Work,Power and Energy Quiz-11
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.
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Q1. A ball moving with velocity 2 m/s. collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be?
Solution
By conservation of linear momentum 2m=mv1+2mv2⇒v1+2v2=2 By definition of e,e=1/2=(v2-v1)/(2-0) ⇒v2-v1=1⇒v1=0 and v2=1ms-1
By conservation of linear momentum 2m=mv1+2mv2⇒v1+2v2=2 By definition of e,e=1/2=(v2-v1)/(2-0) ⇒v2-v1=1⇒v1=0 and v2=1ms-1
Q2. A block of mass 5kg is resting on a smooth surface. At what angle a force of 20N be acted on the body so that it will acquired a kinetic energy of 40J after moving 4m?
Solution
According to work-energy theorem W= Change in kinetic energy FS cosθ=1/2 mv2-1/2 mu2 Substituting the given values, we get 20×4× cosθ=40-0[∵u=0] cosθ=40/80=1/2 θ=cos-1(1/2)=60°
According to work-energy theorem W= Change in kinetic energy FS cosθ=1/2 mv2-1/2 mu2 Substituting the given values, we get 20×4× cosθ=40-0[∵u=0] cosθ=40/80=1/2 θ=cos-1(1/2)=60°
Q3. A 50g bullet moving with a velocity of 10 ms-1 gets embeded into a 950g stationary body. The loss in KE of the system will be ?
Solution
Applying principle of conservation of linear momentum, velocity of the system (v) is m1 v1=(m1+m2 )V,⇒V=(m1 v1)/(m1+m2 )=(50×10)/((50+950))=1/2 ms-1 Initial KE, E1=1/2 m1 v12=1/2×(50/1000)×102=2.5 J Final KE, E2=1/2 (m1+m2 ) v2 =1/2 ((50+950))/1000×1/2=0125 J Percentage loss is KE (E1-E2)/E1 ×100=(2.5-0.125)/2.5=95%
Applying principle of conservation of linear momentum, velocity of the system (v) is m1 v1=(m1+m2 )V,⇒V=(m1 v1)/(m1+m2 )=(50×10)/((50+950))=1/2 ms-1 Initial KE, E1=1/2 m1 v12=1/2×(50/1000)×102=2.5 J Final KE, E2=1/2 (m1+m2 ) v2 =1/2 ((50+950))/1000×1/2=0125 J Percentage loss is KE (E1-E2)/E1 ×100=(2.5-0.125)/2.5=95%
Q4. A long spring, when stretched by x cm has a potential energy U. On increasing the length of spring by stretching to nx cm, the potential energy stored in the spring will be?
Solution
Initially potential energy =1/2 kx2 ⇒ U=1/2 kx2 or 2U=kx2⇒k=2U/x2 When it is stretched to nx cm, then PE = 1/2 kx12=1/2×2U/x2 ×n2 x2=n2 U ∴ Potential energy stored in the spring =n2 U
Initially potential energy =1/2 kx2 ⇒ U=1/2 kx2 or 2U=kx2⇒k=2U/x2 When it is stretched to nx cm, then PE = 1/2 kx12=1/2×2U/x2 ×n2 x2=n2 U ∴ Potential energy stored in the spring =n2 U
Q5. When a force is applied on a moving body, its motion is retarded. Then the work done is ?
Solution
The angle between the displacement and the applied retarded force is 1800 ∴Work done=Fs cos180°-Fs =-Ve
The angle between the displacement and the applied retarded force is 1800 ∴Work done=Fs cos180°-Fs =-Ve
Q6. If a body of mass 3 kg is dropped from the top of a tower of height 25 m, then its kinetic energy after 3 s will be?
Solution
1296 J
1296 J
Q7. A spring of force constant 800 N/m has an extension of 5cm. The work done in extending it from 5 cm to 15 cm is?
Solution
W= 1/2 k(x22-x12 )=1/2×800×(152-52 )×10-4=8J
W= 1/2 k(x22-x12 )=1/2×800×(152-52 )×10-4=8J
Q8. A body moving with a velocity v,breaks up into two equal parts. One of the part retraces back with velocity v.Then, the velocity of the other part is?
Solution
Let Mbe the mass of body moving with velocity v and m be mass of each broken part, velocity of one part which retraces back is v and that of second part is v’. Momentum before breaking=momentum after breaking Mv=m(-v)+mv' Or v'=(Mv+mv)/m Since, M=2m,therefore v'=((2m+m)v)/m=3v .
Let Mbe the mass of body moving with velocity v and m be mass of each broken part, velocity of one part which retraces back is v and that of second part is v’. Momentum before breaking=momentum after breaking Mv=m(-v)+mv' Or v'=(Mv+mv)/m Since, M=2m,therefore v'=((2m+m)v)/m=3v .
Q9. Four particles given, have same momentum. Which has maximum kinetic energy?
Solution
E= P2/2m ∴E∝1/m [If P= constant] i.e., the lightest particle will possess maximum kinetic energy and in the given option mass of electron is minimum
E= P2/2m ∴E∝1/m [If P= constant] i.e., the lightest particle will possess maximum kinetic energy and in the given option mass of electron is minimum
Q10. If reaction is R and coefficient of friction is μ, what is work done against friction in moving a body by distance d?
Solution
As shown a block of mass M is lying over rough horizontal surface. Let μ be the coeeficient of kinetic friction between the two surfaces in contact. The force Of friction between the block and horizontal surface is given by F=μR=μMg (∵R=Mg) To move the block without acceleration, the force (P)required will be just equal to the force of friction , ie , P=F=μR If d is the distance moved , then work done is given by W=P×d=μRd
As shown a block of mass M is lying over rough horizontal surface. Let μ be the coeeficient of kinetic friction between the two surfaces in contact. The force Of friction between the block and horizontal surface is given by F=μR=μMg (∵R=Mg) To move the block without acceleration, the force (P)required will be just equal to the force of friction , ie , P=F=μR If d is the distance moved , then work done is given by W=P×d=μRd
