Home
Class 11
PHYSICS
Figure shows the graph of elastic potent...

Figure shows the graph of elastic potential energy `(U)` stored versus extension, for a steel wire `(Y = 2xx10^(11) Pa)` of volume `200 c c`. If area of cross- section `A` and original length L, then

A

`4m`

B

`2m`

C

`4cm`

D

`2cm`

Text Solution

Verified by Experts

The correct Answer is:
B
`U = (1)/(2) kx^(2)`. It is parabola symmetric about `U-` axis
At `x = 0.2mm, U = 0.2J` from the figure
`:. 0.2 = (1)/(2) k (2xx10^(-4))^(2)`
`rArr k = 10^(7) Nm^(-1)`
`k = (YA)/(L)`
`rArr (A)/(L) = (k)/(Y) = (10^(7))/(2xx10^(11)) = 5xx10^(-5) m rarr (i)`
`AL` = volume `= 200xx10^(-6) m^(3) rarr (ii)`
On solving equaction (i) and (ii), we get `A = 100^(-4) m^(2)` and `L = 2m`
Promotional Banner

Topper's Solved these Questions

  • MECHANICAL PROPERTIES OF SOLIDS

    NARAYNA|Exercise MULTIPLE ANSWER TYPE|10 Videos

  • MECHANICAL PROPERTIES OF SOLIDS

    NARAYNA|Exercise COMPREHENSION TYPE|11 Videos

  • MECHANICAL PROPERTIES OF SOLIDS

    NARAYNA|Exercise NCERT BASED QUESTIONS|21 Videos

  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - III|30 Videos

  • MOTION IN A PLANE

    NARAYNA|Exercise Level-II(H.W)|31 Videos

Similar Questions

Explore conceptually related problems

In CGS system, the Young's modulusm of a steel wire is 2 xx 10^(12) . To double the length of a wire of unit cross-section area, the force

Select and write the corrcet answer : The elastic potential energy per unit volume in a copper wire (Y = 1.2 xx 10^(11) Pa) stretched by 0.5 % of its length is

The area of cross section of a steel wire (Y=2.0 xx 10^(11) N//m^(2)) is 0.1 cm^(2) . The force required to double is length will be

Calculate the work done in stretching a steel wire of Young's modulus of 2 xx 10^(11) Nm^(-2) , mass of 40 kg, length of 200 cm and area of cross-section is 0.06 cm^(2)~ slowly applied without the elastic limit being reached.

Calculate the work done in streching a steel wire of Young's modulus of 2xx10^(11) Nm^(-2) ,length of 200 cm and area of cross-section is 0.06 cm^(2) slowly applied without the elastic limit being reached.

A block of mass M is suspended from a wire of length L, area of cross-section A and Young's modulus Y. The elastic potential energy stored in the wire is

Rigidity modulus of steel is eta and its Young's modulus is Y. A piece of steel of cross sectional area A is chaged into a wire of length L and area (A)/(10) then :

A 20 N stone is suspended from a wire and its length changes by 1% . If the Young's modulus of the material of wire is 2xx10^(11)N//m^(2) , then the area of cross-section of the wire is 2xx10^(11)N//m^(2) , then the area of cross-section of the wire will be

What is the force requiredto stretch a steel wire of 1 cm^(2) cross-section to 1.1 times its length ? (Y = 2 xx 10^(11) N//m^(2))