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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. The unit of Wien’s constant b is
•  Wm-2K-4
•  m-1K-1
•  Wm2
•  MK
Solution
(d) According to Wien’s law the product of wavelength corresponding to maximum intensity of radiation and temperature of body (in Kelvin) is constant ie,λ_m T=b=constant, where b is Wien’s constant and has value 2.89×10-3 m-K.

Q2.The SI unit of gravitational potential is
•  J
•  Jkg-1
•  Jkg
•  Jkg2
Solution
(b) Gravitational potential =work/mass Hence, SI unit gravitational potential = (unit of work)/(unit of mass) = J/kg=Jkg^(-1) or ms^(-2)

Q3.  A physical quantity is measured and its value is found to be nu where n= numerical value and u= unit. Then which of the following relations is true
•  n∝u2
•  n∝u
•  n∝√u
•  n∝1/u
Solution
(d) . P=nu∴n∝ 1/u

Q4. Frequency is the function of density (ρ), length (a) and surface tension (T). Then its value is
•  1/2a3/2/√T
•  3/2a3/2/√T
•  1/2a3/2/T3/4
•  None of these
Solution
(a) Let n=kρaabTc where [ρ]=[ML-3 ],[a]=[L] and [T]=[MT-2] Comparing dimensions both sides we get a=(-1)/2,b=(-3)/2 and c=1/2 ∴η=kρ-1/2a-3/2T-1/2 =(K√T)/(ρ1/2a3/2 )

Q5.The units of modulus rigidity are
•  N - m
•  N / m
•  N - m2
•  N / m2
Solution
(c)

Q6.The circular divisions of shown screw gauge are 50. It moves 0.5 mm on main scale in one rotation. The diameter of the ball is
•  2.25mm
•  2.20 mm
• 1.20 mm
•  1.25mm
Solution
(c) Zero error =5×0.5/50=0.05 mm Actual measurement =2×0.5 mm+25×0.5/50-0.05 mm =1 mm+0.25 mm-0.05 mm=1.20 mm

Q7.Which of the following has same dimensions
•  Current Density and charge density
•  Angular Momentum and Momentum
•  Spring constant and surface energy
•  Force and torque
Solution
(c) Spring constant =F/l=[ML0T-2]. Surface energy =Energy/Area=[ML0T-2]

Q8.The Vander Waal’s equation of state for real gases is given as (P+a/V2)(V-b)=nRT which of the following terms has dimensions different from that of energy
•  PV
•  Ta / v2
•  ab/V2
•  bP
Solution
(b) PV=[energy] Vander Waal’s equation is (P+a/V2 )(V-b)=nRT The dimensions of a/V2 should be that of P and b is that of volume Work done (or energy) should have the dimensions of PV ∴ [a/V2 ×b]= [Energy] [bP]= [Energy] [a/V2 ]=[P] is having dimensions different from energy

Q9.In the relation p=α/β e(-αz/kθ),p is the pressure, z the distance, k is Boltzmann constant and θ is the temperature, the dimensional formula of β will be
•  [M0L2T0]
•  [ML2T ]
•  [ML0T-1 ]
•  ML2T-1
Solution
(a) In given equation, az/kθ should be dimensionless α=kθ/z ⇒ [α]=[ML2T-2K-1×K]/[L] =[MLT-2] And p=α/β ⇒ [β]=[α/p]=[MLT-2) ]/[ML-1T-2] =[M0L2T0]

Q10. A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity η flowing per second through a tube of radius r and length l and having pressure p across its end, is
•  V= (πpr4)/8ηl
•  V= πηl/(8pr4)
•  V= 8pηl/(πr4)
• V= πpη/(8lr4)
Solution
(a) Formula for viscosity η=(πpr4)/8Vl⇒V=(πpr4)/8ηl #### Written by: AUTHORNAME

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