NLM Quiz 17

NLM Quiz-17

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 force of 20N is applied on a body of mass 5 kg resting on a horizontal plane. The body gains a kinetic energy of 10 joule after it moves a distance 2 m. The frictional force is:

•  10 N
•  15 N
•  20 N
•  30 N

Solution
Kinetic energy =10 J ⇒1/2 mv²=10⇒v²=4 From third equation of motion v²=u²+2as 4=0+2×a×2 ⇒a=1 m/s² ∵F_s=F-ma=20-5×1=15 N

Q2. The resultant force of 5 N and 10 N can not be

•  12 N
•  8 N
•  4 N
•  5 N

Solution
F_max=5+10=15N and F_min=10-5=5 N Range of resultant 5≤F≤15

Q3. A boy of mass 100 g is sliding from an inclined plane of inclination 30°. What is the frictional force experienced if μ=1.7

•  1.7×√2×1/√3 N
•  1.7×√3×1/2 N
•  1.7×√3 N
•  1.7×√2×1/3 N

Solution
F_k=μ_k R=μ_k mg cos⁡θ F_k=1.7×0.1×10×cos⁡30°=1.7×√3/2 N

Q4. The coefficient of static friction μ_s between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pully are assumed to be smooth and massless (g=10 ms^-2))

•  2.0 KG
•  4.0 KG
•  0.2 KG
•  0.4 KG

Solution
0.4 KG

Q5.  Work done by a frictional force is.

•  Negative
•  Positive
•  Zero
•  All of the above

Solution
 Work done by friction can be positive, negative and zero depending upon the situation

Q6. A force of 19.6 N when applied parallel to the surface just moves a body of mass 10 kg kept on a horizontal surface. If a 5 kg mass is kept on a horizontal surface. If a 5 kg mass is kept on the first mass, the force applied parallel to the surface to just move the combined body is?

•  29.4 N(
•  39.2 N
•  18.6 N
•  42.6 N

Solution
F_l∝R ∴F_l∝mi.e. limiting friction depends upon the mass of body. So, ⇒(F₁ )^'=3/2×F_l=3/2×19.6=29.4 N .

Q7. Which of the four arrangements in the figure correctly shows the vector addition of two forces (F₁ ) ⃗and (F₂ ) ⃗ to yield the third force (F₃ ) ⃗

•  
  
  
•  
  
  
•  
  
  
•  
  
  

Solution

Q8. A smooth block is released at rest on a 45° incline and then slides a distanced. The time taken to slide is n times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

•  μ_k=1-1/n²
•  μ_k=√(1-1/n² )
•  μ_s=1-1/n²
•  μ_s=√(1-1/n² )

Solution
When friction absent a₁=g sin⁡θ ∴ s₁=1/2 a₂ t₂² ….(i) When friction in present a₂=g sin⁡θ-μ_k g cos⁡θ ∴ s₂=1/2 a₂ t₂² ….(ii) From Eqs. (i) and (ii), we have 1/2 a₁ t₁²=1/2 a₂ t₂² or a₁ t₁²=a₂ (nt₁ )² (∵ t₂=nt₂) or a₁=n² a₂ or a₂/a₁ =(g sin⁡θ-μ_k g cos⁡θ )/(g sin⁡θ )=1/n² or (g sin⁡45°-μ_k g cos⁡45° )/(g sin⁡45° )=1/n² or 1-μ_k=1/n² or μ_k=1-1/n²

Q9. Refer to the system shown in figure. The acceleration of the masses is

•  g/3
•  g/6
•  g/9
•  g/12

Solution
5g-T₂=5a….(i) T₂-T₁-3g=3a…(ii) T₁-g=a…(iii) 

Adding Eqs (i) and (iii), -T₂+T₁+4g=6a Adding this to Eq. (ii), we get g=9a or a=g/9 

Q10. A 1000 kg lift is supported by a cable that can support 2000 kg. The shortest distance in which the lift can be stopped when it is descending with a speed of 2.5 ms^-1 is [Take g=10 ms^-2)]:

•  1 m
•  2 m
•  5/32m
•  5/16m

Solution
T=mg+ma 2000g =1000g+1000a or a=g Direction is upward Now, 0²-2.5²=-2×10×s or s=(2.5×2.5)/20 =625/(100×20)=25/80 m= 5/16 m