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Question 131

JEE Main 2025• Physics Section A

A river is flowing from west to east direction with speed of 9 km h^–1. If a boat capable of moving at a maximum speed of 27 km h^–1 in still water, crosses the river in half a minute, while moving with maximum speed at an angle of 150° to direction of river flow, then the width of the river is :

AOption (a)
300 m
BOption (b)
112.5 m
COption (c)
75 m
DOption (d)
112.5 x $\sqrt{3} , \text{m}$

View answer & explanation

Correct answer

112.5 m

Why this answer is correct

Calculation:

Time taken to cross = 30 seconds = 30 / 3600 hr = 1 / 120 hr

Perpendicular velocity, V_⊥ = 27 / 2 km/hr

⇒ Width of river = V_⊥ × time

⇒ Width = (27 / 2) × (1 / 120) km

⇒ Width = (27 × 1000) / (2 × 120) m

⇒ Width = 112.5 m

∴ Width of the river is 112.5 m

Question 132

JEE Main 2025• Physics Section A

A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is :

AOption (a)
(4d/3, 0, 0)
BOption (b)
(d/4, 0, 0)
COption (c)
(3d/4, 0, 0)
DOption (d)
(d/3, 0, 0)

View answer & explanation

Correct answer

(d/4, 0, 0)

Why this answer is correct

Calculation:

Let the position where electric field is zero be at distance x from charge +q at origin.

Then, distance from +9q is (d − x)

Equating magnitudes of electric fields,

⇒ (kq) / x² = (k × 9q) / (d − x)²

⇒ 1 / x² = 9 / (d − x)²

⇒ (d − x)² = 9x²

⇒ d − x = 3x

⇒ x = d / 4

∴ The coordinates of point P are (d / 4, 0, 0)

Question 133

JEE Main 2025• Physics Section A

The battery of a mobile phone is rated as 4.2 V, 5800 mAh. How much energy is stored in it when fully charged ?

AOption (a)
43.8 kJ
BOption (b)
48.7 kJ
COption (c)
87.7 kJ
DOption (d)
24.4 kJ

View answer & explanation

Correct answer

87.7 kJ

Why this answer is correct

Calculation:

Given, V = 4.2 V

Battery capacity = 5800 mAh = 5800 × 3600 × 10⁻³ C

⇒ Q = 5800 × 3.6 = 20880 C

⇒ Energy = V × Q = 4.2 × 20880 = 87700 J

⇒ Energy = 87.7 kJ

∴ Energy stored in the battery when fully charged is 87.7 kJ

Question 134

JEE Main 2025• Physics Section A

A particle is subjected two simple harmonic motions as

$x_1 = \sqrt{7} \sin 5t , \text{cm} \ \text{and } x_2 = 2\sqrt{7} \sin \left( 5t + \frac{\pi}{3} \right) , \text{cm}$

where x is displacement and t is time in seconds. The maximum acceleration of the particle is x × 10^–2 ms^–2. The value of x is :

AOption (a)
175
BOption (b)
$25\sqrt{7}$
COption (c)
$5\sqrt{7}$
DOption (d)
125

View answer & explanation

Correct answer

175

Why this answer is correct

Calculation:

Given:

x₁ = √7 sin 5t cm

x₂ = 2√7 sin(5t + π/3) cm

Angular frequency, ω = 5 rad/s

Using phasor diagram:

Angle between x₁ and x₂ = 60°

Resultant amplitude A:

A = √[ (√7)² + (2√7)² + 2 × √7 × 2√7 × cos 60° ]

⇒ A = √[ 7 + 4 × 7 + 2 × √7 × 2√7 × (1/2) ]

⇒ A = √49 = 7 cm

Maximum acceleration:

a_max = A × ω²

⇒ a_max = 7 × 25 cm/s²

⇒ a_max = 175 × 10^-2 m/s²

∴ The value of x is 175

Question 135

JEE Main 2025• Physics Section A

The relationship between the magnetic $(\chi)$ susceptibility and the magnetic permeability $(\mu)$ is given by :

( $(\mu)$ ₀ is the permeability of free space and $(\mu)$ _r is relative permeability)

AOption (a)
$\chi = \frac{\mu}{\mu_0} - 1 \qquad$
BOption (b)
$\chi = \frac{\mu_r}{\mu_0} + 1$
COption (c)
$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

$\chi = \mu_r + 1 \qquad$
DOption (d)
$\chi = 1 - \frac{\mu}{\mu_0}$

View answer & explanation

Correct answer

$\chi = \frac{\mu}{\mu_0} - 1 \qquad$

Why this answer is correct

Calculation:

We are given: μ_r = 1 + χ

⇒ χ = μ_r - 1

Also, μ = μ₀ μ_r

⇒ μ_r = μ / μ₀

Substitute into the equation for χ:

⇒ χ = (μ / μ₀) - 1

∴ χ = (μ / μ₀) - 1

Question 136

JEE Main 2025• Physics Section A

A zener diode with 5V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a 400  resistor connected in series, the zener current is found to be 4 times load current. The load current (I_L) and load resistance (R_L) are :

