paper 2 questions & solutions
TRANSCRIPT
Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph.No. : 0744-2777777, 0744-2777700 | Toll Free : 1800 258 5555 | FAX No. : +91-022-39167222 | TToo KKnnooww mmoorree :: ssmmss RREESSOO aatt 5566667777
Website : www.resonance.ac.in | E-mail : ccoonnttaacctt@@rreessoonnaannccee..aacc..iinn | CCIINN :: UU8800330022RRJJ22000077PPLLCC002244002299
TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((AADDVVAANNCCEEDD)) 22002211 SSoolluuttiioonn ppoorrttaall
22-05-2016JEE (Advanced) 2016
®
DATE: 03-10-2021
JEE (ADVANCED) 2021
PAPER 2
Questions & Solutions
Time : 2:30 pm – 05:30 pm | Duration : 3 Hrs. | Total Marks : 180
SSUUBBJJEECCTT:: PPHHYYSSIICCSS
Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph.No. : 0744-2777777, 0744-2777700 | Toll Free : 1800 258 5555 | FAX No. : +91-022-39167222 | TToo KKnnooww mmoorree :: ssmmss RREESSOO aatt 5566667777
Website : www.resonance.ac.in | E-mail : ccoonnttaacctt@@rreessoonnaannccee..aacc..iinn | CCIINN :: UU8800330022RRJJ22000077PPLLCC002244002299
TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((AADDVVAANNCCEEDD)) 22002211 SSoolluuttiioonn ppoorrttaall
PAPER-2 : INSTRUCTIONS TO CANDIDATES
Question Paper-2 has three (03) parts: Physics, Chemistry and Mathematics.
Each part has a total Nineteen (19) questions divided into four (04) sections (Section-1, Section-2, Section-3 & Section-4)
Total number of questions in Question Paper-2 are Fifty Seven (57) and Maximum Marks are One Hundred and Eighty (180).
Type of Questions and Marking Schemes SECTION-1
• This section contains SIX (06) questions. • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is (are) correct answer(s). • For each question, choose the option(s) corresponding to (all) the correct answer(s). • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +4 If only (all) the correct option(s) is(are) chosen; Partial Marks : +3 If all the four options are correct but ONLY three options are chosen; Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of
which are correct; Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct
option; Zero Marks : 0 If unanswered; Negative Marks : −2 In all other cases. • For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1 mark; choosing ONLY (B) will get +1 mark; choosing ONLY (D) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get −2 marks.
SECTION 2 • This section contains THREE (03) question stems. • There are TWO (02) questions corresponding to each question stem. • The answer to each question is a NUMERICAL VALUE. • For each question, enter the correct numerical value corresponding to the answer in the designated place using the mouse and the on-screen virtual numeric keypad. • If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal
places. • Answer to each question will be evaluated according to the following marking scheme: Full Marks : +2 If ONLY the correct numerical value is entered at the designated place; Zero Marks : 0 In all other cases
SECTION 3 • This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02) questions. • Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct
answer. • For each question, choose the option corresponding to the correct answer. • Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −1 In all other cases.
Resonance Eduventures Limited REGISTERED & CORPORATE OFFICE : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph.No. : 0744-2777777, 0744-2777700 | Toll Free : 1800 258 5555 | FAX No. : +91-022-39167222 | TToo KKnnooww mmoorree :: ssmmss RREESSOO aatt 5566667777
Website : www.resonance.ac.in | E-mail : ccoonnttaacctt@@rreessoonnaannccee..aacc..iinn | CCIINN :: UU8800330022RRJJ22000077PPLLCC002244002299
TThhiiss ssoolluuttiioonn wwaass ddoowwnnllooaadd ffrroomm RReessoonnaannccee JJEEEE ((AADDVVAANNCCEEDD)) 22002211 SSoolluuttiioonn ppoorrttaall
SECTION 4
• This section contains THREE (03) questions.
• The answer to each question is a NON-NEGATIVE INTEGER.
