8. Two infinite planes each with uniform surface charge density +σ C/m² are kept in such a way that the angle between them is 30^o. The electric field in the region shown between them is given by:

1. a.
   \(\frac{\sigma }{2\epsilon _{0}}\left [ (1-\frac{\sqrt{3}}{2})\hat{y}-\frac{1}{2}\hat{x}) \right ]\)
2. b.
   \(\frac{\sigma }{2\epsilon _{0}}\left [ (1+\frac{\sqrt{3}}{2})\hat{y}-\frac{1}{2}\hat{x}) \right ]\)
3. c.
   \(\frac{\sigma }{2\epsilon _{0}}\left [ (1-\frac{\sqrt{3}}{2})\hat{y}+\frac{1}{2}\hat{x}) \right ]\)
4. d.
   \(\frac{\sigma }{2\epsilon _{0}}\left [ (1+\frac{\sqrt{3}}{2})\hat{y}+\frac{1}{2}\hat{x}) \right ]\)

9. If the magnetic field in a plane electromagnetic wave is given by B˙ = 3 × 10⁻⁸ sin(1.6 × 10³x + 48 × 10¹⁰ t)ˆj T then what will be expression for electric field?

1. a. E˙ = 3 × 10⁻⁸sin(1.6 × 10³x + 48 × 10¹⁰t)ˆi V/m
2. b. E˙ = 3 × 10⁻⁸sin(1.6 × 10³x + 48 × 10¹⁰t)ˆj V/m
3. c. E˙ = 60 sin(1.6 × 10³x + 48 × 10¹⁰t)kˆ V/m
4. d. E˙ = 9 sin(1.6 × 10³x + 48 × 10¹⁰t)kˆ V/m

Solution:

1. We know that,

   \(\left | \frac{E_{0}}{B_{0}} \right |=c\)

   B₀ = 3 × 10⁻⁸

   =⇒ E₀ = B₀ × c = 3 × 10⁻⁸ × 3 × 10⁸

   = 9 N/C

   ∴ E = E_osin(ωt − kx + φ)kˆ = 9 sin(ωt − kx + φ)kˆ

   E˙ = 9 sin (1.6 × 10³x + 48 × 10¹⁰t)kˆ V/m

   Answer: (d)

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10. The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10⁻¹⁶s. The frequency of revolution of the electron in its first excited state (in s⁻¹) is

1. a. 6.2 × 10¹⁵
2. b. 7.8 × 10¹⁴
3. c. 1.6 × 10¹⁴
4. d. 5.6 × 10¹²

Solution:

1. Time period is proportional to n³ /Z²

   Let T₁ be the time period in ground state and T₂ be the time period in its first excited state.

   T_{1 =} n³ /2²

   (Where, n = excitation level and 2 is atomic no.)

   (T₁/T₂) = (n₁/n₂)³

   

   Given,

   T₁ = 1.6 × 10⁻¹⁶s

   So,

   (1.6 × 10⁻¹⁶) / T₂= (1/2)³

   T₂ = 12.8 x 10^-16 s

   Frequency is given by f = 1/T

   f = 1/ (12.8 x 10^-16) Hz

   f = 7.8128 × 10¹⁴ Hz

   Answer: (b)

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11. A LCR circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having damping constant ’b’, the correct equivalence will be

1. a. L ↔1/b, C ↔1/m, R ↔1/k
2. b. L ↔ k, C ↔ b, R ↔ m
3. c. L ↔ m, C ↔ k, R ↔ b
4. d. L ↔ m, C ↔ 1/k , R ↔ b

Solution:

1. For damped oscillator by Newton’s second law

   − kx − bv = ma

   kx + bv + ma = 0

   kx + b(dx/dt) + m(d²x/dt²) =0

   For LCR circuit by KVL

   − IR − L (dI/dt)–q/c = 0

   ⇒ IR + L (dI/dt) + q/c = 0

   ⇒ q/c + R(dq/dt) + L(d²q/dt²) = 0

   By comparing

   R ⇒ b

   c ⇒ 1/k

   m ⇒ L

   Answer: (d)

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12. Visible light of wavelength 6000 × 10⁻⁸ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minima is at 60^o from the central maxima. If the first minimum is produced at θ₁, then θ₁ is close to

1. a. 20^o
2. b. 45^o
3. c. 30^o
4. d. 25^o

Solution:

1. For single slit diffraction experiment: Angle of minima are given by

   sin θ_n = nλ/d (sin θ_n ≠ θ_n as θ is large )

   sin θ₂ = sin 60^◦ = √3/2 = 2λ/ d = (2 × 6000 × 10⁻¹⁰)/d ......1

   sin θ₁ = λ/d= (6000 x 10⁻¹⁰)/d ......2

   Dividing (1) and (2)

   ⇒

   \(\frac{\frac{\sqrt{3}}{2}}{sin \theta _{1}} = 2\)

   ⇒ sin θ₁ = √3/4 = 0.43

   As, the value is coming less than 30^o the only available option are 20^o and 25^o but by using approximation we get θ₁ = 25^◦

   Answer: (d)

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13. The radius of gyration of a uniform rod of length l about an axis passing through a point l/4 away from the center of the rod, and perpendicular to it, is

1. a. √ (7/48)l
2. b. √ (3/8) l
3. c. (1/4) l
4. d. (1/8) l

Solution:

1. 

   Moment of inertia of rod about axis perpendicular to it passing through its centre is given by

   I = Ml²/12 + M(l/4)²

   I = (3Ml² + 4Ml²)/48

   Now, comparing with I = Mk² where k is the radius of gyration

   k=√7l²/48

   k=l√7/48

   Answer: (a)

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14. A satellite of mass m is launched vertically upward with an initial speed u from the surface of the earth. After it reaches height R(R = radius of earth), it ejects a rocket of mass m/10 so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G = gravitational constant; M is the mass of earth)

1. a.
   \(5m\left[u^{2}-\frac{119}{200}\frac{GM}{R}\right]\)
2. b.
   \(\frac{m}{20}\left [ u-\sqrt{\frac{2GM}{3R}} \right ]^{2}\)
3. c.
   \(\frac{3m}{8}\left [ u+\sqrt{\frac{5GM}{6R}} \right ]^{2}\)
4. d.
   \(\frac{m}{20}\left [ u^{2}+\frac{113}{200}{\frac{GM}{R}} \right ]\)

Solution:

1. T.E_ground= T.ER

   (1/2)mu² + (-GMm/R) = (1/2)mv² +(-GMm/2R)

   (1/2)mv² = (1/2)mu²+(-GMm/2R)

   v² = u²+(-GMm/R)

   \(v=\sqrt{u^{2}+(\frac{-GM}{R})}\)

   The rocket splits at height R. Since, separation of rocket is impulsive therefore conservation of momentum in both radial and tangential direction can be applied.

   \(\frac{m}{10}V_{r}=\frac{9m}{10}\sqrt{\frac{GM}{2R}}\)
   \(\frac{m}{10}V_{r}=m\sqrt{u^{2}-\frac{GM}{R}}\)

   

   =

   \(5m\left[u^{2}-\frac{119}{200}\frac{GM}{R}\right]\)

   Answer: (a)

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15. Three point particles of mass 1 kg, 1.5 kg and 2.5 kg are placed at three corners of a right triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The centre of mass of the system is at the point:

1. a. 0.9 cm right and 2.0 cm above 1 kg mass
2. b. 2.0 cm right and 0.9 cm above 1 kg mass
3. c. 1.5 cm right and 1.2 cm above 1 kg mass
4. d. 0.6 cm right and 2.0 cm above 1 kg mass

Solution:

1. 

   Taking 1 kg as the origin

   x_com= (m₁x₁ + m₂x₂ + m₃x₃)/(m₁+m₂+m₃)

   = (1 × 0 + 1.5 × 3 + 2.5 × 0)/5

   x_com = 0.9

   y_com= (m₁y₁ + m₂y₂ + m₃y₃)/ (m₁ + m₂+m₃)

   y_com= (1 × 0 + 1.5 × 0 + 2.5 × 4)/5

   ycom = 2

   Centre of mass is at (0.9, 2)

   Answer: (a)

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16. If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece should be close to:

1. a. 22 mm
2. b. 12 mm
3. c. 2 mm
4. d. 33 mm

Solution:

