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Question 1. Three identical solid spheres each have mass ‘m’ and diameter ‘d’ are touching each other as shown in the figure. Calculate ratio of moment of inertia about the axis perpendicular to plane of paper and passing through point P and B as shown in the figure. Given P is centroid of the triangle.
Solution:
Distance of centroid from (point P) centre of sphere = (2/3)×(√3d/2) = d/√3
By Parallel axis theorem,
Moment of Inertia about P = 3[ (2/5)M(d/2)^{2} + M(d/√3)^{2}] = (13/10) Md^{2}
Moment of Inertia about B = 2[ (2/5)M(d/2)^{2} + M(d)^{2}] + (2/5)M(d/2)^{2} = (23/10)Md^{2}
Now ratio = 13/23
Answer:(a)
Question 2. A solid sphere having a radius R and uniform charge density ρ. If a sphere of radius R/2 is carved out of it as shown in the figure. Find the ratio of the magnitude of electric field at point A and B
Solution:
For solid sphere,
Field inside sphere, E = ρr/3ε_{0}
And field outside sphere, E = ρr^{3}/3r^{2}ε_{0}
where r is distance from centre and R is radius of sphere.
Electric field at A due to sphere of radius R (sphere 1) is zero and therefore, net electric field will be because of sphere of radius R/2 (sphere 2) having charge density (-ρ)
E_{A }= - ρR/2(3 ε_{0})
ǀE_{A}ǀ = ρR/6 ε_{0}
Similarly, Electric field at point B = E_{B} = E_{1B }+ E_{2B}
E_{1B} = Electric Field due to solid sphere of radius R = ρr/3ε_{0}
E_{2B} = Electric Field due to solid sphere of radius R/2 which having charge density (-ρ)
=
= -ρR/54ε_{0}
E_{B} = E_{1A} + E_{2A} = (ρR/3ε_{0}) - (ρR/54ε_{0})
= 17 ρR/54ε_{0}
Answer: (c)
Question 3. Consider an infinitely long current carrying cylindrical straight wire having radius 'a'. Then the ratio of magnetic field due to wire at distance a/3 and 2a, respectively from axis of wire is
Solution:
B_{A }=
B_{B }= μ_{0}i(2a)/2(2a)^{2} = μ_{0}i /4a
B_{A}/B_{B }= 4/6 = 2/3
Answer: (b)
Question 4. Particle moves from point 𝐴 to point 𝐵 along the line shown in figure under the action of force
Solution:
= ½ + ½
= 1 J
Answer: (a)
Question 5. For the given P-V graph of an ideal gas, chose the correct V-T graph. Process BC is adiabatic. (Graphs are schematic and not to scale)
Solution:
For process 3 - 1; Volume is constant;
For process 1 - 2, PV^{γ} Constant , and PV = nRT therefore TV^{γ-1} = Constant ;
therefore as V increases T decreases and also relation is non linear, so curve will not be a straight line.
For process 2-3; pressure is constant , therefore V = kT
From above, correct answer is option a.
Answer: (a)
Question 6.An electric dipole of moment
Solution:
The electric dipole of moment p = q.a where a is distance between charge.
Electric field E at position r is given by
along tangential direction , where
Since already in question,
This means field is along tangential direction and dipole is also perpendicular to radius vector.
Since electric field and dipole are along same line, we can write
From option, on putting = -1× 10^{29} , we get,
Answer: (c)
Question 7. A body A of mass m is revolving around a planet in a circular orbit of radius R. At the instant the particle B has velocity , another particle of mass m/2 moving at velocity of V/2, collides perfectly inelastically with the first particle. Then, the combined body
Solution:
(m/2)(V/2) + mV = [m + (m/2)]V_{f}
V_{f} = 5V/6
Where V is orbital velocity
Escape velocity will be √2V and at velocity less than escape velocity but greater than orbital velocity (V), the path will be elliptical. At orbital velocity (𝑉), path will be circular. At velocity less than orbital velocity path will remain part of ellipse and it will either orbit in elliptical path whose length of semi major axis will be less than radius of circular orbit or start falling down and collide with the planet but it will not fall vertically down as path will remain part of ellipse. Hence the resultant mass will start moving in an elliptical orbit around the planet.
Answer: (d)
Question 8. Two particles of equal mass m have respective initial velocities
Solution:
Let v_{1} and v_{2} be the final velocities after collision in x and y direction respectively.
