JEE ADVANCED 2021 P1 - entrancekart.com

JEE ADVANCED 2021 P1

PHYSICS SOLUTIONS

SECTION 1 (maximum marks: 12)

This section contains FOUR (4) questions

Each question has FOUR options. 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

Negative marks: −1 in all other cases

Kalyan’s Physics Challenge / JEE ADVANCED 2020 P1

1. The smallest division on the main scale of a Vernier calipers is 0.1 cm. Ten divisions of the Vernier scale correspond to nine divisions of the main scale. The figure below on the left shows the reading of this calipers with no gap between its two jaws. The figure on the right shows the reading with a solid sphere held between the jaws. The correct diameter of the sphere is

(A) 3.07 cm (B) 3.11 cm (C) 3.15 cm (D) 3.17 cm

Solution:

Smallest division on the main scale of Vernier Calipers: 0.1 cm

10 divisions on Vernier scale correspond to 9 divisions on main scale

Least count of the Vernier calipers: LC = 1 MSD – 1 VSD = 1 MSD –9

10MSD = 1 −

9

10MSD =

1

10x 0.1 = 0.01 cm

With no gap between the jaws, 6th division of Vernier scale coincides with one division on main scale

As 0 of Vernier scale is to the left of 0 of mains scale, the zero error is negative:

Zero error: − 10 − 6 𝐿𝐶 = −4 x 0.01 = −0.04 cm

Reading of the Vernier calipers: MSR + VC (LC) = 3.1 + 1 (0.01) = 3.1 + 0.01 = 3.11 cm

Correct diameter of the sphere: d = 3.11 + zero correction

d = 3.11 + 0.04 = 3.15 cm


2. An ideal gas undergoes a four step cycle as shown in P – V diagram below. During this cycle, heat is absorbed by the gas in

(A) steps 1 and 2 (B) steps 1 and 3 (C) steps 1 and 4 (D) steps 2 and 4

Solution:

Step 1: isobaric expansion: dQ = dW + dU (work done and change in internal energy are positive)

Heat is supplied to the gas

Step 2: isochoric process: dW = 0; dU is negative as the temperature decreases: dQ is negative

Heat is rejected by the gas

Step 3: isobaric compression: work done is negative and internal energy decreases as the temperature decreases

Heat is rejected by the gas

Step 4: isochoric process: dW = 0; dU is positive as the temperature increases: dQ is positive

Heat is absorbed by the gas


3. An extended object is placed at point O, 10 cm in front of a convex lens L1 and a concave lens L2 is placed 10 cm behind it, as shown in figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm. The refractive index of both lenses is 1.5. The total magnification of this lens system is

(A) 0.4 (B) 0.8 (C) 1.3 (D) 1.6

Solution:

Focal length of convex lens: 1

𝑓1= (n – 1)

2

𝑅= (1.5 – 1)

2

20= 0.05 → 𝑓1 = 20 cm

Focal length of concave lens: 1

𝑓2= (n – 1) −

2

𝑅= (1.5 – 1) −

2

20= −0.05 → 𝑓2 = −20 cm

For convex lens: 1

𝑣−

1

𝑢=

1

𝑓1→

1

𝑣−

1

−10=

1

20→

1

𝑣= −

1

20→ 𝑣 = −20 cm

Magnitude of magnification provided by convex lens: m1 = 𝑣

𝑢= 20

10= 2

For concave lens: 1

𝑣−

1

𝑢=

1

𝑓2→

1

𝑣−

1

−30=

1

−20→

1

𝑣= −

5

60= −

1

12→ 𝑣 = −12 cm

Magnitude of magnification provided by concave lens: m2 = 𝑣

𝑢= 12

30= 2

5

Total magnification of the lens system: m = m1m2 = 2 x 2

5= 4

5= 0.8


4. A heavy nucleus Q of half life 20 min undergoes alpha decay with probability of 60% and beta decay with probability of 40%. Initially, the number of Q nuclei is 1000. The number of alpha decays of Q in the first one hour is

(A) 50 (B) 75 (C) 350 (D) 525

Solution:

