Q. No.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 rubberband by \(L\) is:
1. \(\frac{1}{2}(aL^2+bL^3)\)
2. \(\frac{aL^2}{2}+\frac{bL^3}{3}\)
3. \(\frac{1}{2}(\frac{aL^2}{2}+\frac{bL^3}{3})\)
4. \(aL^2+bL^3\)
A particle of mass \(m\) moving in the \(x\) direction with speed \(2v\) is hit by another particle of mass \(2m\) moving in the \(y\) direction with speed \(v\). If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:
1. \(44\%\)
2. \(50\%\)
3. \(56\%\)
4. \(62\%\)
1. \(0.3~\text{J}\)
2. \(0.6~\text{J}\)
3. \(1.5~\text{J}\)
4. \(0.8~\text{J}\)
(take potential energy \({=0}\) for \({r = 0)}\)
1. \({1\over 2}{\alpha r^3}\)
2. \({5\over 6}{\alpha r^3}\)
3. \({4\over 3}{\alpha r^3}\)
4. \({\alpha r^3}\)
1. \({mv^2\over 2(L-nr)}\)
2. \({mv^2\over L-2nr}\)
3. \({mv^2\over L-nr}\)
4. zero
A point particle of mass \(m\), moves along the uniformly rough track \(PQR\) as shown in the figure. The coefficient of friction, between the particle and the rough track equals \(\mu\). The particle is released, from rest, from the point \(P\) and it comes to rest at a point \(R\). The energies, lost by the ball, over the parts, \(PQ\) and \(QR\), of the track, are equal to each other, and no energy is lost when the particle changes direction from \(PQ\) to \(QR\).
The values of the coefficient of friction \(\mu\) and the distance \(x~(=QR)\), are, respectively close to:
1. \(0.2~\text{and}~6.5~\text{m}\)
2. \(0.2~\text{and}~3.5~\text{m}\)
3. \(0.29~\text{and}~3.5~\text{m}\)
4. \(0.29~\text{and}~6.5~\text{m}\)
A person trying to lose weight by burning fat lifts a mass of \(10~\text{kg}\) upto a height of \(1~\text{m}\) \(1000\) times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies \(3.8\times 10^7~\text{J}\) of energy per kg which is converted to mechanical energy with a \(20\%\) efficiency rate. Take \(g= 9.8~\text{ms}^{-2}\):
1. \(2.45\times 10^{-3}~\text{kg}\)
2. \(6.45\times 10^{-3}~\text{kg}\)
3. \(9.89\times 10^{-3}~\text{kg}\)
4. \(12.89\times 10^{-3}~\text{kg}\)
A time dependent force \(F = 6t \) acts on a particle of mass \(1~\text{kg}.\) If the particle starts from rest, the work done by the force during the first \(1~\text s\) will be:
1. \(4.5~\text{J}\)
2. \(22~\text{J}\)
3. \(9~\text{J}\)
4. \(18~\text{J}\)
1. \(\frac{5}{3}{h}\)
2. \(\infty\)
3. \(\frac{8}{3}{h}\)
4. \(3{h}\)
1. \(\tan\theta=\dfrac{\sqrt3+\sqrt2}{1-\sqrt2}\)
2. \(\tan\theta=\dfrac{1-\sqrt3}{1+\sqrt2}\)
3. \(\tan\theta=\dfrac{\sqrt3-\sqrt2}{1-\sqrt2}\)
4. \(\tan\theta=\dfrac{1-\sqrt3}{\sqrt2(1+\sqrt3)}\)
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