## NLM Quiz-15

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. In the figure, the ball A is released from rest when the spring is at its natural length. For the block B of mass M to leave contact with the ground at same stage, the minimum mass of A must be:

•  2M
•  M
•  M/2
•  A function of M and the force constant of the spring
Solution
For minimum mass of m, mass M breaks off contact when elongation in spring is maximum At the time of break off, block A is at lowest position and its speed is zero. At an instant t1 mg-kx=ma v dv/dx=(mg-kx)/m

Where x0 is maximum elongation is spring 0=gx0-(kx02)/2m x=2mg/k At the time of break off of block B Mg=kx0 Mg=2mg m=M/2

Q2. The engine of a jet aircraft applies a thrust force of 105 N during take off and causes the plane to attain a velocity of 1 km/sec in 10 sec. The mass of the plane is
•  102 kg
•  103 kg
•  104 kg
•  105 kg
Solution
Acceleration produced in jet =(Change in velocity)/Time a=((103-0))/10=100 m/s2 ∴Mass= Force/Acceleration=105/102 =103 kg

Q3. A block is kept on an inclined plane of inclination Î¸ and length l. The velocity of particle at the bottom of incline is (the coefficient of friction is Î¼)
•  √(2gl(Î¼ cos⁡Î¸-sin⁡Î¸) )
•  √(2gl(sin⁡Î¸-Î¼ cosÎ¸) )
•  √(2gl(sin⁡Î¸+Î¼ cosÎ¸))
•  √(2gl(cos⁡Î¸-Î¼ sin⁡Î¸))
Solution
The various forces acting on the block are as shown From Newton’s law

mg sin⁡Î¸-f=ma ….(i) Where f is frictional force and a the acceleration downwards. Since, there is no motion perpendicular to surface, we have R-mg cos⁡Î¸=0 R=mg cos⁡Î¸ ….(ii) Also, f=Î¼R=Î¼ mg cos⁡Î¸ Putting the value in Eq. (i) we get mg sin⁡Î¸-Î¼ mg cos⁡Î¸=ma a=g sin⁡Î¸-Î¼ g cos⁡Î¸ Now, velocity at bottom v2=u2-2as Since, v=0 ∴u=√2as Given,s=l,a=g sin⁡Î¸-gÎ¼ cos⁡Î¸ ∴ u=√(2l(g sin⁡Î¸-gÎ¼ cos⁡Î¸ )) u=√(2gl(sin⁡Î¸-Î¼ cos⁡Î¸ ))

Q4. A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N. The friction between wall and body is 0.3. Then the frictional force is equal to
•  6 N
•  20 N
•  600 N
•  700 N
Solution
Weight of body =2×10=20 N This force has the tendency to move the block, so friction force = 20 N.

Q5.  A blumb bob is hung from the ceiling of a train compartment. The train moves on an inclined track of inclination 30° with horizontal. Acceleration of train up the plane is a=9/2. The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is.
•  30°
•  tan-1⁡(2/√3)
•  tan-1⁡(√3/2)
•  tan-1⁡(2)
Solution

T sin⁡Î¸-mg sin⁡Î¸=ma T sin⁡Î¸=mg sin⁡Î¸+mg/2 …(i) T cos⁡Î¸=mg cos⁡Î¸ …(ii) Dividing Eq. (i) by Eq. (ii), we get tan⁡Î¸=2/√3

Q6. Which of the following quantities measured from different inertial reference frames are same?
•  Force(
•  Velocity
•  Displacement
•  Kinetic energy
Solution
Kinetic energy being a scalar quantity, hence measured from different inertial frame gives the same value, while the other three being vector quantities their values vary.

Q7. A block of mass ‘m’ is connected to another block of mass ‘M’ by a spring (massless) of spring constant ‘K’. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force ‘F’ starts acting on the block of mass ‘M’ to pull it. Find the force on the block of mass ‘m’
•  mF/M
•  ((M+mF))/m
•  mF/((m+M))
•  MF/((m+M))
Solution
Acceleration of the system =F/(M+m) and

Force on the block m=Kx=ma=mF/(m+M)

Q8. If in a stationary lift, a man is standing with a bucket full of water, having a hole at its bottom. The rate of flow of water through this hole is R0. If the lift starts to move up and down with same acceleration and then that rates of flow of water are Ru and Rd, then
•  R0>Ru>Rd
•  Ru>R0>Rd
•  Rd>R0>Ru
•  Ru>Rd>R0
Solution
Rate of flow will be more when lift will move in upward direction with some acceleration because the net downward pull will be more and vice-versa Fupward=m(g+a) and Fdownward=m(g-a)

Q9. The acceleration of the 500 g block in figure is

•  6g/13 downwards
•  7g/13 downwards
•  8g/13 downwards
•  9g/13 upwards
Solution
500g -T=500a T-100g sin⁡30°-T'=100a or T-T'-50g=100a

Again, T'-50g=50a From Eqs. (ii) and (iii) T-100g=150a Adding Eqs. (i) and (iv), 400g = 650 a or a=400g/650=8g/13 This acceleration is downwards

Q10. Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force F applied on the upper string produces an acceleration of 2 m/s2 in the upward direction in both the blocks. If T and T' be the tensions in the two parts of the string, then ( g=9.8 m/s2):

•  T=70.8 N and T'=47.2 N
•  T=58.8 N and T'=47.2 N
•  T=70.8 N and T'=58.8 N
• T=70.8 N and T'=0
Solution
FBD of mass 2 kgFBD of mass 4kg

TT'-19.6=4…(i) T'-39.2=8…(ii) From (ii), T'=47.2 N And substituting T' in (i), we get T=4+19.6+47.2⇒T=70.8 N

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