NLM Quiz 16

NLM Quiz-16

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 takes time t to reach the bottom of an inclined plane of angle θ with the horizontal. If the plane is made rough, time taken now is 2t. The coefficient of the friction of the rough surface is

•  3/4 tan⁡θ
•  2/3 tan⁡θ
•  1/4 tan⁡θ
•  1/2 tan⁡θ

Solution
μ= tan⁡θ (1-1/n² )=tan⁡θ (1-1/2² )=3/4 tan⁡θ

Q2. A body of mass M at rest explodes into three pieces, two of which of mass M/4 each are thrown off in perpendicular directions with velocities of 3 m/s and 4 m/s respectively. The third piece will be thrown off with a velocity of

•  1.5 m/s
•  2.0 m/s
•  2.5 m/s
•  3.0 m/s

Solution
Momentum of one piece =M/4×3 Momentum of the other piece =M/4×4 ∴ Resultant momentum =√((9M²)/16+M² )=5M/4 The third piece should also have the same momentum Let its velocity be v, then 5M/4=M/2×v ⇒v=5/2=2.5 m/sec

Q3. A block at rest slides down a smooth inclined plane which makes an angle 60° with the vertical and it reaches the ground in t₁ second. Another block is dropped vertically from the same point and reaches the ground in t₂ second. Then the ratio of t₁:t₂ is

•  1:2
•  2:1
•  1:3
•  1:√2

Solution
l= 1/2 g cos⁡60° t¹² ….(i) lcosθ= 1/2 gt₂² …(ii) t₁²/t₂² =1/cos² 60° =1/4 t₁:t₂=2:1 

Q4. The time period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with an acceleration g/3, the time period is

•  T√3
•  T√3/2
•  T/√3
•  T/3

Solution
T=2π√(l/g) and T^'=2π√(l/(4g/3)) [As g^'=g+a=g+g/3=4g/3] ∴T^'=√3/2 T

Q5. Two masses A and B of 15 kg and 10 kg are connected with a string passing over a frictionless pulley fixed at the corner of a table (as shown in figure). The coefficient of friction between the table and block is 0.4. The minimum mass of C, that may be placed on A to prevent it from moving is

•  10 kg
•  5 kg
•  Zero
•  15 kg

Solution
 The following free body diagram shows the various forces acting on the system. Let m be the minimum mass of block C and f_s be the maximum value of static friction.

For block A R=m+m_Ag,f_s=T ∴ μm+m_Ag=T…(i) For block B T=m_B g …(ii) From Eqs. (i) and (ii), we get m=(m_B-μm_A)/μ m= (10-0.4×15)/0.4=10 kg

Q6. A 60 kg man stands on a spring scale in the lift. At some instant he finds, scale reading has changed from 60 kg to 50 kg for a while and then comes back to the original mark. What should we conclude

•  The lift was in constant motion upwards
•  The lift was in constant motion downwards
• The lift while in constant motion upwards, is stopped suddenly
•  The lift while in constant motion downwards, is suddenly stopped

Solution
For upward acceleration apparent weight =m(g+a) If lift suddenly stops during upward motion then apparent weight =m(g-a) because instead of acceleration, we will consider retardation In the problem it is given that scale reading initially was 60 kg and due to sudden jerk reading decreasing and finally comes back to the original mark i.e. , 60 kg So, we can conclude that lift was moving upward with constant speed and suddenly stops

Q7. The mass of ship is 2×10⁷ kg. On applying a force of 25×10⁵ N , it is displaced through 25 m. After the displacement, the velocity acquired by the ship will be

•  12.5 m/s
•  5 m/s
•  3.7 m/s
•  2.5 m/s

Solution
Here : Mass of ship m=2×10⁷kg, Force F=25×10⁵ N Displacement s=25 m According to the Newton’s second law of motion F=ma ⇒a=F/m=25×10⁵/(2×10⁷=12.5×10^-2) m/s² The relation for final velocity is v²=u²+2as ⇒v²=0+2×12.5×10^-2 )×25 ⇒v=√6.25=2.5 m/s

Q8. 300 joule of work is done in sliding up a 2 kg block on an inclined plane to a height of 10 metres. Taking value of acceleration due to gravity ‘g’ to be 10 m/s², work done against friction is

•  100 J
•  200 J
•  300 J
•  Zero

Solution
Work done against gravity =mgh=2×10×10=200 J Work done against friction =(Total work done-work done against gravity)=300-200=100 J

Q9. In the figure shown, a block of weight 10 N is resting on a horizontal surface. The coefficient of static friction between the block and the surface μ_s=0.4. A force of 3.5 N will keep the block in uniform motion, once it has been set in motion. A horizontal force of 3N is applied to the block then the block will

•  Move over the surface with constant velocity
•  Move having accelerated motion over the surface
•  Not move
•  First it will move with a constant velocity for some time and then will have accelerated motion

Solution
F_l=μ_s R=0.4×mg=0.4×10=4Ni.e. minimum 4N force is required to start the motion of a body. But applied force is only 3N. So the block will not move

Q10. Rocket engines lift a rocket from the earth surface because hot gas with high velocity

•  Push against the earth
•  Push against the air
•  React against the rocket and push it up
•  Heat up the air which lifts the rocket

Solution
It works on the principle of conservation of momentum