## DUAL NATURE OF RADIATION AND MATTER quiz-14

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. An electron of mass m and charge e initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time t ignoring relativistic effects is
•  -h/eEt2
•  -e Et/E
•  -mh/e Et2
•  -h/e E
Solution
Here, u=0;a=eE/m;v=?;t=t ∴ v=u+at=0+eE/m t de-Broglie wavelength, Î»=h/mv=h/(m(eEt/m))=h/eEt Rate of change of de-Broglie wavelength dÎ»/dt=h/eE (-1/t2 )=-h)/eEt2

Q2.One electron and one proton is accelerated by equal potential. Ratio in their de-Broglie wavelength is
•  1
•  me/mp
•  √(mp/me )
•  √(me/mp )
Solution
If a charge particle of mass m and charge q is accelerated through a potential difference V and E is the energy acquired by the particle , then E=qV If v is velocity of particle, then E=1/2 mv2 Or v=√(2E/m)  Now, de-Broglie wavelength of particle is Î»=h/mv=h/(m√(2E /m) Substituting the value of E, we get Î»=h/√2mqV For electron, Î»e=h/√(2me eV) For proton, Î»p=h/√(2mp eV) ∴ Î»e/Î»p =√(mp/me )

Q3. A 1Î¼A beam of protons with a cross-sectional area of 0.5 sq.mm is moving with a velocity of 3×104 ms-1. Then charge density of beam is
•   6.6×10-4 C/m3
•  6.6×10-5 C/m3
•  6.6×10-6 C/m3
•  None of these
Solution
For one second, distance = Velocity =3×104 m/sec and Q=i×1=10-6 C.Charge density =Charge/Volume =10-6/(3×104×0.5×10-6 )=6.6×10-5 C/m3

Q4. The dependence of the short wavelength limit Î»min on the accelerating potential V is represented by the curve of figure

•  A
•  B
•  C
•  None of these
Solution
Î»min=hc/eV⇒log⁡.Î»min =log⁡ (hc/e)-log⁡V ⇒log⁡ Î»min =-log⁡ V+log⁡ (hc/e) This is the equation of straight line having slope (-1) and intercept log⁡(hc/e) on+loge⁡Î»min axis

Q5. X-rays are produced due to
•  Break up of molecules
•  Changing in atomic energy level
•  Changing in nuclear elergy level
Solution
Changing in atomic energy level

Q6. In Millikan’s oil drop experiment, a charged drop of mass 1.8×10-14 kg is stationary between the plates. The distance between the plates 0.9 cm and potential difference between the plates is 2000 V. The number of electrons on the oil drop is
•  10
•  5
• 50
•  20
Solution
In equilibrium, qE=mg Also, E=V/d and q=ne ∴ ne×V/d=mg ⇒ n=mgd/eV Given, m=1.8×10-14 kg, g=10ms-2, d=0.9 cm=0.9×10-2 m, e=1.6×10-19 C, V=2000 V ∴ n=(1.8×10-14×10×0.9×10-2)/(1.6×10-19×2000) n=81/16≈5 electrons

Q7.In Millikan’s oil drop experiment, a charged drop falls with terminal velocity V. if an electric field E is applied in vertically upward direction then it starts moving in upward direction with terminal velocity 2V. if magnitude of electric field is decreased to E/2, then terminal velocity will becomes
•  V/2
•  V
•  3V/2
•  2V
Solution
In the absence of electric field (i.e.E=0) mg=6Ï€Î·rv

…(i) In the presence of Electric field mg+QE=6Ï€Î·r(2v)

…(ii) When electric field to reduced to E/2 mg+Q(E/2)=6Ï€Î·r(v'

…(iii) After solving (i), (ii) and (iii) We get v'=3/2 v

Q8. The wavelength of de-Broglie wave is 2Î¼m, then its momentum is (h=6.63× 10-34 Js)
•  3.315×10-28 kg-ms-1
•  1.66×10-28 kg-ms-1
•  4.97×10-28 kg-ms-1
•  9.9×10-28 kg-ms-1
Solution
According to de-Broglie hypothesis Î»=h/p Where, h is Planck’s constant. ⇒ p=h/Î» Given, h=6.63×10-34 J-s Î»=2Î¼m=2×10-6 m∴ ∴ p=(6.63×10-34)/(2×10-6) =3.315×10-28 kg-ms-1

Q9.When a monochromatic point source of light is at a distance 0.2 m from a photoelectric cell, the saturation current and cut-off voltage are 12.0 mA and 0.5 V. If the same source is placed 0.4 m away from the photoelectric cell, then the saturation current and the stopping potential respectively are
•  4 mA and 1V
•  12 mA and 1V
•  3 mA and 1V
•  3 mA and 0.5 V
Solution
The cut-off voltage or stopping potential measure maximum kinetic energy of the electron. It depends on the frequency of incident light whereas the current depends on the number of photons incident. Hence, cut-off voltage will be 0.5 V. Now by inverse square law, 12∝ 1/(0.2)2 or I∝1/(0.4)2 ∴ I/12=(0.2)2/(0.4)2 =1/4 or I=12/4=3mA

Q10. The momentum of a charged particle moving in a perpendicular magnetic field depends on
•  Its charge
•  The strength of magnetic field
• All of the above
Solution
q v B=mv2/r or mv=q r B.

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