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Paper 1 - Physics

Question 1: A current carrying wire heats a metal rod. The wire provides a constant power P to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as T(t) = T₀ (1 + βt^1/4) where β is a constant with appropriate dimension of temperature. The heat capacity of metal is:

a)
$\frac{4P(T(t) - T_0)^3}{\beta^4 T_o^4}$
b)
$\frac{4P(T(t) - T_0)^2}{\beta^4 T_o^3}$
c)
$\frac{4P(T(t) - T_0)^4}{\beta^4 T_o^5}$
d)
$\frac{4P(T(t) - T_0)}{\beta^4 T_o^2}$

Solution:

Heat capacity = dQ/dt
H = dQ/dT ⇒ dQ/dt = H dT/dt
P = H dT/dt
= (HT₀/4 ) β t^-3/4
⇒ 4P/T₀ β = Ht^-3/4
⇒ t^-3/4 =
$(\frac{T-T_0}{T_0 \beta})^3$

⇒ H =
$\frac{4P(T-T_0)^3}{T_0^4 \beta^4}$

Question 2: In a capillary tube of radius 0.2 mm the water rises up to height of 7.5 cm with angle of contact equal to zero. If another capillary with same radius but of different material dipped in the same liquid. The height of water raised in capillary will be, if angle of contact becomes 60°.

a) 7.5 cm
b) 15 cm
c) 3.75 cm
d) 30 cm

Solution:

T = Rhρg/2cosθ

h/cosθ = constant

7.5/1 = h’/(1/2)

⇒ h’ = 3.75 cm

Question 3: A sample of ₁₉K⁴⁰ disintegrates into two nuclei Ca & Ar with decay constant λ_Ca = 4.5 x 10^-10 S^-1 and λ_Ar = 0.5 x 10^-10 S^-1 respectively. The time after which 99% of ₁₉K⁴⁰ gets decayed is:

a) 6.2 × 10⁹ sec
b) 9.2 × 10⁹ sec
c) 7.2 × 10⁹ sec
d) 4.2 × 10⁹sec

Solution:

λ = λ

(1/100) N₀ = N₀ e^-λt

ln(1/100) = -( λ₁ + λ₂)t + ln 100 = +( λ₁ + λ₂)t²

⇒ (2.303x2)/(5x10^-10) = t

Or t = 9.2 x 10⁹ sec

Question 4: Consider a spherical gaseous cloud of mass density ρ(r) in a free space where r is the radial distance from its centre. The gaseous cloud is made of particles of equal mass m moving in circular orbits about their common centre with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If ρ(r) is constant with time. The particle number density n(r) = ρ(r)/m is:
(g = universal gravitational constant)

a) 3K/πr²m²G
b) K/2πr²m²G
c) K/πr²m²G
d) K/6πr²m²G

Solution:

GMm/r = mv²/r = 2/r x ½ x mv² = 2k/r

M = 2kr/Gm

4πr² dr ρ = 2k dr/Gm

Or ρ = k/2πGmr²

Question 5: A thin spherical insulating shell of radius R caries a uniformly distributed charge such that the potential at its surface is V₀. A hole with small area α4πR² (α <<<1) is made in the shell without effecting the rest of the shell. Which one of the following is correct.

a) The magnitude of Vector E at a point located on a line passing through the hole and shell’s centre on a distance 2R from the centre of spherical shell will be reduced by αV₀/2R
b) Potential at the centre of shell is reduced by 2αV₀
c) The magnitude of vector E at the centre of shell reduced by αV₀/2R
d) The ratio of potential at the centre of the shell to that of the point at (1/2) R from centre towards the hole will be (1-α)/(1-2α)

Solution:

dq = Q/4πR² - dA = Qα

V_c = KQ/R - KαQ/R = V₀(1- α)

V_B = KQ/R - KαQ/[R/2] = V₀(1 - 2α)

Therefore, V_C / V_B = 1 - α / 1 - 2α

Question 6: A charged shell of radius R carries a total charge Q. Given ϕ as the flux of electric field through a closed cylindrical surface of height h, radius r & with its center same as that of the shell. Here center of cylinder is a point on the axis of the cylinder which is equidistant from its top & bottom surfaces. Which of the following are correct.

a) If h > 2R and r > R then ϕ = Q/Є_o
b) If h > 2R and r = 4R/5 then ϕ = Q/5Є_o
c) If h < 8R/5 and r = 3R/5 then ϕ = 0
d) If h > 2R and r = 3R/5 then ϕ = Q/5Є_o

Solution:

a, b and d

a.
$h > 2Rr> R$

<img class="img-responsive" src="https://cdn1.byjus.com/wp-content/uploads/2021/06/JEE-Advanced-2019-Physics-Paper-1-Question-6.jpg" alt="JEE Main 2019 Jan Shift 2 Physics Solutions" />

