PAPER-1 (B.E./B.TECH.)
JEE MAIN 2019 Computer Based Test
Solutions of Memory Based Questions
Date: 10th April 2019 (Shift-2)
Time: 02:30 P.M. to 05:30 P.M.
Durations: 3 Hours | Max. Marks: 360
Subject: Physics
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JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS 1. A current of 10 amp. flows along side of an equilateral triangle of side '1m'. What is the strength of magnetic field at the centroid of triangle? (µo is magnetic permeability of space)
(A) 45
πµo T
(B) 15
πµo T
(C) 15√3
πµo T
(D) 5√3
πµo T
Solution: (A)
The perpendicular distance of centroid from any side,
1
2√3 m
Magnetic field due to one side.
B1 =µoI
4πr[cosθ1 + cosθ2]
=µ0(10)
4πr (1
2√3)[cos30° + cos30°]
=5√3
π µ0 [
√3
2+
√3
2] =
15µ0π
Net magnetic field B = 3B1 =45
πµ0 T
2. Find the maximum value of force that can be applied on 3kg block for which 1kg block does not slide.
(A) 6 N (B) 8 N (C) 10 N (D) 16 N Solution: (D)
Maximum possible acceleration of 1 kg block a =µmg
m= µg = 2
m
s2
For 1 g block not to slide the maximum acceleration of both blocks should be
Fmax − µmg = mamax Fmax − 0.2 × 4 × 10 = 4 × 2 Fmax − 8 = 8
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS Fmax = 16 N
3. A beam of light of intensity 25 w
cm2 has photons of energy 1 mJ , Find the number of photons per second incident on
the surface of area 25 cm2
(A) 6.25 × 105 s−1 (B) 8.25 × 105 s−1 (C) 6.25 × 104 s−1
(D) 5.25 × 105 s−1 Solution: (A) Power = no of photons per sencond x energy of each photon I. A.= n. Eλ
n =IA
Eλ=
25 × 25
10−3= 6.25 × 105 s−1
4. A particle of mass 'm' and charge 'q' is suspended in an uniform electric field 'E' with the help of an insulating wire of length 'l' as shown in figure. Find the time period of oscillation.
(A) 2π√l
√g2+(qE
m)2
(B) 2π√l
√g+qE
m
(C) 2π√l
√g−qE
m
(D) 2π√l
g
Solution: (A)
Downward Force = √(qE)2 + (mg)2
geff = √g2 + (qE
m)2
Time period (T) = 2π√l
geff
= 2π √
l
√g2 + (qEm
)2
5. Li++ is in ground state. When radiation of wavelength λo is incident on it, it emits 6 different radiations during de-excitation. FInd λo
(A) 1230 Å
(B) 520 Å
(C) 1080 Å
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS (D) 1480 Å Solution: (C) Number of photons emitted when atom de-excites form state n is nC2
∴ nC2 = 6 ⇒n(n − 1)
2= 6 ⇒ n = 4
1
λ= R z2 (
1
12−
1
42) = 1.097 × 107 × 9 × (1 −
1
16)
⇒ λ = 1080 Å 6. A fixed charge P and another charge Q have same charge 1 µc and same mass 4 × 10−6kg are initially kept at distance 1mm. Find the velocity of charge Q when seperation between P & Q become 9 mm. (A) 103 m/s (B) 2 × 103 m s⁄ (C) 4 × 103 m s⁄ (D) 8 × 103 m s⁄ Solution: (B) Form conservation in mechanical energy: −ΔP. E. = ΔK. E.
K(1 × 106)2 [1
10−3−
1
9 × 10−3] =
1
2× 4 × 10−6 × V2
9 × 109 × 10−12 [8
9 × 10−3] = 2 × 10−6 × V2
V2 = 4 × 106 V = 2 × 103 m/s 7. A sound source is moving towards a fixed observer with speed 50 m/s . The frequency observed by observer is 1000 𝐻𝑧. Find apparent frequency observed by observer when source is moving away with same speed. (A) 350 Hz (B) 550 Hz (C) 750 Hz (D) 950 Hz Solution: (C)
When source approaches f1 = f0v
v−vs
When source recedes f2 = f0v
v+vs
⇒ f2
f1=
v−vs
v+vs
⇒ f2 = 1000 ×350−50
350+50 Hz = 750 Hz
8. What is the torque required to rotate a coin of mass 1 Kg , radius r = 1 , 25 times in 5 second starting from rest (A) 5π × 10−4 N − m (B) π × 10−4 N − m (C) 5π N − m (D) 3π × 10−4 N − m Solution: (A)
Angular displacement, θ = 2π × 25 rad
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS Here θ =
1
2αt2 ⇒ α =
2θ
t2=
2×2π×25
52 = 4π rad/s2
The torque required is
τ = I. α = (mr2
4+ mr2) . α
=5
4× 1 × (1 × 10−2) × 4π N − m
= 5𝜋 × 10−4 N − m 9. A particle is projected vertically upward with a speed v0. If the drag force acting on the particle is fdrag = mϑV
2,
what is the time when the particle reach the maximum height?
