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CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 14 - MCQExams.com
CBSE
Class 11 Engineering Physics
Work,Energy And Power
Quiz 14
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 massive disc of radius $$R$$ is moving with a constant velocity $$u$$ on an frictionless table. Another small disc collides with it elastically with a speed of $$v_{0} = 0.3\ m/s$$, the velocities of the discs are parallel. The distance $$d$$ shown in the figure is equal to $$R/2$$, friction between the discs is negligible.
For which $$u$$ (in m/s) will the small disc move perpendicular to its original motion after the collision?
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$$1$$
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$$0.1$$
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$$2$$
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$$0.2$$
A simple pendulum has length l. The minimum velocity to be pendulum so that it performs vertical circular motion is
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$$\sqrt {gl} $$
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$$\sqrt2{gl} $$
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$$\sqrt5{gl} $$
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$$\sqrt {\frac{1}{g}} $$
A man raises 1 kg wt. to a height of 100 cm and holds it there for 30 minutes. How much work has he performed?
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$$1\times 9.8J$$
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$$1\times 9.8\times 30\times 60J$$
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$$1\times 9.8\times 30J$$
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$$1\times 9.8\times 30erg$$
Explanation
Given
$$m=1\ kg\\h=100\ cm=1\ m \\t=30\ mit$$
Work dont $$W=$$ Force $$\cdot $$ Displacement
Force apply to lift the mass = Weight of the body= $$1\times 9.8$$ N
From above equation,
$$W=1\times 9.8\times 1\ J$$
$$W=1\times 9.8 \ J$$
Option A
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
A bullet of mass $$m$$ and speed $$v$$ hits a pendulum bob of mass $$M$$ at time $$t _ { 1 }$$ and passes completely through the bob. The bullet emerges at time $$t _ { 2 }$$ with a speed of $$v / 2$$. The pendulum bob is suspended by a stiff rod of length $$l$$ and negligible mass. After the collision, the bob can barely swing through a complete vertical circle.At time $$t _ { 3 }$$ , the bob reaches the highest position.What quantities are conserved in this process?
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Total kinetic energy of the bob and the bullet during the time interval $$\Delta t = t _ { 2 } - t _ { 1 }$$
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Total momentum of the bob and the bullet during the time interval $$\Delta t = t _ { 2 } - t _ { 1 } $$
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Total mechanical energy of the bob and the bullet during the time interval $$t _ { 3 } - t _ { 1 } $$
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Momentum of the bob after $$t _ { 2 }$$
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.
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%
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50%
0%
75%
0%
100%
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
Fig. shows a smooth curved track terminatingin a smooth horizontal part. A spring ofspring constant $$400\mathrm { Nm }$$ is attached at one end to the wedge fixed rigidly with the horizontal part. $$A\ 40 \mathrm { g }$$ mass is released from rest at a height of $$5\mathrm { m }$$ on the curved track. The maximum compression of the spring is $$\left( g = 10 \mathrm { m } / \mathrm { s } ^ { 2 } \right)$$
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$$10\mathrm { cm }$$
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$$5\mathrm { cm }$$
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$$12\mathrm { cm }$$
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$$8\mathrm { cm }$$
A block of mass m is connected rigidly with a smooth plank by a light spring of stiffness K. If the plank is moved with constant velocity $${ \upsilon }_{ 0 }$$, find the work done by the external agent till the maximum compression of the spring.
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$$\dfrac { 1 }{ 2 } { mu }_{ 0 }^{ 2 }$$
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$$\frac { 3 }{ 4 } { mu }_{ 0 }^{ 2 }$$
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$${ mu }_{ 0 }^{ 2 }$$
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$$\dfrac { 3 }{ 2 } { mu }_{ 0 }^{ 2 }$$
Explanation
Let work done by spring and external agent be $$W_{s}$$ and $$W_{e} $$
respectively.
At the time of maximum compression, the velocity of block will be equal to velocity of plank i.e $$v_0$$.