$\phi = \frac{Q}{\varepsilon _{o}} \: clearly \: from \: Gauss's \: Law$

b. Suppose,
$h = \frac{8R}{5}r = \frac{3R}{5}$

<img class="img-responsive" src="https://cdn1.byjus.com/wp-content/uploads/2021/06/JEE-Advanced-2019-Physics-Paper-1-Question-6-2.jpg" alt="JEE Main 2019 Jan Shift 1 Physics Solutions" />

$\phi = 0$

So for
$h < \frac{8R}{5}, \;\phi = 0$

c. For h = 2R

r = 4R / 5

<img class="img-responsive" src="https://cdn1.byjus.com/wp-content/uploads/2021/06/JEE-Advanced-2019-Physics-Paper-1-Question-6-3.jpg" alt="JEE Main 2019 Jan Shift 1 Physics Solutions" />

Shaded charge =
$2\pi\frac{1 - cos53^{o}\times (Q)}{4\pi }$

∴ = Q/5

∴
$q_{enclosed} = \frac{2Q}{5}$

∴
$\phi = \frac{2Q}{5_{\varepsilon_{o} }}$

For h > 2Rr = 4R / 5

∴
$\phi = \frac{2Q}{5_{\varepsilon_{o} }}$

d. Similar to option c for h = 2R

r = 3R / 5

$q_{enclosed} = 2 \times 2\pi (1 - cos 37^{o})\frac{Q}{4\pi } = \frac{Q}{5}$

∴
$\phi = \frac{Q}{5_{\varepsilon_{o} }}$

Question 7: Which statements is/are corect:
JEE Advanced Question Paper Physics Paper 1

a) At time t = 0, the S1 is closed instantaneous current in the closed circuit will be 25 mA
b) The key S₁ is kept closed for long time such that capacitors are fully charged. Now key S₂ is closed at this time the instantaneous current across 30Ω resistor between P & Q will be 0.2A.
c) If key S₁ is kept closed for long time such that capacitors are fully charged the voltage across C₁ will be 4V.
d) If S₁ is kept closed for long time such that capacitors are fully charged the voltage difference between P & Q will be 10V.

Solution:

When S₁ is closed then charge on capacitor is zero.

While replacing all capacitors with wire, we have

i = t/[70+100+30] = 5/200 = 25 mA

When circuit is in steady state, q/10 + q/80 + q/80 - 5 = 0

or q = 40 μC

Therefore, voltage across C₁ = 40/10 = 4 volt

Just after closing of S₂, charge on capacitor remain same, KVL

-10+30x+40/10+y×70=0

or 30x + 70y = 6

Again, -40/80 + 5 + (x-y)30 - 40/80 + (x - y) 100 - 10 + 30x = 0

⇒ 160x - 130y - 6 = 0

Solving both the equations, we have x = 0.05 amp and y = 96/1510

Option (b) and (c) are correct.

Question 8: A galvanometer of resistance 10 ohm and maximum current of 2μA is converted into voltmeter of range 100 mV and when converted into ammeter then range is 1mA. When these voltmeter and ammeter are connected by a (ideal) battery in series with a resistance of R = 1000 Ω, then

a) Measured value of R is between 978Ω and 996Ω
b) Resistance of voltmeter 105Ω
c) Shunt resistance is 20mΩ
d) If the ideal battery is replaced by non-ideal battery with internal resistance of 5Ω then R will be > 1000 Ω

Solution:

Question 9: Conducting wire of parabolic shape, initially y = x² is moving with velocity

$\vec v = v_0 \hat i$

in a non-uniform magnetic field

$\vec B = B_0(1 + (y/L)^{\beta}) \hat k$

as shown in figure. If V0, B0, L and B are +ve constants and Δϕ is potential difference develop between the ends of wire, then correct statement(s) is/are
JEE Advanced Solved Question Paper Physics

JEE Advanced Solved Question Paper Physics

a) |Δφ| = (1/2) B₀V₀L for β = 0
b) |Δφ| = (4/3) B₀V₀L for β = 2
c) |Δφ| is proportional to the length of wire projected on y-axis.
d) |Δφ| remains same if the parabolic wire is replaced by a straight wire, y = x, initially of length √2l

Solution:

B, C is correct

D is also correct because projection of wire on y axis is same.

Question 10: If in a hypothetical system if the angular momentum and mass are dimensionless. Then which of the following is true.

a) The linear momentum varies as L^-1
b) The energy varies as L^-2
c) The power varies as L^-4
d) The force varies as L^-5

Solution:

[M] = [M⁰L⁰T⁰]

[J] = [ML²T^-1]

[ML²T^-1] = [M⁰L⁰T⁰]

⇒ [L²] = [T]

Momentum:

[P] = [MLT^-2. LT^-1] = [ML²T^-3] = [L²L^-6] = [L^-4]

[E] = [MLT^-2. L] = [L²L^-4] = [L^-2]

[F] = [LL^-4] = [L^-3]

Linear Momentum = MLT-1

Mo LL-2 = Lsup>-1

Options A,B,C are correct.