• For each question, enter the correct integer corresponding to the answer using the mouse and the on-
screen virtual numeric keypad in the place designated to enter the answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If ONLY the correct integer is entered;
Zero Marks : 0 In all other cases.
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 1
7340010333
PART-I : PHYSICS
SECTION 1
• This section contains SIX (06) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four
option(s) is (are) correct answer(s).
• For each question, choose the option(s) corresponding to (all) the correct answer(s).
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is(are) chosen;
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen;
Partial Marks : +2 If three or more options are correct but ONLY two options are chosen, both of
which are correct;
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct
option;
Zero Marks : 0 If unanswered;
Negative Marks : −2 In all other cases. • For example, in a question, if (A), (B) and (D) are the ONLY three options corresponding to correct
answers, then choosing ONLY (A), (B) and (D) will get +4 marks; choosing ONLY (A) and (B) will get +2 marks; choosing ONLY (A) and (D) will get +2marks; choosing ONLY (B) and (D) will get +2 marks; choosing ONLY (A) will get +1 mark; choosing ONLY (B) will get +1 mark; choosing ONLY (D) will get +1 mark; choosing no option(s) (i.e. the question is unanswered) will get 0 marks and choosing any other option(s) will get −2 marks.
1. One end of a horizontal uniform beam of weight W and length L is hinged on a vertical wall at point O
and its other end is supported by a light inextensible rope. The other end of the rope is fixed at point Q,
at a height L above the hinge at point O. A block of weight 𝛼W is attached at the point P of the beam, as
shown in the figure (not to scale). The rope can sustain a maximum tension of ( 22 )W. Which of the
following statement(s) is(are) correct?
(A) The vertical component of reaction force at O does not depend on 𝛼
(B)The horizontal component of reaction force at O is equal to W for 𝛼 = 0.5
(C) The tension in the rope is 2W for 𝛼 = 0.5
(D) The rope breaks if 𝛼 > 1.5
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 2
7340010333
Ans. (ABD)
Sol.
Ny 45°
W W
Nx
T
W
2W45sinT o
W2
W
2T
2
2
1WT
(D) To break the string T > W22
2
2
1W > W22
2
3
(C) = 0.5 = 1/2 ,
2
12
2
1WT = W2
(B) ox 45cosTN = W2 ×
2
1 Nx = W
(A) Nv + Tsin45° = W+W Nv + 2
W+ W = W + W Nv =
2
W It does not depend on .
2. A source, approaching with speed u towards the open end of a stationary pipe of length 𝐿, is emitting a
sound of frequency fs. The farther end of the pipe is closed. The speed of sound in air is v and f0 is the fundamental frequency of the pipe. For which of the following combination(s) of u and fs, will the sound reaching the pipe lead to a resonance? (A) u = 0.8 v and fs = f0 (B) u = 0.8 v and fs = 2 f0 (C) u = 0.8 v and fs = 0.5 f0 (D) u = 0.5 v and fs = 1.5 f0
Ans. (AD)
Sol.
u
fs
Source
The appeared frequency of the sound will be
uv
0vf'f s , and for resonance, it should match with any
of the natural frequency of the closed organ pipe. fundamental frequency of the closed organ pipe is f0, then its natural frequencies will be f0, 3f0, 5f0…….(2n – 1)f0 for resonance
0s f)1n2(uv
vf
0s f)1n2(v
u1f
(A, B, C) u = 0.8 v fs = (1 – 0.8)(2n – 1)f0
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 3
7340010333
fs = 0f5
1n2
n = 1, fs = 5
f0
n = 2, fs = 5
f3 0
n = 3, fs = 00 f
5
f5
n = 4, fs = 5
f7 0
n = 5, fs = 5
f9 0
n = 6, fs = 5
f11 0
So option (A) is correct, (B) and (C) are in correct (D) u = 0.5 v fs = (1 – 0.5)(2n – 1)f0
fs = 0f2
1n2
n = 1 fs = 2
f0
n = 2 fs = 00 f5.1
2
f3
n = 3 fs = 2
f5 0
3. For a prism of prism angle = 60°, the refractive indices of the left half and the right half are,
respectively, n1 and n2 (n2 ≥ n1) as shown in the figure. The angle of incidence i is chosen such that the
incident light rays will have minimum deviation if n1 = n2 = n = 1.5. For the case of unequal refractive
indices, n1 = n and n2 = n + ∆ n (where ∆n << n1), the angle of emergence e = i + ∆e. Which of the
following statement(s) is(are) correct?