So, $$\Delta K = \dfrac{1}{2}mv_0^2$$ (because plank's velocity did not change)
Now, from the reference of plank
Work done by Spring $$(W_s)$$= Change in Kinetic Energy of block
$$ W_s = 0- \dfrac{1}{2}mv_0^2$$
$$\Rightarrow W_s = -\dfrac{1}{2}mv_0^2$$
On Applying Work-Energy theorem on whole system from the reference of ground, we have
$$W_{e}+ W_s = \Delta K$$
$$\Rightarrow W_{e} -\dfrac{1}{2}mv_0^2 = \dfrac{1}{2}mv_0^2-0$$
$$ \Rightarrow W_{e} = \dfrac{1}{2}mv_0^2 + \dfrac{1}{2}mv_0^2$$
$$ = mv_0^2$$
So, the correct option will be $$(C)$$
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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$$(-1,-1)$$
A bucket tied at the end of a $$1.6m$$ long string is whirled in a vertical circle with a constant speed. What should be the minimum speed so that the water from the bucket does not spill when the bucket is at the highest position, (Take g=10m/s^2$$)
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$$16m/s$$
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$$6.25m/s$$
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$$4m/s$$
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$$2m/s$$
A heavy particle hanging from a fixed point by a light inextensible string of length l h projected horizontally with a speed of $$\sqrt { { g }^{ l } } $$. Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion, when the tension in the string is equal to the weight of the particle.
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$$\sqrt { gl } $$
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$$\sqrt { 2gl } $$
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$$\sqrt { 3gl } $$
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$$\sqrt { \frac { gl }{ 3 } } $$
The distance x moved by a body of mass 0.5 kg by a force varies with time as $$ x = 3t^2 + 4t + 5 $$, x is expressed in meter and t is seconds. The work done by force in first 2 seconds.
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75 J
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50 J
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60 J
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100 J
A strign of length $$1=1m$$ is fixed at one end and carries a mass of $$100gm$$ at other end. The string makes $$\sqrt { 5 } \pi$$ revolutions per second about a vertical axis passing through its second end. What is the angle of inclination of the string with the vertical?
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$$30^0$$
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$$45^0$$
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$$60^0$$
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$$75^0$$
A body of mass m is moved up the plane of varying slope by a tangential force up to height h.The coefficient of friction between the surface and the block is $$\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}
$$
A ball of mass 2 kg hits a floor with a speed of 4 m/s at an angle of $$ 60 ^o $$ with the normal.
If (e=1/2); then the change in the kinetic energy of the ball is
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-6J
0%
-12J
0%
-9J
0%
-3J
In a conical pendulum, the centripetal force $$\left[ \dfrac { { mv }^{ 2 } }{ r } \right] $$ acting on the bob is given by
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$$\dfrac { mgr }{ \sqrt { { L }^{ 2 }-{ r }^{ 2 } } } $$
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$$\dfrac { mgr }{ { L }^{ 2 }-{ r }^{ 2 } } $$
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$$\dfrac { { L }^{ 2 }-{ r }^{ 2 } }{ mgt } $$
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$$\dfrac { mgL }{ \left[ { L }^{ 2 }-{ r }^{ 2 } \right] ^{ 1/2 } } $$
A stone tied to a piece of string is shrilled in a vertical cirlce with uniform speed. In what position of the stone is the tension in the string greatest
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in the highest position of stone
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in the lowest position of stone
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in the position when string is horizontal
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is same for all positions of stone
A force acts on a 2$$\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$$
A body of mass 200 g moving on a test has final K.E. of 50 J after travelling a distance of 10 cm. Assuming 90% loss of energy due to friction. The initial speed of the body is
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10 $$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}$$
In head on elastic collision of two bodies of equal masses, it is not possible :
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the velocities are interchanged
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the speeds are interchanged
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the momenta are interchanged
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the faster body speeds up and the slower body shows down
The maximum and minimum tensions in a string of length 0.1 m with a stone 5 g tied to it and whirled in a vertical circle are in the ratio 13 :The velocity of the body at highest point is
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$$0.7\ ms^{-1}$$
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$$0.7\sqrt 3\ ms^{-1}$$
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$$\sqrt 3\ ms^{-1}$$
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$$7\sqrt 3 ms^{-1}$$
Potential energy v/s displacement curve for one dimensional conservative field is shown. Force at A and B is respectively -
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Positive, Positive
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Positive, Negative
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Negative, Positive
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Negative, Negative
A circular road of radius 1000 m has banking angle $$45^{\circ}$$ The maximum safe speed $$\left ( in ms^{-1} \right )$$ of a car having a mass 2000 kg will be (if the coefficient of the friction between tyre and road is 0.5)
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0%
172
0%
124
0%
99
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86
An escalator is moving downwards with a uniform speed u. A man of mass 'm' is running upwards or it at a uniform speed v. If the height of the escalator is h, the work done by the man in going up the escalator is
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zero
0%
mgh
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$$\frac{mghu}{(v-u)}$$
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$$\frac{mghv}{(v-u)}$$
Figure shows (x, t), (y, t) diagram of a particle in two-dimensions.
