7. Which of the following will NOT be observed when a multimeter (operating in resistance measuring mode) probes connected across a component, are just reversed?

1. 1) Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is metal wire.
2. 2) Multimeter shows a deflection, accompanied by a splash of light out of connected component in one direction and NO deflection on reversing the probes if the chosen component is LED.
3. 3) Multimeter shows an equal deflection in both cases i.e. before and after reversing the probes if the chosen component is resistor.
4. 4) Multimeter shows NO deflection in both cases i.e. before and after reversing the probes if the chosen component is capacitor.

8. A uniform rod of length ‘l’ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) (ml²/12) ω²sinθcosθ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces F_H and F_v about the CM. The value of θ is then such that:

1. 1) cos θ = 2g/3l ω²
2. 2) cos θ = 3g/2l ω²
3. 3) cos θ = g/2l ω²
4. 4) cos θ = g/I ω²

Solution:

1. Answer: 2

   

   

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9. Two resistors 400 Ω and 800 Ω are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 k Ω across 400 Ω resistors is close to:

1. 1) 2.05 V
2. 2) 2 V
3. 3) 1.95 V
4. 4) 1.8 V

Solution:

1. Answer: 3

   

   \(i=\frac{6}{800+\frac{400\times 10000}{400+10000}}\)

   I = 6/1184.61 = 0.00506

   V_v = 6-800 x 0.00506 = 6 – 4.05 = 1.95

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10. A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m/s² . Assume there is no rotational motion and loss of energy after the collision is negligible.]

1. 1) 23 J
2. 2) 21 J
3. 3) 20 J
4. 4) 19 J

Solution:

1. Answer: 2

   

   Applying law of conservation of linear momentum

   0.1×20 = (1.9+0.1)V

   2 = 2 V

   V = 1 m/sec

   KE= (1/2) mv² = (1/2) 2(1)² = 1J

   TE = KE + mgh = 1+2×10×1 = 21J

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11. A metallic sphere cools from 50°C to 40°C in 300 s. If atmospheric temperature around is 20°C, then the sphere’s temperature after the next 5 minutes will be close to :

1. 1) 35°C
2. 2) 31°C
3. 3) 33°C
4. 4) 28°C

Solution:

1. Answer: 3

   \(\frac{\Delta T}{\Delta t}=k\left [ \frac{T_{f}+T_{i}}{2}-T_{0} \right ]\)
   \(\frac{50-40}{300}=k\left [ \frac{90}{2}-20 \right ]\)
   \(\frac{40-T}{300}=k\left [ \frac{40+T}{2}-20 \right ]\)
   \(\frac{10}{40-T}=\left [ \frac{25\times 2}{40+T-40} \right ]\)

   T = 200 -5T

   6T =200

   T = 33⁰C

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12. To raise the temperature of a certain mass of gas by 50°C at a constant pressure, 160 calories of heat is required. When the same mass of gas is cooled by 100°C at constant volume, 240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

1. 1) 6
2. 2) 7
3. 3) 5
4. 4) 3

Solution:

1. Answer: 1

   Q =nC_p ∆T

   160 = nC_p (50)

   240 = nC _v100

   (16/24) = (C_p/2C_v)

   ϒ=(4/3)

   ϒ= (1+ (2/f))

   (4/3) – 1=(2/f)

   f=6

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13. The radius R of a nucleus of mass number A can be estimated by the formula R = (1.3 × 10^–15)A ^1/3 m. It follows that the mass density of a nucleus is of the order of: (Mprot. = Mneut =1.67 × 10^–27 kg)

1. 1) 10¹⁷ kg m^–3
2. 2) 10¹⁰ kg m^–3
3. 3) 10²⁴ kg m^–3
4. 4) 10³ kg m^–3

Solution:

1. Answer: 1

   \(\rho _{nucleus} = \frac{mass}{volume} = \frac{A}{(\frac{4}{3})\pi r_{0}^{3}A} = \frac{3}{4 \pi r_{0}^{3}}\)
   \(= 2.3 \times 10^{17} \; kg/m^{3}\)

