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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Alternating CurrentTelegram @unacademyplusdiscounts67R 30 1As, power factor, cosφ= = = = cos 60°Z 60 2⇒ f = 60°, i.e., current lags behind the emf.So, we can conclude that the circuit is a series LR.12. (c) : Quality factor of an L-C-R circuit is given by,LQ = 1R C1 15 .Q 1 =20 −635 × 101 25 .Q 2 = ×25−645 × 101 35 .Q 3 =15 −630 × 103= 50 × = 10.35705= 40 × = 943 .90100 35= = 22.7715 31 15 . 40Q 4 = × = = 730 .25−645 × 10 30Clearly Q 3 is maximum of Q 1 , Q 2 , Q 3 , and Q 4 .Hence, option (c) should be selected for better tuning ofan L-C-R circuit.13. (d) : Current through resistor,i = Current in the circuitV0V0= =2 2 2 2R + X R + ( 1/ωC)CVoltage across capacitor, V = iX CV01=×2 2R + ( 1/ωC)ω C= V02 2 2R ω C + 1As C a < C b\ i a < i b and V a > V b14. (b) : Here, R = 3 W, X L = 3 WThe phase difference between the applied voltage and thecurrent in the circuit isXL 3 Ω−1 πtanφ= = = 1 or φ = tan ( 1 ) =R 3 Ω415. (b) : Here, Resistance, R = 30 WInductive reactance, X L = 20 W at 50 HzQ X L = 2puLυ′ = ′ × = ⎛ ⎝ ⎜ 100 ⎞XLXL⎠⎟ × 20 Ω=40 Ωυ 502 2 2 2Impedance, Z = R + ( X L ′ ) = ( 30) + ( 40 ) = 50 WVCurrent in thecoil, I = VAZ= 20050= 4Ω16. (b) :As V L = V C = 300 V, therefore the given series LCR circuitis in resonance.\ V R = V = 220 V, Z = R = 100 WVCurrent, I = V. AZ= 220100= 22ΩHence, the reading of the voltmeter V 3 is 220 V and thereading of ammeter A is 2.2 A.17. (d) : In series LCR, current is maximum at resonance.∴ Resonant frequency, ω =1LC∴2ω = or, L =LC2ω CGiven w = 1000 s –1 and C = 10 mF1∴ L == 01 . H = 100 mH.−61000 × 1000 × 10 × 101XC − X18. (d) : tanφ=or tan π − ωL⎛ ⎞ ωR ⎝⎜4 ⎠⎟ = CR111R = −ω L or ( R+ 2πfL)= or C =ωC2πfC2πf( R+2πfL)LR19. (a, b) : Quality factor, Q = ωSince ω 2 1= LC1∴ Quality factor, Q =ωRC20. (b) : For resonance condition, the impedance willbe minimum and the current will be maximum. This isonly possible when X L = X C .XL+ XCThereforetanφ== 0 or φ=0.R21. (a) : Impedance Z in an ac circuit is2 2Z = R + ( XC− XL) ; where X C = capacitive reactanceand X L = inductive reactance.1Also XC= and XL= ωLωC22 ⎛ 1−3⎞∴ Z = ( 50)+− × ×⎝⎜314 20 10−6× ×⎠⎟314 100 10or Z = 56 WVThe power loss in the circuit isrmsPav= ⎛ 2R⎝ ⎜ ⎞Z ⎠⎟∴ = ⎛ 2P 10 ⎞av⎝ ⎜ ⎠⎟256× 50 = 079 . W( )22. (c) : Here, V R = 80 V, V C = 40 V, V L = 100 VPowerfactor, cosφ== R Z = VR= VRV 2 2VR + ( VL −VC)80 80== = 08 .2 2( 80) + ( 100 −40)100

Telegram @unacademyplusdiscounts68 NEET-AIPMT Chapterwise Topicwise Solutions Physics23. (c) : Here, L = 20 mH = 20 × 10 –3 H,C = 50 mF = 50 × 10 –6 F, R = 40 W,V = 10sin 340t = V 0 sinwtw = 340 rad s –1 , V 0 = 10 VX L = wL = 340 × 20 × 10 –3 = 6.8 W4XC = 1C= 1 10=−340 × 50 × 10 34 × 5= 58. 82 Ωω6Z = R 2 + ( XC− XL) 2 = ( 40) 2 + ( 58. 82 −68. )22 2= ( 40) + ( 52. 02) = 65.62 ΩThe peak current in the circuit isVRI0= 0 10Z= 40A,cosφ65 62= Z= ⎛ ⎝ ⎜ ⎞.65.62 ⎠⎟Power loss in A.C. circuit,= V I cos φ 1= VI cos φrms rms2 0 01= × × × =2 10 10 40046 . W65. 62 65.6224. (c) : When L is removed, the phase differencebetween the voltage and current istanφ 1 = X CRπ XCtan =3 Ror XC= Rtan60° or XC= 3RWhen C is removed, the phase difference between thevoltage and current istanφ 2 = XL π XorLtan =R 3 Ror XL= Rtan60°=3RAs X L = X C , the series LCR circuit is in resonance.Impedance of the circuit,2 2Z = R + ( XL− XC)= RR RPower factor, cosφ= = =Z R 1125. (d) : Given: i= sin( 100πt)ampere2Compare it with i = i 0 sin(wt), we get1i 0 = A21 ⎛ π ⎞Given: e= sin t+volt⎝⎜100π⎠⎟23Compare it with e = e 0 sin(wt + f), we get1e 0 = V, φ =π2 3i0 1e01∴ irms= = Aand erms= = V2 2 2 2Average power consumed in the circuit,P = i rms e rms cosf= ⎛ ⎝ ⎜ 1 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ 1 ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ 1 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ 1 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ 1 ⎞2 2 3 2 2 2 ⎠ ⎟ = 1cos πW826. (d) : Average power, P = E r.m.s I r.m.s cosf2 2 RZ = R + ( XL− XC) ,cosφ =ZEr.m.s2 RBut Ir.m.s= ∴ P = Er.m.s⋅Z2ZR22∴ P = Er.m.s2 2{ R + ( XL−XC)} = ε R⎡2⎛ ⎞⎤2 1⎢R+ −⎝⎜Lω⎠⎟ ⎥⎣Cω⎦27. (a) : Averagepower = EI 0 02 cosφ28. (d) : X L = 31 W, X C = 25 W, R = 8 WImpedance of series LCR is2 2Z = ( R ) + ( XL− XC)2 2= () 8 + ( 31− 25)= 64+ 36 = 10 Ω .R 8Power factor, cos φ= = = 08 .Z 1029. (c) : The impedance Z of a series LCR circuit isgiven by, Z R2= + X L − X2( C)1where X L = wL and X C = , w is angular frequency.ωCAt resonance, X L = X C , hence Z = R.\ V R = V (supply voltage)VRV∴ r.m.s. current, I = =R R\ Power loss = I 2 R = V 2 /R30. (b) : The dissipation of power in an a.c. circuit is (P)= V × I × cosq. Therefore current flowing in the circuitdepends upon the phase angle between voltage (V) andcurrent (I) of the a.c. circuit.31. (b) : Current (I) = 5 sin (100t – p/2) and voltage (V)= 200 sin (100t). Comparing the given equation, with thestandard equation, we find that phase between currentπand voltage is φ = = 90 °2Power consumption P = I rms V rms cosf = I rms V rms cos90° = 032. (d) : In case of oscillatory discharge of a capacitorthrough an inductor, charge at instant t is given byq = q 0 coswt; where, ω=1 LC q CV2V2∴ cos ωt = = = ( Q q=CV)q0CV1V1Current through the inductordqI = ddt= (dt q 0cos ω t ) =−q0ωsinωt12 1| I| = CV1[ −cos t] / 21 ωLC( )⎡= − ⎛ ⎞⎤ ⎡ − ⎤VC 2V12 /CV V⎢⎝ ⎜ 2 ⎥⎠⎟= ⎢ 1 2 2 2 12 /⎥1 1L ⎣⎢V1⎦⎥⎣ L ⎦

Telegram @unacademyplusdiscounts

68 NEET-AIPMT Chapterwise Topicwise Solutions Physics

23. (c) : Here, L = 20 mH = 20 × 10 –3 H,

C = 50 mF = 50 × 10 –6 F, R = 40 W,

V = 10sin 340t = V 0 sinwt

w = 340 rad s –1 , V 0 = 10 V

X L = wL = 340 × 20 × 10 –3 = 6.8 W

4

XC = 1

C

= 1 10

=

−

340 × 50 × 10 34 × 5

= 58. 82 Ω

ω

6

Z = R 2 + ( XC

− XL) 2 = ( 40) 2 + ( 58. 82 −68

. )

2

2 2

= ( 40) + ( 52. 02) = 65.

