PAPER-1 (B.E./B.TECH.) JEE MAIN 2019€¦ · PAPER-1 (B.E./B.TECH.) JEE MAIN 2019 Computer Based Test Solutions of Memory Based Questions Date: 8th April 2019 (Shift-2) Time: 02:30

PAPER-1 (B.E./B.TECH.)

JEE MAIN 2019 Computer Based Test

Solutions of Memory Based Questions

Date: 8th 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 8 APRIL 2019 SHIFT-2 PHYSICS

1. In the following figure the shaded portion is removed. Find the COM of remaining part.

(A) (

5

12a,

5

12b)

(B) (a

2,b

2)

(C) (a

2,b

4)

(D) (a

3,b

3)

Solution: (A)

If the total mass of object mass 'M' then the mass removed part will be M

4.

XCM =M ·

a

2−

M

4(3a

4)

M −M

4

=

Ma

2−

3Mb

163M

4

=5

12a

XCM =M ·

b

2−

M

4(3b

4)

M −M

4

=

Mb

2−

3Ma

163M

4

=5

12b

2. A variable resistance 'R' is connected to a battery with internal resistance 'r'. what is the value of 'R' to get maximum power delivered in resistor.


(A) R = r (B) R = 2r (C) R = r/2 (D) r = 0 Solution: p = i2R

(E

R + r)2

R

For maximum power: 𝑑𝑃

𝑑𝑅= 0

−(E

R + r)R (

E

R + r)2

+ (E

R + r)2

= 0

1

R + r−

2R

(R + r)2= 0

1 =2R

R + r

R = r 3. A solid sphere and a cylinder are given same velocity at the fact of the inclined plane as shown in figure. Assume pure rolling throughout the motion and find the ratio of the maximum heights climbed on inclined plane

(ℎsphere

ℎcylinder)

(A) 2: 5 (B) 7: 15 (C) 14: 15 (D) 14: 30 Solution: (C) From conservation of mechanical energy

For solid sphere 1

2𝑚3𝑉

2 +1

2𝐼3𝜔

2 = 𝑚𝑠𝑔ℎ𝑠


𝑉2

2+

𝑉2

5= 𝑔ℎ𝑠

7𝑉2

10= 𝑔ℎ𝑠 …(i)

For cylinder 1

2𝑚𝑐𝑉

2 +1

2𝐼𝑐𝜔

2 = 𝑚𝑐𝑔ℎ𝑐

𝑉2

2+

𝑉2

4= 𝑔ℎ𝑐

3𝑉2

4= 𝑔ℎ𝑐 …(ii)

From (i) and (ii) ℎ𝑠

ℎ𝑐=

14

15

4. Two magnetic dipoles are lying as shown in figure. The separation between them is very large. Find the force on a charge Q which is projected with speed V as shown.

(A) √2

μ

π

MQV

d3

(B) Zero

(C) μ0

4π

MQV

d3

(D) 2√2 μ0MQV

πd3

Solution: ()

B1 = 2(μ0

4π)

M

d 2⁄ along x-axis

B2 =μ0

4π

(2M)

(d 2⁄ ) along y-axis

B1 = B2, Hence Bnet will be at 45o from x-axis. Now, velocity and magnetic field are in same direction. Hence no force will act on the particle.

5. In a region electric field is given by E⃗⃗ = (Ax + B)𝑖 where A = 20 unit and B = 10 unit of electric potential at X = 1 m is V1 and at X = −5 m is V2. Then V1 − V2 is equal to (A) 180 V (B) 150 V (C) 200 V (D) 100 V Solution: (A)

V2 − V1 = −∫E. dx

= − ∫(ax + b) dx

1

−5

= −[ax2

2+ bx]

−5

1

= − [(a

2+ b) − (

289

2− 5b)]


= − [20

2+ 10 − 250 + 50]

= 180 V 6. A simple pendulum undergoes 20 oscillation in 30 second of the length of pendulum is 55 cm, then find the percentage error in acceleration due to gravity. (least count of measurement of time is 1 second and for length 1 mm) (P8268) (A) 6.8% (B) 5.0% (C) 6.6% (D) 1.0% Solution: (A)

g =4π2l

T2

∆g

g=

∆l

l+

2∆T

T

=0.1

55+ 2. (

1

30)

= 0.06848 Percentage error ≈ 6.85% 7. Find reading in voltmeter if potential gradient on wire PQ is 0.01V m⁄ .

(A) 2 mV (B) 3 mV (C) 5 mV (D) 8 mV Solution: (C) Reading = Potential gradient × length = 0.01 × 0.5 = 0.005 V = 5 mV 8. Find the ratio of density of nuclei of O16 and Ca40 (A) 1: 1 (B) 1: 2 (C) 2: 1 (D) 1: 4 Solution: (A) Nucleous density is independent of mass number. So, the ratio of density will be 1: 1


9. From the pressure of volume graph which of the following option relevant in order (isochoric, isobaric, isothermal, adiabatic)

(A) a, b, c, d (B) d, a, b, c (C) b, c, a, d (D) d, c, b, a Solution: (A) a → Isobaric process b → Isothermal process c → Adiabatic process d → Isochoric process 10. When a voltage source e = e0 sin(100t) is connected by a circuit. It was found that the phase difference between voltage and current was 45o . Identify the circuit. (A) R = L circuit with R = 1 kΩ, L = 1 mH (B) R-C circuit with R = 100 Ω, C = 10 μF (C) R-L circuit with R = 1Ω, L = 10H (D) R-C circuit with R = 1 kΩ, C = 10 μF Solution: () For circuit (A): R = 103Ω and XL = Lω = (10−3)(100) = 0.1

tan θ =XL

R≠ 1

For circuit (B): R = 100Ω; XC =1

ωC=

1

(100)(10×10−6)= 1000Ω

tan θ =XC

R≠ 1

For circuit (C): R = 1Ω; XL = ωL = (100)(10) = 1000 Ω

tan θ =XL

R≠ 1

For circuit (D): R = 1 kΩ = 1000Ω; XC =1

ωC=

1

(100)(10×10−6)= 1000

tan θ =XC

R= 1; θ = 45o

11. In the given figure, find total current drawn from the battery:


(A)

32

9

(B) 9

32

(C) 4

3

(D) 14

23

Solution: (B)

Req = 160 3⁄

i =15

160 3⁄=

9

32 Amp

12. A parallel plate capacitor 1 μF. Two plates are given charge 2μc and 4 μc. Then find the potential difference between plates (A) 1 volt (B) 2 volts (C) 3 volts (D) 4 volts Solution: (A)

Charge inside capacitor = 1μC

V =Q

C=

1×10−6

1×10−6 = 1 volt

13. An object starts moving with uniform acceleration from rest. Initially it was at origin. Then which of the following graphs are correct.


(A)

(B)

(C)

(D) Solution: (D) Options: 1 → ABD 2 → ABC 3 → BC 4 → ∆ 1 Acceleration = constant; a v → t graph will be constant line parallel to time axis. Option (A) is correct. 2 v = u + at; velocity vs time graph will be straight line passing through origin. Option (B) is correct. u = 0; initially at rest v = at

3 X =1

2at2 (particle was initially at origin)

X v → t graph will be parabola passing through origin. Option (D) is correct.

14. Two elastic wire 𝐴 and 𝐵 having length 𝑙𝐴 = 2 𝑚 and 𝑙𝐵 = 1.5 𝑚 have Young’s Modules ratio 𝑌𝐴

𝑌𝐵=

7

4 of 𝑟𝐵 =

2 mm then find the radius of 𝐴 for which change in length in both wire will be same for the application of same force. (A) 1.7 mm (B) 1.5 mm


(C) 3.4 mm (D) 2.0 mm Solution: (A) Strain = ∆𝑙 𝑙⁄ Stress = F A⁄

Young’s modulus =Stress

Strain=

F A⁄

(∆l l⁄ )=

F

A (

l

∆l)

YA

YB= (

lAAA

)(AB

lB) =

7

4

(2

πrA2)(

πrB2

1.5) =

7

4

rA2 =

2

1.5 × 7× (2 mm)2

rA = 1.7 mm

15. A⃗⃗ 1 and A⃗⃗ 2 are two vectors such that their magnitude is 3 and 5 unit and sum of these vectors is also 5 unit. Then find the value of (2A1 + 3A2). (2A1 − 2A2) (A) 123 (B) −123

(C) 291

2

(D) 591

2

Solution: (B)

|A⃗⃗ 1 + A⃗⃗ 2| = 5

|A⃗⃗ 1 + A⃗⃗ 2|2= 25

|A⃗⃗ 1|2+ |A⃗⃗ 2|

2+ 2A⃗⃗ 1A⃗⃗ 2 = 25

9 + 25 + 2A⃗⃗ 1. A⃗⃗ 2 = 25

A⃗⃗ 1. A⃗⃗ 2 = −92⁄

Now, (2A⃗⃗ 1 + 3A⃗⃗ 2). (2A⃗⃗ 1 − 2A⃗⃗ 2) = 4|A⃗⃗ 1|2− 4A⃗⃗ 1A⃗⃗ 2 + 6A⃗⃗ 1A⃗⃗ 2 − 6|A⃗⃗ 2|

2

= 4. |A⃗⃗ 1|2+ 2A⃗⃗ 1A⃗⃗ 2 − 6|A⃗⃗ 2|

2

= 4 × 9 + 2(−9

2) − 6 × 25

= 36 − 9 − 150 = −123 16. Two infinite wire are carrying current 𝑖 along 𝑥 and 𝑦 axis as shown. Find the magnetic field at point 𝑃.

