Chapterwise - Topicwise

PHYSICS

•

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Index NEET Solved Paper 2018

2018-

1-2018-12

1.

Physical World, Units and Measurements

1-7

2.

Motion in a Straight Line

3.

Motion in a Plane

18-30

4.

Laws of Motion

31-42

5.

Work, Energy and Power

43-56

6.

System of Particles and Rotational Motion

57-73

7.

Gravitation

74-85

8.

Mechanical Properties of Solids

86-87

9.

Mechanical Properties of Fluids

88-90

8-17

10. Thermal Properties of Matter

91-100

11. Thermodynamics

101-110

12. Kinetic Theory

111-116

13. Oscillations

117-128

14. Waves

129-144 (iii)

15. Electric Charges and Fields

145-152

16. Electrostatic Potential and Capacitance

153-164

17. Current Electricity

165-187

18. Moving Charges and Magnetism

188-205

19. Magnetism and Matter

206-210

20. Electromagnetic Induction

211-217

21. Alternating Current

218-226

22. Electromagnetic Waves

227-231

23. Ray Optics and Optical Instruments

232-247

24. Wave Optics

248-254

25. Dual Nature of Radiation and Matter

255-268

26. Atoms

269-276

27. Nuclei

277-291

28. Semiconductor Electronics : Materials, Devices and Simple Circuits

292-308

(iv)

NEET Solved Paper 2018 1.

2.

3.

4.

5.

A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27°C two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz, the velocity of sound in air at 27°C is (1) 330 m/s (2) 339 m/s (3) 300 m/s (4) 350 m/s An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is (1) smaller (2) 5 times greater (3) equal (4) 10 times greater A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s2 at a distance of 5 m from the mean position. The time period of oscillation is (1) 2p s (2) p s (3) 1 s (4) 2 s The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is (1) independent of the distance between the plates (2) linearly proportional to the distance between the plates (3) inversely proportional to the distance between the plates (4) proportional to the square root of the distance between the plates Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular

6.

7.

8.

9.

deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is (1) 40 W (2) 25 W (3) 500 W (4) 250 W A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from (1) the current source (2) the magnetic field (3) the induced electric field due to the changing magnetic field (4) the lattice structure of the material of the rod An inductor 20 mH, a capacitor 100 mF and a resistor 50W are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is (1) 0.79 W (2) 0.43 W (3) 1.13 W (4) 2.74 W A metallic rod of mass per unit length 0.5 kg m–1 is lying horizontally on a smooth inclined plane which makes an angle of 30°with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is (1) 7.14 A (2) 5.98 A (3) 11.32 A (4) 14.76 A A carbon resistor of (47 ± 4.7) kW is to be marked with rin gs of differen t colours for its identification. The colour code sequence will be (1) Violet – Yellow – Orange – Silver (2) Yellow – Violet – Orange – Silver (3) Green – Orange – Violet – Gold (4) Yellow – Green – Violet – Gold

PHYSICS

2018- 2

10.

11.

A set of 'n' equal resistors, of value 'R' each,are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of'n' is (1) 10 (2) 11 (3) 9 (4) 20 A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n? I

O

n

(4) O

14.

n

I

(3)

13.

15.

n

O

n

In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength l of the light used is 5896 Å and distance D between the screen and slits is100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same l and D) the separation between the slits needs to be changed to (1) 1.8 mm (2) 1.9 mm (3) 1.7 mm (4) 2.1 mm An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of (1) small focal length and large diameter (2) large focal length and small diameter (3) small focal length and small diameter (4) large focal length and large diameter Unpolarised light is incident from air on a plane surface of a material of refractive index 'm'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation? (1) Reflected light is polarised with its electric vector parallel to the plane of incidence

17.

18.

19.

20.

æ1ö i = tan –1 ç ÷ èmø

æ1ö i = sin –1 ç ÷ èmø An em wave is propagating in a medium with a r velocity V = Viˆ . The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along (1) –z direction (2) +z direction (3) –x direction (4) –y direction The refractive index of the material of a prism is 2 and the angle of the prism is 30°.One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is (1) 60° (2) 45° (3) Zero (4) 30° An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm.If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be (1) 30 cm away from the mirror (2) 36 cm away from the mirror (3) 36 cm towards the mirror (4) 30 cm towards the mirror The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance (1) 0.138 H (2) 138.88 H (3) 13.89 H (4) 1.389 H For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is (1) 20 (2) 10 (3) 15 (4) 30 The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is (1) 1 : 1 (2) 1 : –1 (3) 1 : –2 (4) 2 : –1

(4)

16.

(2)

I

12.

(3)

I

(1) O

(2) Reflected light is polarised with its electric vector perpendicular to the plane of incidence

NEET Solved Paper 2018 21.

2018- 3

An electron of mass m with an initial velocity r V = V0iˆ (V 0 > 0) enters an electric field r E = – E0iˆ (E0 = constant > 0) at t = 0. If l0 is its

25.

de-Broglie wavelength initially, then its deBroglie wavelength at time t is (1)

22.

23.

l0 æ eE0 ç1 + mV 0 è

ö t÷ ø

(2)

æ eE ö l0 ç1 + 0 t ÷ è mV0 ø

26.

(3) l0 (4) l0t When the light of frequency 2n0 (where n0 is threshold frequency), is incident on a metalplate, the maximum velocity of electronsemitted is v1. When the frequency of the incident radiation is increased to 5n0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is (1) 1 : 2 (2) 1 : 4 (3) 2 : 1 (4) 4 : 1 In the combination of the following gates the output Y can be written in terms of inputs A and B as

27.

(3) IB = 40 mA, IC = 5 mA, b = 125 (4) IB = 20 mA, IC = 5 mA, b = 250 In a p-n junction diode, change in temperature due to heating (1) Affects only reverse resistance (2) Affects only forward resistance (3) Affects the overall V-I characteristics of p-n junction (4) Does not affect resistance of p-n junction A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same.Which of the following physical quantities would remain constant for the sphere? (1) Angular velocity (2) Moment of inertia (3) Angular momentum (4) Rotational kinetic energy The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

B

A B

(1) 24.

A

Y

A×B

(2)

A×B+A×B

(3) A + B (4) A × B + A × B In the circuit shown in the figure, the input voltage Vi is 20 V, VBE = 0 and VCE = 0. The values of IB, IC and b are given by

28.

20 V Rc 4 kW Vi

RB 500 kW B

C E

(1) IB = 40 mA, IC = 10 mA, b = 250 (2) IB = 25 mA, IC = 5 mA, b = 200

29.

S

C

(1) KA < KB < KC (2) KA > KB > KC (3) KB > KA > KC (4) KB < KA < KC If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct? (1) Raindrops will fall faster (2) Walking on the ground would become more difficult (3) ‘g’ on the Earth will not change (4) Time period of a simple pendulum on the Earth would decrease A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (Kt) as well as rotational kinetic energy (Kr) simultaneously. The ratio Kt : (Kt + Kr) for the sphere is (1) 7 : 10 (2) 5 : 7 (3) 2 : 5 (4) 10 : 7

PHYSICS

2018- 4

30.

31.

32.

33.

34.

35.

A small sphere of radius ‘r’ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to (1) r3 (2) r2 4 (3) r (4) r5 A sample of 0.1 g of water at 100°C and normal pressure (1.013 × 105 Nm–2) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is (1) 104.3 J (2) 208.7 J (3) 84.5 J (4) 42.2 J Two wires are made of the same material and have the same volume. The first wire has crosssectional area A and the second wire has crosssectional area 3A. If the length of the first wire is increased by Dl on applying a force F, how much force is needed to stretch the second wire by the same amount? (1) 9 F (2) 6 F (3) F (4) 4 F The power radiated by a black body is P and it radiates maximum energy at wavelength, l0. If the temperature of the black body is now changed so that it radiates maximum energy at 3 wavelength l0 , the power radiated by it 4 becomes nP. The value of n is 4 3 (1) (2) 3 4 81 256 (3) (4) 256 81 At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere? (Given : Mass of oxygen molecule (m) = 2.76 × 10–26 kg Boltzmann’s constant kB = 1.38 × 10–23 JK–1) (1) 2.508 × 104 K (2) 8.360 × 104 K (3) 1.254 × 104 K (4) 5.016 × 104 K The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

V B A

36.

37.

38.

O T 2 2 (1) (2) 5 3 2 1 (3) (4) 7 3 The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is (1) 13.2 cm (2) 8 cm (3) 16 cm (4) 12.5 cm The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is (1) 26.8% (2) 20% (3) 12.5% (4) 6.25% A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

h

B A

3 D (2) D 2 7 5 D D (3) (4) 5 4 Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed w about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation (1) WC > WB > WA (2) WA > WB > WC (3) WA > WC > WB (4) WB > WA > WC

(1)

39.

vL

NEET Solved Paper 2018 40.

41.

42.

2018- 5

Which one of the following statements is incorrect? (1) Rolling friction is smaller than sliding friction. (2) Limiting value of static friction is directly proportional to normal reaction. (3) Coefficient of sliding friction h as dimensions of length. (4) Frictional force opposes the relative motion. A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be (1) 0.5 (2) 0.25 (3) 0.4 (4) 0.8 A block of mass m is placed on a smooth inclined wedge ABC of inclination q as shown in the figure. The wedge is given an acceleration ‘a’ towards the right. The relation between a and q for the block to remain stationary on the wedge is A m

43.

44.

45.

a q C

g g (2) a = cosec q sin q (3) a = g tan q (4) a = g cos q A toy car with charge q moves on a frictionless horizontal plane surface under the influence of r aruniform electric field E . Due to the force q E , its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively (1) 2 m/s, 4 m/s (2) 1 m/s, 3 m/s (3) 1.5 m/s, 3 m/s (4) 1 m/s, 3.5 m/s r The moment of the force, F = 4iˆ + 5 ˆj – 6kˆ at (2, 0, –3), about the point (2, –2, –2), is given by (1) –8iˆ – 4 ˆj – 7kˆ (2) –4iˆ – ˆj – 8kˆ

(1)

B

a=

(3) –7iˆ – 4 ˆj – 8kˆ (4) –7iˆ – 8 ˆj – 4kˆ A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is (1) 0.521 cm (2) 0.525 cm (3) 0.529 cm (4) 0.053 cm

ANSWER KEY 1

2

6

1

11

1

16

2

21

1

26

3

31

2

36

1

41

2

2

1

7

1

12

2

17

2

22

1

27

2

32

1

37

1

42

3

3

2

8

3

13

4

18

3

23

2

28

3

33

4

38

3

43

2

4

1

9

2

14

2

19

1

24

3

29

2

34

2

39

1

44

3

5

4

10

1

15

2

20

2

25

3

30

4

35

1

40

3

45

3

PHYSICS

2018- 6

Hints & Solutions 1.

(2) Two successive resonance are produced at 20 cm and 73 cm of column length \

l = (73 – 20) × 10–2 m 2

Clearly, F is independent of the distance between plates. 5.

(4) Current sensitivity of moving coil galvanometer

Þ l = 2 × (73 – 20) × 10–2 Velocity of sound, v = nl = 339.2 ms–1

Vs =

(1) As we know, F = qE = ma Þ a=

\

1qE 2 t 2m

t=

2hm qE

RG =

(1) Rod gains gravitational potential energy which comes from energy of current source.

7.

(1) Power dissipated in an LCR series circuit connected to an a.c. source of emf E P = Erms irms cos f =

(2) From question, acceleration, a = 20 m/s2, and displacement, y = 5m =

Þ w = 2 rad/s Time period of pendulum, 2p 2p = = ps w 2

Q2 2 Ae 0

For isolated capacitor Q = constant

Z2

=

2 Erms R

1 ö æ R 2 + ç wL – ÷ C wø è

2

æ 10 ö ç ÷ ´ 50 è 2ø 1 æ ö (50)2 + ç 314 ´ 20 ´10 –3 – ÷ 314 ´ 100 ´ 10–6 ø è

2

Solving we get, P = 0.79 W 8.

(3) From figure, for equilibrium, mg sin 30° = I/B cos 30°

(1) Electrostatic force between the metal plates Fplate =

2 Erms R

2

w2y

Þ 20 = w2(5)

4.

Is 5 ´1 5000 = = = 250 W –3 Vs 20 ´10 20

6.

Since, electron has smaller mass so it will take smaller time.

T=

...(ii)

Resistance of galvanometer

h=

|a| =

NBA CRG

Dividing eqn. (i) by (ii)

qE m

i.e., time t µ m as ‘q’ is same for electron and proton.

3.

...(i)

Voltage sensitivity of moving coil galvanometer,

= 2 × 320 [73 – 20] × 10–2 2.

NBA C

Is =

Þ I= =

mg tan 30° lB

0.5 ´ 9.8 0.25 ´ 3

= 11.32 A

° 30 is n g m 30°

B

s co

° 30

B Il 30° IlB

NEET Solved Paper 2018 9.

10.

2018- 7

(2) Colour code for carbon resistor 0

Black

1

Brown

2

Red

± 10% Silver

3

Orange

± 20% No colour

4 5 6 7 8 9

Yellow Green Blue Violet Grey White

E nR + R

l d

So, 0.20° =

l d Now putting the value of l

Again, 0.21° =

0.20°´ 2mm 0.21° d = 1.9 mm

d=

\ 13.

(4) For telescope, angular magnification =

Angular resolution =

14. ...(ii)

Plane polarised reflected light

(n + 1) R 10 = 1 æ ö ç + 1÷ R n è ø Solving we get, n = 10 (1) Short circuited current,

I=

ne e = nr r

I

O

n So, I is independent of n and I is constant.

D So, D should 1.22l

be large. So, objective lens of refracting telescope should have large focal length (f0) and large diameter D for larger angular magnification. (2) When reflected light rays and refracted rays are perpendicular, reflected light is polarised with electric field vector perpendicular to the plane of incidence.

Dividing eq. (ii) by (i),

11.

f0 fE

So, focal length of objective lens should be large.

...(i)

E 10 I = R +R n

(2) Angular width = l 2mm Þ l = 0.20° × 2

Tolerance : ± 5% Gold

(47 ± 4.7) kW = 47 × 103 ± 10% \ Yellow - Violet - Orange - Silver (1) In series grouping equivalent resistance Rseries = nR In parallel grouping equivalent resistance R Rparallel = n I=

12.

i m

15.

90° Partial polarised refracted light

Also, tan i = m (i = Brewester angle) (2) As we know, r r r E´B = V r ( Ejˆ) ´ ( B) = Viˆ (Q Electric field vector is along +y axis) r So, B = Bkˆ i.e., direction of magnetic field vector is along +z direction.

PHYSICS

2018- 8

16.

(2) For retracing the path, light ray should be normally incident on silvered face. A = r + O Þ r = 30°

19.

Applying Snell’s law at point M, sin i 2 = sin 30° 1 2´

Þ sin i =

1

or, sin i =

2

i

M

30°

1 2

N 150 1 = = N0 600 4

60° 30°

m=

N = N0e–lt Þ ln

2

i.e., i = 45°

O

(2)

2.303 ´ T 1

1 1 1 using mirror formula, f = v + u 1 1 1 1 = + 15 v1 u

Þ

1 1 1 = + v1 –15 40

v1 = –24 cm When object is displaced by 20 cm towards mirror Now, u2 = –20 1 1 1 = + f v2 u2

kze2 2rn

Total energy, E =

21.

l0 =

ur E0

18.

v2 = –60 cm

h mV0

25 × 10–3 =

1 ´ L ´ (60 ´ 10–3 )2 2

...(i)

F

–

V0

Acceleration of electron

Therefore image shifts away from mirror by = 60 – 24 = 36 cm (3) From question energy stored in inductor, U = 25 × 10–3 J Current, I = 60 mA Energy stored in inductor U =

– kze2 2rn

So, Kinetic energy : total energy = 1 : –1 (1) Initial de-Brogile wavelength

1 1 1 1 1 1 = – Þ v = 20 – 15 –15 v2 20 2

\

N0 N

2.303 ´10 log10 4 0.693 (2) In a Bohr orbit of the hydrogen atom Kinetic energy,

k=

\

So,

0.693

log10

=

20.

