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CBSE Questions for Class 11 Medical Physics Work, Energy And Power Quiz 14 - MCQExams.com
CBSE
Class 11 Medical Physics
Work, Energy And Power
Quiz 14
A force acts on a $$30$$ gm particle in such a way that the position of the particle as a function of time is given by $$x = 3t - { 4t }^{ 2 } + { t }^{ 3 }$$, where $$x$$ is in meters and $$t$$ is in seconds. The work done during the first $$4$$ second is :-
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$$5.28\ J$$
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$$450\ mJ$$
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$$490\ mJ$$
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$$530\ mJ$$
Under the action of variable force, at any instant the displacement is
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assumed to be infinitesimally small so that the force is assumed to be constant.
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constant whatever be the magnitude of force.
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half the magnitude of force.
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assumed to be infinitesimally small so that the force is assumed to be perpendicular to the displacement.
If the net external force acting on the system of particles is if the then which of the following may vary?
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Momentum of the system
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Kinetic energy of the system
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Velocity of centre of mass
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Position of centre of mass
A particle of mass m moves on a straight line with its velocity varying with the distance travelled according to the equation $$v = a\sqrt{x}$$, where a is constant. Find the total work done by all the forces during a displacement from x = 0, to x = d.
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$$ 5 ma^2 \ d/2$$
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$$ ma^3 \ d/2$$
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ma d/2
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$$ ma^2 \ d/2$$
A force $$\overrightarrow { F } =\left( 3t\hat { i } +5\hat { j } \right) N$$ acts on a body and its displacement varies as $$\overrightarrow { S } =\left( 2{ t }^{ 2 }\hat { i } -5\hat { j } \right). $$ work done by this force in $$t=0$$ to $$2sec$$ is
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$$23 J$$
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$$32 J$$
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$$Zero$$
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$$can't$$ $$obtained$$
A particle of mass m moves along the quarter section of the circular path whose centre is at the origin. the radius of the circular path is a.. A force $$ \overrightarrow F = y \hat i - x\hat j $$ newton acts on the particle, where x. y denote the coordinates of position of the particle. calculate the work done by this force in taking the particle from point A (a, 0) to point B(0, a) along the circular path.
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$$ \frac { \pi a^2}{4} J $$
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$$ \frac { \pi a^2}{2} J $$
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$$ -\frac { \pi a^2}{2} J $$
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$$ -\frac { \pi a^2}{4} J $$
A particle is displaced from (1, 2) m to (0,0) m along the path $$y = 2x^3$$. Work done by a force $$\vec F = (x^3 \hat j + y \hat i)$$ N acting on the particle, during this displacement, is
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-1.5 J
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1.5 J
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2.5 J
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-2.5 J
For a particle in rectilinear motion under constant acceleration the change in kinetic energy of a particle is equal to the work done on it by the net force.
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True
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False
The total mechanical energy of a spring-mass system in simple harmonic motion is $$E$$. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $$A$$ remains the same. The new mechanical energy will
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Become $$2E$$
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Become $$\dfrac{E}{2}$$
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Become$$\sqrt 2E$$
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Remain $$E$$
A ball of mass m is moving with a speed $$20\ m/sec$$. It strikes an identical ball which is at rest. After collusion each ball moves making an angle of $${ 45 }^{ \circ }$$ with the original of motion. Calculate their speeds after collision.
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$$10\sqrt { 3} m/sec\quad$$ for each ball
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$$20\sqrt { 2} m/sec\quad $$ for each ball
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$$30\sqrt { 2 } m/sec\quad $$ for each ball
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$$10\sqrt { 2 } m/sec\quad $$ for each ball
A force of $$\overrightarrow { F } =2x\hat { x } +2\hat { y } +3z^{ 2 }\hat{k}N$$ is acting on a particle. Find the work done by this force in displacing the body from $$(1,2,3)m$$ to $$(3,6,1)m$$.
