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JEE Main 2018 April 15 Shift 1 Paper & Solutions - JEE MCQ


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30 Questions MCQ Test - JEE Main 2018 April 15 Shift 1 Paper & Solutions

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JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 1

A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius R/2, and the other mass, in a circular orbit of radius 3R/2. The difference between the final and initial total energies is 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 1






JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 2

In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25 cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 2

Given,  
5 complete rotations of screw = 0.25cm
So 1 rotation of screw = 0.05
Hence, 1 main scale division = 0.05 cm
and 1 circular scale = 0.05/100 division = 5 × 10−4 cm.
Now Reading is 4 main scale and 30 circular scale divisions 
So , thickness  = 4 × 0.05 + 30 × 5 × 10−4
= 0.2150 cm.

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JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 3

The B-H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 1000 turns/ cm. The current that should be passed in the solenoid to demagnetise the ferromagnet completely is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 3

From given diagram
Coercivity = −100 (A/m)
Hence 
Reverse magnetic field required to demagnetize the substance 
= µo H
= µo 100
so µo 100 = µo.N.I

I = 1mA

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 4

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ1 and λ2, their de Broglie wavelength in the frame of reference attached to their centre of mass is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 4




JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 5

Light of wavelength 550 nm falls normally on a slit of width 22.0 x 10-5cm. The angular position of the second minima from the central maximum will be (in radians): 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 5

b sin θ = nλ
b sin θ = 2λ

sin θ = 1/2
θ = π/6

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 6

A uniform rod AB is suspended from a point X, at a variable distance x from A, as shown. To make the rod horizontal, a mass in is suspended from its end A. A set of (m, x) values is recorded. The appropriate variables that give a straight line, when plotted, are: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 6


= y = mx + c

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 7

A Helmholtz coil has a pair of loops, each with N turns and radius R. They are placed coaxially at distance R and the same current I flows through the loops in the same direction. The magnitude of magnetic field at P, midway between the centres A and C, is given by [Refer to figure given below]: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 7



JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 8

An ideal capacitor of capacitance 0.2 µF is charged to a potential difference of 10 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self-inductance 0.5 mH. The current at a time when the potential difference across the capacitor is 5 V, is:  

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 8

Qo = 0.2 × 10 µC = 2µC

Vo = 10V
Ei of capacitor = 1/2 × 0.2 µf × (10v)2
= 10µJ
= Ef of capacitor = 1/2 × 0.2µf × (5v)2
= 2.5 µj
⇒ Einductor  = 7.5 µJ = 1/2 Li2
⇒ 7.5 × 10−6 = 1/2 × 0.5 × 10−3 × i2
⇒ 30 × 10−3 = i2
⇒ i = √3/10 = 0.17 A

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 9

Take the mean distance of the moon and the sun from the earth to be 0.4 x 106 km and 150 x 106 km respectively. Their masses are 8 x 1022 kg and 2 x 1030 kg respectively. The radius of the earth is 6400 km. Let ΔF1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2 be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1/ΔF2 is:

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 9






=  (375)3 (4 × 10−8)
= (0.37 × 103)3(4 × 10−8)
= 1.64

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 10

An automobile, travelling at 40 km/h, can be stopped at a distance of 40m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding):  

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 10

u = 40 Km/hr
μ = 100/9 m/s
V2 − u2 = 2a × 40
a = −1.54 m  m/s2
Now. 

S = 160m

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 11

A charge Q is placed at a distance a/2 above the centre of the square surface of edge a as shown in the figure  

The electric flux through the square surface is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 11

Through whole cube = Q/∈o
Through one face = Q/6∈o

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 12

A force of 40 N acts on a point B at the end of an L-shaped object, as shown in the figure. The angle θ that will produce maximum moment of the force about point A is given by 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 12


τA = f cosθ(4) + Fsinθ(2)
τA = 2F(sinθ + 2cosθ
A/dθ = 2F[cosθ − 2sinθ] = 0
Tan θ = 1/2

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 13

The equivalent capacitance between A and B in the circuit given below, is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 13



JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 14

A monochromatic beam of light has a frequency v = 3/2π × 1012 Hz and is propagating along the direction  It is polarized along the k̂ direction. The acceptable form for the magnetic field is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 14

Wave propagation vector should be along 

So, b is the only option which satisfies the above condition.

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 15

A particle is oscillating on the X-axis with an amplitude 2 cm about the point x0 = 10 cm, with a frequency ω. A concave mirror of focal length 5 cm is placed at the origin (see figure). 

Identify the correct statements.
(A)The image executes periodic motion.
(B)The image executes non-periodic motion.
(C)The turning points of the image are asymmetric w.r.t. the image of the point at x =10 cm.
(D)The distance between the turning points of the oscillation of the image is 100/21 Cm.

