QBManager

Mechanical Properties of Solids

140954 Two wires made of same material are clamped rigidly at one end and pulled by the same force on the other end. The length and the radius of the first wire are three times those of the second wire. If $x$ is the increase in the length of the first wire, then the increase in the length of the second wire is

1

\frac{1}{3} x

2

3 x

3

9 x

4

\sqrt{3} x

Explanation:

B Given, $l_{1} = 3 l_{2}$
$r_{1} = 3 r_{2}$
$Δ l = x$
We know that,
$Y = \frac{F \cdot l_{1}}{A_{1} Δ l_{1}}$
$Y = \frac{F \cdot l_{1}}{π r_{1}^{2} Δ l_{1}}$
$Y = \frac{F \cdot 3 l_{2}}{π {(3 r_{2})}^{2} x}$
$Y = \frac{F \cdot 3 l_{2}}{π 9 r_{2}^{2} \cdot x} = \frac{F l_{2}}{π 3 r_{2}^{2} \times x}$
and, $Y = \frac{F l_{2}}{π r_{2}^{2} \times Δ l_{2}}$
since, wires made of same material form (i) \& (ii),
$Y_{1} = Y_{2}$
$\frac{{Fl}_{2}}{π r_{2}^{2} Δ l_{2}} = \frac{{Fl}_{2}}{π 3 r_{2}^{2} \cdot x}$
$Δ l_{2} = 3 x$

Shift-I]

Mechanical Properties of Solids

140955 A rubber band catapult has initial length $2 cm$ and cross-sectional area $5 {mm}^{2}$ . It is stretched to $2 c m$ and then released to project a stone of mass of $20 g$ . The velocity of projected stone is (Young's modulus of rubber $= 5 \times 10^{8} {Nm}^{- 2}$ )

1

20 {ms}^{- 1}

2

50 {ms}^{- 1}

3

100 {ms}^{- 1}

4

250 {ms}^{- 1}

Explanation:

B Given that,
$A = 5 {mm}^{2}, l = 2 cm$ , mass $(m) = 20 g = 20 \times 10^{- 3} kg$ Stiffness of rubber band-
$K = \frac{YA}{l} = \frac{(5 \times 10^{8}) \times 5 \times 10^{- 6}}{0.02}$
$= \frac{25}{2} \times 10^{4} \frac{N}{m}$
Conservation of mechanical energy
Elastic potential energy of band = kinetic energy of stone
$\frac{1}{2} K l^{2} = \frac{1}{2} {mv}^{2}$
$v = \sqrt{\frac{K}{m} \times l^{2}}$
$= \sqrt{\frac{\frac{25}{2} \times 10^{4}}{20 \times 10^{- 3}} \times {(2 \times 10^{- 2})}^{2}}$
$= \frac{5}{2} \times 10^{3} \times 2 \times 10^{- 2} = 50 m / s$

Shift-I]

Mechanical Properties of Solids

140956 The elastic limit of a metal is $\frac{400}{π} MPa$ . If a rod of this metal is to support a $484 N$ load without exceeding its elastic limit, the minimum diameter of the rod is

1

2.2 mm

2

1.2 mm

3

2 mm

4

16 mm

Explanation:

A Given that,
$(Stress)_{max} = \frac{400}{π} \times 10^{6} N / m^{2}, F_{max} = 484 N$
Stress $= \frac{Force}{Area} = \frac{F}{A} = \frac{F \times 4}{π d^{2}}$
$\frac{400}{π} \times 10^{6} = \frac{484 \times 4}{π \times d^{2}}$
$d^{2} = \frac{484 \times 4}{400 \times 10^{6}} = 4.84 \times 10^{- 6}$
$d = 2.2 mm$

Shift-I]

Mechanical Properties of Solids

140957 A metal cube has an edge length of $90 cm$ . If $2 \times 10^{9} N / m^{2}$ of pressure (or stress) is required to reduce the edge length to $89.5 cm$ , then the bulk modulus of the metal is

1

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

2

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

3

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

4

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

Explanation:

D Given that,
$l_{1} = 90 cm$
$V_{1} = {(l_{1})}^{3} = (90)^{3} = 729000 {cm}^{3}$
$l_{2} = 89.5 cm$
$V_{2} = {(l_{2})}^{3} = (89.5)^{3} = 716917.375 {cm}^{3}$
$P = 2 \times 10^{9} N / m^{2}$
We know that,
$B = \frac{P}{\frac{Δ V}{V}} = \frac{2 \times 10^{9}}{\frac{12082.625}{729000}} = \frac{2 \times 10^{9}}{0.01657} = 120.7 \times 10^{9}$

Shift-II]