–

2

Þ t=

40 cm

N0 = lt N

1 N0 ln l N

Þ t=

f = 15 cm

17.

25 ´ 2 ´ 106 ´10 –3 = 13.89 H 3600 (1) Number of nuclei remaining, N = 600 – 450 = 150 after time ‘t’

L=

1 2 LI 2

eE0 (Q F = ma = eE0) m Velocity after time ‘t’

a=

eE0 ö æ t V = ç V0 + m ÷ø è

So, l =

h = mV

h eE ö æ m çV0 + 0 t ÷ m ø è

NEET Solved Paper 2018 =

h é eE ù mV0 ê1 + 0 t ú ë mV0 û

2018- 9

=

l0 é eE0 ù tú ê1 + ë mV0 û

20 V

...(ii)

IC

Dividing eqs. (ii) by (i), de-Broglie wavelength l =

22.

Vi

l0

IC =

1 2 mv1 2

1 2 mv2 2

1 2 mv2 2

Þ IB =

23.

(2)

A B

...(ii)

26.

20 500 ´ 103

= 40 mA

IC 25 ´ 10 –3 = = 125 Ib 40 ´ 10–6 (3) On heating, number of electron-hole pairs increases, so overall resistance of diode will change. Hence forward biasing and reversed biasing both are changed. (3) Angular momentum, remains conserved until the torque acting on sphere remains zero. tex = 0 dL =0 dt i.e., angular momentum L = constant. (2) Speed of the planet will be maximum when its distance from the sun is minimum as mvr = constant.

So,

A B

A× B Y

A B

A×B

Y = (A × B + A × B) 24.

= 5 × 10–3 = 5 mA

b=

25.

1 v12 = 4 v22 v1 1 = v2 2

4 ´ 103

Þ 20 = IB × 500 × 103

Dividing eqn. (i) by (ii),

\

(20 – 0)

or, Vi = 0 + IBRB

1 2 mv1 ...(i) 2 when light of frequency, 5n0 is incident on a metal plate

4hn0 =

Vb

500 kW

Vi = VBE + IBRB

hn0 =

h(5n0) = hn0 +

Ib

é eE0 ù tú ê1 + ë mV0 û

(1) Using Einstein’s photoelectric equation, E = W0 + kmax When light of frequency, 2n0 is incident on a metal plate, h(2n0) = hn0 +

RB

RC = 4 kW

(3) From question, VBE = = 0, Vi = 20 V VCE = 0 Vb = 0 (earthed) IB = ?, IC = ?, b = ?

27.

Perihelion A

B

VC C

S VA

Aphelion

Point A is perihelion and C is aphelion. Clearly, VA > VB > VC So, KA > KB > KC

PHYSICS

2018- 10

28.

(3) If universal gravitational constant becomes ten times, then G¢ = 10 G Acceleration due to gravity, g =

29.

æ F ö æ F¢ Dl = ç ÷ 3l = ç AY è ø è 3 AY

GM

R2 So, acceleration due to gravity increases. (2) In rolling motion, rotational kinetic energy. Kt =

From equation (i) & (ii),

33.

And, Kt + Kr =

Þ l0 T =

1 2 1 2 mv + l w 2 2

30.

31.

32.

34.

2 r 2 (r – s) g h 9 Power µ r5

T = 8.360 × 104 K 35.