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$$-10J$$
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$$100J$$
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$$10J$$
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$$1J$$
A skater of weight $$30 \ kg$$ has initial speed $$32m/s$$ and second one of weight $$40 \ kg$$ has $$5m/s$$. After the collision, they stick together and have a speed $$5m/s$$. Then the loss in KE is
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$$48J$$
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$$96J$$
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Zero
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None of these
Explanation
Given:
Weight of the skater 1 is $$m_1=30 \ kg$$
His initial speed $$u_1=32 \ m/s$$
Weight of the skater 2 is $$m_2=40 \ kg$$
His initial speed $$u_2=5 \ m/s$$
After collision they stick together
Both of their speed $$v_1=v_2=v=5 \ m/s$$
We know that kinetic energy $$=\dfrac{1}{2}mv^2$$
Initial kinetic energy $$=KE_{skater 1}+KE_{skater 2}$$
$$=\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2$$
$$=\dfrac{1}{2}(30)(32)^2+\dfrac{1}{2}(40)(5)^2$$
$$=\dfrac{1}{2}(30)(1024)+\dfrac{1}{2}(40)(25)$$
$$=15360+500=15860 \ J$$
Final kinetic energy
$$=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2$$
$$=\dfrac{1}{2}m_1v^2+\dfrac{1}{2}m_2v^2=\dfrac{1}{2}(m_1+m_2)v^2$$
$$=\dfrac{1}{2}(30+40)(5)^2$$
$$=\dfrac{1}{2}(70)(25)=875 \ J$$
Loss in kinetic energy $$\Delta KE=15860-875$$
$$=14985 \ J$$
The answer is none of the given options. So option D is the answer.
The figure shows the force-time graph for a body of mass $$10\, kg$$ initially at rest. The velocity gained by the body in $$6 \sec$$ will be
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Zero
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$$6\, m/s$$
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$$3\, m/s$$
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$$4\, m/s$$
A ball is dropped from a height of 10m. It strikes the ground and rebounds up to a height of 2.5m. During the collision the percent loss in kinetic energy is?
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25%
0%
50%
0%
75%
0%
100%
A perticle which is experiencing a force , given by $$\overset { \rightarrow }{ F } =\overset { \rightarrow }{ 31 } -12\overset { \rightarrow }{ J }$$, undergoes a displacement $$\overset { \rightarrow }{ d } = 4\overset { \rightarrow }{ i } $$. If the particle had a kinetic energy of $$3\vec{ j}$$ at the beginning of the displacement , what is its kinetic energy at the end of the displacement ?
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$$9J$$
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$$12J$$
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$$15J$$
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$$10J$$
A particle moves along x-axis under the action of a position dependent force $$ F= (5x^2-2x) N $$. work done by forces on the particle when it moves from origin to x=3 m is
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45J
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36j
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32J
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42J
A particle of mass 2kg is moving in a circular path of radius 1m with a speed that varies with time as $$v=(t^2-t)m/s$$. The force acting on the particle at t=2s is
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8N
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6N
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10N
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5N
A particle is moving in a force field given by F $$=$$ $${y^2}i - {x^2}j$$. starting from A, the particle has to reach C either along ABC or ADC. Let the work done along the two paths be $${W_1}$$ and $${W_2}$$ respectively. Then ($${W_1}$$,$${W_2}$$) are
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$$(-1,1)$$
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$$(1,0)$$
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$$(1,1)$$
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$$(-1,-1)$$
The distance x moved by a body of mass 0.5 kg by a force varies with time as $$ x = 3t^2 + 4t + 5 $$, x is expressed in meter and t is seconds. The work done by force in first 2 seconds.
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75 J
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50 J
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60 J
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100 J
A bullet of mass $$20 g$$ leaves a riffle at an initial speed $$100 m/s$$ and strikes a target at the same level with speed $$50 m/s$$ . The amount of work done by the resistance of air will be
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0%
$$100 J$$
0%
$$25 J$$
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$$75 J$$
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$$50 J$$
A body of mass m is moved up the plane of varying slope by a tangential force up to height h.The coefficient of friction between the surface and the block is $$\mu $$.
1.The work done on the block.
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by gravity depends upon the height h
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by gravity depends upon the force F
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by friction depends upon the speed of block.
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by friction is $$\mu $$mgx, when moved slowly.