Mechanical Properties of Solids

140954 Two wires made of same material are clamped rigidly at one end and pulled by the same force on the other end. The length and the radius of the first wire are three times those of the second wire. If $x$ is the increase in the length of the first wire, then the increase in the length of the second wire is

1

\frac{1}{3} x

2

3 x

3

9 x

4

\sqrt{3} x

Explanation:

B Given, $l_{1} = 3 l_{2}$
$r_{1} = 3 r_{2}$
$Δ l = x$
We know that,
$Y = \frac{F \cdot l_{1}}{A_{1} Δ l_{1}}$
$Y = \frac{F \cdot l_{1}}{π r_{1}^{2} Δ l_{1}}$
$Y = \frac{F \cdot 3 l_{2}}{π {(3 r_{2})}^{2} x}$
$Y = \frac{F \cdot 3 l_{2}}{π 9 r_{2}^{2} \cdot x} = \frac{F l_{2}}{π 3 r_{2}^{2} \times x}$
and, $Y = \frac{F l_{2}}{π r_{2}^{2} \times Δ l_{2}}$
since, wires made of same material form (i) \& (ii),
$Y_{1} = Y_{2}$
$\frac{{Fl}_{2}}{π r_{2}^{2} Δ l_{2}} = \frac{{Fl}_{2}}{π 3 r_{2}^{2} \cdot x}$
$Δ l_{2} = 3 x$

Shift-I]

Mechanical Properties of Solids

140955 A rubber band catapult has initial length $2 cm$ and cross-sectional area $5 {mm}^{2}$ . It is stretched to $2 c m$ and then released to project a stone of mass of $20 g$ . The velocity of projected stone is (Young's modulus of rubber $= 5 \times 10^{8} {Nm}^{- 2}$ )

1

20 {ms}^{- 1}

2

50 {ms}^{- 1}

3

100 {ms}^{- 1}

4

250 {ms}^{- 1}

Explanation:

B Given that,
$A = 5 {mm}^{2}, l = 2 cm$ , mass $(m) = 20 g = 20 \times 10^{- 3} kg$ Stiffness of rubber band-
$K = \frac{YA}{l} = \frac{(5 \times 10^{8}) \times 5 \times 10^{- 6}}{0.02}$
$= \frac{25}{2} \times 10^{4} \frac{N}{m}$
Conservation of mechanical energy
Elastic potential energy of band = kinetic energy of stone
$\frac{1}{2} K l^{2} = \frac{1}{2} {mv}^{2}$
$v = \sqrt{\frac{K}{m} \times l^{2}}$
$= \sqrt{\frac{\frac{25}{2} \times 10^{4}}{20 \times 10^{- 3}} \times {(2 \times 10^{- 2})}^{2}}$
$= \frac{5}{2} \times 10^{3} \times 2 \times 10^{- 2} = 50 m / s$

Shift-I]

Mechanical Properties of Solids

140956 The elastic limit of a metal is $\frac{400}{π} MPa$ . If a rod of this metal is to support a $484 N$ load without exceeding its elastic limit, the minimum diameter of the rod is

1

2.2 mm

2

1.2 mm

3

2 mm

4

16 mm

Explanation:

A Given that,
$(Stress)_{max} = \frac{400}{π} \times 10^{6} N / m^{2}, F_{max} = 484 N$
Stress $= \frac{Force}{Area} = \frac{F}{A} = \frac{F \times 4}{π d^{2}}$
$\frac{400}{π} \times 10^{6} = \frac{484 \times 4}{π \times d^{2}}$
$d^{2} = \frac{484 \times 4}{400 \times 10^{6}} = 4.84 \times 10^{- 6}$
$d = 2.2 mm$

Shift-I]

Mechanical Properties of Solids

140957 A metal cube has an edge length of $90 cm$ . If $2 \times 10^{9} N / m^{2}$ of pressure (or stress) is required to reduce the edge length to $89.5 cm$ , then the bulk modulus of the metal is

1

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

2

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

3

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

4

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

Explanation:

D Given that,
$l_{1} = 90 cm$
$V_{1} = {(l_{1})}^{3} = (90)^{3} = 729000 {cm}^{3}$
$l_{2} = 89.5 cm$
$V_{2} = {(l_{2})}^{3} = (89.5)^{3} = 716917.375 {cm}^{3}$
$P = 2 \times 10^{9} N / m^{2}$
We know that,
$B = \frac{P}{\frac{Δ V}{V}} = \frac{2 \times 10^{9}}{\frac{12082.625}{729000}} = \frac{2 \times 10^{9}}{0.01657} = 120.7 \times 10^{9}$

Shift-II]

