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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 particle of mass m collides with another stationary particle of mass M. If the particle m stops just after collision, the coefficient of restitution of collision is equal to?
•  1
•  m/M
•  (M-m)/(M+m)
•  m/(M+m)
Solution
As net horizontal force acting on the system is zero, hence momentum must remain conserved. Hence mu+0=0+mv2⇒v2=mu/M As per definition, e=-(v1-v2 ) /((u2-u1))=(v2-0)/(o-u)=v2/u=(mu/M)/u=m/M

Q2. A string is wound round the rim of a mounted fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord. Neglecting friction and mass of the string, the angular acceleration of the wheel is?
•  50 s-2
•  25 s-2
•  12.5 s-2
•  6.25 s-2
Solution
Here, M=20kg,R=20cm=1/5 m Moment of inertia of flywheel about its axis is
I= 1/2 MR2=1/2×20kg×(1/5 m)2 =0.4kgm2 As τ=Iα Where α is the angular acceleration ∴α=τ/I=FR/I=(25×1/5)/0.4=5Nm/(0.4kgm2 )=12.5s-2

Q3. A man weighing 80 kg is standing in a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a speed of 1 m/s, then after 4 sec his displacement relative to the ground will be?
•  5 m
•  4.8 m
•  3.2 m
•  3.0 m
Solution
Displacement of the man with respect to trolley in 4 sec xmT=4m⇒xmT=xm+xT⇒xT=4-xm Position of centre of mass remain constant ⇒(4-xm )320=xm×80⇒xm=16/5=3.2 m
Alternatively: If the man starts walking on the trolley in the forward direction then whole system will move in backward direction with same momentum. Momentum of man in forward direction = Momentum of system (man + trolley) in backward direction ⇒80×1=(80+320)×v⇒v=0.2 m/s So the velocity of man w.r.t. ground 1.0-0.2=0.8 m/s ∴Displacement of man w.r.t. ground =0.8×4=3.2 m

Q4. 2 bodies of different masses of 2 kg and 4 kg are moving with velocities 20 m/s and 10 m/s towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass?
•  5 m/s
•  6 m/s
•  8 m/s
•  Zero
Solution
m1=2kg,m2=4kg,v 1=20m/s,v 2=-10m/s v cm=(m1 v 1+m2 v 2)/(m1+m2 )=(2×20-4×10)/(2+4)=0 m/s

Q5. The moment of inertia of a uniform ring of mass M and radius r about a tangent lying in its own plane is ?
•  2Mr2
•  3/2 Mr2
•  Mr2
•  1/2 Mr2
Solution
3/2 Mr2

Q6. A man of 50 kg mass is standing in a gravity free space at a height of 10m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2m/s. When the stone reaches the floor, the distance of the man above the floor will be?
•  20 m
•  9.9 m
•  10.1 m
•  10 m
Solution
Let distance of man from the floor be (10+x)m. As centre of mass of system remains at 10m above the floor So 50(x)=0.5(10)⇒x=0.1m ⇒ distance of the man above the floor =10+0.1 =10.1m

Q7. Two balls each of mass m are placed on the vertices A and B of an equilateral triangle ABC of side 1m. A ball of mass 2m is placed at vertex C. The centre of mass of this system from vertex A (located at origin) is?
•  (1/2 m,1/2 m)
•  (1/2 m,√3 m)
•  (1/2 m,√3/4 m)
•  (√3/4 m,√3/4 m)
Solution
The centre of mass is given by
x ̅=(m1 x1+m2 x2+m3 x3)/(m1+m2+m3 ) x ̅=(m×0+m×1+2m×(1/2))/(m+m+2m) x ̅=2m/4m=1/2 m y ̅=(m1 y1+m2 y2+m3 y3)/(m1+m2+m3 ) y ̅=(m×0+m×0+2m×√3⁄2)/(m+m+2m) =√3/4m ∴ Centre of mass is (1/2 m,√3/4 m).

Q8. One quarter of the disc of mass m is removed. If r be the radius of the disc, the new moment of inertia is?
•  3/2 mr2
•  (mr2)/2
•  3/8 mr2
•  None of these
Solution
Moment of inertia of whole disc about an axis through centre of disc and perpendicular to its plane is I=1/2 mr2 As one quarter of disc is removed, new mass, m'=3/4 m ∴I'=1/2 (3/4 m) r2=3/8 mr2

Q9. Three identical spheres, each of mass 1 kg are kept as shown in figure below, touching each other, with their centers on a straight line. If their centres are marked P,Q, R respectively, the distance of centre of mass of the system from P is?
•  (PQ+PR+QR)/3
•  (PQ+PR)/3
•  (PQ+QR)/3
•  (PR+QR)/3
Solution
r1=0,r2=PQ,r3=PR Distance of centre of mass from P is r=(r1+r2+r3)/3=(0+PQ+PR)/3=(PQ+PR)/3

Q10. If kinetic energy of a body remains constant, then momentum-mass graph is?
Solution
K= p2/2m p2=2Km This is an equation of parabola. . ## Want to know more

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET & IIT JEE COACHING: Rotational Motion Quiz 15
Rotational Motion Quiz 15