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14. A perfectly diamagnetic sphere has a small spherical cavity at its centre, which is filled with a paramagnetic substance. The whole system is placed in a uniform magnetic field B. Then the field inside the paramagnetic substance is:

1. 1) much large than
   \(\left | \vec{B} \right |\)
   and parallel to
   \(\vec{B}\)
2. 2) zero
3. 3)
   \(\vec{B}\)
4. 4) much large than
   \(\left | \vec{B} \right |\)
   but opposite to
   \(\vec{B}\)

Solution:

1. Answer: 2

   

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15. Concentric metallic hollow spheres of radii R and 4R hold charges Q₁ and Q₂ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is:

1. 1)
   \(\frac{3Q_{2}}{4\pi \varepsilon _{0}R}\)
2. 2)
   \(\frac{3Q_{1}}{4\pi \varepsilon _{0}R}\)
3. 3)
   \(\frac{3Q_{1}}{16\pi \varepsilon _{0}R}\)
4. 4)
   \(\frac{Q_{1}}{4\pi \varepsilon _{0}R}\)

Solution:

1. Answer: 3

   

   \(\sigma =\frac{Q_{1}}{4\pi R^{2}}=\frac{Q_{2}}{4\pi 16R^{2}}\)

   16Q₁=Q₂

   \(V_{R}-V_{4R}=\frac{KQ_{1}}{R}+\frac{KQ_{2}}{4R}-\frac{KQ_{1}}{4R}-\frac{KQ_{2}}{4R}\)
   \(=\frac{3KQ_{1}}{4R}=\frac{3Q_{1}}{16\pi \varepsilon _{0}R}\)

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16. The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is

\(\vec{E}=E_{0}\hat{j}cos(\omega t-kx)\)

.The magnetic field

\(\vec{B}\)

, at the moment t = 0 is:

Solution:

1. 
2. 
3. 

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17. A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dB/dt = 0.032 Ts^–1. The induced current in the loop is close to (Resistivity of the metal wire is 1.23 × 10^–8 Ωm)

1. 1) 0.53 A
2. 2) 0.61 A
3. 3) 0.34 A
4. 4) 0.43 A

Solution:

1. Answer: 2

   

   

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18. Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to:

1. 1) 2: 1
2. 2) 1: 2
3. 3) 5: 7
4. 4) 10: 7

Solution:

1. Answer: 1

   

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19. Two light waves having the same wavelength λ in vacuum are in phase initially. Then the first wave travels a path L₁ through a medium of refractive index n₁ while the second wave travels a path of length L₂ through a medium of refractive index n₂. After this the phase difference between the two waves is:

1. 1)
   \(\frac{2\pi }{\lambda }(n_{1}L_{1}-n_{2}L_{2})\)
2. 2)
   \(\frac{2\pi }{\lambda }(\frac{L_{1}}{n_{1}}-\frac{L_{2}}{n_{2}})\)
3. 3)
   \(\frac{2\pi }{\lambda }(\frac{L_{2}}{n_{1}}-\frac{L_{1}}{n_{2}})\)
4. 4)
   \(\frac{2\pi }{\lambda }(n_{2}L_{1}-n_{1}L_{2})\)

Solution:

1. Answer: 1

   \(\lambda _{n1}=\frac{\lambda }{n_{1}}\)
   \(\lambda _{n2}=\frac{\lambda }{n_{2}}\)
   \((\Delta \phi )_{1}=\frac{2\pi}{\lambda _{n1}}L_{1}\)
   \((\Delta \phi )_{2}=\frac{2\pi}{\lambda _{n2}}L_{2}\)
   \(\Delta \phi _{1}-\Delta \phi _{2}=\frac{2\pi}{\lambda }(n_{1}L_{1}-n_{2}L_{2})\)

