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Question 141

JEE Main 2025• Physics Section A

Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is ′α′.'\alpha'. . Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :

  1. AOption (a)

    α\alpha

  2. BOption (b)

    α\alpha/4

  3. COption (c)

    α\alpha/8

  4. DOption (d)

    α\alpha/2

View answer & explanation

Correct answer

Answer: (b)B.

α\alpha/4

Why this answer is correct

Calculation:

 

Before cutting

α=Mℓ212⋯(i)\alpha = \frac{M\ell^2}{12} \quad \cdots (i)

After cutting

For each half:

⇒ Mass = M / 2

⇒ Length = ℓ / 2

Moment of inertia for each half, α' (single part) = (M/2) × (ℓ/2)2 / 12

⇒ α' = (M/2) × (ℓ2 / 4) / 12 = (M × ℓ2) / 96

Total moment of inertia, α' = 2 × (M × ℓ2 / 96) = (M × ℓ2) / 48

⇒ α' = α / 4

∴ Correct option is (2).

Question 142

JEE Main 2025• Physics Section A

A spherical surface separates two media of refractive indices 1 and 1.5 as shown in figure. Distance of the image of an object 'O', is : (C is the center of curvature of the spherical surface and R is the radius of curvature)

  1. AOption (a)

    0.24 m right to the spherical surface

  2. BOption (b)

    0.4 m left to the spherical surface

  3. COption (c)

    0.24 m left to the spherical surface

  4. DOption (d)

    0.4 m right to the spherical surface  

View answer & explanation

Correct answer

Answer: (b)B.

0.4 m left to the spherical surface

Why this answer is correct

Calculation:

⇒ (1.5 / v) - (1 / -0.2) = (1.5 - 1) / 0.4

⇒ (1.5 / v) = 1.25 - 5 = -3.75

⇒ v = 1.5 / (-3.75) = -0.4 m

∴ v = -1.2 m.

Question 143

JEE Main 2025• Physics Section A

Match List–I with List–II. 

List–IList–II
(A) Coefficient of viscosity(I) [ML0T–3]
(B) Intensity of wave(II) [ML–2T–2]
(C) Pressure gradient(III) [M–1LT2]
(D) Compressibility(IV) [ML–1T–1]

Choose the correct answer from the options given below : 

  1. AOption (a)

    (A)–(I), (B)–(IV), (C)–(III), (D)–(II)  

  2. BOption (b)

    (A)–(IV), (B)–(I), (C)–(II), (D)–(III) 

  3. COption (c)

    (A)–(IV), (B)–(II), (C)–(I), (D)–(III)   

  4. DOption (d)

    (A)–(II), (B)–(III), (C)–(IV), (D)–(I)

View answer & explanation

Correct answer

Answer: (b)B.

(A)–(IV), (B)–(I), (C)–(II), (D)–(III) 

Why this answer is correct

Explanation:

Coefficient of viscosity (η):

Expressed as force per unit area per velocity gradient.

⇒ Dimensional formula: [M1 L-1 T-1].

Intensity (I):

Power delivered per unit area perpendicular to wave propagation.

SI unit is watt per square meter (W/m2).

⇒ Dimensional formula: [M1 L0 T-3].

Pressure gradient:

Represents the rate of change of pressure with distance.

Its unit is pascal per meter (Pa/m).

⇒ Dimensional formula: [M L-2 T-2].

Compressibility (K):

Defined as the fractional change in volume per unit increase in pressure.

Inverse of bulk modulus (elastic modulus of volume).

⇒ Dimensional formula: [M-1 L1 T2].

Question 144

JEE Main 2025• Physics Section A

A small bob of mass 100 mg and charge +10 μ\mu C is connected to an insulating string of length 1 m. It is brought near to an infinitely long non conducting sheet of charge density 'σ\sigma ' as shown in figure. If string subtends an angle of 45° with the sheet at equilibrium the charge density of sheet will be :

(Given, ϵ0\epsilon_0 = 8.85 × 10–12 Fm\frac{\text{F}}{\text{m}} and acceleration due to gravity, g = 10 m/s2) 

  1. AOption (a)

    0.885 nC/m2 

  2. BOption (b)

    17.7 nC/m2

  3. COption (c)

    885 nC/m2

  4. DOption (d)

    1.77 nC/m2

View answer & explanation

Correct answer

Answer: (d)D.

1.77 nC/m2

Why this answer is correct

Calculation:

The FBD is shown below

From equilibrium condition:

⇒ q × (σ / 2 ε0) = mg

⇒ σ = (2 ε0 mg) / q

⇒ σ = (2 × 8.85 × 10-12 × 100 × 10-6 × 10) / (10 × 10-6)

⇒ σ = 17.7 × 10-10 C/m2

⇒ σ = 1.77 nC/m2

∴ Surface charge density σ = 1.77 nC/m2.

Question 145

JEE Main 2025• Physics Section A

A monochromatic light is incident on a metallic plate having work function ϕ . An electron, emitted normally to the plate from a point A with maximum kinetic energy, enters a constant magnetic field, perpendicular to the initial velocity of electron. The electron passes through a curve and hits back the plate at a point B. The distance between A and B is :

(Given : The magnitude of charge of an electron is e and mass is m, h is Planck's constant and c is velocity of light. Take the magnetic field exists throughout the path of electron)

  1. AOption (a)

    2m(hcλ−ϕ)/eB\sqrt{2m \left( \frac{hc}{\lambda} - \phi \right) / eB} \qquad

  2. BOption (b)

    m(hcλ−ϕ)/eB \sqrt{m \left( \frac{hc}{\lambda} - \phi \right) / eB} \

  3. COption (c)

    8m(hcλ−ϕ)/eB\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right) / eB} \qquad

  4. DOption (d)

    2m(hcλ−ϕ)/eB2 \sqrt{m \left( \frac{hc}{\lambda} - \phi \right) / eB}

View answer & explanation

Correct answer

Answer: (c)C.