Conserving linear momentum
By equating i and j
v_{1 }= 3u/4
v_{2} = u/4
Initial K.E = (mv^{2}/2) + (m/2)×(u/√2)^{2} = 3mu^{2}/4
Final K.E = (2m/2)×(u√10/4)^{2} = 5mu^{2}/8
Change in K.E = (3mu^{2}/4) - (5mu^{2}/8) = mu^{2}/8
Answer: (d)
Question 9.Three harmonic waves of same frequency (v) and intensity (I_{0}) having initial phase angles 0, /4, -/4 rad respectively. When they are superimposed, the resultant intensity is close to;
Solution:
Amplitudes can be added using vector addition A_{resultant }=(√2+1)A
Since, I ∝A^{2}, Where 𝐼 is intensity.
Therefore, I_{res }= (√2+1)^{2} 𝐼_{0} = 5.8 𝐼_{0} (approx.)
Answer: (a)
Question 10. An ideal liquid (water) flowing through a tube of non-uniform cross-sectional area, where area at A and B are 40 cm^{2 }and 20 cm^{2 }respectively. If pressure difference between A & B is 700 N/m^{2}, then volume flow rate is (density of water = 1000 kgm^{−3})
Solution:
Using equation of continuity
V_{A} ×Area_{A}= V_{B} ×Area_{B }
40V_{A}=20V_{B}
2V_{A} = V_{B}
Using Bernoulli’s equation
P_{A }+ ½ V_{A}^{2} = 𝑃_{𝐵} + ½ 𝜌V_{B}^{2}
P_{A }- P_{B} = ½ 𝜌 (V_{B}^{2} - V_{A}^{2} )
P = ½ 1000( V_{B}^{2} - V_{B}^{2}/4
= 500×3V_{B}^{2}/4
V_{B }= √(P×4/1500)
= √(700×4/1500)
= √(28/15)m/s
Volume flow rate = V_{B}×Area_{B}
= 20 × 100 ×√(28/15)cm^{3}/s
= 2732.5 cm^{3}/s
So, answer comes nearly 2720 cm^{3}/s
Answer: (a)
Question 11. A screw gauge advances by 3 mm on main scale in 6 rotations. There are 50 divisions on circular scale. Find least count of screw gauge?
Solution:
Pitch = 3/6 = 0.5 mm
Least count =Pitch / number of divisions
= 0.5mm/ 50
= 1100mm
= 0.01 mm
= 0.001 cm
Answer: (b)
Question 12. A telescope of aperture diameter 5 m is used to observe the moon from the earth. Distance between the moon and earth is 4 × 10^{5} km. The minimum distance between two points on the moon's surface which can be resolved using this telescope is close to (Wavelength of light is 5500 Å )
Solution:
Minimum angle for clear resolution,
= 1.22 /a
d = 1.22 (/a)d
r = 4×10^{8 }m
Aperture, a = 5
= 5500 A^{0}
Distance = r d
= 4×10^{8 }×1.22 (5500×10^{-10}/5)
= 53.68
Nearest option is 60m.
Answer: (a)
Question 13. Radiation with wavelength 6561 Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3×10^{-4} T. If the radius of largest circular path followed by electron is 10 mm, the work function of metal is close to;
Solution:
From photoelectric equation,
hc/λ=W+K.E _{max }
Where, hc = 12400 eV Å
⇒12400/6561=W+K.E _{max }
⇒1.89 eV=W+K.E_{ max }…(1)
Radius of charged particle moving in a magnetic field is given by
r = mv/qB
½ mv^{2} = K.E_{max} = eV
⇒V=0.8 V
So, K.E_{max}=0.8 eV
Substituting in (1),
1.89 =W+0.8
i.e. W=1.1 eV (approx)
Answer: (c)
Question 14. Kinetic energy of the particle is 𝐸 and it's de–Broglie wavelength is 𝜆. On increasing its K.E by Δ𝐸, it's new de–Broglie wavelength becomes /2. Then Δ𝐸 is
Solution:
𝜆 = ℎ/𝑚𝑣
= ℎ√(2𝑚(𝐾𝐸))
⇒𝜆 ∝ 1√𝐾𝐸
𝜆/𝜆/2 = √(𝐾𝐸_{𝑓} / 𝐾𝐸_{𝑖}
4𝐾𝐸𝑖 = 𝐾𝐸𝑓
⇒Δ𝐸 = 𝐾𝐸_{𝑓} -𝐾𝐸_{𝑖}
= 4𝐾𝐸_{𝑖}−𝐾𝐸_{𝑖}
= 3𝐾𝐸_{𝑖}
= 3𝐸
Answer: (a)
Question 15. A quantity is given by f = √(hc^{5}/G) ^{where c is speed of light, G is universal gravitational constant and h is the Planck’s constant. Dimension of f is that of;}
Solution:
hc = E
[E] = [ML^{2}T^{-2}]
Therefore [hc] = [ML^{3}T^{-2}]
[c] = [LT^{-1}]
[G] = [M^{-1} L^{3}T^{-2}]
√(hc^{5}/G) ^{= [ML2T-2]}
The above dimension is of energy.
