## Work Power and Energy Quiz-7

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 billiards player hits a stationary ball by an identical ball to pocket the target ball in a corner pocket that is at an angle of 35° with respect to the direction of motion of the first ball. Assuming the collision as elastic and that friction and rotational motion are not important, the angle made by the target ball with respect to the incoming ball is
•  35°
•  50°
•  55°
•  60°
Solution
(C)

Q2.A machine which is 75% efficient uses 12 J of energy in lifting up a 1 kg mass through a certain distance. The mass is then allowed to fall through, that distance. The velocity of the ball at the end of its fall is
•  √24ms-1
•  √32 ms-1
•  √18 ms-1
•  3ms-1
Solution
(c) Useful work =75/100×12 J=9J Now, 1/2×1×v2=9 or v=√18 ms-1

Q3.  A car manufacturer claims that his car can be accelerated from rest to a velocity of 10 ms-1 in 5 s. If the total mass of the car and its occupants is 1000 kg, then the average horse power developed by the engine is
•  103/746
•  104/746
•  105/746
•  8
Solution
(b) a=(10-0)/5 ms-2=2ms-2; F=ma or F=1000×2 N = 2000 N Average velocity =(0+10)/2 ms-1=5ms-1 Average power =2000×5 W=104 W Required horse power is 104/746

Q4. A body of mass M moves with velocity v and collides elastically with a another body of mass m(M>>m) at rest then the velocity of body of mass m is
•  V
•  2V
•  V/2
•  Zero
Solution
(b) When target is very light and at rest then after head on elastic collision it moves with double speed of projectile i.e. the velocity of body of mass m will be 2v

Q5.A particle, initially at rest on a frictionless horizontal surface, is acted upon by a horizontal force which is constant in size and direction. A graph is plotted between the work done (W) on the particle, against the speed of the particle, (v). If there are no other horizontal forces acting on the particle the graph would look like
Solution
(d) Work done = change in kinetic energy W=1/2 mv2 ∴W∝v2 graph will be parabolic in nature

Q6. A body of mass 50 kg is projected vertically upwards with velocity of 100 m/sec. 5 seconds after this body breaks into 20 kgand 30 kg. If 20 kg piece travels upwards with 150 m/sec, then the velocity of the block will be
•  15 m/sec downwards
•  15 m/sec upward
• 51 m/sec downwards
•  51 m/sec upward
Solution
(a) Velocity of 50 kg mass after 5 sec of projection v=u-gt=100-9.8×5=51 m/s At this instant momentum of body is an upward direction P_initial=50×51=2550 kg-m/s After breaking 20 kg piece travels upwards with 150 m/s let the speed of 30 kg mass is V P_final=20×150+30×V By the law of conservation of momentum Pinitial=Pfinal ⇒2550=20×150+30×V⇒V=-15 m/s i.e. it moves in downward direction

Q7.A man, by working a hand pump fixed to a well, pumps out 10 m3water in 1 s. If the water in the well is 10 m below the ground level, then the work done by the man is
•  103J
•  104J
•  105J
•  106J
Solution
(d) Mass to be lifted = 10 ×102 kg [∴ density of water = 103kgm-3] Height, h=10 m Work done = 104×10×10=106 J

Q8.A body of mass m_1 is moving with a velocity V. It collides with another stationary body of mass m_2. They get embedded. At the point of collision, the velocity of the system
•  increases
•  Decreases but not become zero
•  Remain same
•  zero
Solution
(b) By momentum conservation before and after collision m1 V+m2×0=(m1+m2)v⇒v=m1/(m1+m2) V i.e. Velocity of system is less than V

Q9. An intense stream of water of cross-sectional area A strikes a wall at an angle θ with the normal to the wall and returns back elastically. If the density of water is ρ and its velocity is v,then the force exerted in the wall will be
•  2Avρ cos⁡θ
•  2Av2ρcos⁡θ
•  2Av2ρ
•  2Avρ
Solution
(b) Linear momentum of water striking per second to the wall P1=mv=Avρ v=Av2 ρ, similarly linear momentum of reflected water per second Pr=Av2ρ
Now making components of momentum along x- axes and y-axes. Change in momentum of water per second =Picos⁡θ+Prcos⁡θ =2Av2ρcos⁡θ By definition of force, force exerted on the Wall =2Av2ρcos⁡θ

Q10. Two solid rubber balls A and B having masses 200 and 400 g respectively are moving in opposite directions with velocity of A equal to 0.3 m/s. After collision the two balls come to rest, then the velocity of B is
•  0.15 m/sec
•  1.5 m/sec
•  -0.15 m/sec
• None
Solution
(C) Initial linear momentum of system =mAv->A+mBv->B =0.2×0.3+0.4×vB
Finally both balls come to rest ∴ final linear momentum =0 By the law of conservation of linear momentum 0.2×0.3+0.4×vB=0 ∴vBB=-(0.2×0.3)/0.4=-0.15 m/s #### Written by: AUTHORNAME

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