## Work Power and Energy QUIZ-4

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. A bomb is kept stationary at a point. It suddenly explodes into two fragments of masses 1gand 3g. The total K.E. of the fragments is 6.4×104 J. What is the K.E. of the smaller fragment

•  2.5x104J
•  3.5x104J
•  4.8x104J
•  5.2x104J
Solution
(c)

As the momentum of both fragments are equal therefore E1/E22 =m2/m11 =3/1 i.e., E11=3E2 …(i) According to problem E11+E2=6.4×104 J …(ii) By solving equation (i) and (ii), we get E11=4.8×104 J and E22=1.6 ×104 J

Q2.If the heart pushes 1 cc of blood in 1 s under pressure 20000Nm-2 ,the power of heart is
•  0.02 W
•  400 W
•  5×10-10 W
•  0.2 W
Solution
(a) Given, pressure=20000Nm-2 Volume=1cc=10-6 m3 ∵Power=pressure×volume per second ∴Power=20000×10-6 p=0.02 w

Q3.  The energy required to accelerate a car from 10 m/s to 20 m/s is how many times the energy required to accelerate the car from rest to 10 m/s
•  Equal
•  4 times
•  2 times
•  3 times
Solution
(d) Kinetic energy for first condition =1/2 m(v22-v12 )=1/2 m(2022-102 )=150 mJ K.E. for second condition =1/2 m(〖10〗2-02 )=50mJ ∴(K.E.)I/(K.E.)II=150m/50m=3

Q4. A uniform chain of length L and mass Mis 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
•  MgL
•  MgL/3
•  MgL/9
•  MgL/18
Solution
(d)

W= MgL/(2n2)=MgL/(2(3)2 )=MgL/18 [n=3 Given ]

Q5.When a 1.0 kg mass hangs attached to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition,
•  1.5 joule
•  2.0 joule
•  2.5 joule
•  3 joule
Solution
(b) Force constant of a spring k=F/x=mg/x=(1×10)/(2×10-2) )⇒k=500 N/m Increment in the length =60-50=10 cm U=1/2 kx2=1/2 500(10×10-2)2=2.5 J

Q6. A mass m is attached to the end of a rod of length l. The mass goes around a vertical circular path with the other end hinged at the centre. What should be the minimum velocity of mass at the bottom of the circle, so that the mass complete the circle?
•  √4gl
•  √3gl
• √5gl
•  √gl
Solution
(c) When a particle is moved in a circle under the action of a torque then such motion is non-uniform circular motion. Applying principle of conservation of energy, total mechanical energy at L =total mechanical energy at H

∴ 1/2 mvLL2=1/2 mvH2+MG(2l) But vHH2=gl ∴1/2 mvLL2=1/2 m(gl)+2mgl Or vL^2=5gl Or vL=√5gl Hence for looping the vertical loop, the minimum velocity at the lowest point L IS √(5gl.)

Q7.A bullet of mass a and velocity b is fired into a large block of mass c. The final velocity of the system is
•  (c/(a+b)).b
•  (a/(a+c)).b
•  ((a+b)/c).a
•  ((a+c)/a).b
Solution
(a)

Initially bullet moves with velocity b and after collision bullet get embedded in block and both move together with common velocity By the conservation of momentum ⇒a×b+0=(a+c)V⇒V=ab/(a+c)

Q8.A bomb of mass M at rest explodes into two fragments of masses m1and m2. The total energy released in the explosion is E. If E1 and E2 represent the energies carried by masses m11 and m22 respectively, then which of the following is correct?
•  (E11=m2/M )E
•  (E1=m1/m2 )E
•  (E1=m1/M )E
•  (E1=m2/m1 )E
Solution
(A)

Q9. Consider elastic collision of a particle of mass m moving with a velocity u with another particle of the same mass at rest. After the collision the projectile and the struck particle move in directions making angles Î¸11 and Î¸22 respectively with the initial direction of motion. The sum of the angles Î¸11+Î¸2, is
•  45°
•  90°
•  135°
•  180°
Solution
(b)

Q10. The force F acting on a particle moving in a straight line is shown in figure. What is the work done by the force on the particle in the 1st meter of the trajectory

•  5J
•  10J
•  15 J
• 2.5J
Solution
(d) Work done (W)= Area under curve of F-x graph = Area of triangle OAB=1/2×5×1=2.5 J

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