Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Work Power and Energy question

2014 · Shift 0 · Q71
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Work Power and Energy
  5. /2014 · Shift 0 · Q71

Work Power and Energy question

2014 · Shift 0 · Q71

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
When a rubber-band is stretched by a distance xxx, it exerts restoring force of magnitude F=ax+bx2F = ax + b{x^2}F=ax+bx2 where aaa and bbb are constants. The work done in stretching the unstretched rubber-band by LLL is :
  1. A
    aL2+bL3a{L^2} + b{L^3}aL2+bL3
  2. B
    12(aL2+bL3){1 \over 2}\left( {a{L^2} + b{L^3}} \right)21​(aL2+bL3)
  3. C
    aL22+bL33{{a{L^2}} \over 2} + {{b{L^3}} \over 3}2aL2​+3bL3​
  4. D
    12(aL22+bL33){1 \over 2}\left( {{{a{L^2}} \over 2} + {{b{L^3}} \over 3}} \right)21​(2aL2​+3bL3​)
View written solutionFree

Correct answer: C

  1. Given restoring force

    When the rubber band is stretched by distance xxx, the restoring force magnitude is F(x)=ax+bx2.F(x)=ax+bx^2.F(x)=ax+bx2.

    To stretch it slowly from x=0x=0x=0 to x=Lx=Lx=L, the external force applied at each stage must balance this restoring force.

  2. Work done in stretching

    The work done by the external agent is W=∫0LF(x) dx.W=\int_0^L F(x)\,dx.W=∫0L​F(x)dx.

    Substituting F(x)=ax+bx2F(x)=ax+bx^2F(x)=ax+bx2, W=∫0L(ax+bx2) dx.W=\int_0^L (ax+bx^2)\,dx.W=∫0L​(ax+bx2)dx.

  3. Integrate term by term

    W=a∫0Lx dx+b∫0Lx2 dx.W=a\int_0^L x\,dx+b\int_0^L x^2\,dx.W=a∫0L​xdx+b∫0L​x2dx.

    Using standard integrals, ∫0Lx dx=[x22]0L=L22,\int_0^L x\,dx=\left[\frac{x^2}{2}\right]_0^L=\frac{L^2}{2},∫0L​xdx=[2x2​]0L​=2L2​, ∫0Lx2 dx=[x33]0L=L33.\int_0^L x^2\,dx=\left[\frac{x^3}{3}\right]_0^L=\frac{L^3}{3}.∫0L​x2dx=[3x3​]0L​=3L3​.

    Therefore, W=a(L22)+b(L33).W=a\left(\frac{L^2}{2}\right)+b\left(\frac{L^3}{3}\right).W=a(2L2​)+b(3L3​).

  4. Final expression

    W=aL22+bL33\boxed{W=\frac{aL^2}{2}+\frac{bL^3}{3}}W=2aL2​+3bL3​​

  5. Match with options

    This corresponds to Option C.

  6. Comparison with stored answer

    Stored correct answer: C

    Our derived answer: C

    Hence, they agree.

PreviousNext

More from Work Power and Energy

  • This question has Statement 1 and Statement 2. Of the four choices given after the Statements, choose the one that best describes the two Statements. If two springs S1​ and S2​ of force constants k1​ and k2​,…2012 · MCQ
  • The potential energy function for the force between two atoms in a diatomic molecule is approximately given by U(x)=x12a​−x6b​, where a and b are constants and x is the distance…2010 · MCQ
  • An athlete in the olympic games covers a distance of 100m in 10s. His kinetic energy can be estimated to be in the range2008 · MCQ
  • A 2kg block slides on a horizontal floor with a speed of 4m/s. It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is 15N and spring constant is 10,000N/m. The spring…2007 · MCQ
  • A particle is projected at 60∘ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is2007 · MCQ
  • A mass of Mkg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45∘ with the initial vertical direction is2006 · MCQ
  • A ball of mass 0.2kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2m while applying the force and the ball goes upto 2m height further, find the magnitude of the force. (consider g=10m/s2).2006 · MCQ
  • A particle of mass 100g is thrown vertically upwards with a speed of 5m/s. The work done by the force of gravity during the time the particle goes up is2006 · MCQ