Mechanical Properties of Solids

140954 Two wires made of same material are clamped rigidly at one end and pulled by the same force on the other end. The length and the radius of the first wire are three times those of the second wire. If $x$ is the increase in the length of the first wire, then the increase in the length of the second wire is

1

\frac{1}{3} x

2

3 x

3

9 x

4

\sqrt{3} x

Explanation:

B Given, $l_{1} = 3 l_{2}$
$r_{1} = 3 r_{2}$
$Δ l = x$
We know that,
$Y = \frac{F \cdot l_{1}}{A_{1} Δ l_{1}}$
$Y = \frac{F \cdot l_{1}}{π r_{1}^{2} Δ l_{1}}$
$Y = \frac{F \cdot 3 l_{2}}{π {(3 r_{2})}^{2} x}$
$Y = \frac{F \cdot 3 l_{2}}{π 9 r_{2}^{2} \cdot x} = \frac{F l_{2}}{π 3 r_{2}^{2} \times x}$
and, $Y = \frac{F l_{2}}{π r_{2}^{2} \times Δ l_{2}}$
since, wires made of same material form (i) \& (ii),
$Y_{1} = Y_{2}$
$\frac{{Fl}_{2}}{π r_{2}^{2} Δ l_{2}} = \frac{{Fl}_{2}}{π 3 r_{2}^{2} \cdot x}$
$Δ l_{2} = 3 x$

Shift-I]

Mechanical Properties of Solids

140955 A rubber band catapult has initial length $2 cm$ and cross-sectional area $5 {mm}^{2}$ . It is stretched to $2 c m$ and then released to project a stone of mass of $20 g$ . The velocity of projected stone is (Young's modulus of rubber $= 5 \times 10^{8} {Nm}^{- 2}$ )

1

20 {ms}^{- 1}

2

50 {ms}^{- 1}

3

100 {ms}^{- 1}

4

250 {ms}^{- 1}

Explanation:

B Given that,
$A = 5 {mm}^{2}, l = 2 cm$ , mass $(m) = 20 g = 20 \times 10^{- 3} kg$ Stiffness of rubber band-
$K = \frac{YA}{l} = \frac{(5 \times 10^{8}) \times 5 \times 10^{- 6}}{0.02}$
$= \frac{25}{2} \times 10^{4} \frac{N}{m}$
Conservation of mechanical energy
Elastic potential energy of band = kinetic energy of stone
$\frac{1}{2} K l^{2} = \frac{1}{2} {mv}^{2}$
$v = \sqrt{\frac{K}{m} \times l^{2}}$
$= \sqrt{\frac{\frac{25}{2} \times 10^{4}}{20 \times 10^{- 3}} \times {(2 \times 10^{- 2})}^{2}}$
$= \frac{5}{2} \times 10^{3} \times 2 \times 10^{- 2} = 50 m / s$

Shift-I]

Mechanical Properties of Solids

140956 The elastic limit of a metal is $\frac{400}{π} MPa$ . If a rod of this metal is to support a $484 N$ load without exceeding its elastic limit, the minimum diameter of the rod is

1

2.2 mm

2

1.2 mm

3

2 mm

4

16 mm

Explanation:

A Given that,
$(Stress)_{max} = \frac{400}{π} \times 10^{6} N / m^{2}, F_{max} = 484 N$
Stress $= \frac{Force}{Area} = \frac{F}{A} = \frac{F \times 4}{π d^{2}}$
$\frac{400}{π} \times 10^{6} = \frac{484 \times 4}{π \times d^{2}}$
$d^{2} = \frac{484 \times 4}{400 \times 10^{6}} = 4.84 \times 10^{- 6}$
$d = 2.2 mm$

Shift-I]

Mechanical Properties of Solids

140957 A metal cube has an edge length of $90 cm$ . If $2 \times 10^{9} N / m^{2}$ of pressure (or stress) is required to reduce the edge length to $89.5 cm$ , then the bulk modulus of the metal is

1

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

2

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

3

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

4

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

Explanation:

D Given that,
$l_{1} = 90 cm$
$V_{1} = {(l_{1})}^{3} = (90)^{3} = 729000 {cm}^{3}$
$l_{2} = 89.5 cm$
$V_{2} = {(l_{2})}^{3} = (89.5)^{3} = 716917.375 {cm}^{3}$
$P = 2 \times 10^{9} N / m^{2}$
We know that,
$B = \frac{P}{\frac{Δ V}{V}} = \frac{2 \times 10^{9}}{\frac{12082.625}{729000}} = \frac{2 \times 10^{9}}{0.01657} = 120.7 \times 10^{9}$

Shift-II]

