Test: Single Correct MCQs: Work, Energy and Power | JEE Advanced
21 Questions MCQ Test Physics For JEE | Test: Single Correct MCQs: Work, Energy and Power | JEE Advanced
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If a machine is lubricated with oil (1980)
Mechanical efficiency =
The output work will increase because the friction becomes less. Thus the mechanical efficiency increases.
Two masses of 1 gm and 4 gm are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac = k2rt2 where k is a constant. The power delivered to the particles by the force acting on it is:
The centripetal acceleration
A spring of force-constant k is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of
A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to
A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = –kx + ax3. Here k and a are positive constants. For x≥ 0 , the functional form of the potential energy U(x) of the particle is
we have potential energy zero twice (out of which one is at origin).
Also, when we put x = 0 in the given function,
should be zero. These characteristics are represented by (d).
An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end.
The mass is released with the spring initially unstretched.
Then the maximum extension in the spring is
The above situation can also be looked upon as the decrease in the gravitational potential energy of spring mass system is equal to the gain in spring elastic potential energy.
If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1,2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation between W1, W2 and W3
Note : In a conservative field work done does not depend on the path. The gravitational field is a conservative field.
A particle is acted by a force F = kx, where k is a +ve constant. Its potential energy at x = 0 is zero. Which curve correctly represents the variation of potential energy of the block with respect t o x
We know that ΔU = – W for conservative forces
A block (B) is attached to two unstretched springs S1 and S2 with spring constants k and 4k, respectively (see fig. I).
The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (figure II) and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B. The ratio y/x is –
When the block B is displaced towards wall 1, only spring S1 is compressed and S2 is in its natural state.
This happens because the other end of S2 is not attached to the wall but is free. Therefore the energy stored in the system = 1/2
k1 x2 . When the block is released, it will come back to the equilibrium position, gain momentum, overshoot to equilibrium position and move towards wall 2. As this happens, the spring S1 comes to its natural length and S2 gets compressed. As there are no frictional forces involved, the P.E. stored in the spring S1 gets stored as the P.E. of spring S2 when the block B reaches its extreme position after compressing S2 by y.
Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?
Let the radius of the circle be r. Then the two distance travelled by the two particles before first collision is 2πr. Therefore
2v ´ t + v ´ t =2pr
where t is the time taken for first collision to occur.
Distance travelled by particle with velocity v is equal to 
Therefore the collision occurs at B.
As the collision is elastic and the particles have equal masses, the velocities will interchange as shown in the figure. According to the same reasoning as above, the 2nd collision will take place at C and the velocities will again interchange.
With the same reasoning the 3rd collision will occur at the point A. Thus there will be two elastic collisions before the particles again reach at A.
A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is
The forces acting on the bead as seen by the observer in the accelerated frame are : (a) N ; (b) mg ; (c) ma (pseudo force).
Let θ is the angle which the tangent at P makes with the X- axis. As the bead is in equilibrium with respect to the wire, therefore N sinθ = ma and N cos θ = mg
From (i) & (ii)
A block of mass 2 kg is free to move along the x-axis. It is at rest and from 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 the figure. The kinetic energy of the block after 4.5 seconds is
Area under F – t graph gives the impulse or the change in the linear momentum of the body. As the initial velocity (and therefore the initial linear momentum) of the body is zero, the area under F – t graph gives the final linear momentum of the body.
Area of ΔAOB
The work done on a particle of mass m by a force,
(K being a constant of appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x – y plane is
Let us consider a point on the circle The equation of circle is x2 + y2 = a2 The force is
The force acts radially outwards as shown in the figure and the displacement is tangential to the circular path.
Therefore the angle between the force and displacement is 90° and W = 0 option (d) is correct.
A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface.
The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figure are only illustrative and not to the scale.
First the kinetic energy will increase as per eq (1). As the balls touches the ground it starts deforming and loses its K.E. (K.E. converting into elastic potential energy). When the deformation is maximum, K.E. = 0.
The ball then again regain its shape when its elastic potential energy changes into K.E. As the ball moves up it loses K.E. and gain gravitational potential energy.
These characteristics are according to graph (b).
A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to
By integrating, we get
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 work required to pull the hanging part on to the table is
The hanging part of the chain which is to be pulled up can be considered as a point mass situated at the centre of the hanging part. The equivalent diagram is drawn.
Note : The work done in bringing the mass up will be equal to the change in potential energy of the mass.
W = Change in potential energy
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle.
The motion of the particle takes place in a plane. It follows that :
When the force is perpendicular to the velocity and constant in magnitude then the force acts as a centripetal force, and the body moves in a circular path. The force is constant in magnitude, this show the speed is not changing and hence kinetic energy will remain constant.
Note : The velocity changes continuously due to change in the direction. The acceleration also changes continuously due to change in direction.
A force F = - K 
(where K is a positive constant) acts on a particle moving in the xy plane. Starting from the origin, the particle is taken along the positive x axis to the point (a, 0), and then parallel to the y axis to the point (a, a), The total work done by the force F on the particle is
The expression of work done by the variable force F on the particle is given by
In going from (0, 0) to (a, 0), the coordinate of x varies from 0 to 'a', while that of y remains zero. Hence, the work done along this path is :
In going from (a, 0) to (a, a) the coordinate of x remains constant (= a) while that of y changes from 0 to 'a'.
Hence, the work done along this path is
Hence, W = W1 + W2 = – ka2
A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is
Applying the principle of conservation of energy (K.E.)B + (P.E.)B = (K.E.)A + (P.E.)A,
we get
Change in velocity =
A small ball starts moving from A over a fixed track as shown in the figure. Surface AB has friction. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies of the ball at A, B and C, respectively.
Then
At point A, potential energy of the ball = mghA
At point B, potential energy of the ball = 0
At point C, potential energy of the ball = mghC
Total energy at point A, EA = KA + mghA
Total energy at point B, EB = KB
Total energy at point C, EC = KC + mghC
According to the law of conservation of energy.
EA = EB = EC ... (i)
EA = EB ⇒ EC > KC ...(ii)
EA = EC
KA + mghA = KB + mghC
Option (b) is correct
From (i),(ii) and (iv), we get hA> hC; KB > KC.Option (a) is correct.
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