A steel wire is rigidly fixed at both ends its length mass and cross sectional area are 1m, 0.1 kg and `10^(6)m^(2)` respectively then the temperature of the wire is lowered by `20^(@)C` if the transverse waves are setup by plucking the wire at 0.25m from one end and assuming that wire vibrates with minimum number of loops possible for such a case. Find the frequency of vibration (in Hz). [coefficient of linear expansion of steel `=1.21xx10^(-6)//.^(@)C` and young's modulus `=2xx10^(11)//N//m^(2)`]

A steel wire is rigidly fixed at both ends. Its length mass and cross-sectionl area are 1m, 0.1kg and 10^(-8)m^(2) respectively. Tension in the wire is produced by lowering the temperature by 20^(@)C . If the transerverse waves are some up by plucking the wire at 0.25m from one end and assuming that the wire viberates with minimum number of loops possble for such a case. The frequency of viberation (in Hz) is found to be. 1.11. Find the value of K. Given alpha=1.21xx10^(-5).^(@)C^(-1). Y-2xx10^(11)N//m^(2)

A steel wire of length lm, mass 0.1 kg and uniform cross-sectional area 10^(-6)m^(2) is rigidly fixed at both ends. The temperature of the wire is lowered by 20^(0)C . If the transverse waves are set up by plucking the string in the middle, calculate the frequency of the fundamental mode of vibration [Y=2xx10^(11)N//m^(2) and alpha=1.21xx10^(-5)l^(0)C]

A steel wire of length im, mass 0.1kg and uniform cross sectional area 10^(-6) m^(2) is rigidly fixed at both ends. The temperature of the wire is lowered by 20^(@)C . If the transverse waves are set up plucking the string in the middle, calculate the frequency of the fundamental mode of vibration. (Y=2xx10^(11)N//m^(2), a=1.21xx10^(-5)//^(@)C)

A steel wire of length 1m , mass 0.1kg and uniform cross-sectional area 10^(-6)m^(2) is rigidly fixed at both ends. The temperature of the wire is lowered by 20^(@)C . If transverse waves are set up by plucking the string in the middle.Calculate the frequency of the fundamental mode of vibration. Given for steel Y = 2 xx 10^(11)N//m^(2) alpha = 1.21 xx 10^(-5) per ^(@)C

A steel wire of length 1m , mass 0.1kg and uniform cross-sectional area 10^(-6)m^(2) is rigidly fixed at both ends. The temperature of the wire is lowered by 20^(@)C . If transverse calculate the frequency of the fundamental mode of vibration. Given for steel Y = 2 xx 10^(11)N//m^(2) alpha = 1.21 xx 10^(-5) per ^(@)C

In a stretched wire under tension and fixed at both ends, the area of cross section of the wire is halved and the tension is doubled. The frequency of the wire will be