## UNITS AND MEASUREMENTS-QUIZ 15

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. Which of the following is not represented in correct unit
•  Stress/Strain=N/m2
•  Surface tension =N/m
•  Energy =kg-m/sec
•  Pressure =N/m2
Solution
(c) Unit of energy will be kg- m2/sec2

Q2.If 1 g cm s-1= x newton-second, then the number x is equal to
•  1×10-3
•  3.6×10-3
•   1×10-5
•  6×10-4
Solution
(c) x=(1gcms-1/T2 =(1gcms-1/(1kg×1ms-1×1s) =(1gcms-1/(103 g×〖102 cms2×1s)=10-5

Q3. The time taken by an electron to go from ground state to excited state is one
shake (one shake = 10-8s). this time in nanosecond will be
•   10 ns
•  4 ns
•  2 ns
•  25 ns
Solution
(a) n1 u1=n2 u2 n2=(1 shake)/(1 ns) =(10-8 s)/(10-9 s) ∴ n2=10

Q4. The dimensional formula of angular velocity is
•  M0 L0 T-1
•  MLT-1
•  M0 L0 T1
•  ML0 T-2
Solution
(a) Angular velocity = Î¸/t,[Ï‰]=([M0 L0 T0])/([T])=[T-1]

Q5.The physical quantity which has the dimensional formula M1 T-3 is
•  Surface tension
•  Solar constant
•  Density
•  Compressibility
Solution
(b) Solar constant is energy received per unit area per unit time i.e.
([ML2 T-2])/([L2 ][T])=[M1T-3]

Q6. The dimensions of e2/4Ï€Îµ0 hc, where e,Îµ0,h and c are electronic charge, electric permittivity,
Planck’s constant and velocity of light in vacuum respectively
•  [M0 L0 T0]
•  [M1 L0 T0]
• [M0 L1 T0]
•  [M0 L0 T1]
Solution
(a) [e]=[AT],∈0=[M-1 L-3 T4 A2 ],[h]=[ML2 T-1]
And [c]=[LT-1]
∴[e2/(4Ï€Ïµ0 hc)]=[(A2 T2)/(M-1 L-3 T4 A2×ML2 T-1×LT-1 )]
=[M0 L0 T0]

Q7.Farad is not equivalent to
•  q/V
•  qv2
•  q2/J
•  J/V2
Solution
(b) Capacitance C=Charge/potential=q/V Also potential =work/charge
(∵V=W/q) ∴ C=q2/J as well as C=J/V2 .
Thus, (a), (c), (d) are equivalent to farad but (b) is not equivalent to farad.

Q8.Given X=(Gh/c2 )(1/2), where G,h and c are gravitational constant, Planck’s constant and the velocity of light
respectively. Dimensions of X are the same as those of
•  Mass
•  Time
•  Length
•  Acceleration
• Solution
(c) [X]=[(M-1 L3 T-2×ML2 T-1/L3 T-3 ](-1/2)=[L]

Q9.The velocity of a particle v at an instant t is given by v=at+bt2 the dimension of b is (Given that radius of earth = 6400 km)
•  [L]
•  [LT-1 ]
•  [LT-2 ]
•  [LT-3]
Solution
(d) Given, v=at+bt2 Applying the law of homogeneity [v]=[bt2 ]
Or [LT-1 ]=[bT2 ]
Or [b]=[LT-3]

Q10. Dimensions of luminous flux are
•  ML2T-2
•  ML2 T-3
•  ML2 T-1
•  MLT-2
Solution
(B)ML2 T-3

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