Answer: (b)
Question 16. A vessel of depth 2h is half filled with a liquid of refractive index √2 in upper half and with a liquid of refractive index 2√2 in lower half. The liquids are immiscible. The apparent depth of inner surface of the bottom of the vessel will be;
Solution:
Assume, air is present outside container
Apparent height as seen from liquid 1 (having refractive index 𝜇_{1} = √2 ) to liquid 2 (refractive index 𝜇_{2} = 2√2)
D = ℎ𝜇_{1}/𝜇_{2} = ℎ/2
Now, actual height perceived from air, h + ℎ/2 = 3ℎ/2
Therefore, apparent depth of bottom surface of the container (apparent depth as seen from air (having refractive index 𝜇_{0} = 1) to liquid 1(having refractive index 𝜇_{1}= √2 )
= (3ℎ/2)×(𝜇_{0}/ 𝜇_{1} )
= (3ℎ/2)×(1/√2)
= 3ℎ/2√2
= 3√2ℎ/4
Answer: (a)
Question 17. In the given circuit diagram, a wire is joining point B & C. Find the current in this wire;
Solution:
Since resistance 1 Ω and 4 Ω are in parallel
∴𝑅’ = 4×1/(4+1) = 4/5
Similarly we can find equivalent resistance (𝑅′′) for resistances 2 Ω and 3 Ω
⇒𝑅’’ = 6/5
And 𝑅′ and 𝑅′′ are in series
∴ 𝑅_{𝑒𝑓𝑓 }= (4/5) + (6/5) = 2 Ω
So total current flowing in the circuit ‘𝑖’ can be given as
𝑖 = 𝑉/𝑅_{𝑒𝑓𝑓 }= 20/2 = 10A
Current will distribute in ratio opposite to resistance.
So, distribution will be as
So current in the branch BC will be 𝐼 = (4𝑖/5) - (3𝑖/ 5)
= 𝑖/5
= 10/ 5
= 2 A
Answer (b)
Question 18. Two plane electromagnetic waves are moving in vacuum in whose electric field vectors are given by
Solution:
Given that the magnetic field vectors are:
Since, the variation of vectors E_{1} and E_{2} are along x and y respectively.
Therefore, direction of propagation of vectors E_{1} and E_{2} will be along x and y respectively. Since
So, the magnetic field vectors of the electromagnetic wave are given by
Answer (b)
Question 19.Consider two ideal diatomic gases A and B at some temperature T. Molecules of the gas A are rigid, and have a mass m. Molecules of the gas B have an additional vibration mode and have a mass m/4 . The ratio of molar specific heat at constant volume of gas A and B is;
Solution:
We know that,
Molar heat capacity at constant volume, 𝐶_{𝑉 }= 𝑓𝑅/2 (Where f is degree of freedom)
Since, A is diatomic and rigid, degree of freedom for A is 5
Therefore, Molar heat capacity of A at constant volume (C_{v})_{A } = 5R/2
Since, B is diatomic and has extra degree of freedom because of vibration; degree of freedom for B is 5 + 2×1= 7 (1 vibration for each atom).
Therefore, Molar heat capacity of B at constant volume (C_{V})_{B} = 7R/2
Ratio of molar specific heat of A and B = (C_{V})_{A }/ (C_{V})_{B} = 5/7
Answer: (d)
Question 20. A charged particle of mass ‘m’ and charge ‘q’ is moving under the influence of uniform electric field
A. Magnitude of electric field
B. Rate of work done by electric field at P is
C. Rate of work done by both fields at Q is zero.
D. The difference between the magnitude of angular momentum of the particle at P and Q is 2mva.
Solution:
Considering statement A
Let, Net work done by magnetic field be W_{B} and net work done by electric field be W_{E}.
