrm{~Pa}$

Shift-I]

Mechanical Properties of Solids

140961 A horizontal metal rod of diameter $4 \mathrm{~cm}$ projects $6 \mathrm{~cm}$ from a wall. The shear modulus of the metal is $3 \times 10^{10} \mathrm{~N} \mathrm{~m}^{-2}$. An object of mass $1100 \mathrm{~kg}$ is suspended from the free end of the rod. The free end of the rod moves down by a distance of (Acceleration due to gravity $=10$ $\mathrm{ms}^{-2}$, neglect the mass of the rod)

1 $2.5 \times 10^{-6} \mathrm{~m}$

2 $6.75 \times 10^{-5} \mathrm{~m}$

3 $2.25 \times 10^{-6} \mathrm{~m}$

4 $1.75 \times 10^{-5} \mathrm{~m}$

Shift-II]

Mechanical Properties of Solids

140958 A copper wire of length $2.4 \mathrm{~m}$ and an aluminum wire of length $0.7 \mathrm{~m}$. Both having diameter $2 \mathrm{~mm}$ are connected end to end. When stretched by a load, the obtained elongation is found to be $0.6 \mathrm{~mm}$. The applied load is
(Young's modulus of copper $=1.2 \times 10^{11} \mathrm{Nm}^{-2}$ and Young's modulus of aluminum $=0.7 \times 10^{11}$ $\mathrm{Nm}^{-2}$ )

1 $12 \pi \mathrm{N}$

2 $24 \pi \mathrm{N}$

3 $20 \pi \mathrm{N}$

4 $80 \pi \mathrm{N}$

Explanation:

C Given,
$\mathrm{L}_{\mathrm{C}}=2.4 \mathrm{~m} \quad, \quad \mathrm{Y}_{\mathrm{C}}=1.2 \times 10^{11} \mathrm{Nm}^{-2}$
$\mathrm{L}_{\mathrm{A}}=0.7 \mathrm{~m} \quad, \quad \mathrm{Y}_{\mathrm{A}}=0.7 \times 10^{11} \mathrm{Nm}^{-2}$
Load $=\mathrm{W}$
Diameter $=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$
So radius $=1 \times 10^{-3} \mathrm{~m}$
Total elongation $=0.6 \mathrm{~mm}=0.6 \times 10^{-3} \mathrm{~m}$ As we know,
Stress $=$ Young's modulus $\times$ strain
$\frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y} \times \frac{\Delta \mathrm{L}}{\mathrm{L}}$
$\therefore \quad \frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y}_{\mathrm{C}} \times \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{C}}}=\mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}$
or, $\quad \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\Delta \mathrm{L}_{\mathrm{A}}}=\frac{\mathrm{Y}_{\mathrm{A}}}{\mathrm{Y}_{\mathrm{C}}} \times \frac{\mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{A}}}$
$=\frac{0.7 \times 10^{11}}{1.2 \times 10^{11}} \times \frac{2.4}{0.7}$
$\frac{\Delta \mathrm{L}_{\mathrm{C}}}{\Delta \mathrm{L}_{\mathrm{A}}}=\frac{2}{1}$
Net elongation $=0.6 \times 10^{-3} \mathrm{~m}$
$\Delta \mathrm{L}_{\mathrm{C}}+\Delta \mathrm{L}_{\mathrm{A}}=0.6 \times 10^{-3} \mathrm{~m}$
From equation (ii) and (iii),
$\Delta \mathrm{L}_{\mathrm{C}} =0.4 \times 10^{-3} \mathrm{~m}$
$\Delta \mathrm{L}_{\mathrm{A}} =0.2 \times 10^{-3} \mathrm{~m}$
From equation (i), we get
$\frac{\mathrm{W}}{\mathrm{A}}=\mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}=\mathrm{Y}_{\mathrm{C}} \times \frac{\Delta \mathrm{L}_{\mathrm{C}}}{\mathrm{L}_{\mathrm{C}}}$
$\mathrm{W}=\mathrm{A} \times \mathrm{Y}_{\mathrm{A}} \times \frac{\Delta \mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{A}}}$
Put the value of $\triangle \mathrm{L}_{\mathrm{A}}$ and $\triangle \mathrm{L}_{\mathrm{C}}$, we get
$\mathrm{W}=\pi \mathrm{r}^{2} \times 0.7 \times 10^{11} \times \frac{0.2 \times 10^{-3}}{0.7}$
$\mathrm{~W}=\pi \times\left(1 \times 10^{-3}\right)^{2} \times 0.7 \times 10^{11} \times \frac{0.2 \times 10^{-3}}{0.7}$
$\mathrm{~W}=20 \pi$

Shift-II]

Mechanical Properties of Solids

140959 A metallic rod breaks when strain produced is $0.2 \%$. The Young's modulus of the material of the rod is $7 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$. The area of cross section of support a load of $10^{4} \mathrm{~N}$ is

1 $7.1 \times 10^{-4} \mathrm{~m}^{2}$

2 $7.1 \times 10^{-2} \mathrm{~m}^{2}$

3 $7.1 \times 10^{-8} \mathrm{~m}^{2}$

4 $7.1 \times 10^{-6} \mathrm{~m}^{2}$

Karnataka CET-2022

Mechanical Properties of Solids

140960 A metal cube has an edge length $0.1 \mathrm{~m}$. The bulk modulus of the metal is $1.4 \times 10^{11} \mathrm{Nm}^{-2}$. When subjected to hydraulic pressure, the volume contraction of the cube is $5 \times 10^{-2} \mathrm{~cm}^{3}$. The applied pressure is

1 $2 \times 10^{6} \mathrm{~Pa}$

2 $1.4 \times 10^{7} \mathrm{~Pa}$

3 $7 \times 10^{6} \mathrm{~Pa}$

4 $5 \times 10^{5} \mathrm{~Pa}$

Shift-I]

Mechanical Properties of Solids

140961 A horizontal metal rod of diameter $4 \mathrm{~cm}$ projects $6 \mathrm{~cm}$ from a wall. The shear modulus of the metal is $3 \times 10^{10} \mathrm{~N} \mathrm{~m}^{-2}$. An object of mass $1100 \mathrm{~kg}$ is suspended from the free end of the rod. The free end of the rod moves down by a distance of (Acceleration due to gravity $=10$ $\mathrm{ms}^{-2}$, neglect the mass of the rod)

1 $2.5 \times 10^{-6} \mathrm{~m}$

2 $6.75 \times 10^{-5} \mathrm{~m}$

3 $2.25 \times 10^{-6} \mathrm{~m}$

4 $1.75 \times 10^{-5} \mathrm{~m}$

Shift-II]

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