AOption (a)
I_L = 20 mA; R_L = 250 $\Omega$
BOption (b)
I_L = 10 A; R_L = 0.5 $\Omega$
COption (c)
I_L = 0.02 mA; R_L = 250 $\Omega$
DOption (d)
I_L = 10 mA; R_L = 500 $\Omega$

View answer & explanation

Correct answer

I_L = 10 mA; R_L = 500 $\Omega$

Why this answer is correct

Calculation:

From the circuit diagram,

⇒ 5i = 20 / 400

⇒ 5i = 1 / 20

⇒ i = 1 / 100 = 0.01 A = 10 mA (Load current)

Given load voltage, V_L = 5 V

⇒ Load resistance, R_L = V_L / i

⇒ R_L = 5 / (10 × 10^-3) = 500 Ω

∴ Load current i = 10 mA and Load resistance R_L = 500 Ω

Question 137

JEE Main 2025• Physics Section A

In an adiabatic process, which of the following statements is true ?

AOption (a)
The molar heat capacity is infinite
BOption (b)
Work done by the gas equals the increase in internal energy
COption (c)
The molar heat capacity is zero
DOption (d)
The internal energy of the gas decreases as the temperature increases

View answer & explanation

Correct answer

The molar heat capacity is zero

Why this answer is correct

Calculation:

For adiabatic process,

⇒ dQ = 0

⇒ Molar heat capacity = 0

⇒ From first law, dQ = dU + dW

⇒ 0 = dU + dW ⇒ dU = -dW

Also,

⇒ dU = (f / 2) × nR × dT

∴ Only option (3) is correct.

Question 138

JEE Main 2025• Physics Section A

A square Lamina OABC of length 10 cm is pivoted at 'O'. Forces act at Lamina as shown in figure. If Lamina remains stationary, then the magnitude of F is :

AOption (a)
20 N
BOption (b)
0 (zero)
COption (c)
10 N
DOption (d)
$10\sqrt{2} , \text{N}$

View answer & explanation

Correct answer

10 N

Why this answer is correct

Concept

For rotational equilibrium, the net torque about the pivot must be zero.
Torque = Force × perpendicular distance from pivot.

Explanation

The square lamina of side $10,cm$ is pivoted at point $O$ .

Forces of $10,N$ act at different points producing clockwise and anticlockwise torques about $O$ .

Calculations

Taking moments about point $O$ :

The forces at A and B produce clockwise moments.
The forces at C and the unknown force $F$ produce anticlockwise moments.

Balancing torques for equilibrium:

$F \times 10 = 10 \times 10$

$\Rightarrow F = 10,N$

Thus, the magnitude of the force $F$ required for equilibrium is $10,N$ .

Question 139

JEE Main 2025• Physics Section A

Let B1 be the magnitude of magnetic field at center of a circular coil of radius R carrying current I. Let B2 be the magnitude of magnetic field at an axial distance 'x' from the center. For x $R = 3 : 4, \frac{B_2}{B_1} \text{ is :}$

AOption (a)
4 : 5
BOption (b)
16 : 25
COption (c)
64 : 125
DOption (d)
25 : 16

View answer & explanation

Correct answer

64 : 125

Why this answer is correct

Calculation:

Given:

⇒ B₁ = (μ₀ i) / (2R)

⇒ B₂ = B₁ × sin³ θ

⇒ Therefore, B₂ / B₁ = sin³ θ

⇒ sin θ = 4 / 5

⇒ (sin θ)³ = (4 / 5)³ = 64 / 125

∴ The ratio B₂ / B₁ = 64 / 125.

Question 140

JEE Main 2025• Physics Section A

Considering Bohr's atomic model for hydrogen atom :

(A) the energy of H atom in ground state is same as energy of He⁺ ion in its first excited state.

(B) the energy of H atom in ground state is same as that for Li⁺⁺ ion in its second excited state.

(D) the energy of He⁺ ion in its first excited state is same as that for Li⁺⁺ ion in its ground state

Choose the correct answer from the options given below :

AOption (a)
(B), (D) only
BOption (b)
(A), (B) only
COption (c)
(A), (D) only
DOption (d)
(A), (C) only

View answer & explanation

Correct answer

(A), (B) only

Why this answer is correct

Concept Used:

Energy levels in hydrogen-like atoms are proportional to the atomic number divided by the square of the principal quantum number

E = k × Z² / n² (where k is a constant)

Excited states correspond to n values greater than 1

Explanation:

From the given data:

⇒ For 1^st excited state, n = 2

⇒ For 2^nd excited state, n = 3

∴ Only statements A and B are correct.

Page 14

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Search Previous Year Questions

Question 131

Question 132

Question 133

Question 134

Question 135

Question 136

Question 137

Question 138

Question 139

Question 140

SSC Exams

Banking

Railways

Defence & Police

Teaching

State & Civil Services

Entrance Exams

Other Govt Exams

Quantitative Aptitude

Logical Reasoning

English Language

General Awareness

SSC CGL

Short sets.Sharper decisions.

Reasoning

Quantitative Aptitude

English Comprehension

General Awareness

Daily Current Affairs

Revisit earlier editions

Put your knowledge into practice

Search Previous Year Questions

Question 131

Question 132

Question 133

Question 134

Question 135

Question 136

Question 137

Question 138

Question 139

Question 140

Short sets.
Sharper decisions.