Mechanical Properties of Solids

140954 Two wires made of same material are clamped rigidly at one end and pulled by the same force on the other end. The length and the radius of the first wire are three times those of the second wire. If $x$ is the increase in the length of the first wire, then the increase in the length of the second wire is

1

\frac{1}{3} x

2

3 x

3

9 x

4

\sqrt{3} x

Explanation:

B Given, $l_{1} = 3 l_{2}$
$r_{1} = 3 r_{2}$
$Δ l = x$
We know that,
$Y = \frac{F \cdot l_{1}}{A_{1} Δ l_{1}}$
$Y = \frac{F \cdot l_{1}}{π r_{1}^{2} Δ l_{1}}$
$Y = \frac{F \cdot 3 l_{2}}{π {(3 r_{2})}^{2} x}$
$Y = \frac{F \cdot 3 l_{2}}{π 9 r_{2}^{2} \cdot x} = \frac{F l_{2}}{π 3 r_{2}^{2} \times x}$
and, $Y = \frac{F l_{2}}{π r_{2}^{2} \times Δ l_{2}}$
since, wires made of same material form (i) \& (ii),
$Y_{1} = Y_{2}$
$\frac{{Fl}_{2}}{π r_{2}^{2} Δ l_{2}} = \frac{{Fl}_{2}}{π 3 r_{2}^{2} \cdot x}$
$Δ l_{2} = 3 x$

Shift-I]

Mechanical Properties of Solids

140955 A rubber band catapult has initial length $2 cm$ and cross-sectional area $5 {mm}^{2}$ . It is stretched to $2 c m$ and then released to project a stone of mass of $20 g$ . The velocity of projected stone is (Young's modulus of rubber $= 5 \times 10^{8} {Nm}^{- 2}$ )

1

20 {ms}^{- 1}

2

50 {ms}^{- 1}

3

100 {ms}^{- 1}

4

250 {ms}^{- 1}

Explanation:

B Given that,
$A = 5 {mm}^{2}, l = 2 cm$ , mass $(m) = 20 g = 20 \times 10^{- 3} kg$ Stiffness of rubber band-
$K = \frac{YA}{l} = \frac{(5 \times 10^{8}) \times 5 \times 10^{- 6}}{0.02}$
$= \frac{25}{2} \times 10^{4} \frac{N}{m}$
Conservation of mechanical energy
Elastic potential energy of band = kinetic energy of stone
$\frac{1}{2} K l^{2} = \frac{1}{2} {mv}^{2}$
$v = \sqrt{\frac{K}{m} \times l^{2}}$
$= \sqrt{\frac{\frac{25}{2} \times 10^{4}}{20 \times 10^{- 3}} \times {(2 \times 10^{- 2})}^{2}}$
$= \frac{5}{2} \times 10^{3} \times 2 \times 10^{- 2} = 50 m / s$

Shift-I]

Mechanical Properties of Solids

140956 The elastic limit of a metal is $\frac{400}{π} MPa$ . If a rod of this metal is to support a $484 N$ load without exceeding its elastic limit, the minimum diameter of the rod is

1

2.2 mm

2

1.2 mm

3

2 mm

4

16 mm

Explanation:

A Given that,
$(Stress)_{max} = \frac{400}{π} \times 10^{6} N / m^{2}, F_{max} = 484 N$
Stress $= \frac{Force}{Area} = \frac{F}{A} = \frac{F \times 4}{π d^{2}}$
$\frac{400}{π} \times 10^{6} = \frac{484 \times 4}{π \times d^{2}}$
$d^{2} = \frac{484 \times 4}{400 \times 10^{6}} = 4.84 \times 10^{- 6}$
$d = 2.2 mm$

Shift-I]

Mechanical Properties of Solids

140957 A metal cube has an edge length of $90 cm$ . If $2 \times 10^{9} N / m^{2}$ of pressure (or stress) is required to reduce the edge length to $89.5 cm$ , then the bulk modulus of the metal is

1

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

2

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

3

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

4

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

Explanation:

D Given that,
$l_{1} = 90 cm$
$V_{1} = {(l_{1})}^{3} = (90)^{3} = 729000 {cm}^{3}$
$l_{2} = 89.5 cm$
$V_{2} = {(l_{2})}^{3} = (89.5)^{3} = 716917.375 {cm}^{3}$
$P = 2 \times 10^{9} N / m^{2}$
We know that,
$B = \frac{P}{\frac{Δ V}{V}} = \frac{2 \times 10^{9}}{\frac{12082.625}{729000}} = \frac{2 \times 10^{9}}{0.01657} = 120.7 \times 10^{9}$

Shift-II]

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: