JEE Main is an examination conducted for admission into some of the reputed colleges in India. Students can find the solved paper of JEE Main 2018 Physics Jan 10 Shift 1 on this page. The subject experts at BYJU’S have solved all these questions. These questions are answered in the easiest possible method. Students can also download these solutions from our website in PDF format. Test
1. It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional oss of its energy is p_{d}; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p_{C}. The values of p_{d} and p_{c} are respectively:
Solution:
L.M.C
mV_{1} + 2m V_{2} = mV_{0}
V_{1} + 2V_{2} = V_{0} …(1)
V_{2} – V_{1} = V_{0}…(2)
On solving
V_{2} = 2V_{0}/3 and V_{1}= V_{0}/3
Final KE of neutron= (1/2)mV_{1}^{2}= (1/2)m(V_{0}/3)^{2}
=
Loss of K.E =
Fractional loss P_{d} = (8/9) = 0.89
Similarly collision between N and C
mV_{1} + 12 m.V_{2} = mV_{0}
V_{1} + 12V_{2} = V_{0} …(1)
V_{1}-V_{2} = V_{0} …(2)
On Solving
V_{2} = 2V_{0}/13
And V_{1} = 11V_{0}/13
Final K.E =
Loss of KE =
Fractional loss P_{c} = (48/169) = 0.28
2. The mass of a hydrogen molecule is 3.32 x 10^{-27} kg. If 10^{23} hydrogen molecules strike, per second, a fixed wall of area 2 cm^{2} at an angle of 45^{0} to the normal, and rebound elastically with a speed of 10^{3} m/s, then the pressure on the wall is nearly:
Solution:
Change in momentum of one molecule
Force
Where n -> no. of molecules incident per unit time
Pressure P = Force/Area
Answer: (a)
3. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A mass less piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere,(dr/r) is:
Solution:
Bulk modulus
From eq.(1) and (2)
Answer: (a)
4. Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The internal resistances of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between.
Solution:
5. A particle is moving in a circular path of radius a under the action of an attractive potential U = -k/2r^{2}. Its total energy is
Solution:
F= -k/r^{3}
Centripetal force
Total Energy E = K + U = 0
Answer: (a)
6. Two masses m_{1} = 5 kg and m_{2} = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is:
Solution:
At equilibrium f_{r} = T = m_{1}g
f_{rmax} =(m_{2} + m)g
(10+m)g =5g
10 +m = 5/0.15
10+m = 100/3
m = (70/3)kg = 23.3 kg.
The minimum weight in the options is 27.3 kg.
Answer: (d)
7. If the series limit frequency of the Lyman series is v_{L}, then the series limit frequency of the Pfund series is:
Solution:
Series limit is
Lyman : ∞
→ 1
P fund :∞
→ 5
v_{Lyman} = RC
v_{Pfund} = RC/25
8. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. the intensity of light beyond B is found to be I/2. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I/8. The angle between polarizer A and C is:
Solution:
Unpolarized light passes the A
I _{after A} = I/2
I _{after B} = I/2 (given)
The angle between A and B is 90^{0}
Let A and C have angle θ
I _{after c} = (I/2)cos^{2} θ
I _{after B} = (I/2)cos^{2} θ cos^{2}(90 - θ )= I/8, therefore, [cos θ sin θ]^{2}= ¼
Sin 2 θ = 1
Therefore, θ = 45^{0}
Answer: (a)
9. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let λ_{n}, λ_{g} be the de Broglie wavelength of the electron in the n^{th} state and the ground state respectively. Let A_{n} be the wavelength of the emitted photon in the transition from the n^{th} state to the ground state. For large n, (A, B are constants)
Solution:
mvr = (nh/2 π)
λ_{de Broglie }= (h/p) = (2 πr/n)
λ_{n }=(2 πr_{n}/n)
_{}
Answer: (c)
10. The reading of the ammeter for a silicon diode in the given circuit is:
Solution:
Knowledge based
Si diode has forward bias resistance
200 at 2 V
400 at 1 V
⇒ here R_{fb} < 200 … at 3V
I = (V-ΔV)/R = (3-0.7)/200 = 2.3/200 = 11.5 mA
Answer: (a)
11. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii r_{e}, r_{p}, r_{α} respectively in a uniform magnetic field B. The relation between r_{e}, r_{p}, r_{α} is:
Solution:
r = mv/qB = p/qB
K = (1/2)mv^{2}-------same
= (p^{2}/2m)
Therefore, p ∞ √m
B same
r∝(p/q) or √m/q
q_{p} =q_{e}
m_{p} = 183 m_{e}
q_{α} = 2 q_{p}
m _{α}= 4 m_{p}
Mass of electron is least and charge qe = e
So, r _{e} < r _{p} = r
Answer: (d)
12. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material of dielectric constant k = 5/3 is inserted between the plates, the magnitude of the induced charge will be:
Solution:
Battery remains connected as dielectric is introduced
So E, V unchanged
q_{0} = C_{0}V
q = kC_{0}V
Induced charge q = q q_{0}
= C_{0}V(k-1)
= 90 x 10^{-12} x 20((5/3)-1) = 1.2 nC
Answer: (c)
13. For an RLC circuit driven with voltage of amplitude V_{m} and frequency
Solution:
Quantity factor
Answer: (c)
14. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized for transmission. How
Solution:
No of channels =(carrier frequencyx0.1)/channel bandwidth
= (0.1 x 10 x 10^{9})/(5 x 10^{3})= 2 x 10^{5}
Answer: (a)
15. A granite rod of 60 cm length is clamed at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 x 10^{3} kg/m^{3} and its Young’s modulus is 9.27 x 10^{10} Pa. What will be the fundamental frequency of the longitudinal vibrations?
Solution:
On substituting the value, we get the required answer.
Answer: (c)
16. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is:
Solution:
M.I. about origin
Answer: (b)
17. Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities + σ, -σ and +σ respectively. The potential of shell B is:
Solution:
Answer: (d)
18. In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.
Solution:
When switch s is opened
E = λL_{1}----------(I)
Where λ is potential gradient
when switch is closed E - ir = λL_{2} … (II)
Replace i = E/(R+r)
Answer: (d)
19. An EM wave from air enters a medium. The electric fields are
Solution:
Answer: (a)
20. The angular width of the central maximum in a single slit diffraction pattern is 60^{0}. The width ofthe slit is 1μm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e., distance between the cenrtres of each slit.)
Solution:
2α = 60^{0}
a=1 μm
D = 50 cm
(Cond. for minima/Path diff Δx = a sin θ = n λ)
a=1 μm and θ =30^{0} and n=1
λ = 0.5 μm
If same setup is used for YDSE
Fringe width β = λD/d = 1cm
d=0.5 x 10^{-6} x 0.5/0.01 = 25 μm
d = 25 μm
Answer: (c)
21. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10^{12}/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avagadro number = 6.02 x 10^{23} gm mole^{-1}
Solution:
Frequency
K =m(2 πf))^{2}
Mass of 1 atom
m=18 x 10^{-26} Kg
k = 18 x 10^{-26}(2π x 10^{12})^{2} = 4 π^{2} x 18 x 10^{-2}
k = 7.1 N/m (π^{2} =10)
Answer: (d)
22. From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:
Solution:
Answer: (c)
23. In a collinear collision, a particle with an initial speed v_{0} strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is:
Solution:
L.M.C
mv_{1}+mv_{2}= mv_{0}
V_{1}+V_{2}=V_{0}----------(I)
Initial KE = (1/2)mV_{0}^{2}
Final KE = (1/2)mV_{1}^{2 }+ (1/2)mV_{2}^{2}
final kE = (3/2)initial kE
(I)^{2} – (II)
2V_{1}V_{2} = -(1/2) V_{0}^{2}----------(III)
(V_{1}-V_{2})^{2} = (V_{1}+V_{2})^{2} - 4V_{1}V_{2}
(V_{1}-V_{2})^{2} = V_{0}^{2}+V_{0}^{2} =2V_{0}^{2}
Answer: (d)
24. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B_{1}. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is B_{2}. The ratio B_{1}/B_{2} is:
Solution:
Dipole moment μ = niA
μ = ix πR^{2}
If dipole moment is doubled keeping current const. R_{2} = R_{1}√2
Magnetic Field at center of loop
B= μ_{0}i/2R
Answer: (a)
25. The density of the material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:
Solution:
= 1.5 + 3 x 1 = 4.5
Answer: (a)
26. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k Ω. How much was the resistance on the left slot before interchanging the resistances?
Solution:
Let balancing length be L
If R_{1} and R_{2} are interchanged balancing length is (L - 10)
From equ (1) and (2)
Hence, L^{2}-10L = 110 x 100 +L^{2} – 210L
Therefore, 200L = 110 x 100
L = 55 cm
R_{1} + R_{2} = 1000 Ω
On Solving
R_{1} = 550 Ω
R_{2} = 450 Ω
Answer: (a)
27. In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t and i = 20 sin(30 t – (π/4)). In one cycle of a.c., the average power consumed by the circuit and the wattles current are, respectively:
Solution:
Answer: (d)
28. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.
Solution:
If Vs t is a straight line with -ve slope acc = - ve const.
Displacement V_{s} time is a parabola opening downward
Only incorrect option is (4)
Correct distance Vs time graph is
Answer: (4)
29. Two moles of an ideal monoatomic gas occupies a volume V at 27^{0}C. The gas expands adiabatically to a volume 2 V. Calculate (a) the final temperature of the gas and (b) change in its internal energy.
Solution:
ΔU = -2.76 kJ
Answer: (1)
30. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then:
Solution:
T ∝ R^{(n+1)/2}
Answer: (a)