(A) The value of ∆e (in radians) is greater than that of ∆n
(B) ∆e is proportional to Δn
(C) Δe lies between 2.0 and 3.0 milliradians, if ∆n = 2.8 × 10−3
(D) Δe lies between 1.0 and 1.6 milliradians, if ∆n = 2.8 × 10−3
Ans. (BC)
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 4
7340010333
Sol.
ie
A=60°
rm=30° rm=30°
n1 n2
n1 = n2 = n = 2
3 for minimum deviation
302
60
2
Arm
and i = e =
2
Asin
2
Asin
n
min
4
7)cos(,
4
3)sin(
2
60sin
sin
2
3
If n1 = n = 2
3and n2 = n + n
e = +
(1) sin = n sin30°
(n + n) sin30° = (1) sin( + )
Solving 2
1 (n) = sin( + ) – sin
sin)sin(
2
n
cosd
)(sind
cos2
n
4
7
2
n
n
2
64.2
2
7
134.1n
n , so option (A) is incorrect
n = (1.34) n option (B) is correct
2
7108.2
2
7n 3
33
108.077
106.5
= (2.64 × 0.8) × 10–3 = 2.11 × 10–3 rad
option (C) is correct
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 5
7340010333
4. A physical quantity S
is defined as
0
BES
, where E
is electric field, B
is magnetic field and 𝜇0 is the
permeability of free space. The dimensions of S
are the same as the dimensions of which of the
following quantity(ies) ?
(A) CurrenteargCh
Energy
(2)
TimeLength
Force
(3)
volume
Energy (4)
Area
Pow er
Ans. (BD)
Sol. 0
BEs
0
BE
is pointing vector and its dimension is same as that of intensity (w/m2)
Dimension = 2
22
TL
TML MT
–3
Alternate Solutions
Dimension of Electric field (E) , [E] = MA–1
LT–3
Dimensions of Magnetic field (B); [B] = MA–1
T–2
Dimensions of magnetic permeability (0) ; [0] = MA–2
T–2
L
[S] = LTMA
)TMA)(LTMA(EB22
2131
0
= MT–3
5. A heavy nucleus N, at rest, undergoes fission N → P + Q, where P and Q are two lighter nuclei. Let 𝛿 = MN – MP – MQ, where MP, MQ and MN are the masses of P, Q and N, respectively. EP and EQ are
the kinetic energies of P and Q, respectively. The speeds of P and Q are vP and vQ, respectively. If c is
the speed of light, which of the following statement(s) is(are) correct?
(A) EP + EQ = c2 (B)
2
QP
PP c
MM
ME (C)
P
Q
Q
P
M
M
v
v
(D) The magnitude of momentum for P as well as Q is 2c , where 𝜇 = )MM(
MM
QP
QP
Ans. (ACD)
Sol.
N P P
P Q
(Ptotal) = 0 (Ptotal)f = 0
So momentum of one nucleus is P in forward direction then momentum of the other nucleus will be P in
backwards direction
Energy released = (m)C2 = C2 = KEP + KEQ
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 6
7340010333
P
2
Pm2
PKE ,
Q
2
Qm2
PKE
QPQP
m
1:
m
1KE:KE = mQ : mP
2
QP
QP C
mm
mKE
2
QP
PQ C
mm
mKE
KEP + KEQ = C2
2
Q
2
P
2
Cm2
P
m2
P
QP
QP
mm
mm2CP
6. Two concentric circular loops, one of radius R and the other of radius 2R, lie in the xy-plane with the
origin as their common center, as shown in the figure. The smaller loop carries current 1 in the anti-
clockwise direction and the larger loop carries current 2 in the clockwise direction, with 2 > 21. )y,x(B
denotes the magnetic field at a point (x, y) in the xy-plane. Which of the following statement(s) is(are) correct?
(A) )y,x(B
is perpendicular to the xy-plane at any point in the plane
(B) | )y,x(B
| depends on x and y only through the radial distance 22 yxr
(C) | )y,x(B
| is non-zero at all points for r < R
(D) )y,x(B
points normally outward from the xy-plane for all the points between the two loops
Ans. (AB)
Sol.
x
y
i1,R
i2, 2R
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 7
7340010333
(A) Magnetic field at the plane of the ring is perpendicular to the plane. However they bend as they
move forward.
(B) By symmetry, we can say that B
will be same at all the points having the same radial distance. so B
(x,y) will depend only the radial distance r = 22 yx
(C) (Bnet)centre = 1201020 i2iR4R2
i
)R2(2
i
Since i2 > 2i1 so Bnet at the centre will be non zero in
direction. But at some other point, Bnet may be zero.
From the graph , it is clear that Bnet = 0 at for r (O, R). So option (C) is incorrect
=
i1,R
i2, 2R
= –
ve
ve
B
r=0 r=R r=2R r
(D) For the graph , it is clear that B = –Ve indirection for r(R to 2R), So option D is also incorrect.
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 8
7340010333
SECTION 2
• This section contains THREE (03) question stems.
• There are TWO (02) questions corresponding to each question stem.
• The answer to each question is a NUMERICAL VALUE.
• For each question, enter the correct numerical value corresponding to the answer in the designated place
using the mouse and the on-screen virtual numeric keypad.
• If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal
places.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +2 If ONLY the correct numerical value is entered at the designated place;
Zero Marks : 0 In all other cases
Question Stem for Question Nos. 7 and 8
Question Stem
A soft plastic bottle, filled with water of density 1 gm/cc, carries an in be with some airverted glass test-tu
(ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm made of a thick, and it is
glass of density 2.5 gm/cc. Initially the bottle is sealed at atmospheric pressure 𝑝0 0= 1 5 Pa so that the
volume of the trapped air is v0 side at constant= 3.3 cc. When the bottle is squeezed from out
temperature, the pressure inside rises and the volume of the trapped air reduces. It und that the testis fo
tube begins to sink at pressure p0 + e, the volume ofp without changing its orientation. At this pressur
the trapped air is v0 − v.
Let v = X cc and p = Y × 103 Pa.
7. The value of X is .
Ans.
8.
Ans.
0.3
The value of Y is .
10.00
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222 To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 9
7340010333
Sol. As soon as, the upwards and downward force on the tube are balanced, the tube will just start sinking.
(w) (Vtube + Vair) g = (mtube) g
(2cm3 + Vair) g = (5 gm) g
mg
B = (w) (Vtube + Vair) g
Vair = 3 cm, while the initial volume of the air was 3.3 cm3 so the decrease in the volume
V = 3.3 – 3 = 0.3 cm3
Vtube = 3
tube
tube
cm/gm5.2
gm5m
Vtube = 2 cm3
Since the temperature of the tapped air remains constant, so PV = constant
(P+P) (V – V) = PV
P = 10 × 103 = Y × 103
Y = 10
Question Stem for Question Nos. 9 and 10
Question Stem
A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length
L = 1.0 m. It is suspended from a fixed point at height H = 0.9 m above a frictionless horizontal floor.
Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A
horizontal impulse P = 0.2kg-m/s is imparted to the bob at some instant. After the bob slides for some
distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular
momentum of the pendulum about the point of suspension just before the bob lifts off is J kg-m2/s. The
kinetic energy of the pendulum just after the lift- off is K Joules.
9. The value of J is .
Ans. 0.18
10. The value of K is .
Ans. 0.16
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 10
7340010333
Sol.
1m
J V = 2m/s
0.9m
Angular momentum about top point = mv × 0.9
= 0.2 × 0.9
0.18 kg m2/s
When string get taut a Impulse act by string on particle.