1. Magnification of compound microscope for least distance of distinct vision setting (strained eye)

   M = (L/f₀)(1 + (D/f_e))

   where L is the tube length

   f₀ is the focal length of objective

   D is the least distance of distinct vision = 25 cm

   \(375 = \frac{150\times 10^{-3}}{5\times 10^{-3}}(1+\frac{25\times 10^{-2}}{f_{e}})\)
   \(12.5 = (1+\frac{25\times 10^{-2}}{f_{e}})\)
   \(\frac{25\times 10^{-2}}{f_{e}}=11.5\)

   ∴ fe ≈ 21.7 × 10⁻³m = 22 mm

   Answer: (a)

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17. Speed of transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.0 mm²) is 90m/s. If the Young’s modulus of wire is 16 × 10¹¹Nm⁻², the extension of wire over its natural length is

1. a. 0.03 mm
2. b. 0.04 mm
3. c. 0.02 mm
4. d. 0.01 mm

Solution:

1. Given, M = 6 grams = 6 × 10⁻³ kg

   L = 60 cm = 0.6 m

   A = 1 mm² = 1 × 10⁻⁶m²

   Using the relation, v² = T/ µ

   ⇒ T = µv²

   = V² x M/L

   As Young’s modulus, Y = stress/strain

   strain = stress/Y= T/AY

   strain =∆L/L = (V²(M/L))/AY = V²(M/AYL)

   ∆L = V²M/AY

   ∆L =(8100 × 6 × 10⁻³)/(1 × 10⁻⁶ × 16 × 10¹¹)

   ∆L = 0.03 mm

   Answer: (a)

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18.1 liter of dry air at STP expands adiabatically to a volume of 3 litres. If γ = 1.4, the work done by air is (31.4 = 4.655) (take air to be an ideal gas)

1. a. 48 J
2. b. 100.8 J
3. c. 90.5 J
4. d. 60.7 J

Solution:

1. Given, P₁ = 1 atm, T₁ = 273 K (At STP)

   In adiabatic process,

   P₁V₁^γ = P₂V₂^γ

   \(P_{2}=P_{1}\left [ \frac{V_{1}}{V_{2}} \right ]^{\gamma }\)

   P₂ = 1 × (1/3)^1.4

   P₂ = 0.2164 atm

   Work done in adiabatic process is given by

   W =( P₁V₁ − P₂V₂)/(γ – 1)

   W = ((1 × 1 − 3 × .2164)/ 0.4) x 101.325

   Since, 1 atm = 101.325 kPa and 1 Liter = 10⁻³ m³

   W=89.87J

   Closest answer is 90.5 J

   Answer: (c)

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19. A bob of mass m is tied by a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from the rest, the bob starts falling vertically. When it has covered a distance h, the angular speed of the wheel will be:

1. a. r√3/4gh
2. b. (1/r)√4gh/3
3. c. ( r√3)/2gh
4. d. (1/r)√2gh/3

Solution:

1. By energy conservation,

   mgh = (1/2) mv² + (1/2)Iω

   ⇒gh = v²/2 +ω²r²/4 ------------(1)

   Since the rope is inextensible and also it is not slipping,

   ∴ v = rω--------(2)

   from eq. (1) and (2)

   gh = ω²r²/2 +ω²r²/4

   gh = (3/4) ω²r²

   ⇒ ω² = 4gh/3r²

   ω= (1/r) )√4gh/3

   Answer: (b)

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20. A parallel plate capacitor has plates of area A separated by distance ‘d’ between them. It is filled with a dielectric which has a dielectric constant varies as k (x) = k(1 + αx), where ‘x’ is the distance measured from one of the plates. If (αd <<1), the total capacitance of the system is best given by the expression:

1. a.
   \(\frac{A\epsilon _{0}k}{d}\left [ 1+(\frac{\alpha d}{2})^{2}\right ]\)
2. b.
   \(\frac{A\epsilon _{0}k}{d}\left [ 1+(\frac{\alpha d}{2})\right ]\)
3. c.
   \(\frac{A\epsilon _{0}k}{d}\left [ 1+(\frac{\alpha^{2} d}{2})\right ]\)
4. d.
   \(\frac{A\epsilon _{0}k}{d}\left [ 1+{\alpha d})\right ]\)

Solution:

1. Given, k(x) = k (1 + αx)

   dC = Aε₀k/dx

   Since all capacitance are in series, we can apply

   

   On putting the limits from 0 to d

   = ln (1 + αd)/ kε₀Aα

   Using expression ln(1 + x) = x – x²/2 +-----

   And putting x = αd where, x approaches to 0.

