A composite wire consists of a steel wire of length 1 . 5 m and a copper wire of length 2 . 0 m , with a uniform cross-section area of 2 . 5 × 10 - 5 m 2 . It is loaded with a mass of 200 kg . Find the extension produced. Young's modulus of copper is 1 . 0 × 10 11 m - 2 and that of steel is 2 . 0 × 10 11 Nm - 2 . Take g = 9 . 8 ms - 2

Given: L s t e e l = 1 . 5 m L c o p p e r = 2 . 0 m Area of cross-section A = 2 . 5 × 10 - 5 m 2 Load m = 200 kg Y s t e e l = 2 . 0 × 10 11 N / m 2 Y c o p p e r = 1 . 0 × 10 11 N / m 2 g = 9 . 8 ms - 2 To find, Tension due to load The tension due to the load will remain the same in the composite wire. So the tensile stress in both wire is the same. Tensile stress = L o a d A r e a o f s e c t i o n = m g A Tensile stress = 200 × 9 . 8 2 . 5 × 10 - 5 stress = 7 . 84 × 10 7 N / m 2 . . . . . . . . . . . . . . i For steel wire Y s = T e n s i l e s t r e s s s t r a i n Y s = T e n s i l e s t r e s s ∆ l s l s ∆ l s = Change in length of steel wire 2 . 0 × 10 11 = 7 . 84 × 10 7 ∆ l s 1 . 5 From i ∆ l s = 7 . 84 × 10 7 × 1 . 5 2 . 0 × 10 11 = 5 . 88 × 10 - 4 m ∆ l s = 5 . 88 × 10 - 14 m . . . . . . . . . . . . . . . . . ii For copper wire Y c = T e n s i l e s t r e s s s t r a i n Y c = T e n s i l e s t r e s s ∆ l c l c ∆ l c = Change in length of copper wire 1 . 0 × 10 11 = 7 . 84 × 10 7 ∆ l c 2 ∆ l c = 7 . 84 × 10 7 ∆ l c 2 ∆ l c = 7 . 84 × 10 7 × 2 1 . 0 × 10 11 = 1 . 508 × 10 - 3 m ∆ l c = 1 . 508 × 10 - 3 m So extension in composite root = ∆ l s + ∆ l c = 5 . 88 × 10 - 14 + 1 . 508 × 10 - 3 = 2 . 156 × 10 - 3 m or 2 . 156 mm Tension due to load = 2 . 156 mm

A substance breaks down by a stress of 10 9 N m - 2 . If the density of the substance be 3 × 10 3 kg m - 3 , find that length of the wire made of the same substance, by which it will break under its own weight when suspended.

Given: Breaking stress = 10 9 N / m 2 Density of substance = 3 × 10 3 kg / m 3 = ρ To find, Find that length of the wire by which it will break under its own weight Breaking stress = F A = m g A Breaking stress = v ρ g A [ m = volume × density] Breaking stress = A L ρ g A [Volume = Area × Length] v = A × L Put up the values 10 9 = A × L × 3 × 10 3 × 9 . 8 A L = 10 9 3 × 10 3 × 9 . 8 = 3 . 4 × 10 4 m Length L = 3 . 4 × 10 4 m

EASY

11th ICSE

IMPORTANT

Earn 100

If Young's modulus of steel is $2.0 \times 10^{11} N / m^{2}$ then how much weight be suspended from a steel wire of length $2.0 m$ and diameter $1.0 mm$ so that the length of wire be increased by $1.0 mm$ . $(g = 9.8 m / s^{2})$

Important Questions on Elasticity

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 18 - Elasticity>NUMERICALS>Q 9

A composite wire consists of a steel wire of length

1.5 m

and a copper wire of length

2.0 m

, with a uniform cross-section area of

2.5 \times 10^{- 5} m^{2}

. It is loaded with a mass of

200 kg

. Find the extension produced. Young's modulus of copper is

1.0 \times 10^{11} m^{- 2}

and that of steel is

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

. Take

g = 9.8 {ms}^{- 2}

HARD

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 18 - Elasticity>NUMERICALS>Q 10

A rod

A D

consisting of three segments

A B, B C

and

C D

joined together is hanging vertically from a fixed support.

A

. The segments are respectively

0.1 m

0.2 m

and

0.15 m

. The cross-section of the rod is uniformly

10^{- 4} m^{2}

. A weight of

10 kg

is hung from

D

. Calculate the displacements of point

B, C

and

D

. Neglect the weight of the rod. Given,

Y_{A B} = 2.5 \times 10^{10} {Nm}^{- 2}

Y_{A B} = 4.0 \times 10^{10} {Nm}^{2}

and

Y_{C D} = 1.0 \times 10^{10} {Nm}^{- 2}

EASY

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 18 - Elasticity>NUMERICALS>Q 11

A piece of metal of

2 kg

weight is suspended from one end of a vertical wire whose other end is fixed and the piece is fully immersed in an oil of density

0.7 \times 10^{3} kg m^{- 3}

. The length of the wire increases by

1 mm

. If the diameter of the wire is

0.6 mm

, Young's modulus is

2.0 \times 10^{11} N m^{- 2}

and the volume of the metal-piece is

800 {cm}^{3}

, then calculate the initial length of the wire.

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 18 - Elasticity>NUMERICALS>Q 12

A weight of

20 kg

is suspended from a wire. The area of cross-section of the wire is

1 {mm}^{2}

and when stretched, the length is exactly

6 m

. When the weight is removed, the length reduces to

5.995 m

. Calculate the Young's modulus

(Y)

of material of the wire.

EASY

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 18 - Elasticity>NUMERICALS>Q 13

The length of a rubber cord increases to

60 cm

under a weight of

100 g

suspended from it and to

70 cm

under a weight of

120 g

. Find the initial length of the cord. Also calculate the weight under which the length of the cord will become

74 cm

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 18 - Elasticity>NUMERICALS>Q 14

A substance breaks down by a stress of

10^{9} N m^{- 2}

. If the density of the substance be

3 \times 10^{3} kg m^{- 3}

, find that length of the wire made of the same substance, by which it will break under its own weight when suspended.

MEDIUM

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 18 - Elasticity>NUMERICALS>Q 15

For steel the breaking stress is

8.0 \times 10^{6} N m^{2}

and the density is

8.0 \times 10^{3} kg m^{- 3}

. Find the maximum length of a steel wire which can be suspended without breaking under its own weight.

(g = 10 m s^{- 2})

EASY

11th ICSE

IMPORTANT

ISC Physics Class XI Part 1>Chapter 18 - Elasticity>NUMERICALS>Q 16

A body of

10 kg

tied at one end of

0.3 m

wire is rotated in a horizontal circle. The area of cross-section of wire is

10^{- 6} m^{2}

and the breaking stress is

4.8 \times 10^{7} N m^{2}

. By how much maximum angular speed can the body be rotated?

If Young's modulus of steel is 2.0×1011 N/m2 then how much weight be suspended from a steel wire of length 2.0 m and diameter 1.0 mm so that the length of wire be increased by 1.0 mm. g=9.8 m/s2

Important Questions on Elasticity

Learn and Prepare for any exam you want !

If Young's modulus of steel is $2.0 \times 10^{11} N / m^{2}$ then how much weight be suspended from a steel wire of length $2.0 m$ and diameter $1.0 mm$ so that the length of wire be increased by $1.0 mm$ . $(g = 9.8 m / s^{2})$