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 15



JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 16

A tuning fork vibrates with frequency 256 Hz and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is 340 ms-1

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 16

Third normal mode of frequency in open pipe, 
f = 3vs/2l
Where, Vs = 340m/s 
Get L = 2m
Or
L = 200 cm

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 17

A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ρ1 and ρ21> ρ2), fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 17


h1ρ1 − 1 = h2ρ2
(R Cos θ + R sinθ)ρ2 = (R cosθ − R sinθ)ρ1
1 + ρ2) Sin θ = (ρ1 − ρ2 )Cosθ

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 18

In a common emitter configuration with suitable bias, it is given that RL is the load resistance and RBE is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by:  
β is current gain, IB, IC and IE are respectively base, collector and emitter currents. 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 18

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 19

A given object takes n times more time to slide down a 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline Is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 19



So

Sin θ = n2 sinθ − µn2 cosθ
1 = n2 − µn2
1 − n2 = −µn2
n2 − 1 = −µn2
µ = 1 − 1/n2

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 20

One mole of an ideal monatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 27°C. The work done on the gas will be: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 20



= 300R ln 2

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 21

A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 21

Given, 
Q2 = 500 cal.
ηCarnot = 1 - T2/T1
= 1/6
and (COP)HP = 1/ηCarnot = 6
(Cop)HP − (Cop)Ref = 1
So, (cop)Ref = 5 = Q2/w
so w = 100 cal
Or w = 420 J

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 22

In the given circuit all resistances are of value R ohm each. The equivalent resistance between A and B is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 22


Due to short circuit current will flow along
B-C-D-E-F-A
So
R eq = R + R 
R eq = 2R 
 

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 23

A solution containing active cobalt 60/27 Co having activity of 0.8 µCi and decay constant λ is injected in an animal's body. If 1 cm3 of blood is drawn from the animal's body after 10 hrs of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body? (1 Ci = 3.7 x 1010 decays per second and at t = 10 hrs e –λt = 0.84) 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 23

(dn/dt) = −λNo
0.8 µ Ci = −λNo
N1 = No(0.84)
v → N1
v cm3 → N1/v


v = 3.7/5 × 0.84 × 0.84 × 104 cm3
= 0.5 × 104 cm3 = 5 × 103cm3 = 5 lit

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 24

In a meter bridge, as shown in the figure, it is given that resistance Y =12.5 Ω and that the balance is obtained at a distance 39.5 cm from end A (by Jockey J). After interchanging the resistances X and Y, a new balance point is found at a distance I2 from end A. What are the values of X and l2

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 24

For balanced wheat stone bridge 
x(100 − l1) = y × l1
So x(100 − 39.5) = 12.5(39.5)
x = 8.16 Ω
If x and y inter changed
y(100 − l2) = xl2
12.5(100 − l2) = 8.16 l2
get l2 = 60.5 cm
 

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 25

The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 25

E1 = ionization energy of ionized He
E2 = 2. 2E2
E1 = 13.6 ev
E2 = 13.6/2.2 = 6.18 ev
Total = E1 + E2 = 13.6 ev + 6.18 ev
= 20 ev 

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 26

The number of amplitude modulated broadcast stations that can be accomodated in a 300 kHz band width for the highest modulating frequency 15 kHz will be: 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 26

The station will require a band width of 30 khz 
So
No. of stations = 300/30
= 10

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 27

The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively. 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 27


s1 = 1/2 × 3 × (15)2 = 1.5 × 225
= 225 + 112.5
= 337.5 m
s2 = v2t
s2 = 15 × 30 = 450 m ⟹ s2 − s1 = 112.5 m
For catching up ⟹ s1 = s2
30t = 1/2 × 3 × t2
20 = t 

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 28

A body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x direction about its equilibrium position, taken to be at x = 0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct? 

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 28

Energy at the extreme 
= 1/2 kA2 = T. E
After switching on electric field 
New mean position ⟹ kxo = qE 
xo = qE/k
So entreme position also shifts by qE/k

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 29

A planoconvex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from left to right as shown in Fig-A. 
If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig. B, it acts like an optical system of focal length 10 cm. The refractive index of the material of lens is:  

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 29

When plane surface is silvered 
Focal length f1 = 
and f1 = 28 cm
When curved surface is silvered 
Focal length f2 = R/2μ ... (ii)


µ = 1.55

JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 30

The relative error in the determination of the surface area of a sphere is α. Then the relative error in the determination of its volume is:  

Detailed Solution for JEE Main 2018 April 15 Shift 1 Paper & Solutions - Question 30

Area of sphere (A) = 4πR2
taking log
ln A = In(4π) + 2In(R)
differentiating both sides



now, similarly

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