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Work Power and Energy Quiz-8

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 space craft of mass M and moving with velocity v suddenly breaks in two pieces of same mass m. After the explosion one of the mass m becomes stationary. What is the velocity of the other part of craft?
  •  Mv/(M-m)
  •  v
  •  Mv/m
  •  (M-m)/m v
Solution
(C) (c) Let velocity of masses after explosion be v1 and v2, then from law of conservation of momentum, we have Momentum before explosion = Momentum after explosion MV=m1 v1+m2 v2 Given m1=m2=m,v2=0 , ∴ Mv=mv1+m×0 ⇒ v1=Mv/m.

Q2.A force F=-K(yi+xj) (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a,0) and then parallel to the y-axis to the point (a,a). The total work done by the force F on the particles is
  •  -2 Ka2
  •  2 Ka2
  •  -Ka2
  •  Ka2
Solution
 (c)
 While moving from (0,0) to (a,0) Along positive x-axis, y=0 ∴F ⃗=-kxj ̂ i.e., force is in negative y- direction while displacement is in positive x-direction ∴W_1=0 Because force is perpendicular to displacement Then particle moves from (a,0) to (a.a) along a line parallel to y-axis (x=+a) during this F ⃗=-k(yi ̂+aj ̂) The first component of force, -kyi ̂ will not contribute any work because this component is along negative x-direction (-i ̂) while displacement is in positive y-direction (a,0)to (a,a). The second component of force i.e.-kaj ̂ will perform negative work ∴W_2=(-kaj ̂ )(aj ̂ )=(-ka)(a)=-ka2 So net work done on the particle W=W1+W2 =0+(-ka2 )=-ka2

Q3.  A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It slides down a smooth surface to the ground, then climbs up another hill of height 30 m and finally slides down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is
  •  10 m/s
  •  10√30 m/s
  •  40 m/s
  •  20 m/s
Solution


Q4. Adjacent figure shows the force-displacement graph of a moving body, the work done in displacing body from x=0 to x=35 m is equal to


  •  50 J
  •  25 J
  •  287.5 J
  •  200J
Solution

Q5.A bomb is kept stationary at a point. It suddenly explodes into two fragments of masses 1g and 3g. The total KE of the fragments is 6.4×104J. What is the KE of the smaller fragment?
  •  2.5×104 J
  •  3.5×104 J
  •  4.8×104 J
  •  5.2×104 J
Solution
  (c) Let m_1,m_2be the masses of first and second fragments respectively and v1,v2be their velocities after explosion. From conservation of momentum Mv=m1,m2+m2 v2 Where, M is mass of bomb before explosion and vits velocity. Since, bomb is stationary, hence v=0 Given ,m1=1g=1×〖10〗^(-3) kg=0.001kg m2=3g=3×〖10〗^(-3) kg=0.003kg and Ek=6.4×〖10〗^4 J ∴0=m1 v1+m2 v2 or 0=0.001v1+0.003v2 Or v2=-v1/3 .............(i) Total kinetic energy is EK=1/2 m1 v1^2+1/2 m2 v2^2 Ek=1/2×(0.001) v12+1/2×(0.003) v22…(ii) ∴Ek=1/2×(0.001)v12+1/2 (0.003)×(-(v12)/3)2 Ek=1/2×(0.001)(v12+3×(v12)/9) Ek=1/2×(0.001)×(4v11^2)/3=((0.002) v12)/3 …(iii) ∴ 6.4×104=((0.002) v12)/3 Or v12=(3×6.4×〖104)/0.002 Or v12=(3×6.4×104)/0.002 orv12=96×〖106=9.6×〖107 ms-1 Hence, kinetic energy of smaller fragment is E1K'=1/2 m11 v11^2 E1k'=1/2×(0.001)×9.6×〖10〗^7 E1k' =4.8×〖10〗^4 J.

Q6. What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of 10 cm


  •  0.6 m/s
  •  1.4 m/s
  • 1.8 m/s
  •  2.2 m/s
Solution
(b) v=√2gh=√(2×9.8×0.1)=√1.96=1.4 m/s

Q7.Which of the following statements are incorrect? (i)If there were no friction, Work need to be done to move a body up an inclined plane is zero. (ii)If there were no friction, moving vehicles could not be stopped even by locking the brakes. (iii)As the angle of inclination is increased, the normal reaction on the body placed on it increases. (iv)A duster weighing 0.5 kg is pressed against a vertical board with a force of 11 N. If the coefficient of friction is 0.5, the work done in rubbing it upward through a distance of 10 cm is 0.55J.
  •  (i)and(ii)
  •  (i),(ii),(iv)
  •  (i),(iii),and(iv)
  •  All of these
Solution
(c) The explanation are given below (i)If a body is moved up in inclined plane, then the work done against friction force is zero as there is no friction. But a work has to be done against the gravity.


 So, this statement is incorrect. (ii)If there were no friction, moving vehicles could not be stopped by locking the brakes. Vehicles are stopped by air friction only. So , This Statement is correct. (iii) In this situation the normal reaction is given R=mg cos⁡α …(i) Ifα increase then the value of cos α also decreases. So, this Statement is incorrect. (iv)When the duster is rubbing upward then an external force is applied and its value is 


 F’=0.5g+μR ∴ F^'=0.5g+0.5×11 Or F'=(0.5×10+5.5)N (Here R=11 N) Or F'=10.5N Hence, work done in rubbing the duster through a distance of 10 cm. W=F'×d ⇒ W=10.5×10/100 J Or F'=10.5J

Q8.238U nucleus decays by emitting an alpha particle of speed v ms-1. The recoil speed of the residual nucleus is (in ms^(-1))
  •  -4v/234
  •  V/4
  •  -4v/238
  •  4v/238
Solution
(a) Initially (238U)nucleus was at rest and after decay its part moves in opposite direction According to conservation of momentum 4v+234V=238×0⇒V=-4v/234



Q9. A body of mass 5 kg is placed at the origin, and can move only on the x-axis. A force of 10 N is acting on it in a direction making an angle of 60° with the x-axis and displaces it along the x-axis by 4 metres. The work done by the force is
  •  2.5 J
  •  7.25 J
  •  40 J
  •  20 J
Solution
(d) W=FS cos⁡θ=10×4×cos⁡60°=20 Joule

Q10. 80. The potential energy of a 1 kg particle free to move along the x-axis is given by V(x)=(x4/4-x2/4)J The total mechanical energy of the particle is 2J. Then, the maximum speed (in m/s) is
  •  √2
  •  1/√2
  •  2
  • 3/√2
Solution



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