By Work-Energy theorem
W_{𝐵}+𝑊_{𝐸 }= (1/2)𝑚(2𝑣)^{2 }− (1/2) mv^{2}
⇒ 0+𝑞𝐸_{𝑜}2𝑎 = (3/2) 𝑚𝑣^{2}
𝐸_{𝑜} = (3/4) 𝑚𝑣^{2 }/𝑞𝑎
So, statement A is correct
Now, considering statement B
Rate of work done at P = Power of electric force
= 𝑞𝐸_{𝑜}𝑣
= (3/4) 𝑚𝑣^{3 }/𝑎
So, statement B is correct
Now, considering statement C
At Q,
Vector E perpendicular to vector v.
Vector B perpendicular to vector v.
So, dw/dt = 0 for both forces
So, statement C is correct
Now, considering statement D
Angular momentum should be defined about a point which is not given in question but let’s find angular momentum about origin.
Change in magnitude of angular momentum of the particle at P and Q about origin
Hence, Δ𝐿 = 3mva
So, statement D is wrong.
Answer: (b)
Question 21. In a fluorescent lamp choke (a small transformer) 100 V of reversible voltage is produced when choke changes current in from 0.25 A to 0 A in 0.025 ms. The self-inductance of choke (in mH) is estimated to be;
Solution:
Fluorescent lamp choke will behave as an inductor
By using faraday law to write induced emf,
∈ -L dI/dt = 0
100 = L(0.25)×10^{3}/0.025
L = 100×10^{-4} H
= 10 mH
Answer: (10)
Question 22.A wire of length l = 0.3 m and area of cross section 10^{–2} cm^{2} and breaking stress 4.8×10^{7} N/m^{2} is attached with block of mass 10 kg. Find the maximum possible value of angular velocity (𝑟𝑎𝑑/𝑠) with which block can be moved in a circle with string fixed at one end.
Solution:
Breaking stress 𝜎 = T/A
T = 𝑚𝜔^{2}l
⇒ 𝜎 = 𝑚𝜔^{2}l/𝐴
⇒ 𝜔^{2 }= A/mI
= (4.8×10^{7}×10^{−6})/ 10×0.3 = 16
⇒𝜔 = 4 𝑟𝑎𝑑/𝑠
Answer: (4)
Question 23. The distance x covered by a particle in one dimension motion varies as with time 𝑡 as x^{2} = at^{2} + 2bt + c, where a, b, c are constants. Acceleration of particle depend on x as x^{–n} , the value of n is;
Solution:
Let v be velocity, be the acceleration then,
x^{2} = at^{2} + 2bt + c
2xv = 2at + 2b
xv = at + b ..(i)
v = (at + b)/x
Now, differentiating equation (i)
v^{2} + ax = a
ax = a - (at+b)^{2}/x^{2}
= a(at^{2} +2bt+c)-(at+b^{2})/x^{3}
= ac - b^{2}/x^{3}
∝ x^{-3}
Answer: (3)
Question 24. A rod of length 1 m pivoted at one end is released from rest when it makes 30° from the horizontal as shown in the figure below.
If of rod is √𝑛 at the moment it hits the ground, then find n.
Solution:
By using conservation of energy,
mg (𝑙/2)sin30^{∘ }= (½)(ml^{2}/3)^{2}
On solving ^{2} = 15
𝜔=√15
Therefore, n = 15.
Answer: (15)
Question 25. In the given circuit both diodes are ideal having zero forward resistance and built-in potential of 0.7 V. Find the potential of point E in volts.
Solution:
We have to apply nodal analysis on both left and right side and check what can be voltage at E. For nodal analysis, voltage at B, F and G will be 0 volts and voltage at A will be 12.7 volt and voltage at H will be 4 volts.
If, we apply Nodal from right side, voltage at E will be 12 volt (diode between A and E will be forward biased). Now voltage at E is 12 volt and voltage at H is 4 volt and since, diode between E and H is reversed biased and any difference of voltage is possible across reverse biased. So, this is possible.
If, we apply Nodal from left side, voltage at E will be 3.3 volt (diode between E and H will be forward biased). Now voltage at E is 3.3 volt and voltage at A is 12 volt and since, diode between E and A is forward biased and in forward biased difference of voltage of 0.7 volt is allowable. So, this case is not possible. Therefore current will also not flow through GH. Hence, V_{E} = 12 V.
Answer (12)