62 Ω

The peak current in the circuit is

V

R

I0

= 0 10

Z

= 40

A,cosφ

65 62

= Z

= ⎛ ⎝ ⎜ ⎞

.

65.

62 ⎠

⎟

Power loss in A.C. circuit,

= V I cos φ 1

= VI cos φ

rms rms

2 0 0

1

= × × × =

2 10 10 40

046 . W

65. 62 65.

62

24. (c) : When L is removed, the phase difference

between the voltage and current is

tanφ 1 = X C

R

π XC

tan =

3 R

or XC

= Rtan60° or XC

= 3R

When C is removed, the phase difference between the

voltage and current is

tanφ 2 = XL π X

or

L

tan =

R 3 R

or XL

= Rtan60°=

3R

As X L = X C , the series LCR circuit is in resonance.

Impedance of the circuit,

2 2

Z = R + ( XL

− XC)

= R

R R

Power factor, cosφ= = =

Z R 1

1

25. (d) : Given: i= sin( 100πt)

ampere

2

Compare it with i = i 0 sin(wt), we get

1

i 0 = A

2

1 ⎛ π ⎞

Given: e= sin t+

volt

⎝

⎜100π

⎠

⎟

2

3

Compare it with e = e 0 sin(wt + f), we get

1

e 0 = V, φ =

π

2 3

i0 1

e0

1

∴ irms

= = Aand erms

= = V

2 2 2 2

Average power consumed in the circuit,

P = i rms e rms cosf

= ⎛ ⎝ ⎜ 1 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ 1 ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ 1 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ 1 ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ 1 ⎞

2 2 3 2 2 2 ⎠ ⎟ = 1

cos π

W

8

26. (d) : Average power, P = E r.m.s I r.m.s cosf

2 2 R

Z = R + ( XL

− XC) ,cosφ =

Z

Er.m.s

2 R

But Ir.m.s

= ∴ P = Er.m.s

⋅

Z

2

Z

R

2

2

∴ P = Er.m.s

2 2

{ R + ( XL

−XC)} = ε R

⎡

2

⎛ ⎞

⎤

2 1

⎢R

+ −

⎝

⎜Lω

⎠

⎟ ⎥

⎣

Cω

⎦

27. (a) : Averagepower = EI 0 0

2 cosφ

28. (d) : X L = 31 W, X C = 25 W, R = 8 W

Impedance of series LCR is

2 2

Z = ( R ) + ( XL

− XC)

2 2

= () 8 + ( 31− 25)

= 64+ 36 = 10 Ω .

R 8

Power factor, cos φ= = = 08 .

Z 10

29. (c) : The impedance Z of a series LCR circuit is

given by, Z R

2

= + X L − X

2

( C)

1

where X L = wL and X C = , w is angular frequency.

ωC

At resonance, X L = X C , hence Z = R.

\ V R = V (supply voltage)

VR

V

∴ r.m.s. current, I = =

R R

\ Power loss = I 2 R = V 2 /R

30. (b) : The dissipation of power in an a.c. circuit is (P)

= V × I × cosq. Therefore current flowing in the circuit

depends upon the phase angle between voltage (V) and

current (I) of the a.c. circuit.

31. (b) : Current (I) = 5 sin (100t – p/2) and voltage (V)

= 200 sin (100t). Comparing the given equation, with the

standard equation, we find that phase between current

π

and voltage is φ = = 90 °

2

Power consumption P = I rms V rms cosf = I rms V rms cos90° = 0

32. (d) : In case of oscillatory discharge of a capacitor

through an inductor, charge at instant t is given by

q = q 0 coswt; where, ω=

1 LC

q CV2

V2

∴ cos ωt = = = ( Q q=

CV)

q0

CV1

V1

Current through the inductor

dq

I = d

dt

= (

dt q 0cos ω t ) =−q

0ωsinω

t

1

2 1

| I| = CV1

[ −cos t] / 2

1 ω

LC

( )

⎡

= − ⎛ ⎞

⎤ ⎡ − ⎤

V

C 2

V

12 /

CV V

⎢

⎝ ⎜ 2 ⎥

⎠

⎟

= ⎢ 1 2 2 2 12 /

⎥

1 1

L ⎣⎢

V1

⎦⎥

⎣ L ⎦

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