Half life of the radioactive nucleus: T = 20 min

Probability of alpha decay: 60%

Probability of beta decay: 40%

Initial number of nuclei: N0 = 1000

60% of 1000 = 600 radioactive nuclei may undergo alpha decay

Number of nuceli remaining after 1 hour: Nα = 𝑁0(𝛼)1

2

𝑡/𝑇= 600

1

2

60/20= 600

8= 75

Number of alpha decays of Q in the first 1 hour: 600 – 75 = 525



This section contains THREE (3) question stems

There are TWO (2) questions corresponding to each question stem

The answer to each equation is a Numerical Value

For each question, enter the correct numerical value corresponding to the answer in the designated place

If the numerical value has more than two decimal places, truncate/round-off the value to Two decimal places


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 Numbers: 5 & 6

A projectile is thrown from a point O on the ground at an angle 450 from the vertical and with a speed 5 2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 sec after thesplitting. The other part, t seconds after the splitting, falls to the ground at a distance 𝑥 m from the point O. The acceleration due to gravity g = 10 m/s2

5. Th value of t is 6. The value of 𝑥 is

Solution:

Maximum height attained by the projectile: H = 𝑢2 𝑠𝑖𝑛2 𝜃

2𝑔= 50 × 𝑠𝑖𝑛2 (450)

2(10)= 2.5 x

1

2= 1.25 m

Time taken by the freely falling part to reach ground: t = 2𝐻

𝑔=

2(1.25)

10= 0.5 sec

Range of the original projectile: R = 𝑢2 sin 2𝜃

𝑔= 50 × sin 90

10= 5 m

Consrvation of momentum at the highest point of trajectory: mu cos θ = 𝑚

2v1 +

𝑚

2v2 → 5 2 cos 450 =

1

20 +

1

2v2 → v2 = 10 m/s

Range of horizontal projectile (second part): r = v2

2𝐻

𝑔= 10

2(1.25)

10= 10 x 0.5 = 5 m

Distance of the second part from point of projection: 𝑥 = 𝑅

2+ 𝑟 = 2.5 + 5 = 7.5 m


Question stem for Question Numbers: 7 & 8

In the circuit shown below, the switch S is connected to position P for a long time so that the charge on the capacitor becomes q1 𝜇C. Then S is switched to position Q. After a long time, the charge on the capacitor is q2 𝜇C

7. The magnitude of q1 is 8. The magnitude of q2 is

Solution:

When the switch is connected to position P, after a long time the capacitor gets fully charged.

A fully charged capacitor behaves as open circuit

Apply KVL to the loop: – 1(i) – 1 + 2 – 2 (i) = 0 → i = 1

3A

Apply KVL to the loop: VA – 1 (i) – 1 = VB → VA – VB = 1 + 1 1

3= 4

3volt

Potential difference across the capacitor: VAB = 4

3volt

Charge on the capacitor: q1 = CVAB = 1 x 4

3= 1.33 𝝁C


Solution:

When the switch is connected to position Q, after a long time the capacitor gets fully charged.

A fully charged capacitor behaves as open circuit

Apply KVL to the loop: – 1(i) + 2 – 2 (i) = 0 → i = 2

3A

Apply KVL to the loop: VA – 1 (i) = VB → VA – VB = 1 x 2

3= 2

3volt

Potential difference across the capacitor: VAB = 2

3volt

Charge on the capacitor: q1 = CVAB = 1 x 2

3= 0.67 𝝁C


Question stem for Question Numbers: 9 & 10

Two point charges −Q and +𝑄

3are placed in XY plane at the origin (0, 0) and a point (2, 0) respectively, as shown in the figure.

This results in an equipotential circle of radius R and potential V = 0 in the XY plane with its centre at (b, 0). All the lengths are measured in meters

9. The value of R is10. The value of b is

Solution:

Consider a point P (x, y) in XY plane which is at a distance r1 from A and r2 from B as shown in figure

Potential at point P: V = −𝑘𝑄

𝑟1+

𝑘𝑄

3𝑟2= 𝑘𝑄 −

1

𝑥2+𝑦2+

1

3

1

(𝑥−2)2+𝑦2

0 = −1

𝑥2+𝑦2+

1

3

1

(𝑥−2)2+𝑦2→

1

𝑥2+𝑦2=

1

3

1

(𝑥−2)2+𝑦2→

1

𝑥2+𝑦2= 1

3

1

((𝑥−2)2+𝑦2)→ 𝑥2 + 𝑦2 = 3 ((𝑥 − 2)2+𝑦2)

