Work-Power-and-Energy-Quiz-2

Work Power and Energy Quiz-2

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. 

Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. 

Q1. The potential energy of a particle in a force field is U=A/r² -B/r, where A and B are positive constants and r is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is

•  B/2A
•  2A/B
•  A/B
•  B/A

Solution
(b) For equilibrium dU/dr=0 ⇒(-2A)/r³+B/r² =0 r=2A/B For stable equilibrium (d²U)/(dr²) should be positive for the value of r Here (d²U)/(dr²)=6A/r⁴-2B/r³ is +ve value for r=2A/B

Q2.Consider the following statements. A and B and identify the correct answer given below. I. Body initially at rest is acted upon by a constant force. The rate of change of its kinetic energy varies linearly with time. II. When a body is at rest, it must be in equilibrium.

•  A and B are correct
•  A and B are wrong
•  A is Correct B is wrong
•  A is Wrong B is Correct

Solution
(c) KE=1/2 mv²=1/2 m(at)²=1/2 ma²t² Rate of change of KE, dk/dt=d/dt (1/2 ma² t²2 )=ma²t ∵ dk/dt∝t So, statement Ais correct. When the body is at rest then it may be or may not be in equilibrium, so statement Bis wrong.

Q3.  A body at rest breaks into two pieces with unequal mass

•  Both of them have equal speeds
•  Both of them move along a same line with unequal speeds
•  Sum of their momentum is non zero
•  They move along different lines with different speeds

Solution
(b)

Q4. A mass of 50 kg is raised through a certain height by a machine whose efficiency is 90%, the energy is 5000 J. If the mass is now released, its KE on hitting the ground shall be

•  5000
•  4500
•  4000
•  5500

Solution
(b) Because the efficiency of machine is 90%, hence, potential energy gained by the mass =90/100× energy spend =90/100×5000=4500 J When the mass is released now, gain in KE on hitting the ground = Loss of potential energy =4500 J

Q5.A body of mass 2 kg is projected at 20 m/s at an angle of 60°above the horizontal. Power on the block due to the gravitational force at its highest point is

•  200W
•  100√3 W
•  50W
•  None of these

Solution
 (d) P=v cos⁡θ=mg v cos⁡90°=0

Q6. An engine pumps water through a hose pipe. Water passes through the pipe and leaves it with a velocity of 2 m/s. The mass per unit length of water in the pipe is 100 kg/m. What is the power of the engine

•  800 W
•  400 W
• 200 W
•  100 W

Solution
(a) Power =Fv=v(m/t)v=v² (ρAv) =ρAv³=(100) (2)³=800 W

Q7.A particle free to move along the x-axis has potential energy given by U(x)=k[1-exp⁡(-x²)]for -∞≤x≤+∞, where k is a positive constant of appropriate dimensions. Then

•  At point away from the origin, the particle is in unstable equilibrium
•  For any finite non-zero value of x, there is a force directed away from the origin
•  If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin
•  For small displacements from x=0, the motion is simple harmonic

Solution
(d)

Q8.The relation between the displacement X of an object produced by the application of the variable force F is represented by a graph shown in the figure. If the object undergoes a displacement from X=0.5 m to X=2.5 m the work done will be approximately equal to

•  16J
•  32J
•  1.6J
•  8J

Solution
(a) Work done = Area under curve and displacement axis = Area of trapezium =1/2×(sum of two parallel lines)×distance between them =1/2 (10+4)×(2.5-0.5)=1/2 14×2=14 J As the area actually is not trapezium so work done will be more than 14 Ji.e. approximately 16 J

Q9.Two rectangular blocks A and B of masses 2kg and 3 kg respectively are connected by spring of spring constant 10.8 Nm^-1and are placed on a frictionless horizontal surface. The block A was given an initial velocity of 0.15 ms^-1in the direction shown in the figure. The maximum compression of the spring during the motion is

•  0.01m
•  0.02m
•  0.05m
•  0.03m

Solution
(c) As the block A moves with velocity with velocity 0.15 ms^-1, it compresses the spring Which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, i.e. 0.15 ms^-1. Let this velocity be v. Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage. According to the law of conservation of linear momentum, 

we get m_Au=(m_A+m_B)v Or v=(m_A u)/(m_A+m_B ) (2×0.15)/(2+3)=0.06ms^-1 According to the law of conservation of energy 1/2 m_A u²=1/2 (m_A+m_B) V²+1/2 kx² 1/2 m_A u²-1/2 (m_A+m_B) v²=1/2 kx² 1/2×2×(0.15)²-1/2 (2+3) (0.06)^2=1/2 kx² 0.0225-0.009= 1/2 kx² or 0.0135= 1/2 kx² Or x=√(0.0027/k)=√(0.0027/10.8) =0.05m

Q10. A force of 5N, making an angle θ with the horizontal, acting on an object displaces it by 0.4m along the horizontal direction. If the object gains kinetic energy of 1J, the horizontal component of the force is

•  1.5N
•  2.5N
•  3.5N
• 4.5N

Solution
(b) Work done on the body = K.E. gained by the body Fs cos⁡θ=1⇒F cos⁡θ=1/s=1/0.4=2.5N