33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020
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33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG. All previous year questions , and these are downloaded from the sources in the internet. I don't own these resources and these are copyrighted by MTG.

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Current ElectricityTelegram @unacademyplusdiscountsHints & Explanations271. (a) : Flow of electrons, n t = 107 /secondTherefore, current( I)= q ne nt= t= t×e= 10 7 × (1.6 × 10 –19 ) = 1.6 × 10 –12 A2. (b) : The resistance of a wire of length l and area Aand resistivity r is given aslR = ρAGiven, l′ = nlAs the volume of the wire remains constantAl Al A\ A′l′ = Al or A′= = or A′=l ′ nl n22∴3.ρl′R′=ρnln ρlor R′= = =nRA′A Anl(b) : Resistance of a wire, R = ρ = 4 ΩA...(i)When wire is stretched twice, its new length be l′. Thenl′ = 2lOn stretching volume of the wire remains constant.\ lA = l′A′ where A′ is the new cross-sectional areaorlA′ = A = =l ′ l A A2 2\ Resistance of the stretched wire isR ′′ = lA′ = 2lA= l4( / 2)A= 4(4 W) = 16 W (Using (i))∆l4. (d) : = 01 . ∴ l = 11 .lBut the area also decreases by 0.1.Mass = rlA = Vr, ln l + ln A = ln mass.∴ ∆l+ ∆ A= ⇒ ∆ l= −∆A0l A l ALength increases by 0.1, resistance increases, areadecreases by 0.1, then also resistance will increase. Totalincrease in resistance is approximately 1.2 times, dueto increase in length and decrease in area. But specificresistance does not change.5. (b) : Resistance of wire =ρ l Al lR ∝ =A π2rWhen length and radius are both doubled2l1R1 ∝ ⇒RR2 1 ∝π( 2r)2The specific resistance of wire is independent of geometryof the wire, it only depends on the material of the wire.6. (c) : According to given parameters in questionl3 ρ 100R = ρ ⇒100Ω= ρ ⇒ = ⋅A A A 3Thus total resistance of 50 cm wire isρR1A l 100 50= = × 05 . = Ω.3 3The total current in the wire is I = 6100 A.Therefore potential difference across the two points onthe wire separated by a distance of 50 m is50 6V = IR1= × =3 100 1 V7. (a) : Three wires of lengths and cross-sectional areas= (l, A), (2l, A/2) and (l/2, 2A).lResistance of awire R ∝AFor I st wire, R 1 ∝ l/A = Rnd2lForII wire,R2∝ =4RA / 2rdl / 2 RFor III wire,R3∝ =2A4Therefore resistance of the wire will be minimum for III rdwire.8. (b) : Length (l) = 50 cm = 0.5 m;Area (A) = 1 mm 2 = 1 × 10 –6 m 2 ;Current (I) = 4A and voltage (V) = 2 volts.Resistance( R) = V.I= 24= 05ΩResistivity( ρ= ) × = . × × −RA 05 1 10 6= 1 × 10 –6 W ml 05 .9. (d) : m = l × area × densityArea ∝ m l2l lR ∝ ∝Area ml1 2 l2 2 l3 2 R1: R2 : R3= : :m1m2m3R1 : R2 : 25 9 1R3= 125 15 11: 3: 5 = : :10. (b) : Here, v d = 7.5 × 10 –4 m/s, E = 3 × 10 –10 V/mMobility, m = vd −475 . × 10=E −103×10m = 2.5 × 10 6

Telegram @unacademyplusdiscounts28 NEET-AIPMT Chapterwise Topicwise Solutions Physics11. (c) : Energy = 2 eV = eEl2 27∴ E = = = 5×10 V/mλ ×−84 1012. (a) 13. (a)14. (a) : For the negative resistance, when we increasethe voltage, the current will decrease. Therefore from thegraph, we find that the current in CD is decreased whenvoltage is increased.15. (d) : The colour code of the given resistor is yellow,violet, brown and gold.According to the colour code digits areYellow - 4Violet - 7Brown - 1and Gold = 5%\ R = 47 × 10 1 ± 5% = 470 W ± 5%16. (b) : (47 ± 4.7) kW = 47 × 10 3 ± 10% W\ Yellow – Violet – Orange – Silver17. (c)18. (c) : For metals, resistivity versus time graph is19. (d)rr 020. (a) : Specific resistance is a property of a materialand it increases with the increase of temperature, but notvary with the dimensions (length, cross-section) of theconductor.21. (a) : For metal specific resistance decreases withdecrease in temperature whereas for semiconductorspecific resistance increases with decrease in temperature.22. (a) : Fuse is an electrical safety device that operatesto provide overcurrent protection to an electrical circuit.23. (c) : Given, Q = at – bt 2dQ∴ I = = a−2btdtAt t = 0, Q = 0 ⇒ I = 0Also, I = 0 at t = a/2b\ Total heat produced in resistance R,a/ 2b a/2b22H = I Rdt = R ( a−2bt)dt∫00a/2b2 2 2= R ( a + 4bt − 4abt)dt∫0⎡⎤= R ⎢at+ b t − ab t a b2 2 3 2 / 24 4 ⎥⎣ 3 2 ⎦0∫T⎡2 32 ⎤= R 2 a 4ba 4aba⎢a× + × − × ⎥⎣ 2b3 328b2 4b⎦3 3aR⎡11 1⎤= + −b ⎣⎢⎦⎥ = aR2 6 2 6b24. (b) : Here,Distance between two cities = 150 kmResistance of the wire,R = (0.5 W km –1 )(150 km) = 75 WVoltage drop across the wire,V = (8 V km –1 )(150 km) = 1200 VPower loss in the wire isPV 2 ( 1200 V)2= = = 19200 W=19.2 kWR 75 Ω25. (c) : Power, P V 2=RAs the resistance of the bulb is constant∆P2∆V∴ =P V∆P2∆V%decreaseinpower = × 100 = × 100P V= 2 × 2.5% = 5%26. (c) : Power = 220 V × 4 A = 880 wattsHeat needed to raise the temperature of 1 kg waterthrough 80°C= ms.DT = 1 × 4200 × 80 J = 336 × 10 3 J3336 × 10∴ Time taken =880= 382 s = 6.3 min27. (a) : P = i 2 R or 1 = 25 × R1R = = 004 . Ω25( )28. (c) : In India, PI = 220 2 ; In USA, PRU = ( 110)2RU2 2( 220) ( 110)RAs PI= PU⇒ = ⇒ RU=R RU429. (b) : Fuse wire should have high resistance and lowmelting point.30. (b) : P V 21= or, R ∝RP\ R 40 > R 10031. (a) : H = I 2 Rt = msDTI1 2 ∆T1∆T1 I2 2TI T2 2 = or, ∆ 2 =∆ 2I1 2( 2I1)2∆T2= 5 ×I1 2 = 20°C

Telegram @unacademyplusdiscounts

28 NEET-AIPMT Chapterwise Topicwise Solutions Physics

11. (c) : Energy = 2 eV = eEl

2 2

7

∴ E = = = 5×

10 V/m

λ ×

−8

4 10

12. (a) 13. (a)

14. (a) : For the negative resistance, when we increase

the voltage, the current will decrease. Therefore from the

graph, we find that the current in CD is decreased when

voltage is increased.

15. (d) : The colour code of the given resistor is yellow,

violet, brown and gold.

According to the colour code digits are

Yellow - 4

Violet - 7

Brown - 1

and Gold = 5%

\ R = 47 × 10 1 ± 5% = 470 W ± 5%

16. (b) : (47 ± 4.7) kW = 47 × 10 3 ± 10% W

\ Yellow – Violet – Orange – Silver

17. (c)

18. (c) : For metals, resistivity versus time graph is

19. (d)

r

r 0

20. (a) : Specific resistance is a property of a material

and it increases with the increase of temperature, but not

vary with the dimensions (length, cross-section) of the

conductor.

21. (a) : For metal specific resistance decreases with

decrease in temperature whereas for semiconductor

specific resistance increases with decrease in temperature.

22. (a) : Fuse is an electrical safety device that operates

to provide overcurrent protection to an electrical circuit.

23. (c) : Given, Q = at – bt 2

dQ

∴ I = = a−2bt

dt

At t = 0, Q = 0 ⇒ I = 0

Also, I = 0 at t = a/2b

\ Total heat produced in resistance R,

a/ 2b a/

2b

2

2

H = I Rdt = R ( a−

2bt)

dt

∫

0

0

a/

2b

2 2 2

= R ( a + 4bt − 4abt)

dt

∫

0

⎡

⎤

= R ⎢at+ b t − ab t a b

2 2 3 2 / 2

4 4 ⎥

⎣ 3 2 ⎦0

∫

T

⎡

2 3

2 ⎤

= R 2 a 4b

a 4ab

a

⎢a

× + × − × ⎥

⎣ 2b

3 3

2

8b

2 4b

⎦

3 3

aR⎡1

1 1⎤

= + −

b ⎣

⎢

⎦

⎥ = aR

2 6 2 6b

24. (b) : Here,

Distance between two cities = 150 km

Resistance of the wire,

R = (0.5 W km –1 )(150 km) = 75 W

Voltage drop across the wire,

V = (8 V km –1 )(150 km) = 1200 V

Power loss in the wire is

P

V 2 ( 1200 V)

2

= = = 19200 W=

19.

2 kW

R 75 Ω

25. (c) : Power, P V 2

=

R

As the resistance of the bulb is constant

∆P

2∆V

∴ =

P V

∆P

2∆V

%decreaseinpower = × 100 = × 100

P V

= 2 × 2.5% = 5%

26. (c) : Power = 220 V × 4 A = 880 watts

Heat needed to raise the temperature of 1 kg water

through 80°C

= ms.DT = 1 × 4200 × 80 J = 336 × 10 3 J

3

336 × 10

∴ Time taken =

880

= 382 s = 6.3 min

27. (a) : P = i 2 R or 1 = 25 × R

1

R = = 004 . Ω

25

( )

28. (c) : In India, PI = 220 2 ; In USA, P

R

U = ( 110)

2

RU

2 2

( 220) ( 110)

R

As PI

= PU

⇒ = ⇒ RU

=

R RU

4

29. (b) : Fuse wire should have high resistance and low

melting point.

30. (b) : P V 2

1

= or, R ∝

R

P

\ R 40 > R 100

31. (a) : H = I 2 Rt = msDT

I1 2 ∆T1

∆T1 I2 2

T

I T

2 2 = or, ∆ 2 =

∆ 2

I1 2

( 2I1)

2

∆T2

= 5 ×

I1 2 = 20°

C

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