JEE Main 2014 · 6 Apr · Shift 1 · Physics Q5

← JEE Main 2014 · 6 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 5 of 90

When a rubber-band is stretched by a distance $x$ , it exerts a restoring force of magnitude $F=ax+bx^2$ where $a$ and $b$ are constants. The work done in stretching the unstretched rubber-band by $L$ is :

aL^2+bL^3

\dfrac12\left(aL^2+bL^3\right)

\dfrac{aL^2}{2}+\dfrac{bL^3}{3}

\dfrac12\left(\dfrac{aL^2}{2}+\dfrac{bL^3}{3}\right)

Answer and worked solution

Official answer: C

$W=\int_0^L\left(ax+bx^2\right)dx=\frac{aL^2}2+\frac{bL^3}3.$

Topics:Work by constant and variable forces

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JEE Main 2014 · 6 Apr · Shift 1, Physics question 5