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20. A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. ‘m’ grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of ‘m’ is close to (Latent heat of water = 540 cal g^–1, specific heat of water = 1 cal g^–1 °C^–1)

1. 1) 2
2. 2) 2.6
3. 3) 4
4. 4) 3.2

Solution:

1. Answer: (1)

   

   Heat lost by steam = heat gained by water

   180×1×(31-25)+20×(31-25)= m×540+m×1×(100-31)

   180×6+20×6 = 540m + 100 m - 31m

   1080+120 = 640 m - 31m

   1200 = 609m

   m = 1200/609

   = 1.97

   m = 2

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21. If minimum possible work is done by a refrigerator in converting 100 grams of water at 0°C to ice, how much heat (in calories) is released to the surroundings at temperature 27°C (Latent heat of ice = 80 Cal/gram) to the nearest integer?

Solution:

1. Answer: 8791

   Q₁= m L = 8000cal

   

   Q₂ = W + Q₁

   \(C.O.P =\frac{Q_{1}}{W}=\frac{Q_{1}}{Q_{2}-Q_{1}}=\frac{T_{1}}{T_{2}-T_{1}}\)
   \(\frac{Q_{1}}{W}=\frac{273}{27}\)
   \(W = (27/273)Q_{1}\)
   \(W = (27/273)\times 80 \times 100\)
   \(Q_{2}=\frac{27}{273}\times 80\times 100+80\times 100\)

   = 8791.2 cal

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22. An massless equilateral triangle EFG of side ‘a’ (As shown in figure) has three particles of mass m situated at its vertices. The moment of inertia of the system about the each line EX perpendicular to EG in the plane of EFG is (N/20) ma² where N is an integer. The value of N is _____.

Solution:

1. 

   

   Using parallel axis theorem

   I = moment of inertia of triangle along the median + m(r²)

   r=a/2

   I = ma²+(ma²/4) = (5/4)ma²

   (5/4)ma²= (N/20)ma²

   N =25

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23. A galvanometer coil has 500 turns and each turn has an average area of 3 × 10^–4 m². If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 A is flowing through it, the strength of the field (in T) is ______.

Solution:

1. Answer: 20

   \(\tau =BINAsin\theta\)

   1.5 = B×0.5×500×3×10^-4

   B = (10000/500) = 20 Tesla

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24. A block starts moving up an inclined plane of inclination 30° with an initial velocity of v₀.It comes back to its initial position with velocity v₀/2. . The value of the coefficient of kinetic friction between the block and the inclined plane is close to 1/1000 . The nearest integer to I is____.

Solution:

1. Answer: 346

   

   a= gsin30+μ gcos 30(constant acceleration)

   therefore, by applying kinematic formula

   V₀²= 2ad

   d= V₀²/2a

   from work energy theorem

   W_f = k_f –k_i

   \(-2\mu mgcos30\frac{V_{0}^{2}}{2a}=\frac{1}{2}m\frac{V_{0}^{2}}{4} - \frac{1}{2}mV_{0}^{2}\)
   \(\frac{+2\mu mgcos30}{a}=+\frac{3}{4}\)

   8μgcos30 = 3gsin30+3 μgcos30

   5μgcos30 = 3gsin30

   \(\mu =\frac{3tan30}{5}=\frac{\sqrt{3}}{5}\)
   \(\frac{\sqrt{3}}{5}=\frac{I}{1000}\)

   I = 346

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25. When an object is kept at a distance of 30 cm from a concave mirror, the image is formed at a distance of 10 cm from the mirror. If the object is moved with a speed of 9 cms^–1, the speed (in cms^–1) with which image moves at that instant is ____.

Solution:

1. Answer: 1

   

   \(V_{1}=\frac{v^{2}}{u^{2}}v_{0}\)
   \(V_{1}=\frac{10\times 10}{30\times 30}\times 9\)

   V₁ = 1cm/sec

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Video Lessons - September 3 Shift 2 Physics

JEE Main 2020 Physics Paper With Solutions Shift 2 September 3

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