(A)

μ0I

2d

(B) μ0I

√2 d


(C) Zero

(D) μ0I

d

Solution: () Magnetic field due to wire 1:

B1 =μ0I

2d [outward]

Magnetic field due to wire 2;

B2 =μ0I

2d [inside the plane]

B⃗⃗ net = B⃗⃗ 1 + B⃗⃗ 2 = 0 17. When light of wavelength 𝜆 fall on a metal surface the ejected 𝑒− have linear momentum ‘𝑝’. When light of wavelength 𝜆′ fall the ejected 𝑒− have linear momentum 1.5 𝑝. Considering work function of metal small. Find 𝜆′

(A) 4𝜆

9

(B) 9𝜆

4

(C) 2𝜆

3

(D) 3𝜆

2

Solution: ()

𝐸 =ℎ𝑐

𝜆− 𝜙

𝜙 is very small

𝐸 =ℎ𝑐

𝜆

𝑝2

2 𝑚=

ℎ𝑐

𝜆 …(i)

(1.5 𝑝)2

2 𝑚=

ℎ𝑐

λ′ …(ii)

From equation (i) and equation (ii) 1

(1.5)2=

𝜆′

𝜆

𝜆′ =4

9𝜆

18. An electric dipole is in equilibrium having charge 𝑞 and −𝑞 are placed at distance ‘𝑑’ in uniform electric field. Each charge has mass ‘𝑚’. If dipole is displaced by small angle. Then find its angular frequency of oscillation.

(A) √qE

md

(B) √2qE

md

(C) √qE

2 md

(D) √3 qE

md

Solution: (B) τ = pEsin θ For small angle = pE θ


∝=τ

I=

pE

I θ

(∵ ∝ = ω2θ) Angular frequency = √pE

I

= √(q. d) E

(md2

4+ m

d2

4)

= √2 qE

md

19. Two balls of mass m1 and m2 = 0.5 m1 undergo head on collision as shown.

after collision of V3 = 0.5 V1, then find the value of V4 (A) V4 = V1 + V2 (B) V4 = 2V1 + V2 (C) V4 = 2(V1 + V2) (D) V4 = V1 + 2 V2 Solution: (A) From conservation of linear momentum m1V1 + m2V2 = m1V3 + m2V4 m1V1 + (0.5 m1) V2 = m1(0.5 V1) + (0.5 m1) V4 = 0.5 V1 + 0.5 V2 = 0.5 V4 V4 = V1 + V2 20. A projectile is thrown from earth with minimum energy 𝐸 such that it escapes to infinity. How much energy is required to escape same projectile from surface of moon? Assume density of moon and earth are same and volume of earth is 64 times the volume of moon.

(A) E

16

(B) E

64

(C) E

8

(D) E

32

Solution: (A) Mearth = 64 Mmoon and Rearth = 4 Rmoon

Given: E =G Mem

Re

Now, for moon E′ =G Mmm

Rm

=G

Mearth

64 m

Rearth

4

=G Mearth m

16 Rearth

=E

16


21. The distance between a receiver and transmitter is 50 km. The height of tower required if the height of receiver is 35 m. (Given Rearth = 6400 km) (A) 70 m (B) 60 m (C) 65 m (D) 80 m Solution: (C)

d = √2 RhT + √2 Rhr

50 = √2 × 6400 × hT + √2 × 6400 × (35 × 10−3)

√2 × 6400 × hT = 50 − δ√7

hT =(50 − 2√7)

2

2 × 6400= 0.0649 km

hT ≈ 65 m 22. Frequency of a particle performing damped oscillation is 5 Hz, after every 10 oscillation its amplitude becomes half then find out the time from beginning after which amplitude becomes 1/1000 of its initial amplitude (A) 30 sec (B) 50 sec (C) 20 sec (C) 40 sec Solution: (C) 𝑓 = 5

So, 𝑇 =1

5

10𝑇 =10

5= 2

𝐴0

1000=

𝐴0

(1

2)

𝑡

2

(2)𝑡

2 = 1000

(𝑡

2) log2 = 3

𝑡 =6

log2≈ 20 𝑠

23. Temperature of hydrogen gas at which rms speed of hydrogen molecule is equal to speed of a particle from earth surface is : (A) 102 𝐾 (B) 104 𝐾 (C) 2 × 104 𝐾 (D) 105 𝐾 Solution:(B)

√3𝑅𝑇

𝑚= 11.2 𝑘𝑚 𝑠⁄

3 ×25

3× 𝑇

2 × 10−3= (11.2 × 103)2

𝑇 = 10035.2𝐾 ≈ 104𝐾


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PAPER-1 (B.E./B.TECH.) JEE MAIN 2019€¦ · PAPER-1 (B.E./B.TECH.) JEE MAIN 2019 Computer Based Test Solutions of Memory Based Questions Date: 8th April 2019 (Shift-2) Time: 02:30