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Q. No.A particle of mass m moves on a straight line with its velocity increasing with distance according to the equation \(\begin{equation} v=\alpha \sqrt{x} \end{equation}\), where \(\begin{equation} \alpha \end{equation}\) is a constant. The total work done by all the forces applied on the particle during its displacement from \(x=0\) to \(x=d\), will be :
1. \(\begin{equation} \frac{m}{2 \alpha^2 d} \end{equation}\)
2. \(\begin{equation} \frac{m d}{2 \alpha^2} \end{equation}\)
3. \(\begin{equation} \frac{m \alpha^2 d}{2} \end{equation}\)
4. \(\begin{equation} 2 m \alpha^2 d \end{equation}\)
Subtopic: Work Done by Variable Force |
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A body of
(given, \(R=14\) m, \(g=10\) m/s2 and \(\sqrt{2}=1.4\))
1. \(21.9\) m/s
2. \(10.6\) m/s
3. \(19.8\) m/s
4. \(16.7\) m/s
(given, \(R=14\) m, \(g=10\) m/s2 and \(\sqrt{2}=1.4\))
1. \(21.9\) m/s
2. \(10.6\) m/s
3. \(19.8\) m/s
4. \(16.7\) m/s
Subtopic: Conservation of Mechanical Energy |
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Level 2: 60%+
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A stationary particle breaks into two parts of masses \(\mathrm{m_A}\) and \(\mathrm{m_B}\) which move with velocities \(\mathrm{v_A}\) and \(\mathrm{v_B}\) respectively. The ratio of their kinetic energies \(\mathrm{(K_B:K_A)}\) is:
1. \(\mathrm{m_B:m_A}\)
2. \(1:1\)
3. \(\mathrm{m_Bv_B:m_Av_A}\)
4. \(\mathrm{v_B:v_A}\)
1. \(\mathrm{m_B:m_A}\)
2. \(1:1\)
3. \(\mathrm{m_Bv_B:m_Av_A}\)
4. \(\mathrm{v_B:v_A}\)
Subtopic: Collisions |
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A bullet of mass \(50 ~\text g\) is fired with a speed \(100 ~\text{m/s}\) on a plywood and emerges with \(40 ~\text{m/s}.\) The percentage loss of kinetic energy is :
1. \(44\%\)
2. \(32\%\)
3. \(84\%\)
4. \(16\%\)
1. \(44\%\)
2. \(32\%\)
3. \(84\%\)
4. \(16\%\)
Subtopic: Work Energy Theorem |
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A force acts on a \(2~\text{kg}\) object so that its position is given as a function of times as; \({x=3t^2+5.} \) What is the work done by this force in first \(5\) seconds?
1. \(850 ~\text{J}\)
2. \(950 ~\text{J}\)
3. \(875 ~\text{J}\)
4. \(900 ~\text{J}\)
1. \(850 ~\text{J}\)
2. \(950 ~\text{J}\)
3. \(875 ~\text{J}\)
4. \(900 ~\text{J}\)
Subtopic: Work Energy Theorem |
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Two particles of the same mass \(m\) are moving in circular orbits because of force given by \(F(r) =\left( \frac{-16}{r}-r^3\right). \) The first particle is at a distance \(r = 1 \) and the second at \(r = 4. \) The best estimate for the ratio of kinetic energies of the first and the second particle is closest to:
1. \(3\times 10^-3\)
2. \(6\times 10^2\)
3. \(6\times 10^{-2}\)
4. \( 10^{-1}\)
1. \(3\times 10^-3\)
2. \(6\times 10^2\)
3. \(6\times 10^{-2}\)
4. \( 10^{-1}\)
Subtopic: Concept of Work |
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A body of mass \(m\) starts moving from rest along the \(x\text{-axis}\) so that its velocity varies as \(v = a \sqrt s\) where \(a\) is a constant and \(s\) is the distance covered by the body. The total work done by all the forces acting on the body in the first \(t\) seconds after the start of the motion is:
1. \(8ma^4 t^ 2\)
2. \(\frac{1} {4} ma^4 t^ 2\)
3. \(4ma^4 t ^2\)
4. \(\frac{1} {8} ma^4 t^ 2\)
1. \(8ma^4 t^ 2\)
2. \(\frac{1} {4} ma^4 t^ 2\)
3. \(4ma^4 t ^2\)
4. \(\frac{1} {8} ma^4 t^ 2\)
Subtopic: Concept of Work |
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Three blocks \(A, B~\text{and}~ C\) are lying on a smooth horizontal surface, as shown in the figure. \(A ~\text{and}~B\) have equal masses, \(m\) while \(C\) has mass \(M.\) Block \(A\) is given an initial speed \(v\) towards \(B\) due to which it collides with \(B\) perfectly inelastically. The combined mass collides with \(C,\) also perfectly inelastically if \(\frac 5 6 \text{th}\) of the initial kinetic energy is lost in the whole process. The value of \(({M/m})\) is:
1. \(5\)
2. \(2\)
3. \(4\)
4. \(3\)
1. \(5\)
2. \(2\)
3. \(4\)
4. \(3\)
Subtopic: Collisions |
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Two particles \(A \) and \(B \) of equal mass \({M}\) are moving with the same speed \(v\) as shown in the figure. They collide completely inelastically and move as a single particle \(C.\) The angle \(\theta\) that the path of \(C\) makes with the \(x\text-\)axis is given by:
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. \(\tan\theta=\dfrac{1-\sqrt3}{\sqrt2(1+\sqrt3)}\)
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. \(\tan\theta=\dfrac{1-\sqrt3}{\sqrt2(1+\sqrt3)}\)
Subtopic: Collisions |
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An object is dropped from a height \({h}\) from the ground. Every time it hits the ground it looses \(50\%\) of its kinetic energy. The total distance covered as \({t}\rightarrow\infty\) is:
1. \(\frac{5}{3}{h}\)
2. \(\infty\)
3. \(\frac{8}{3}{h}\)
4. \(3{h}\)
1. \(\frac{5}{3}{h}\)
2. \(\infty\)
3. \(\frac{8}{3}{h}\)
4. \(3{h}\)
Subtopic: Conservation of Mechanical Energy |
Level 3: 35%-60%
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