Work,Power and Energy Quiz 20

Work,Power and Energy Quiz-20

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. A bomb of mass 9 kg explodes into two parts. One part of mass 3 kg moves with velocity 16 m/s ,then the KE of the other part is?

•  162 J
•  150 J
•  192 J .
•  200 J

Solution
From the law of conservation of momentum 3×16+6×v=9×0 Or v=-8 ms^-1 ⇒ v=8ms^-1 (numerically) Therefore, its kinetic energy k=1/2×6×8²=192J

Q2. A body of mass ‘M’ collides against a wall with a velocity v and retraces its path with the same speed. The change in momentum is (take initial direction of velocity as positive)?

•  Zero
•  2Mv
•  Mv
•  -2 Mv

Solution
Change in momentum =mv ₂-mv₁=-mv-mv=-2mv

Q3. The relationship between the force F and position x of a body is as shown in figure. The work done in displacing the body from x=1mto x=5m will be?

•  30 J
•  15 J
•  25 J
•  20 J

Solution
Work done=area enclosed by F-xgraph =area of ABNM + area of CDEN - area of EFGH + area of HIJ 

=1×10+1×5-1×5+ 1/2×1×10 =10+5-5+5=15 J

Q4. A block C of mass mis moving with velocity v₀ and collides elastically with block A of mass m and connected to another block B of mass 2m through spring constant k.What is k if x₀ is compression of spring when velocity of A and B is same ?

•  (mv₀²)/(X₀² )
•  (mv₀²)/(2x₀² )
•  3/2 (mv₀²)/(x₀² )
•  2/3 (mv₀²)/(x₀² )

Solution
Using conservation of linear momentum, we have mv₀=mv+2mv Or v=v₀/3 Using conservation of energy, we have 1/2 mv₀²=1/2 kx₀²+1/2 (3m) v² Where x₀=compression in the spring , ∴mv₀²=kx₀²+(3m) (v₀²)/9 Or kx₀²=mv₀²-(mv₀²)/3 Or kx₀²=(2mv₀²)/3 ∴k=(2mv₀²)/(3x₀² )

Q5. A canon ball is fired with a velocity 200 m/sec at an angle of 60° with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity 100 m/sec, the second one falling vertically downwards with a velocity 100 m/sec. The third fragment will be moving with a velocity ?

•  100 m/s in the horizontal direction
•  300 m/s in the horizontal direction
•  300 m/sin a direction making an angle of 60° with the horizontal
•  200 m/s in a direction making an angle of 60° with the horizontal

Solution
 

Momentum of ball (mass m) before explosion at the highest point =mvi ̂=mu cos⁡60°i ̂ =m×200×1/2 i ̂=100 mi ̂ kgms^-1 

Let the velocity of third part after explosion is V After explosion momentum of system =P ₁+P ₂+P ₃ =m/3×100j ̂-m/3×100j ̂+m/3×Vi ̂ By comparing momentum of system before and after the explosion m/3×100j ̂-m/3×100j ̂+m/3 Vi ̂=100mi⇒V=300m/s

Q6. 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 ?

•  w₁>w₂>w₃
•  w₁=w₂=w₃
•  6ms^-1 w₁ Less than w₂ Less than w₃
•  w₂>w₁>w₃

Solution
Gravitational field is a conservative force field. In a conservative force field work done is path independent. ∴ W₁=W₂=W₃

Q7. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is?

•  250 m/s
•  250√2 m/s
•  400 m/s
•  500 m/s

Solution
R=u√(2h/g)⇒20=V₁ √((2×5)/10) and 100=V₂ √((2×5)/10) ⇒V₁=20 m/s,V₂=100 m/s Applying momentum conservation just before and just after the collision (0.01)(V)=(0.2)(20)+(0.01)(100) V=500 m/s

Q8. A gun of mass 20 kg has bullet of mass 0.1 kg in it. The gun is free to recoil 804 J of recoil energy are released on firing the gun. The speed of bullet (ms^-1) is?

•  √(804×2010)
•  √(2010/804)
•  √(804/2010)
•  √(804×4×10³ )

Solution
Here, m₁=20 kg, m₂=0.1 kg, v₁= velocity of recoil of gun, v₂ = velocity of bullet As m₁ v₁=m₂ v₂ v₁=m₂/m₁ v₂=0.1/20 v₂=v₂/200 Recoil energy of gun=1/2 m₁ v₁² =1/2×20(v₂/200)² 804= (10v₂²)/(4×10⁴ )=(v₂²)/(4×10³ ) v₂=√(804×4×10³ ) ms^-1

Q9. A particle is placed at the origin and a force F=kx is acting on it (where k is positive constant). If U(0)=0, the graph of U(x) versus x will be (where U is the potential energy function)?

•  
  
  
•  
  
  
•  
  
  
•  
  
  

Solution

This is the equation of parabola symmetric to U axis in negative direction

Q10. A wire of length L suspended vertically from a rigid support is made to suffer extension l in its length by applying a force F. The work is?

•  Fl/2
•  Fl
•  2Fl
•  Fl

Solution