Question 11: V – T diagram for n mol monoatomic gas is given below:
JEE Advanced Question Paper Physics 2019

c) Work done in cyclic process is ΔW = nRT₀/2
d) There are only adiabatic and isochoric processes are involved

Solution:

Corresponding PV entraps

(d) There are no adiabatic process is involved.

Only Options (a) and (b) are correct.

Question 12: Apparent depth for point object x in all three cases are H1, H2 & H3 respectively when seen from below given H = 30 cm, n = 1.5 & R = 3m, then
JEE Advanced Past Year Questions 2019

a) 0.8 < H₂ - H₁ < 0.9
b) H₂ > H₁
c) H₂ > H₃
d) H₃ > H₁

Solution:

Case 1:
$\frac{n_2}{n_1} = \frac{d^1}{d}$

⇒ 1/n = d¹/(3Q)

⇒ d¹ = (30/3) x 2 = 20 cm

Case 2:

$\frac{n_2}{v} - \frac{n_1}{u}= \frac{n_2-n_1}{R}$

$\frac{1}{H_2} - \frac{3}{-2 \times 30} = \frac{1-3/2}{-300}$

H₂ = 20.684

Case 3:

$\frac{n_2}{v} - \frac{n_1}{u}= \frac{n_2-n_1}{R}$

$\frac{1}{H_3} - \frac{3}{-2 \times 30} = \frac{1-\frac{3}{2}}{-300}$

$H_{3} = \frac{600}{31} = 19.354 cm$

V₃ = 19.354

Question 13:

Consider the following nuclear fission reaction

Previous Year JEE Advanced Solved Questions 2019

In this fission reaction. Kinetic energy of α-particle emitted is 4.44 MeV. Find the energy emitted as γ – radiation in keV in this reaction.

m(₈₈Ra²²⁶) = 226.005 amu

m(₂He⁴) = 4.000 amu

Solution:

Δm = 0.005 amu

Question 14: N dielectrics are introduced in series in a capacitor of thickness D. Each dielectric have width d = D/N &

dielectric constant of mth dielectric is given by K_m = K(1 + m/N) : [N >> 10³, Area of plates = A]

Net capacitance is given by [KЄ_oA]/[αDln2]. Find value of α.

Solution:

x/m = Δ/N

Integrating we get

C_eq = [KЄ_oA]/[Dln2]

Therefore, α = 1

Question 15: If at angle θ the light takes maximum time to travel in optical fiber. Then the maximum time is x (10^-8), calculate x.

Solution:

1.5 sin θ_C = 1.44 sin 90⁰

sin θ_C = 24/25

and sin θ_C = x/d

⇒ d = 25x/24

Now,

t = L/(C/n₂) = 5 x 10^-8s

Putting: C = 3 x 10⁸, n₂ = 1.5 and L = 10

Question 16: The source S₁ is at rest. The observer and the source S₂ are moving towards S₁ as shown in figure. The roof beats observed by the observer if both sources have frequency 120 Hz and speed of sound 330 m/s in is
JEE Advanced 2019 Solved Questions

Solution:

Question 17: A weight of 100 N is suspended by two wires made by steel and copper as shown in figure length of steel wire is 1 m and copper wire is √3 m. Find ratio of change in length of copper wire (Δ l₀) to change in length of steel wire (Δ l_s). Given Young’s modulus: Y_steel= 2 × 10¹¹ N/m², Y_copper = 1 × 10¹¹ N/m².
JEE Advanced 2019 Year Questions with Solutions

Solution:

Question 18: An optical bench, to measure the focal length of lens, is 1.5 m long and on the bench marks are with spacing 1/4 cm. Now a lens is placed at 75 cm and pin type object is placed at 45 cm marks on the bench. If its image is formed at 135 cm find maximum possible error in calculation of focal length.

Solution:

V = 30 cm and dv = 0.5 cm

V = 60 cm and dv = 0.5 cm

Now, using lens formula,

1/v - 1/u = 1/f

1/60 + 1/30 = 1/f

⇒ f = 20 cm

⇒
$\frac{-dv}{v^2} + \frac{-du}{u^2} = \frac{-df}{f^2}$

⇒
$\frac{df}{f} \times 100 = f[\frac{dv}{v^2} + \frac{du}{u^2}]$

= (20/2)[1/60² + 1/30²]

Multiply by 100 both the sides, we get

df/f = 1.38 and 1.39

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