The displacement time graph of a uniformed accelerating body of mass 2 kg. initially at rest is shown below the magnitude of force acting on the body is
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$$0.5 N$$
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$$1 N$$
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$$2 N$$
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$$4 N$$
Two bodies A and B have masses $$20\ kg$$ and $$5\ kg$$ respectively . Each one is acted upon by a force of $$4\ kg.-wt$$. If they acquire the same kinetic energy in times $$t_{ A }\ and\ t_{ B }$$, then the ratio $$\dfrac { t_{ A } }{ t_{ B } } $$:
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$$\dfrac { 1 }{ 2 } $$
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$$2$$
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$$\dfrac { 2 }{ 5 } $$
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$$\dfrac { 5 }{ 6 } $$
If water is flowing in a pipe at a height 4m from the ground then its potential
energy per unit volume is (Reference is taken at ground, $$ g=10 \mathrm{m} / \mathrm{s}^{2} ) $$
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$$
20 \mathrm{kJ} / \mathrm{m}^{3}
$$
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$$
10 \mathrm{kJ} / \mathrm{m}^{3}
$$
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$$
40 \mathrm{kJ} / \mathrm{m}^{3}
$$
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$$
30 \mathrm{kJ} / \mathrm{m}^{3}
$$
A certain system has potential energy given by the function $$U(x) = - ax^2 + bx^4$$ with constants,a,b>Which of the following is an unstable equilibrium point?
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$$0$$
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$$\sqrt{a/2b}$$
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$$- \sqrt{a/2b}$$
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$$\sqrt{a/b}$$
A ball of mass 2 kg hits a floor with a speed of 4 m/s at an angle of $$ 60 ^o $$ with the normal.
If (e=1/2); then the change in the kinetic energy of the ball is
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-6J
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-12J
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-9J
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-3J
A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the minimum work required to pull the hanging part of the chain on the table is
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$$MgL$$
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$$\dfrac{MgL}{3}$$
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$$\dfrac{MgL}{9}$$
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$$\dfrac{MgL}{18}$$
Explanation
$$\textbf{Step 1 - Draw the FDB -}$$
$$\bullet$$ Mass of hanging chain $$M^{\prime}= \lambda \cdot \dfrac {l}{3}$$
$$= \dfrac {M}{l} \cdot \dfrac {l}{3}$$
$$= \dfrac {M}{3}$$
$$\textbf{Step 2 - By using work-energy theorem}$$
$$W_{net} = \triangle k$$
$$W_{ext} + W_{non-cons} + W_{cons} = \triangle k$$
$$W_{ext} + W_{cons} = \triangle k$$
$$W_{ext} + U_{i} - U_{f} = k_{f} - k_{i}$$....(1)
$$\textbf{Step3:Potential and kinetic energies}$$
If table is taken as reference then
in initial condition potential energy of $$2l/3$$ part is zero as it is placed on the table and potential energy of $$l/3$$ lie at centre of mass.
So $$U_i=M^{\prime}g \dfrac{(\dfrac{l}{3})}{2}$$
$$\Rightarrow U_i=\dfrac{M^{\prime}g l}{6}$$
Final potential energy is zero as whole chain lie on table.
$$U_f=0$$
As chain is slowly pulled on table then kinetic energies remain constant.
$$K_i=K_f=K$$
$$\textbf{Step4:Put values in equation 1}$$
$$W_{ext} - \dfrac{M'gl}{6} - 0 =K-K $$
$$W_{ext} = \dfrac {M}{3} \times g\times \dfrac {l}{6}$$
$$W_{ext} = \dfrac {Mgl}{18}$$ Ans.
Hence option (D) is correct.
A body of mass 200 g moving on a test has final K.E. of 50 J after travelling a distance of 10 cm. Assuming 90% loss of energy due to friction. The initial speed of the body is
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10 $$ms^{-1}$$
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$$10\sqrt{50}ms^{-1}$$
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$$5\sqrt{50}ms^{-1}$$
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$$50\sqrt{50}ms^{-1}$$
A point particle of mass $$m$$, moves along the uniformly rough track $$PQR$$ as shown in 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 particle, 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 :
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0.29 and 6.5 m
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0.2 and 6.5 m
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0.2 and 3.5 m
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0.29 and 3.5 m
A block of mass $$20 kg$$ is being brought down by a chain. If block acquires a speed of $$2 m/s$$ in dropping down $$2 m$$. Find work done by the chain during the process. ($$g= 10m/s^{2}$$ )
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0%
$$-360 J$$
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$$400 J$$
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$$360 J$$
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$$-280 J$$
Potential energy v/s displacement curve for one dimensional conservative field is shown. Force at A and B is respectively -
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Positive, Positive
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Positive, Negative
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Negative, Positive
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Negative, Negative
An electron with $$K E$$ $$6 e V$$ is incident on a
hydrogen atom in its ground state. The collision.