1m
J
vcos 0.9m
90–
vsin
Due to impulse, component of V along string get change but component perpendicular to v remain same
Vcos0.9
V1 = 0 (By string constraint)
New velocity = Vcos = V × 0.9 So new momentum = 0.9 × Pinitial = 0.9 × 0.2 = 0.18
KE = 1.02
)18.0(
m2
P 22
= 0.162 Joule
Question Stem for Question Nos. 11 and 12
Question Stem
n a circuit, a metal filament lamp is connected in series with a capacitor of capacitance C F across a
200 V, 50 Hz supply. The power consumed by the lamp is 500 W while the voltage drop across it is 100
V. Assume that there is no inductive load in the circuit. Take rms values of the voltages. The magnitude
of the phase- angle (in degrees) between the current and the supply voltage is . Assume, 3 5
11. The value of C is ___________
Ans. 100
12. The value of is ___________
Ans. 60
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 11
7340010333
Sol. For lamp
R
VP
2
20500
100100R
i = A520
100
R
V
2c
2 xR
Vi
2
2
C502
1)20(
2005
1600C100
1400
2
1200C100
12
4
22101200
C
1
62
1012
1C 310
12
1C
V = 2R
2c VV
22C 100V200 3100VC
3100
3100
V
V
R
XCtan
R
C
= 60°SECTION 3
• This section contains TWO (02) paragraphs. Based on each paragraph, there are TWO (02) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct
answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : −1 In all other cases.
PARAGRAPH 13 TO 14
A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not
allow any change in flux through itself by inducing a suitable current to generate a compensating flux.
The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a
magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of
radius 𝑎, with its center at the origin. A magnetic dipole of moment m is brought along the axis of this
loop from infinity to a point at distance r (>> a) from the center of the loop with its north pole always
facing the loop, as shown in the figure below.
The magnitude of magnetic field of a dipole m, at a point on its axis at distance r,is 3
0
r
m
2
, where 𝜇 is
the permeability of free space. The magnitude of the force between two magnetic dipoles with moments,
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 12
7340010333
m1 and m2, separated by a distance r on the common axis, with their north poles facing each other, is
4
21
r
mkm, where k is a constant of appropriate dimensions. The direction of this force is along the line
joining the two dipoles.
13. When the dipole 𝑚 is placed at a distance 𝑟 from the center of the loop (as shown in the figure), the
current induced in the loop will be proportional to
(A) 3r
m (B)
2
2
r
m (C)
2r
m (D)
r
m2
Ans. (A)
14. The work done in bringing the dipole from infinity to a distance 𝑟 from the center of the loop by the given
process is proportional to :
(A) 5r
m (B)
5
2
r
m (C)
6
2
r
m (D)
7
2
r
m
Ans. (C)
Sol.
N S r
m
Since it is a super conducting loop, So net flux (self flux + external flux) passing through it will remain constant
(total)i = (total)f
L(0) + (0)(pa2) = Li –
3
0
r
m
2, Here L = self inductance of the super conducting loop.
33
0
r
mi
r
m
L2i
(ii) The current carring superconducting loop will also behave like a magnet, whose magnetic dipole
moment m1 = NiA =(1) 2
3
0 ar
m
L2
31
r
mm
The repulsive force felt be the magnet will be
F = 7
2
4
3
4
21
r
m
r
)m(r
m)K(
r
mKm
Wext = drr
m–Fdr–7
2
Wext 6
2
r
m
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 13
7340010333
PARAGRAPH 15 TO 16
A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle,
as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific
heat at constant volume, CV = 2R. Here, R is the gas constant. Initially, each side has a volume V0 and
temperature T0. The left side has an electric heater, which is turned on at very low power to transfer heat
Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right
side volume to 2
V0 . Consequently, the gas temperatures on the left and the right sides become TL and TR,
respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.
