Advertisements
Advertisements
प्रश्न
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Advertisements
उत्तर १
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m
Diameter of the wires, d = 0.25 m
Hence, the radius of the wires, `r=d/2` = 0.125 cm
Length of the steel wire, L1 = 1.5 m
Length of the brass wire, L2 = 1.0 m
Total force exerted on the steel wire:
F1 = (4 + 6) g = 10 × 9.8 = 98 N
Young’s modulus for steel:
`Y_1 = (F_1/A_1)/(triangleL_1/L_1)`
Where,
ΔL1 = Change in the length of the steel wire
A1 = Area of cross-section of the steel wire = `pir_1^2`
Young’s modulus of steel, Y1 = 2.0 × 1011 Pa
`:.triangleL_1 = (F_1xxL_1)/(A_1xxY_1) = (F_1xxL_1)/(pir_1^2xxY_1)`
`= (98xx1.5)/(pi(0.125 xx 10^(-2))^2 xx 2 xx 10^11) = 1.49 xx 10^(-4)m`
Total force on the brass wire:
F2 = 6 × 9.8 = 58.8 N
Young’s modulus for brass: `Y_2 = (F_2/A_2)/((triangleL_2)/L_2)`
Where
`triangleL_2` = change in length
`A_2` = Area of cross-section of the brass wire
`:.triangleL_2 = (F_2xxL_2)/(a_2xxY_2) = (F_2xx L_2)/(pir_2^2 xx Y_2)`
`= (58.8 xx 1.0)/(pixx(0.125xx10^(-2))^2 xx (0.91xx10^(11))) = 1.3xx10^(-4)m`
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m
उत्तर २
For steel wire total force on steel wire
F1 = 4+6 = 10 kg f = 10 x 9.8 N
l1 = 1.5 m, Δl1 = ?; 2r1 = 0.25 cm
or `r_1 = (0.25)/2 cm = 0.125xx 10^(-2) m`
`Y_1 = 2.0 xx 10^11` Pa
For brass wire, F2 = 6.0 kg f = 6 x 9.8 N
2r2 = 0.25 cm
or `r_2 = (0.25/2) = 0.125 xx 10^(-2) m`
`Y_2 = 0.91 xx 10^11 Pa`, `l_2 = 1.0 m` `trianglel_2 = ?`
Since` Y_1 = (F_1xxl_1)/(A_1xxtrianglel_1) = (F_1xxl_2)/(pir_1^2 xx trianglel_1) => trianglel_1 = (F_1xxl_1)/(pir_1^2xxtrianglel_1)`
or `trianglel_1 = ((10xx9.8)xx1.5xx7)/(22xx(0.125xx10^(-2))^2xx2xx10^11) = 1.49 xx 10^(-4) m`
And `trianglel_2 = (F_2xxl_2)/(pir_2^2 xx Y_2) = ((6xx9.8)xx1xx7)/(22xx(0.125xx10^(-2))^2xx(0.91xx10^11)) = 1.3 xx 10^(-4) m`
संबंधित प्रश्न
The stress-strain graphs for materials A and B are shown in Figure
The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?
Read the following statements below carefully and state, with reasons, if it is true or false
The Young’s modulus of rubber is greater than that of steel;
Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Consider the situation shown in figure. The force F is equal to the m2 g/2. If the area of cross section of the string is A and its Young modulus Y, find the strain developed in it. The string is light and there is no friction anywhere.
A uniform rectangular block of mass of 50 kg is hung horizontally with the help of three wires A, B and C each of length and area of 2m and 10mm2 respectively as shown in the figure. The central wire is passing through the centre of gravity and is made of material of Young's modulus 7.5 x 1010 Nm−2 and the other two wires A and C symmetrically placed on either side of the wire B are of Young's modulus 1011 Nm−2 The tension in the wires A and B will be in the ratio of:
Young's modulus of a perfectly rigid body is ______.
Identical springs of steel and copper are equally stretched. On which, more work will have to be done?
What is the Young’s modulus for a perfect rigid body ?
A steel rod (Y = 2.0 × 1011 Nm–2; and α = 10–50 C–1) of length 1 m and area of cross-section 1 cm2 is heated from 0°C to 200°C, without being allowed to extend or bend. What is the tension produced in the rod?
In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by `(Ypir^4)/(4R) . Y` is the Young’s modulus, r is the radius of the trunk and R is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
A boy's catapult is made of rubber cord which is 42 cm long, with a 6 mm diameter of cross-section and negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms-1. Neglect the change in the area of the cross-section of the cord while stretched. Young's modulus of rubber is closest to ______.
A uniform metal rod of 2 mm2 cross section is heated from 0°C to 20°C. The coefficient of linear expansion of the rod is 12 × 10-6/°C, it's Young's modulus is 1011 N/m2. The energy stored per unit volume of the rod is ______.
If the length of a wire is made double and the radius is halved of its respective values. Then, Young's modules of the material of the wire will ______.
The force required to stretch a wire of cross section 1 cm2 to double its length will be ______.
(Given Young's modulus of the wire = 2 × 1011 N/m2)
What is longitudinal strain?
Which of the following statements about Young's modulus is correct?
In the formula Y = MgL/(πr²l), what does 'l' represent?
The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus, respectively, are 8 × 108 Nm−2 and 2 × 1011 Nm−2, is ______.