...(i)

F¢ 3A, l ...(ii)

(1) Gas is monatomic, so Cp = \

5 R 2

Given process is isobaric dQ = n Cp dT

æ5 ö Þ dQ = n ç R ÷ dT è2 ø dW = P dV = n RdT

F A, 3l

(1) Wire 1:

æ F¢ ö Dl = ç ÷l è 3 AY ø

3k BT mO2 = 11200 m/s

Putting value of KB and mO2 we get,

\ (2) Using first law of thermodynamics equation, DQ = DU + DW Þ 54 × 4.18 = DU + 1.013 × 105(167.1 × 10–6 – 0) (Q DW = PDV) Þ DU = 208.7 J

Wire 2:

4 4 P2 256 æ T¢ ö æ 4ö = n = =ç ÷ So, P ç ÷ = T 3 81 1 è ø è ø (2) Let at temperature T rms speed of oxygen molecules become just sufficient for escaping from the Earth’s atmosphere Vescape = 11200 m/s

Also, Vrms = Vescape =

VT =

æ F ö Dl = ç ÷ 3l è AY ø

3l0 T¢ 4

4 T 3 Power radiated P µ T4

1 2 1 æ 2 2 öæ v ö 7 mv + ç mr ÷ç ÷ = mv 2 2 2è5 10 øè r ø

1 2 mv Kt 5 2 = = \ 7 7 Kt + Kr 2 mv 10 (4) Power = rate of production of heat = F.V = 6phr VT × VT = 6phrVT2 (Q F = 6phVT r stoke’s formula) 2 VT µ r

Q

2

Þ T¢ =

2

=

(4) From Wien’s law lmax T = constant i.e., lmax T1 = lmax T2 1

1 2 mv 2

ö ÷ l or, F¢ = 9 F ø

dW nRdT 2 = = dQ 5 æ5 ö n ç R ÷ dT è2 ø (1) For closed organ pipe, third harmonic

\

36.

Required ratio =

(2 N – 1)V 3V = (Q N = 2) 4l 4l For open organ pipe, fundamental frequenty

n=

NEET Solved Paper 2018 n=

2018- 11

IRing =

NV V = (Q N = 1) 2l 2l ¢

\

3V V = According to question, 4l 2l ¢

Þ l¢ = 37.

40.

f = msN Þ

(1) Efficiency of ideal heat engine, 41.

(2)

æ T2 ö Percentage efficiency, %h = ç 1– T ÷ ´ 100 è 1ø

e=

vL

1 mvL 2 + 0 2

ma cos q

1 2æ 1 ö I w ç DkErot = I w2 ÷ 2 2 è ø or, DW µ I (for same w) 2 1 2 MR 2 , I = MR Disk 5 2

q

mg sin q a

mg

For completing the vertical circle, vL ³ 5gR

Isolid sphere =

1 = 0.25 4 (3) Let the mass of block is m. It will remains stationary if forces acting on it are in equilibrium. i.e., ma cos q = mg sin q Þ a = g tan q

e=

ma

vL2 2 2 Using v – u = 2gh, h = (Q u = 0) 2g

DW =

v 4

v 4 = v

42.

(1) Work done required to bring them rest DW = DKE (work-energy theorem)

Þ v¢ =

Relative velocity of separation Relative velocity of approach

or,

5gR 5 5 = R= D 2g 2 4

4m After Collision

4m Before Collision

Coefficient of restitution,

B

or, h =

m

m

(3) As track is frictionless, so total mechanical energy will remain constant

i.e., 0 + mgh =

v¢

v=0

mv + 4m × 0 = 4 mv¢ + 0

æ 100 ö æ 273 ö = ç 1– ÷ ´100 = ç 373 ÷ ´100 = 26.8% 373 è ø è ø

A

f = [M0L0T0] N

According to law of conservation of linear momentum,

Source temperature, T1 = 0°C = 0 + 273 = 273 K

h

ms = v=0

v

Sink temperature, T2 = 100°C = 100 + 273 = 373 K

39.

WC > WB > WA

(3) Coefficient of friction or sliding friction has no dimension.

4l 2l 2 ´ 20 = = = 13.33 cm 3´ 2 3 3

æ T ö h = ç1 – 2 ÷ è T1 ø

38.

MR2

43.

q

Here ma = Pseudo force on block, mg = weight. t=0 a t = 1 –a t = 2 B –1 (2) vA= 0 v=0 v = 6 ms C t = 3 –1 –a v = –6 ms Acceleration, a =

v –u 6–0 = t 1

= 6 ms–2