Advertisements
Advertisements
प्रश्न
A rigid bar of mass M is supported symmetrically by three wires each of length l. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to ______.
पर्याय
`Y_(copper)/Y_(iron)`
`sqrt((Y_(iron))/(Y_(copper)`
`Y_(iron)^2/Y_(copper)^2`
`Y_(iron)/Y_(copper)^2`
Advertisements
उत्तर
A rigid bar of mass M is supported symmetrically by three wires each of length l. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to `underline(sqrt((Y_(iron))/(Y_(copper))`.
Explanation:
As the bar is supported symmetrically by the three wires, therefore extension in each wire is the same.
Let T be the tension in each wire and the diameter of the wire is D, then Young’s modulus is `Y = "Stress"/"Strain"`
= `(F/A)/((ΔL)/L)`
= `F/A xx L/(ΔL)`
= `F/(pi(D/2)^2) xx L/(ΔL)`
= `(4FL)/(piD^2ΔL)`
⇒ `D^2 = (4FL)/(piΔLY)`
⇒ `D = sqrt((4FL)/(piΔLY)`
As F and `L/(ΔL)` are constants.
Hence, `D ∝ sqrt(1/Y)`
or `D = K/sqrt(Y)` ......(K is the proportionality constant)
Now, we can find ratio as `D_(copper)/D_(iron) = sqrt(Y_(iron)/Y_(copper)`
APPEARS IN
संबंधित 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.
Two wires A and B are made of same material. The wire A has a length l and diameter rwhile the wire B has a length 2l and diameter r/2. If the two wires are stretched by the same force, the elongation in A divided by the elongation in B is
A wire elongates by 1.0 mm when a load W is hung from it. If this wire goes over a a pulley and two weights W each are hung at the two ends, he elongation of he wire will be
The length of a metal wire is l1 when the tension in it T1 and is l2 when the tension is T2. The natural length of the wire is
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.
The temperature of a wire is doubled. The Young’s modulus of elasticity ______.
The temperature of a wire is doubled. The Young’s modulus of elasticity ______.
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?
A truck is pulling a car out of a ditch by means of a steel cable that is 9.1 m long and has a radius of 5 mm. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young’s modulus for steel is 2 × 1011 Nm–2.)
A steel wire of mass µ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains << longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m–3 (Young’s modules Y = 2 × 1011 Nm–2).
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.
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 ______.
Young's modulus is also known as
What does the dimensional formula [L⁻¹M¹T⁻²] represent?
Which of the following statements about Young's modulus is correct?
