Work-Power-And-Energy-Quiz-5

Work Power and Energy Quiz-5

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. Two springs have force constants k₁and k₂. There are extended through the same distance x. If their elastic energies are E₁ and E₂, then E₁/E₂ is equal to

•  k₁:k₂
•  k₂:k₁
•  √(k₁):√(k₂)
•  k₁²:k²²

Solution
(a) E=1/2 kx² ∴ E∝k ∴ E₁/E₂ =k₁/k₂

Q2.A position dependent force F=7-2x+3x² newton acts on a small body of mass 2 kg and displaces it from x=0to x=5m. The work done in joules is

•  70
•  270
•  35
•  135

Solution
    (d)

Q3.  An automobile weighing 1200 kg climbs up a hill that rises 1m in 20s. Neglecting frictional effects. The minimum power developed by the engine is 9000 W. If g=10ms^-2, then the velocity of the automobile is

•  36 km h^-1
•  54 km h^-1
•  72 km h^-1
•  90 km h^-1

Solution
(b) Minimum force mg sin⁡θ, so, minimum power is given by P=mg sin⁡θ v or v=P/(mg sin⁡θ ) or v=(9000×2)/(1200×10×1) ms^-1=15ms^-1 =15×18/5=54 kmh^-1

Q4. A mass of M kg suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45⁰with the initial vertical direction is

•  Mg(√2 +1)
•  Mg√2
•  Mg/√2
•  Mg(√2-1)

Solution
(d) Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using work energy theorem. 

Let us assume that body taken slowly so that its speed doesn’t change, then ∆K=0 =W_F+W_Mg+W_tension (symbols have their usual meanings ) W_F=F×l sin⁡45°, W_Mg=Mg(l-l cos⁡〖45°),W_tension=0 ∴ F=Mg(√2-1)

Q5.A body of mass 10 kg is moving on a horizontal surface by applying a force of 10 N in forward direction. If body moves with constant velocity, the work done by force of fiction for a displacement of 2m is

•  -20J
•  10J
•  20J
•  5J

Solution
 (a) Since body moves with constant velocity, so. Net force on the body is zero. Here, N=mg,F=f ∴W=F ⃗.s ⃗=fs cos⁡180'' =fs=-10×2=-20 J

Q6. The pointer reading v/s load graph for a spring balance is as given in the figure. The spring constant is

•  0.1 kg/cm
•  5 kg/cm
• 0.3 kg/cm
•  1 kg/cm

Solution
(a) Spring constant k=F/x= Slope of curve ∴k=(4-1)/30=3/30=0.1 kg/cm

Q7.A ball dropped from a height of 2m rebounds to a height of 1.5 m after hitting the ground. Then the percentage of energy lost is

•  25
•  30
•  50
•  100

Solution
(a) U₁=mgh₁ and U₂=mgh₂ % energy lost =(U₁-U₂)/U₁ ×100 =(mgh₁-mgh₂)/(mgh₁ ) 100=((h₁-h₂)/h₁)×100 =(2-1.5)/2×100=25%

Q8.

•  2
•  4
•  6
•  8

Solution
(D) 

Q9. Which of the following statements is wrong?

•  KE of a body is independent of the direction of motion
•  In an elastic collision of two bodies ,the momentum and energy of each body is conserved
•  If two protons are brought towards each other the PE of the system decreases.
•  A body cannot have energy without momentum

Solution
(c) If momentum is Zero ie, if p=0,then kinetic energy K=p²/2m=0 But potential energy cannot be zero, thus a body can have energy without momentum.

Q10. Which of the following graphs show variation of potential energy (U) with position x.

•  
  
  
  
•  
  
  
  
•  
  
  
  
• 
  
  

Solution
(c) The variation of potential energy(U) With distance(x)is U=1/2 kx₂ Hence, parabolic graph is obtained.