A film of water is formed between two straight paralel wires each `10 cm` long and at seperation `0.5 cm`. Calculate the work requied to increase `1mm` distance between the wires. Surface tension of water `=72xx10^(-3)N//m`.
Text Solution
Verified by Experts
Initial surface area `=2xx"length"xx"separation"` `=2xx10cmxx0.5cm=10cm^(2)=10xx10^(-4)m^(2)` Final surface area `=2xx10cmxx(0.5+0.1)cm` `=2xx10xx0.6cm^(2)=12xx10^(-4)m^(2)` The required work `WT/_\A=72xx10^(-3)xx(12xx10^(-4)-10xx10^(-4))J` `=72xx10^(-3)xx2xx10^(-4)=144xx10^(-7)J`
A film of water is formed between two straight parallel wires each 10 cm long and at a separatin 0.5 cm . Calculate the work required to increase 1 mm distance between wires. Surface tension = 72 xx 10^(-3) N//m .
A water film is formed between two parallel wires of 10 cm length. The distance of 0.5 cm between the wires is increased by 1mm. What will be the work done ? (Given, surface tension of water of 72 Nm^(-1) )
A film of water is formed between two straight parallel wires of length 10 cm each separated by 0.5cm If their separation is increased by 1mm while still maintaining their parallelism, how much work will have to be done (Surface tension of water =7.2xx10^-2(N)/(m) )
A film of water is formed between two straight parallel wires , each of length 10 cm and separated by 4 mm . How much work should be done to increase their separation by 1 mm , while still maintaining their parallelism ? [ T = 7.5 xx 10^(-2)N//m]
A water film is formed between two stratght parallel wires of length 20 cm each with a separation of 0.1 cm . If the distance between the wires is increased by 0.1 cm , how much work is done ? (T=0.072 N/m)
Calculate the force required to separate the galss plates of area 10^(-2)m^(2) with a film of water 0.05 mm thickness between them (surface tension of waer =70xx10^(-3)N//m) )
A capillary tube of radius 0.5 mm is dipped in a vessel containing water. Calculate the height to which water will rise in the capillary tube. Take, surface tension of water =7.4 xx 10^(-3)" N/m".
Work done is splitting a drop of water of 1 mm radius into 64 droplets is (surface tension of water =72xx10^(-3)j m^(2))
A rectangular frame having dimensions "8 cm "xx " 4 cm " and width of 3 mm is initially placed with its largest face flat on water surface. Calculate the downward force acting on the frame due to surface tension of water ("Surface tension of water "= 7.2 xx 10^(-4)"Ncm"^(-1))
CENGAGE PHYSICS-PROPERTIES OF SOLIDS AND FLUIDS-INTEGER_TYPE
