Question

. A rubber cord has a cross-sectional
area 1 mm² and total unstrectched
length 10.0 cm. It is stretched to
12.5 cm and then released to project
a 'missile of mass 5.0 g. Taking
Young's modulus Y for rubber as
5.0 x 10^8 Nm-2. Calculate is the
velocity of projection.
(1) 5 ms
(2) 35 ms -1
(3)15 ms
0 (4) 25 ms -1​

Answer 1

Answer:

Hey friend,

Use conservation of mechanical energy

Initially the energy stored in stretched wire is given by 1/2*stress*strain*volume. This is completely changed to kinetic energy (1/2m*v^2)

U = KE

1/2 * Y *strain^2 * Area * Original length = 1/2 m*v^2

From this you can find velocity.

strain is change in length/original length

Answer 2

Explanation:

Given  A rubber cord has a cross-sectional  area 1 mm² and total un stretched  length 10.0 cm. It is stretched to  12.5 cm and then released to project  a 'missile of mass 5.0 g. Taking  Young's modulus Y for rubber as

5.0 x 10^8 Nm-2. Calculate is the  velocity of projection.

• Here K.E = 1/2 m v^2
•                = 1/2 x load x extension
• We know that  = F x l / A x Δl
• Or F = YAΔl^2 / l
• Or m v^2 = YAΔl^2 / l l
• Or v = √YA / ml Δl
• Given Y = 5 x 10^8 Nm^-2
• A = 1 mm^2 = 10^-6 m^2
• m = 5 g = 5 x 10^-3 kg
• l = 10 cm = 10 x 10^-2
• So Δl = 12.5 – 10 = 2.5 cm = 2.5 x 10^-2
• Or v = √5 x 10^8 x 10^-6 / 5 x 10^-3 x 10 x 10^-2 x 2.5 x 10^-2
• Or v = √10^6 x 2.5 x 10^-2
• So v = 25 m / s

Reference link will be

https://brainly.in/question/8542042