15. The value of
0
R
T
T is
(A) 2 (B) 3 (C) 2 (D) 3
Ans. (A)
16. The value of 0RT
Q is
(A) 1224 (B) 1224 (C) 125 (D) 125
Ans. (B)
Sol.
gas (1) gas (2)
Left Right
P0, T0, V0, n = 1 P0 , T0, V0, n = 1
TL , f0 P,
2
V3 fR
0 P,T,2
V
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 14
7340010333
R2R2
fCV f = 4,
2
3
4
21
8
21
The gas of the right portion is undergone through an adiabatic process so
1
ff1
ii VTVT
12
3
0R
12
3
002
VTVT
0R T2T 2T
T
0
R
Work done by the gas (2) = – U = – nCV (Tf – Ti)
= – (1) (2R) 00 TT2
= 0RT222
So work done by the gas (1) = + 0RT222
PV = nRT V
TP
for gas (2) T 2 times, volume 2
1times
so 222/1
2P times Pf = 0P22
so pressure of the gas (1) will also be 22 times and volume of gas (1) becomes 2
3 times
PVT )2
322(T
= 23 times
TL = 0T23
U for the left gas = n CV (Tf – Ti)
= (1) (2R) 000 RT226TT23
Heat given to gas (1) = W + U
= 00 RT226RT222
= 122RT4 0
SECTION 4
• This section contains THREE (03) questions.
• The answer to each question is a NON-NEGATIVE INTEGER.
• For each question, enter the correct integer corresponding to the answer using the mouse and the on-
screen virtual numeric keypad in the place designated to enter the answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If ONLY the correct integer is entered;
Zero Marks : 0 In all other cases.
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 15
7340010333
17. In order to measure the internal resistance r1 of a cell of emf E, 𝐸, a meter bridge of wire resistance
R0 = 50, a resistance R0/2, another cell of emf E/2 (internal resistance r) and a galvanometer G are used
in a circuit, as shown in the figure. If the null point is found at =72 cm, then the value of r1 = ___
Ans. 3
Sol.
72 cm
100 m
/236
G
r/2
R0/2=25r1
14
i = 0
O
Resistance of potential wire is R0 = 50
Resistance of 100 m wire = 50
So Resistance of 72 cm wire = 3672100
50
Current
36r
2/
2514
2/
1
r1 = 39 – 36 = 3
18. The distance between two stars of masses 3MS and 6MS is 9R. Here R is the mean distance between
the centers of the Earth and the Sun, and MS is the mass of the Sun. The two stars orbit around their
common center of mass in circular orbits with period nT, where T is the period of Earth’s revolution
around the Sun. The value of n is ___
Ans. 9
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 16
7340010333
Sol. GMs
R2T
2/3
for binary star
)MM(G
r2'T
21
2/3
Put M1 = 3Ms
M2 = 6Ms
r = 9R
)Ms6Ms3(G
R92'T
2/32/3
2/1
2/32/3
9GMs
R92
GMs
9R2 2/3
T' = 9T
value of n = 9
19. In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q, and
R are EP, EQ and ER, respectively, and they are related by EP = 2EQ = 2ER. In this experiment, the
same source of monochromatic light is used for metals P and P while a different source of
monochromatic light is used for the metal R. The work functions for metals P, Q and R are 4.0 eV,
4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in eV, is :
Ans. 6
Sol. ppp0 k4kE ……..(1)
qqq0 k5.4kE ……..(2)
rr' kE
from (1) and (2)
4 + kp = 4.5 + kq
kp = 2kr = 2kq
4 + kp = 4.5 + kq
4 + 2kq = 4.5 + kq
kq = 0.5 eV
kp = 2 × 0.5 = 1eV
kr =2
1
2
kp = 0.5
E = r + kr
= 5.5 + 0.5
= 6.0 eV
| JEE (ADVANCED) 2021 | DATE : 03-10-2021 | PAPER-2 | PHYSICS
Resonance Eduventures Ltd. Reg. Office & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005
Ph. No.: +91-744-2777777, 2777700 | FAX No. : +91-022-39167222
To Know more : sms RESO at 56677 | Website : www.resonance.ac.in | E-mail : [email protected] | CIN : U80302RJ2007PLC024029
This solution was download from Resonance JEE (ADVANCED) 2021 Solution portal
PAGE # 17
7340010333