8m(hcλ−ϕ)/eB\sqrt{8m \left( \frac{hc}{\lambda} - \phi \right) / eB} \qquad

Why this answer is correct

Calculation:

Using KEmax:

⇒ KEmax = (hc / λ) - φ

Momentum of electron:

⇒ p = √(2m KEmax) = √[2m ((hc / λ) - φ)]

Diameter of path:

⇒ dA-B = 2r = 2 (p / qB) = 2 [√(2m ((hc / λ) - φ)) / eB]

Thus,

⇒ dA-B = (2 √(2m ((hc / λ) - φ))) / (eB)

⇒ dA-B = √[ (8m ((hc / λ) - φ)) / (eB) ]

Question 146

JEE Main 2025• Physics Section B

A vessel with square cross–section and height of 6 m is vertically partitioned. A small window of 100 cm2 with hinged door is fitted at a depth of 3 m in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density 1.5 × 103 kg/m3. What force one needs to apply on the hinged door so that it does not get opened ?

​(Acceleration due to gravity = 10 m/s2)

View answer & explanation

Correct answer

Answer: 150Answer: 150

Why this answer is correct

Calculation:

Calculate forces:

⇒ Fℓ = (P0 + ρℓ g h) A

⇒ Fw = (P0 + ρw g h) A

So,

⇒ Fext = Fℓ - Fw

⇒ Fext = [ (P0 + ρℓ g h) - (P0 + ρw g h) ] A

⇒ Fext = (ρℓ - ρw) g h A

⇒ Fext = (1500 - 1000) × 10 × 3 × (100 × 10-4)

⇒ Fext = 150 N

∴ The external force required is 150 N.

Question 147

JEE Main 2025• Physics Section B

A steel wire of length 2 m and Young's modulus 2.0 × 1011 Nm–2 is stretched by a force. If Poisson ratio and transverse strain for the wire are 0.2 and 10–3 respectively, then the elastic potential energy density of the wire is ____ × 105 (in SI units)

View answer & explanation

Correct answer

Answer: 25Answer: 25

Why this answer is correct

Calculation:

From Poisson's ratio formula:

⇒ μ = - (Δr/r) / (Δℓ/ℓ) ⇒ (Δℓ/ℓ) = - (Δr/r) / μ

⇒ (Δℓ/ℓ) = - (10-3) / 0.2 = - 5 × 10-3

Since length increases, take positive value:

⇒ εℓ = 5 × 10-3

⇒ u = (1/2) × Y × εℓ2

⇒ u = (1/2) × 2 × 1011 × (5 × 10-3)2

⇒ u = 1 × 1011 × 25 × 10-6 = 25 × 105 = 25

∴ Elastic potential energy density u = 25 J/m3.

Question 148

JEE Main 2025• Physics Section B

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is 30° in a single slit diffraction pattern recorded using 628 nm light, then the width of the slit is ____ μ\mum.

View answer & explanation

Correct answer

Answer: 6Answer: 6

Why this answer is correct

Calculation:

θ1 = sin-1(2λ/a)

 

θ2 = sin-1(3λ/a)

It is given that:

θ1 + θ2 = 30° = π/6

⇒ sin-1 (2λ/a) + sin-1(3λ/a) = π/6

using sin-1(x) + sin-1(y) = sin-1[x√(1-y2) +y√(1-x2)] 

⇒ (2λ/a) × √(1 - (3λ/a)2) + (3λ/a) × √(1 - (2λ/a)2) = 1/2

Now plug in λ = 628 × 10-9 m and solve numerically to find a.

⇒  a ≈ 6.07 µm

Question 149

JEE Main 2025• Physics Section B

γA\gamma_A is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. γB\gamma_B is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If γAγB=(1+1n) \frac{\gamma_A}{\gamma_B} = \left( 1 + \frac{1}{n} \right) , then the value of n is ______.

View answer & explanation

Correct answer

Answer: 3Answer: 3

Why this answer is correct

Calculation:

We are given a ratio of adiabatic indices (γ):

γA / γB = [(fA + 2) / fA] × [fB / (fB + 2)]

Substitute values:

= (3 + 2)/3 × (6 + 2)/(6 + 2 + 2)

= (5/3) × (8/10)

= 40 / 30

So:

40 / 30 = 1 + (1/n)

⇒ (40 / 30) - 1 = 1/n

⇒ 10 / 30 = 1/n

⇒ n = 3

Question 150

JEE Main 2025• Physics Section B

A person travelling on a straight line moves with a uniform velocity v1 for a distance x and with a uniform velocity v2 for the next 32x\frac{3}{2}x distance. The average velocity in this motion is 507\frac{50}{7}m/s. If v1 is 5 m/s then v2 = _____ m/s.

View answer & explanation

Correct answer

Answer: 10Answer: 10

Why this answer is correct

Calculation:

The average velocity is given by:

vavg = (x1 + x2) / (t1 + t2)

Given: vavg = 50 / 7

Let x be the initial distance:

x1 = x,

x2 = (3x / 2),

t1 = x / 5,

and t2 = (3x / 2) / v2

⇒ 50 / 7 = (x + 3x / 2) / (x / 5 + (3x / 2v2))

⇒ 50 / 7 = (5 / 2) / (1 / 5 + 3 / (2v2))

⇒3 / (2v2) = 7 / 20 - 1 / 5 = (7 - 4) / 20 = 3 / 20

⇒3 / (2v2) = 3 / 20

⇒  v2 = 10 m/s

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