If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.
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8 N along X-axis
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128 N along X-axis
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1 N along Y-axis
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3 N along Y-axis
A rubber ball is bounced on the floor of a room which has its ceiling at a height of $$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$$
A particle of mass $$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 } $$
A block of mass $$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$$
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$$2J$$
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$$3J$$
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$$0J$$
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$$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$$
The potential energy of a particle of mass 1 kg free to move along the x-axis is given by U($$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$$ m/s
A body at rest can have
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Kinetic Energy
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momentum
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both momentum and kinetic Energy
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potential Energy
Explanation
A body at rest can have potential energy because PE is due to position of object but it does not have kinetic energy because KE is due to motion but object is at rest so no KE will be present.Momentum is also not present because momentum is due to velocity and velocity is zero in this case.
For example an object at a roof at rest have no KE but it has gravitational potential energy due to it's height above ground given by $$mgh$$.
A heavy stone hanging from a massless string of length $$15$$m is projected horizontally with speed $$147$$ m/s. The speed of the particle at the point where the tension in the string equals the weight of the particle is?
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$$10$$ m/s
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$$7$$ m/s
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$$12$$ m/s
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None of these
Explanation
Let at angle $$\theta$$ from vertical
speed be $$u$$ m/s and tension equal weight of particle i.e
$$\boxed{T=mg}$$.........(1) [given]
at this location, $$T-mg\cos\theta=\dfrac{mu^2}{l}$$
$$mg-mg\cos\theta=\dfrac{mu^2}{l}$$ [From (1)]
$$\boxed{lg(1-\cos\theta)=u^2}$$
by work energy theorem
$$W_{gravity}=\Delta K.E$$
$$-mgl(1-\cos\theta)=\dfrac{1}{2}mu^2-\dfrac{1}{2}mv^2$$
$$-2gl(1-\cos\theta)=u^2-v^2$$
$$-2u^2=u^2-v^2$$ [From (2)]
$$3u^2=v^2$$
$$u=\dfrac{v}{\sqrt{3}}=\dfrac{147}{\sqrt{3}}=85m/s$$
So answer is $$85m/s$$
A particle originally at rest at the hoghest point of a smooth vertical circle is slightly displaced. it will leave the circle at a vertical distance h below the highest point, such that h is equal to
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R
0%
R/4
0%
R/2
0%
R/3
The graph below show the force acting on a particle moves along the positive $$x-$$axis from the origin to $$x=x_1$$. The force if parallel to the $$x-$$axis and is conservative. The maximum magnitude $$F_1$$ has the same values for all graphs. Rank the solutions according to the change in the potential energy associated with the force. least (or most negative) to greatest (or most positive)
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$$2, 1, 3$$
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$$1, 3, 2$$
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$$2, 3, 1$$
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$$3, 2, 1$$
A stone of mass $$0.3$$ kg attached to a $$1.5$$m long string is whirled around in a horizontal circle at a speed of $$6$$ m/s. The tension in the string is?
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$$10$$N
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$$20$$N
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$$7.2$$N
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None of these
After how much time collision between the blocks will not take place practically:
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$$8\ sec$$
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$$16\ sec$$
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$$12\ sec$$
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$$infinite$$
Two particles A and B having mass m each and charge $$q_1$$ and $$-q_2$$ respectively, are connected at the ends of a non conducting flexible and inextensible string of the length l. The particle A is fixed and B is whirled along a vertical circle with centre at A. If a vertically upward electric field of strength E exists in the space, then for minimum velocity of particle B:
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The tension force in string at lowest point is zero
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The tension force at the lowest point is non-zero
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The tension force at the highest point is zero
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The work done by interaction force between particles A and B is non-zero
When a man increases his speed by $$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$$ to be speed which is a non-negative quantity or in this case, positive.
Option B is correct.
Two perfectly elastic objects $$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$$
0%
$$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$$, the kinetic energy acquired by the block is
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$$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$$.
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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