Work,Power and Energy Quiz 19

Work,Power and Energy Quiz-19

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 body moves from a position r₁= (2i ̂-3j ̂-4k ̂) m to a position, r2= (3i ̂-43j ̂-5k ̂) m under the influence of a constant force F= (4i ̂-4j ̂+5k ̂) N. The work done by the force is?

•  57 J
•  58 J
•  59 J .
•  60 J

Solution
Given,r₁=2i ̂-3j ̂-4k ̂ And r₂=3i ̂-4j ̂+5k ̂ Now, r₂-r₁=i ̂-j ̂+9k ̂ And F=4i ̂+j ̂+6k ̂ ∴ work done = F.r W=(4i ̂+j ̂+6k ̂ ).(i ̂-j ̂+9k ̂) =4-1+54=57 J

Q2. The relationship between force and position is shown in the figure given (in one dimensional case). The work done by the force in displacing a body from x=1 cm to x=5 cm is?

•  20 ergs
•  60 ergs
•  70 ergs
•  700 ergs

Solution
Work done = area between the graph and position axis W=10×1+20×1-20×1+10×1=20 erg

Q3. Choose the incorrect statement?

•  No work is done if the displacement is perpendicular to the direction of the applied force
•  If the angle between the force and displacement vectors is obtuse, then the work done is negative
•  Frictional force in non-conservative
•  All the central forces are non-conservative

Solution
All the central forces are conservative

Q4. An elastic string of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is?

•  1/2 ky²
•  1/2 k(x²+y²)
•  1/2 k(x+y)²
•  1/2 ky(2x+y)

Solution
Elastic force in string is conservative in nature W=-∆V₁ where W= work done by elastic force of string W=-(V_f-V_i )=V_i-V_f or W=1/2 kx²-1/2 k(x+y)² or W=1/2 kx²-1/2 k(x²+y²+2xy) =1/2 kx²-1/2 kx²-1/2 ky²-1/2 k(2xy)=-kxy-1/2 ky² =1/2 ky(-2x-y) The work done against elastic force is W_ext=-W=ky/2(2x+y)

Q5. A ball is dropped from a height h on a floor of coefficient of restitution e.The total distance covered by the ball just before second hit is ?

•  h(1-2e²)
•  h(1+2e²)
•  h(1+e²)
•  he²

Solution
 

Total distance travelled by the ball before its second hit is H=h+2h₁ =h[1+2e² ] (∵h₁=he²)

Q6. A stationary bomb explodes into two parts of masses in the ratio of 1:3 .If the heavier mass moves with a velocity 4ms^-1 , what is the velocity of lighter part ?

•  12ms^-1 opposite to heavier mass
•  12ms^-1 in the direction of heavier mass
•  6ms^-1 opposite to heavier mass
•  6ms^-1 in the direction heavier mass

Solution
The ratio of masses=1:3 Therefore ,m₁=xkg,m₂=3x kg Applying law of conservation of momentum m₁ v₁+m₂ v₂=0 ⇒ x×v₁+3x×4=0 Or v₁=-12ms^-1 Therefore, velocity of lighter mass is opposite to that of heavier mass.

Q7. A spring of spring constant5×10³ N/mis stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5cm is?

•  6.25 N-m
•  12.50 N-m
•  18.75 N-m
•  25.00 N-m

Solution
W= 1/2 k(x₂²-x₁² )=1/2×5×10³ (10²-5² )×10^-4=18.75 J

Q8. Two bodies having same mass 40 kg are moving in opposite directions, one with a velocity of 10 m/s and the other with 7 m/s. If they collide and move as one body, the velocity of the combination is?

•  10 m/s
•  7 m/s
•  3 m/s
•  1.5 m/s

Solution
By the conservation of momentum 40×10+(40)×(-7)=80×v ⇒v=1.5 m/s

Q9. Two masses of 0.25 kg each moves towards each other with speed 3ms^-1 and 1ms^-1 collide and stick together. Find the final velocity?

•  0.5 ms^-1
•  2 ms^-1
•  1 ms^-1
•  0.25 ms^-1

Solution

Here, m=0.25 kg,u₁=3ms^-1,u₂=-1ms^-1 It is an inelastic collision According to conservation of momentum mu₁+mu₂=(m+m)v ⇒v=(mu₁+mu₂)/2m=(u₁+u₂)/2=(3-1)/2=1ms^-1

Q10. A bullet fired from a gun with a velocity of 10⁴ ms^-1 goes through a bag full of straw. If the bullet loses half of its kinetic energy in the bag, its velocity when it comes out of the bag will be?

•  7071.06 ms^-1
•  707 ms^-1
•  70.71 ms^-1
•  707.06 ms^-1

Solution
KE left, 1/2 mv²=1/2 (1/2 mu² ) ∴ velocity left, v=u/√2=10⁴/√2=7071.06 ms^-1