Work-Power-and-Energy-Quiz-3

Work Power and Energy Quiz-3

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 bodies of masses 1 kg and 5 kg are dropped gently from the top of a tower. At a point 20 cm from the ground, both the bodies will have the same

•  Momentum
•  Kinetic Energy
•  Velocity
•  Total Energy

Solution
(c) Velocity of fall is independent of the mass of the falling body

Q2.A 10 H.P. motor pumps out water from a well of depth 20m and fills a water tank of volume 22380 litres at a height of 10 m from the ground. The running time of the motor to fill the empty water tank is

•  5 minutes
•  10 minutes
•  15 minutes
•  20 minutes

Solution
(a) Volume of water to raise =22380 l=22380×〖10〗^-3 m³ P=mgh/t=(V ρgh)/t⇒t=(V ρgh)/P t=(22380×〖10〗^-3×〖10〗³×10×10)/(10×746)=5 min

Q3.  Power of water pump is 2 kW. If g=10 m/sec², the amount of water it can raise in one minute to a height of 10 m is

•  2000 litres
•  1000 litres
•  100 litres
•  1200 litres

Solution
(d) P=mgh/t⇒m=(p×t)/gh=(2×〖10〗³×60)/(10×10)=1200 kg As volume = mass/density⇒v=1200kg/(〖10〗³ kg/m³ )=1.2m^3 Volume =1.2m^3=1.2×〖10〗³ litre=1200litre

Q4. A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing, it rises to1.8 m. The ball loses its velocity on bouncing by a factor of

•  6/25
•  2/5
•  3/5
•  9/25

Solution
(b) If ball falls from height h_1and bounces back up to height h₂ then e=√(h₂/h₁ ) 

 Similarly if the velocity of ball before and after collision are v₁ and v₂ respectively then e=v₂/v₁1 So v₂/v₁1 =√(h₂1/h₁1 )=√(1.8/5)=√(9/25)=3/5 i.e. fractional loss in velocity =1-v₂/v₁1 =1-3/5=2/5

Q5.A coolie 1.5 m tall raises a load of 80 kg in 2 s from the ground to his head and then walks a distance of 40 m in another 2 s. The power developed by the coolie is

•  0.2kW
•  0.4kW
•  0.6kW
•  0.8kW

Solution
 (c) P=mgh/t=(80×10×1.5)/2 =600 W = 0.6 kW

Q6. A bomb at rest explodes into 3 parts of the same mass. The momentum of the 2 parts is -2pI ̂and pj ̂. The momentum of the third part will have a magnitude of

•  P
•  √3p
• p√5
•  Zero

Solution
(c) Momentum of the third part will be equal to the resultant of momentum of two parts. p₃=√(p₁²+p²²) p₃=√((-2p)²+p²) p₃=p√5

Q7.The potential energy of a certain spring when stretched through a distance s is 10 J. The amount of work (in joule) that must be done on this spring to stretch it through additional distance s will be

•  30
•  40
•  10
•  20

Solution
(a) U=1/2 ks²=10 J U^'=1/2 k(s+s)²=4(1/2 ks²)=40 J W=U^'-U=40-10=30 J

Q8.A particle of mass m moving with horizontal speed 6 m/sec as shown in figure. If m<< M than for one dimensional elastic collision, the speed of lighter particle after collision will be

•  2m/sec in original direction
•  2m/sec in opposite to original direction
•  4m/sec in opposite to original direction
•  4m/sec in original direction

Solution
(a) v₁=((m₁-m₂)/(m₁+m₂ )) u₁+(2m₂ u₂)/(m₁+m₂) Substituting m₁=0,V₁1= -u₁+2u₂ ⇒v₁1=-6+2(4)=2 m/s i.e. the lighter particle will move in original direction with the speed of 2 m/s

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 stuck particle move in directions making angles θ₁and θ₂respectively with the initial direction of motion. The sum of the angles θ₁+θ₂

•  45°
•  90°
•  135°
•  180°

Solution
(b)

Q10. A car weighing 1400 kg is moving at a speed of 54 kmh^-1) up a hill when the motor stops. If it is just able to reach the destination which is at a height of 10 m above the point, then the work done against friction (negative of the work done by the friction) is

•  10 kJ
•  15kJ
•  17.5 kJ
• 25kJ

Solution
(c) 1400×10×10+W=1/2×15×15 or W=700×15×15-1400×10×10 or W=700(225-200) J or W=700×25 J =75.5 kJ