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must be elastic
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may be partially elastic
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must be completely inelastic
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may be partially inelastic.
An escalator is moving downwards with a uniform speed u. A man of mass 'm' is running upwards or it at a uniform speed v. If the height of the escalator is h, the work done by the man in going up the escalator is
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zero
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mgh
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$$\frac{mghu}{(v-u)}$$
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$$\frac{mghv}{(v-u)}$$
A proton of mass $$m_p$$ collides with a heavy particle. After collision proton bunches back with 4/9 of its intial kinetic energy. Collision is perfectly elastic. Find mass of heavy particle.
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5 $$m_p$$
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6 $$m_p$$
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3 $$m_p$$
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1.5 $$m_p$$
A block of mass 100$$\mathrm { g }$$ attached to a spring of stiffness 100$$\mathrm { N } / \mathrm { m }$$ is lying on a frictionless floor as shown. block is moved to compress the spring by 10 cm and released. If the collision with the wall is elastic then the time period of oscillations. (in seconds)
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0.133
0%
13.3
0%
0.26
0%
0.3
A force acts on a 2$$\mathrm { kg }$$ object so that itsposition is given as a function of time as $$x = 3 t ^ { 2 } + 5 .$$ What is the work done by this
force in first 5 seconds?
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$$960$$ $$J$$
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$$950 J$$
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$$850$$ $$J$$
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$$875$$ $$J$$
Figure shows (x, t), (y, t) diagram of a particle in two-dimensions.
If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.
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8 N along X-axis
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128 N along X-axis
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1 N along Y-axis
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3 N along Y-axis
In the arrangement show, the pendulum on the left is
pulled aside. It is then released
and allowed to collide with other pendulum which is at rest. A perfectly inelastic collision occurs and the system rises to a height 1/4 h. The ratio of the masses of the pendulum is :
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0%
1
0%
2
0%
3
0%
4
In head on elastic collision of two bodies of equal masses, it is not possible :
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the velocities are interchanged
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the speeds are interchanged
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the momenta are interchanged
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the faster body speeds up and the slower body shows down
Two balls A and B moving in the same direction collide. The mass of B is p times that of A. Before the collision the velocity of A was q times that of B. After the collision A comes to rest. If e be the coefficient of restitution then which of the following conclusion/s is/are correct?
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$$e=\dfrac{p+q}{pq-p}$$
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$$\dfrac{p+q}{pq+q}$$
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$$p>= \dfrac{q}{q-2}$$
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$$p>-1$$
A rubber ball is bounced on the floor of a room which has its ceiling at a height of $$3.2{ m }$$ from the floor. The ball hits the floor with a speed of $$10 m / { s },$$ and rebounds vertically up. If all collisions simply reverse the velocity of the ball, without changing its speed, then how long does it take the ball for a round trip, from the moment it bounces from the floor to the
moment it returns back to it ? Acceleration due to gravity is $$10 m / s ^ { 2 }.$$
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$$4 s$$
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$$2 s$$
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$$0.8 s$$
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$$1.2 s$$
An L shaped tube of uniform cross-sectional area "a" carries a flowing liquid (speed = v m/s). The force requried to keep the tube in equilibrium (p = density, Neglect friction) is
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$$\sqrt{2}a pv^2$$
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$$a p v^2$$
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$$\frac{1}{2}a pv^2$$
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$$2 a pv^2$$
The potential energy of a particle of mass 1 kg free to move along the x-axis is given by U($$x$$) =($$\frac{x^2}{2}$$-$$x$$) joule. If the total mechanical energy of the particle is 2J, then find the maximum speed of the particle. (Assuming only conservative force acts on the particle)
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$$\sqrt5$$ m/s
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$$\sqrt7$$ m/s
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$$\sqrt2$$ m/s
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$$\sqrt6$$ m/s
When a man increases his speed by $$2 m/s$$, he finds that his kinetic energy is doubled, the original speed of the man is
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$$2(\sqrt {2} - 1) m/s$$
0%
$$2(\sqrt {2} + 1) m/s$$
0%
$$4.5\ m/s$$
0%
None of these
Explanation
According to question,
Initial kinetic energy, $$ E_1 = \dfrac{1}{2} mv^2 $$ ....(i)
Then, he increases his speed by $$2 \ m/s$$
2m/s
and K.E. is doubled.