   \(\frac{1}{C}=\frac{d}{k\epsilon _{0}A\alpha }\left [ \alpha d-{\alpha^{2}d}^{2}/2)\right ]\)
   \(C = \frac{A\epsilon _{0}k}{d\left ( 1-\frac{\alpha d}{2} \right )}\)
   \(\Rightarrow C = \frac{\epsilon _{0}kA}{d}\left ( 1+\frac{\alpha d}{2} \right )\)

   Answer: (b)

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21. A non- isotropic solid metal cube has coefficient of linear expansion as 5 × 10⁻⁵/^◦C along the x-axis and 5 × 10⁻⁶/^◦C along y-axis and z-axis. If the coefficient of volumetric expansion of the solid is C × 10⁻⁶/^◦C then the value of C is

Solution:

1. We know that, V = xyz

   ∆v/v = (∆x/x) + (∆y/y) + (∆z/z)

   (1/T)∆v/v = (1/T)((∆x/x)+(∆y/y)+(∆z/z))

   y = α_n + α_y + α_z

   y = (50 × 10⁻⁶/^◦C) + (5 × 10⁻⁶/^◦C) + (5 × 10⁻⁶/^◦C)

   y = 60 × 10⁻⁶/^◦C

   ∴ C = 60

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22. A loop ABCDEFA of straight edges has six corner points A(0,0,0), B(5,0,0), C(5,5,0), D(0,5,0), E(0,5,5), F(0,0,5). The magnetic field in this region is B˙ = (3ˆi + 4kˆ) T. The quantity of flux through the loop ABCDEFA (in Wb) is ---------

Solution:

1. 

   As we know, magnetic flux =

   \(\vec{B}\times \vec{A}\)

   ⇒ (B_x + B_z) • (A_x + A_z)

   ⇒ (3ˆi + 4kˆ) • (25ˆi + 25kˆ)

   ⇒ (75 + 100) Wb

   ⇒ 175 Wb

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23. A carnot engine operates between two reservoirs of temperature 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy in(J ) delivered by the engine to the low temperature reservoir, in a cycle, is

Solution:

1. η = 1 – (T₂/T₁) = 1- (300/900) =2/3

   Given, W = 1200 J

   From conservation of energy

   Q₁ − Q₂ = W

   η = Q₁ − Q₂ = W

   ⇒ Q₁= 1800 J

   ⇒ Q₂ = Q₁ − W = 600 J

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24. A particle of mass 1 kg slides down a frictionless track (AOC) starting from rest at a point A (height 2 m). After reaching C, the particle continues to move freely in air as a projectile. When it reaches its highest point P(height 1 m) the kinetic energy of the particle (in J ) is:
(Figure drawn is schematic and not to scale; take g = 10 m/s²)

Solution:

1. As the particle starts from rest the total energy at point A = mgh = T.EA( where h = 2 m) After reaching point P

   

   T.Ec = K.E. + mgh

   By conservation of energy

   T.E_A = T.Ep

   ⇒ K.E. = mgh = 10 J

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25. A beam of electromagnetic radiation of intensity 6.4× 10⁻⁵ W/cm² is comprised of wavelength, λ = 310 nm. It falls normally on a metal (work function φ = 2 eV) of surface area 1cm². If one in 103 photons ejects an electron, total number of electrons ejected in 1s is 10x (hc = 1240 eV − nm, 1 eV = 1.6 × 10⁻¹⁹ J ), then x is

Solution:

1. P = Intensity × Area

   = 6.4 × 10⁻⁵Wcm⁻² × 1 cm²

   = 6.4 × 10⁻⁵ W

   For photoelectric effect to take place, energy should be greater than work function Now,

   E = (1240/310) = 4eV>2eV

   Therefore, photoelectric effect takes place

   Here n is the number of photons emitted.

   n × E = I × A

   n = IA/E = (6.4 × 10⁻⁵⁾/(6.4 × 10⁻¹⁹)

   n = 10¹⁴

   Where, n is number of incident photon

   Since, 1 out of every 1000 photons are successful in ejecting 1 photoelectron Therefore, the number of photoelectrons emitted is

   = 10¹⁴/10³

   x = 10¹¹

   Answer: (10¹¹)

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Video Lessons - January 7 Shift 1 Physics

JEE Main 2020 Physics Paper January 7 Shift 1

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