𝑥2 + 𝑦2 = 3 𝑥2 + 22 − 4𝑥 + 𝑦2 → 2𝑥2 + 2𝑦2 − 12𝑥 + 12 = 0 → 𝑥2 + 𝑦2 − 6𝑥 + 6 = 0

𝑥2 − 2 3 𝑥 + 6 + 3 − 3 + 𝑦2 = 0 → (𝑥 − 3)2 + 𝑦2 = 32

Radius of the circle: R = 3 = 1.73 m and centre of the circle: (b, 0) = (3, 0)



This section contains SIX (6) questions

Each question has FOUR options. 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)


Full marks: +4 if only all the correct options(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


11. A horizontal force F is applied at the centre of mass of a cylindrical object of mass ‘m’ and radius R, perpendicular to its axis as shown in figure. The coefficient of friction between the object and the ground is μ. The center of mass of the object has an acceleration ‘a’. The acceleration due to gravity is g. Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

(A) For the same F, the value of ‘a’ does not depend on whether the cylinder is solid or hollow(B) For a solid cylinder, the maximum possible value of ‘a’ is 2μg(C) The magnitude of the frictional force on the object due to the ground is always μmg

(D) For a thin walled hollow cylinder, a = 𝑭

𝟐𝒎

Solution:

For translation: Fnet = ma → F – f = ma ---- (1)

For rotation: 𝜏𝑛𝑒𝑡 = Iα → fR = Iα → f = 𝐼𝛼

𝑅---- (2)

Condition for accelerated pure rolling: a = Rα ---- (3)

From the above equations: F −𝐼𝛼

𝑅= ma → F −

𝐼𝑎

𝑅2= ma → F = a 𝑚 +

𝐼

𝑅2→ a =

𝐹

𝑚 +𝐼

𝑅2

(D) For thin walled hollow cylinder: I = mR2

Acceleration of center of mass: a = 𝐹

𝑚 +𝑚𝑅2

𝑅2

= 𝑭

𝟐𝒎


Solution:

(B) For a solid cylinder: I = 1

2mR2

Acceleration of center of mass: a = 𝐹

𝑚 +

12𝑚𝑅2

𝑅2

= 2

3

𝐹

𝑚

From equation (1): F – f = ma →3𝑚𝑎

2− 𝜇𝑁 = 𝑚𝑎 →

1

2𝑚𝑎 = 𝜇𝑚𝑔 → 𝑎 = 2𝜇𝑔

Maximum possible acceleration of the solid cylinder: 𝟐𝝁𝒈


12. A wide slab consisting of two media of refractive indices n1 and n2 is placed in air as shown in the figure. A ray of light is

incident from medium n1 to n2 at an angle θ, where sin θ is slightly larger than 1

𝑛1.

Take refractive index of air as 1. Which of the following statement(s) is(are) correct?

(A) The light ray enters air if n2 = n1

(B) The light ray is finally reflected back into the medium of refractive index n1 if n2 < n1

(C) The light ray is finally reflected back into the medium of refractive index n1 if n2 > n1

(D) The light ray is reflected back into the medium of refractive index n1 if n2 = 1

Solution:

(D) When the angle of incidence is greater than critical angle, the light rays reflect back into denser medium (TIR)

Condition for TIR between n1 and n2: n1 sin θ = n2 sin 90 → n1 sin θ = 1 x 1 → θ = 𝑠𝑖𝑛−11

𝑛1

Since the angle of incidence is greater than critical angle [sin θ is slightly larger than 1/𝑛1], the light ray gets back to n1


Solution:

(C) If n2 > n1; light ray travels from rarer medium to denser medium.

The angle of refraction in n2 is less than the angle of incidence in n1: n1 sin θ = n2 sin r → sin r = 𝑛1

𝑛2sin θ

Since sin θ is slightly larger than 1

𝑛1, r is slightly greater than the crictical angle for n2 and air: 𝑠𝑖𝑛−1

1

𝑛2

Angle of incidence at n2 – air interface: r

Since r is greater than critical angle, the light ray undergoes TIR at n2 – air interface and travels back to n1


13. A particle of mass M = 0.2 kg is initially at rest in the XY plane at a point (𝑥 = −𝑙, 𝑦 = −ℎ), where 𝑙 = 10 m and ℎ = 1 m. The particle is accelerated at time t = 0 with a constant acceleration a = 10 m/s2 along the +ve X direction. Its angular momentum

and torque with respect to the origin, in SI units, are represented by ത𝐿 and ҧ𝜏, respectively. Ƹ𝑖, Ƹ𝑗 and 𝑘 are unit vectors along the +ve

X, Y and Z directions, respectively. If 𝑘 = Ƹ𝑖 x Ƹ𝑗 then which of the following statement(s) is(are) correct?

(A) The particle arrives at the point (𝒙 = 𝒍, 𝒚 = −𝒉) at time t = 2 sec

(B) ത𝝉 = 2 𝒌 when the particle passes through the point (𝒙 = 𝒍, 𝒚 = −𝒉)

(C) ത𝑳 = 4 𝒌 when the particle passes through the point (𝒙 = 𝒍, 𝒚 = −𝒉)

(D) ҧ𝜏 = 𝑘 when the particle passes through the point (𝑥 = 0, 𝑦 = −ℎ)

Solution:

(A) Mass of the particle: M = 0.2 kg

Initial position of the particle: (x, y) = −𝑙,−ℎ = −10,−1

Acceleration of the particle: a = 10 m/s2 (along +ve X direction)

Displacement of the particle: x = 1

2at2 = 0.5 x 10 x 4 = 20 m

Final position of the particle (after 2 sec): (x, y) = (10 m, −1 m)


Solution:

(B) Force acting on the particle: ത𝐹 = Mത𝑎 = 0.2 x 10 Ƹ𝑖 = 2 Ƹ𝑖 N

Position vector of the particle: ҧ𝑟 = 10 Ƹ𝑖 − Ƹ𝑗

Torque acting on the particle: ҧ𝜏 = ҧ𝑟 x ത𝐹 = Ƹ𝑖 Ƹ𝑗 𝑘

10 −1 02 0 0

= Ƹ𝑖 (0 – 0) − Ƹ𝑗 (0 – 0) + 𝑘 (0 + 2) = 2 𝒌

(C) Mass of the particle: M = 0.2 kg

Velocity of the particle: ҧ𝑣 = ത𝑎t = 10 Ƹ𝑖 (2) = 20 Ƹ𝑖

Position vector of the particle: ҧ𝑟 = 10 Ƹ𝑖 − Ƹ𝑗

Angular momentum of the particle: ത𝐿 = m ҧ𝑟 × ҧ𝑣 = 0.2 Ƹ𝑖 Ƹ𝑗 𝑘

10 −1 020 0 0

= 0.2 Ƹ𝑖 (0 – 0) − Ƹ𝑗 (0 – 0) + 𝑘 (0 + 20) = 4 𝒌


14. Which of the following statement(s) is(are) correct about the spectrum of hydrogen atom?

(A) The ratio of the longest wavelength to the shortest wavelength in Balmer series is 𝟗

𝟓(B) There is an overlap between the wavelength ranges of Balmer and Paschen series

(C) The wavelengths of Lyman series are given by 1 +1

𝑚2 λ0 where λ0 is the shortest wavelength of Lyman series and m is an integer

(D) The wavelength ranges of Lyman and Balmer series do not overlap

Solution:

Wavenumber of the emitted photon during electron transition: 1

λ= R

1

𝑛𝑖2 −

1

𝑛𝑓2

Longest wavelength in Balmer series (n = 3 to n = 2): 1

λ𝑙= R

1

4−

1

9= 5𝑅

36

Shortest wavelength in Balmer series (n = ∞ to n = 2): 1

λ𝑠= R

1

4−

1

∞= 𝑅

4

Ratio of longest wavelength to shortest wavelength: λ𝑙

λ𝑠= 36

5𝑅x 𝑅

4= 𝟗

𝟓

Longest wavelength of Lyman series (n = 2 to n = 1): 1

λ𝐿= R 1 −

1

4= 3𝑅

4→ λ𝐿 =

4

3𝑅

Shortest wavelength of Balmer series (n = ∞ to n = 2): 1

λ𝐵= R

1

4−

1

∞= 𝑅

4→ λ𝐵 =

4

𝑅

So, Lyman and Balmer series do not overlap


15. A long straight wire carries a current I = 2 A. A semi circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in figure. Two ends of the semi – circular rod are at distances 1 cm and 4 cm from the wire. At time t = 0, the rod starts moving on the rails with a speed v = 3 m/s (see the figure). A resistor R = 1.4 Ω and a capacitor C0 = 5.0 μF are connected in series between the rails. At time t = 0, C0 is uncharged. Which of the following statement(s) is(are) correct?[μ0 = 4𝜋 x 10−7 SI units, take ln 2 = 0.7]

(A) maximum current through R is 1.2 x 𝟏𝟎−𝟔 A (B) maximum current through R is 3.8 x 10−6 A(C) maximum charge on capacitor C0 is 8.4 x 𝟏𝟎−𝟏𝟐 C (D) maximum charge on capacitor C0 is 2.4 x 10−12 C

Solution:

Magnetic field due to infinitely long current carrying straight conductor: B = 𝜇0

2𝜋

𝐼

𝑥

Consider an element of thickness ‘dx’ in the semi-circular rod at distance ‘x’ from the straight wire

Emf (motional) induced in the element: de = B (dx)v = 𝜇0

2𝜋

𝐼

𝑥v (dx) =

𝜇0𝐼𝑣

2𝜋

1

𝑥𝑑𝑥

Emf induced in the semi-circular rod: e = 𝑑𝑒 = 𝜇0𝐼𝑣

2𝜋14 1

𝑥𝑑𝑥 =

𝜇0𝐼𝑣

2𝜋[ln x] [limits: 1 to 4]

Emf induced in the semi-circular rod: e = 𝜇0𝐼𝑣

2𝜋[ln 4 – ln 1] =

𝜇0𝐼𝑣

2𝜋(2 ln 2) = 2 × 10−7 x 2 x 3 (2 x 0.7) = 16.8 x 10−7 volt

Maximum current through the resistor: imax = 𝑒

𝑅= 16.8 × 10−7

1.4= 1.2 x 𝟏𝟎−𝟔 A

Maximum charge on the capacitor: qmax = eC0 = 16.8 x 10−7 x 5 x 10−6 = 8.4 x 𝟏𝟎−𝟏𝟐 C


16. A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration ‘a’ along a fixed inclined plane with angle θ = 450. P1 and P2 are pressure at points 1 and 2, respectively, located at the base of the

tube. Let β = 𝑃1−𝑃2

𝜌𝑔𝑑, where ρ is density of water, d is the inner diameter of the tube and g is the acceleration due to gravity.

Which of the following statement(s) is(are) correct?

(A) β = 0 when a = g/ 𝟐 (B) β > 0 when a = g/ 2 (C) β = 𝟐−𝟏

𝟐when a = g/2 (D) β =

1

2when a = g/2

Solution:

Acceleration in horizontal direction: ax = a cos 450 = 𝑎

2

Acceleration in vertical direction: ay = g – a sin 450 = g −𝑎

2

Consider a point 3 (vertically above 1) at the same horizontal level of point 2 as shown in figure

Pressure difference between 1 & 3: P1 – P3 = ρayh = ρ 𝑔 −𝑎

2

𝑑

tan 45= ρ 𝑔 −

𝑎

2d

Pressure difference between 2 & 3: P2 – P3 = ρax(d) = ρ𝑎

2d (pressure decreases in the direction of acceleration)

Pressure difference between 1 & 2: P1 – P2 = ρ 𝑔 −𝑎

2d − ρ

𝑎

2d = ρd 𝑔 −

𝑎

2−

𝑎

2= ρd 𝑔 −

2𝑎

2= ρgd 1 −

2𝑎

𝑔

Given: β = 𝑃1−𝑃2

𝜌𝑔𝑑= 1 −

2𝑎

𝑔= 0 → a =

𝒈

𝟐

For a = 𝑔

2; β = 1 −

2

𝑔

𝑔

2= 1 −

1

2=

𝟐−𝟏

𝟐



This section contains THREE (3) questions

The answer to each question is a non negative integer

For each question, enter the correct integer corresponding to the answer


Full marks: +4 if only the correct integer is entered

Zero marks: 0 in all other cases


17. An α particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential V and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the α particle and the sulfur ion move in circular orbits of radii 𝑟𝛼 and 𝑟𝑆,

respectively. The ratio 𝑟𝑆

𝑟𝛼is _____

Solution:

Charge on alpha particle: 𝑞𝛼 = 2 units

Mass of alpha particle: 𝑚𝛼 = 4 amu

Charge on singly charged sulfur ion: 𝑞𝑆 = 1 unit

Mass of singly charged sulfur ion: 𝑚𝑆 = 32 amu

Radius of the circular path followed by the charged particle in uniform magnetic field:

r = 𝑚𝑣

𝐵𝑞=

𝑃

𝐵𝑞=

2𝑚𝑘

𝐵𝑞=

2𝑚𝑞𝑉

𝐵𝑞=

2𝑉

𝐵

𝑚

𝑞

Both the charged particles are accelerated by the same potential difference and both are moving in same magnetic field

Ratio of radii: 𝑟𝑆

𝑟𝛼=

𝑚𝑆

𝑚𝛼

𝑞𝛼

𝑞𝑆=

32

4×

2

1= 4


18. A thin rod of mass M and length ‘a’ is free to rotate in horizontal plane about a fixed vertical axis passing through point O. A thin circular disc of mass M and of radius ‘a/4’ is pivoted on this rod with its center at a distance a/4 from the free end so that it can rotate freely about its vertical axis, as shown in figure. Assume that both the rod and the disc have uniform density andthey remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity Ω and the disc rotating about its vertical axis with angular velocity 4Ω. The total angular momentum of the system about the point O is 𝑀𝑎2Ω

48𝑛. The value of 𝑛 is _____

Solution:

Angular momentum of the rod about O: L1 = Iω = 𝑀𝑎2

3Ω

Angular momentum of the disc about O: L2 = Iω = M 𝑎 −𝑎

4

2Ω =

9𝑀𝑎2

16Ω [for rotation with the rod]

Angular momentum of the disc about its axis: L3 = Iω = 𝑀

𝑎

4

2

24Ω =

𝑀𝑎2

8Ω [for rotation about own axis]

Angular momentum of the system: L = L1 + L2 + L3 = 𝑀𝑎2Ω1

3+

9

16+

1

8=

𝑀𝑎2Ω

4849

The value of n is 49


19. A small object is placed at the centre of a large evacuated hollow spherical container. Assume that the container is maintained at 0 K. At time t = 0, the temperature of the object is 200 K. The temperature of the object becomes 100 K at t = t1 and 50 K at t = t2. Assume the object and the container to be ideal black bodies. The heat capacity of the object

does not depend on temperature. The ratio𝑡2

𝑡1is _____

Solution:

Power radiated by the black body: P = σAT4 ---- (1)

Rate of colling of the black body: 𝑑𝑄

𝑑𝑡= −𝑚𝑠

𝑑𝑇

𝑑𝑡---- (2)

From (1) & (2): σAT4 = −𝑚𝑠𝑑𝑇

𝑑𝑡→

1

𝑇4𝑑𝑇 = −

𝜎𝐴

𝑚𝑠𝑑𝑡 →

1

𝑇4𝑑𝑇 = −k 𝑑𝑡

First time interval: 200100 1

𝑇4𝑑𝑇 = −k 0

𝑡1 𝑑𝑡 →1

3𝑇3(limits: 200 to 100) = kt1 →

1

3

1

1003−

1

2003= kt1 ---- (3)

Second time interval: 20050 1

𝑇4𝑑𝑇 = −k 0

𝑡2 𝑑𝑡 →1

3𝑇3(limits: 200 to 50) = kt2 →

1

3

1

503−

1

2003= kt2 ---- (4)

(4)/(3): 𝑡2

𝑡1=

1

503−

1

2003

1

1003−

1

2003

= 1 −

1

431

23−

1

43

= 1 −

1

641

8−

1

64

= 63/64

7/64= 63

7= 9


Documents

JEE ADVANCED 2021 P1 - entrancekart.com