(A) √𝑔ϑ tan−1 (√𝑔
√vv0)
(B) tan−1(√𝑔ϑ v0)
(C) 1
√𝑔ϑtan−1 (√
𝑔
√v v0)
(D) 1
√𝑔ϑtan−1 (
√ϑ
√𝑔 v0)
Solution: (D) From the drag force, the acceleration of the particle is given by,
a = −(fdrag + W)
m
= −mϑv2+m𝑔
m
= −(ϑv2 + 𝑔)
Therefore, dv
dt= a = −(ϑv2 + 𝑔)
⇒ ∫dv
(ϑv2+𝑔)= −∫ dt
t
0
0
v0
1
ϑ∫
dv
(𝑔ϑ + v
2)= −t
t
v0
∴ t =1
ϑ
1
√𝑔ϑ [
tan−1
(
v
√𝑔ϑ)
]
v0
0
=1
√𝑔ϑtan−1 (
√ϑ
√𝑔 v0)
10. Mercury in a capillary tube of radius R1 has a depression, that is equal to the rise of water in another capillary tube of radius R2 , if the ratio of densities of mercury and water is 13.6, the ratio of surface tension of mercury and water is 7.5 and their angle of contact are θwater = 0° and θHg = 135° in the respective tubes. Thus, the ratio of R1 and R2 is,
(A) 0.2 (B) 0.6 (C) 0.4 (D) 0.8 Solution: (C) Since the depression of mercury in first tube is equal to the rise of water in the second tube
|hHg| = |hwater|
2 SHg cos(θHg)
eHgRHgg=
2Swater cos(θwater)
ewaterRwaterg
Therefore, the ratio of the radii of the respective tubes is,
R1R2
=RHg
Rwater=
ρwaterρHg
SHg
Swater
cos(θHg)
cos(θwater)
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS
=1
13.6× 7.5 ×
1
√2= 0.4
11. He is kept in a rigid contains of volume 67.2 lit. at STP. In order to increase its temperature by 20°C, the heat supplied to the gas is, (A) 780 J (B) 748 J (C) 718 J (D) 680 J Solution: (B) The amount of heat supplied is Q = nCvΔT
= (67.2
22.4) (
3R
2) 20
= 3 ×3
2 × 8.314 × 20
= 748 J 12. A modulating wave of amplitude 100 V and frequency 100 MHz get superimposed with a carrier wave of amplitude 400 V and frequency 300 GHz . Determine respective values of modulating index and difference between the maximum frequency and minimum frequency of modulated wave. (A) 0.25, 2 × 108 Hz (B) 2, 2 × 108 Hz (C) 4, 1 × 108 Hz (D) 2.5 1 × 108 Hz Solution: (A)
Modulating index is given by, m =Am
Ac=
100 V
400 V= 0.25
The difference between the maximum frequency and minimum frequency of modulated wave is given by: 𝛥𝑓 = 𝑓𝑚𝑎𝑥 − 𝑓𝑚𝑖𝑛 = (𝑓𝑐 + 𝑓𝑚) − (𝑓𝑐 − 𝑓𝑚) = 2 𝑓𝑚 = 2(100 MHz) = 2 × 108 Hz 13. A projectile is projected up with speed 2 m/s on an incline plane of inclination 30° at angle 15° from the plane. Find its range on incline plane.
(A) 4
5 (
√3−1
3)
(B) 4
5 (
√3+1
3)
(C) 2
5 (
√3−1
3)
(D) 2
5 (
√3+1
3)
Solution: (A)
Range=2𝑢2𝑠𝑖𝑛𝛼 cos(𝛼+𝛽)
𝑔 cos2 𝛽
β = angle of inclination = 30°
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS α = angle of projection from inclination = 15° u = 2 m/s
𝑅 =2(2)2𝑠𝑖𝑛15° cos(15° + 30°)
g cos2 30°=
8 sin 15° 𝑐𝑜𝑠45°
𝑔 𝑐𝑜𝑠230°
𝑅 =4
5(√3−1
3) [𝑠𝑖𝑛15° =
√3−1
2√2]
14. In YDSE, the ratio of width of slit is 4 ∶ 1, then find the ratio of maximum to minimum intensity. (A) 4: 1 (B) 9: 1 (C) 3: 1 (D) 2: 1 Solution: (B) Intensity ∝ width of slit I1I2
=4
1
ImaxImin
= (√I1 + √I2
√I1 − √I2)
2
=
(
√
I1I2
+ 1
√I1I2
− 1)
2
= (√4 + 1
√4 − 1)
2
= (3
1)2
=9
1
Imax ∶ Imin = 9: 1 15. In the circuit diagram when V0 = 8 Volt , the current through zener diode is 𝑖1 and when V0 = 16 Volt , the corresponding current is 𝑖2 . Find the value of 𝑖2 − 𝑖1.(zener break down voltage Vz = 6 V )
(A) 8 mA (B) Zero (C) 5 mA (D) 1.5 mA Solution: (A) For V0 = 8V
8 𝑉 − 1000 𝑖 = 6 𝑉 i = 2 mA
i0 − i1 =6
4000= 1.5 mA
i1 = 0.5 mA For V0 = 16 V
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS
16 𝑉 − 1000 𝑖0 = 6 𝑉 𝑖0 = 10 𝑚𝐴
𝑖0 − 𝑖2 =6
4000= 1.5 𝑚𝐴
𝑖2 = 8.5 𝑚𝐴 𝑖2 − 𝑖1 = 8.5 − 0.5 = 8 𝑚𝐴 16. What is the time after which the current in the circuit becomes 80% of its maximum current.
(A) ln(5)
100
(B) ln(2)
100
(C) 100 ln(5)
(D) ln(3)
100
Solution: (A) The maximum current through the circuit is:
Is =V
R=
V
0.9+0.1=
V
1
The instantaneous current is:
I = I𝑠 (1 − 𝑒−
𝑅𝑡
𝐿 )
⇒ (80 %) I𝑠 = I𝑠 (1 − 𝑒−
𝑅𝑡𝐿 )
0.8 = (1 − 𝑒−
1×𝑡10×10−3)
0.8 = 1 − 𝑒−100𝑡 Therefore, the time is:
t =ln (
11 − 0.8
)
100=
ln(5)
100
17. A satellite is revolving on a circular path around a planet of mass 8 × 1022 kg and radius 2 × 106 m . Find the number of revolution made by the satellite around the planet in 24 hours. (A) 8 (B) 11 (C) 15 (D) 20 Solution: (B)
T = 2π√R3
GM= 2π √
8×1018
6.67×10−11×8×1022
T ≈ 7700 sec Number of revolution in 24 hour:
𝑛 =86400
7700= 11.22
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS ≈ 11 revolutions 18. A brass rod of length ‘1m’ and cross sectional area 1 mm2 and Young’s modulus 120 × 109 N/m2 is connected with a steel rod of length ‘1m’ and cross sectional area 1 mm2 and Young’s modulus 60 × 109 N/m2 . Find the stress so that the extension of system is 0.2 𝑚𝑚. (A) 8 × 106 N/m2 (B) 16 × 106 N/m2 (C) 4 × 106 N/m2 (D) 106 N/m2 Solution: (A) Stress will be same in both rods
Stress = 𝑌1Δ𝑙1
𝑙= 𝑌2
Δ𝑙2
𝑙
& 𝛥𝑙1 + 𝛥𝑙2 = 0.2 × 10−3 𝑚
Stress× (l
Y1+
l
Y2) = 0.2 × 10−3 m
Stress× (1
120×109+
1
60×109) = 0.2 × 10−3 m
Stress= 8 × 106 N/m2 19. Two radioactive materials have decay constant 5λ and λ . At time 𝑡 = 0 they have same number of nuclei. Find at
what time the ratio of their nuclei will become (1
𝑒)2
.
(A) 1
𝜆
(B) 1
2𝜆
(C) 2
𝜆
(D) 𝜆 Solution: (B)
N = N0e−λt
N1 = N0e−5λt...........(i)
N2 = N0e−λt............(ii)
From equation (i) and (ii)
N1
N2=
e−5λt
e−λt
(1
e)2
= e−4λt
4λt = 2
t =1
2λ
20. Find the change in temperature of 1 mole ideal gas, when its volume changes from 1 V to 2 V . The pressure of the
gas is P = P0 [1 −1
2(V0
V)2
]
(A) P0V
4R+
P0V02
V
(B) P0V0 [V
R+ V]
(C) P0V
R+
P0V02
4VR
(D) P0V
4R+ P0V
Solution: (C) For ideal gas, PV = nRT
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS
Therefore, nRT
V= P0 [1 −
1
2(V0
V)2]
⇒ 𝑇 =𝑃0𝑉
𝑅[1 −
1
2(𝑉0𝑉
)2
]
Therefore, the change in temperature is: 𝛥𝑇 = 𝑇𝑓 − 𝑇𝑖
=𝑃0(2𝑉)
𝑅(1 −
1
2(𝑉02𝑉
)2
) −𝑃0𝑉
𝑅(1 −
𝑉02
2𝑉2)
=P0V
R+
P0V02
4VR
21. The breaking stress of a cylindrical wire is 376 MPa . What is the minimum diameter of the wire if it does not break for a force of 400 N applied on it. (A) 1.1 mm (B) 1.2 m (C) 2.3 mm (D) 1.1 m Solution: (A)
The stress is given by stress =force
area=
F
π(d2
4)
Therefore, the minimum diameter of the wire is:
d = √1
stress×
4F
π
= √1
376 × 106×
4 × 400
π
= 1.1 × 10−3m = 1.1 mm 22. A particle of mass 20 g is moving with a velocity 1 m/s and it penetrates a wooden block of thickness 20 cm with a force of 2.5 × 10−2 N . Find the out speed of the particle when it comes out of the fixed block.
(A) 1
√3 m/s
(B) 1 m/s
(C) 1
√4 m/s
(D) 1
√2 m/s
Solution: (D) The work done is given by:
W = F. 𝑥 =1
2 mVf
2 −1
2 mVi
2
⇒ −2.5 × 10−2 × 0.2 =1
2× 20 × 10−3 (Vf
2 − 12)
Therefore the final velocity of the particles is:
Vf = √(2.5 × 10−2 × 0.2) ×2
20 × 10−3+ 12
= √0.5
=1
√2 m/s
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS 23. A charge ‘q’ is released at a distance R0 from an infinite long wire of linear charge density 𝜆 . Then the velocity of particle will be proportional to
(A) [𝑙𝑛 (𝑅
𝑅0)
1
2]
(B) 𝑙𝑛R
R0
(C) R
(D) R1
2 Solution: (A) Total energy must be constant 𝛥𝐾. 𝐸.+𝛥𝑃. 𝐸.= 0
(1
2mV2 − 0 ) + q(ΔV) = 0
1
2mV2 + q(−2Kλ 𝑙𝑛
R
R0) = 0
V2 =4Kλq
m 𝑙𝑛
R
R0
V = [4Kλq
m 𝑙𝑛
R
R0]1/2
24. The temperature at which rms speed of H2 molecule is equal to escape speed of a particle from earth surface. (A) 103K (B) 104K (C) 105K (D) 106K Solution: (B)
Vrms = √3RT
m= 11.2 Km/s
=3 × 8.314 × T
2 × 10−3= (11.2 × 103)2
T =2 × (11.2)2 × 103
3 × 8.314= 10058.54 K
≈ 104K 25. A particle is performing damped oscillation with frequency 5 Hz. After 10 oscillations its amplitude becomes half.
Find time from beginning after which the amplitude becomes 1
1000
𝑡ℎ of its initial amplitude.
(A) 50 sec (B) 10 sec (C) 20 sec (D) 100 sec Solution: (C) Frequency = 5 Hz
Time period =1
5 second
In 10 oscillations = 10 ×1
5sec = 2 sec. (Amplitude becomes half)
JEE MAIN 10 APRIL 2019 SHIFT-2 PHYSICS
𝐴 = 𝐴0 (1
2)
𝑡2
𝐴01000
= 𝐴0 (1
2)
𝑡2
1000 = 2𝑡2 ⇒
𝑡
2log 2 = log 1000
𝑡 = 2 × (log 1000
log 2)
= 2 ×3
0.301≈ 20 sec
26. If surface tension(s), moment of inertia (I) and Plank's constant (h) are taken as fundamental quantities, then dimension of linear momentum (p) will be given as,
(A) h s1 2⁄ I1 2⁄
(B) h0s1 2⁄ I1 2⁄
(C) h0s−1 2⁄ I1 2⁄
(D) h0s−1 2⁄ I−1 2⁄ Solution: (B)
Let p = hXsYIZ [p] = [h]X[s]Y[I]Z MLT−1 = [ML2T−1]X [MT−2]Y [ML2]Z = MX+Y+ZL2X+2ZT−X−2Y Comparing the power 1 = 𝑋 + 𝑌 + 𝑍 . . . (i) 1 = 2𝑋 + 2𝑍 . . . (ii) −1 = −𝑋 − 2𝑌 . . . (iii)
On solving we will get X = 0 ; Y =1
2 ; Z =
1
2
𝑝 = ℎ0 𝑠1 2⁄ 𝐼1 2⁄