Final kinetic energy $$ E_2=2E_1= \dfrac{1}{2}m(v+2)^2 $$ ....(ii)
Let multiply both sides of (i) by 2,
$$2E_1=mv^2$$ . . . (iii)
Equating (ii) and (iii),
$$mv^2=\dfrac{1}{2}m(v+2)^2$$
$$2v^2=v^2+4v+4$$
Solving this using quadratic formula,
$$ v=(2+2\sqrt{2})\ m/s $$
$$= 2(1+\sqrt{2})\ m/s$$
We reject the other value of $$v$$ as it must be positive because we considered $$v$$ to be speed which is a non-negative quantity or in this case, positive.
Option B is correct.
A particle of mass $$m$$, initially at rest is acted upon by a variable force f varying with time t. It begins to move with a velocity u after the force stops acting.)(curve is semi circle )
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$$u=\frac { \pi f^{ 2 }_{ 0 } }{ 2m } $$
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$$u=\frac { \pi t^{ 2 } }{ 8m } $$
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$$u=\frac { \pi f_{ 0 }T }{ 4m } $$
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$$u=\frac { f_{ 0 }T }{ 2m } $$
In head on elastic collision on two bodies of equal masses, which of the following is possible
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the velocities are interchaned
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the speeds are interchanged
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the momenta are interchanged
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the faster body slows down and the slower body speeds up
A block of mass $$1\,kg$$ is free to move along the X-axis. It is at rest and from time $$t=0$$ onwards it is subjected to a time-dependent force $$F(t )$$ in the X-direction. The force $$F(t )$$ varies with $$t$$ as shown in figure. The kinetic energy of the block at $$t=4s$$ is
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$$1J$$
0%
$$2J$$
0%
$$3J$$
0%
$$0J$$
0%
$$4J$$
Explanation
$$\because$$ Total momentum of block = Area of given graph
$$\therefore P=A_1+A_2+A_3$$
Or $$P=\dfrac{1}{2}\times 1\times 2-\dfrac{1}{2}\times 2\times 2+\dfrac{1}{2}\times 1\times 2$$
Or $$P=0$$ or $$mv=0$$ or $$v=0$$
Hence, kinetic energy at $$(t=4s)=\dfrac{1}{2}mv^2=0$$
Two perfectly elastic objects $$A$$ and $$B$$ of identical mass are moving with velocities $$15\ m/s$$ and $$10\ m/s$$ respectively collide along the direction of line joining them. Their velocities after collision are respectively:
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$$10\ m/s, 15\ m/s$$
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$$20\ m/s, 5\ m/s$$
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$$0\ m/s, 25\ m/s$$
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$$5\ m/s, 20\ m/s$$
Explanation
$$15m+10m=mv_1+mv_2$$
$$25=v_1+v_2$$...............(i)
and $$\dfrac{v_2-v_1}{u_1-u_2}=1$$
$$\Rightarrow \dfrac{v_2-v_1}{15-10}=1$$
$$\Rightarrow v_2-v_1=5$$............(ii)
$$v_1+v_2=25$$
$$\dfrac{v_2-v_1=5}{2v_2=30}$$
$$\therefore v_2=15m/s,v_1=10m/s$$
A constant force of $$5N$$ is applied on a block of mass $$20\ kg$$ for a distance of $$2.0\ m$$, the kinetic energy acquired by the block is
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0%
$$20\ J$$
0%
$$15\ J$$
0%
$$10\ J$$
0%
$$5\ J$$
Explanation
Acceleration $$a = \dfrac {F}{m} =\dfrac {5}{20} = \dfrac {1}{4} m/s^{2}$$
$$ u =0 $$
$$v = \sqrt {(2as)} = \sqrt {\left (\dfrac {2\times 2}{4}\right )} = 1\ m/s$$
$$KE = \dfrac {1}{2} mv^{2} = \dfrac {1}{2} \times 20\times (1)^{2} = 10\ J$$
(or)
Work done, $$W = Fs$$ $$= 5\times 2 = 10\ J$$.
A body at rest can have
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Kinetic Energy
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momentum
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both momentum and kinetic Energy
0%
potential Energy
Explanation
A body at rest can have potential energy because PE is due to position of object but it does not have kinetic energy because KE is due to motion but object is at rest so no KE will be present.Momentum is also not present because momentum is due to velocity and velocity is zero in this case.
For example an object at a roof at rest have no KE but it has gravitational potential energy due to it's height above ground given by $$mgh$$.
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