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Question 61

JEE Main 2026• Physics Section A

Refer the figure given below. μ1 \mu_1 and μ2 \mu_2 are refractive indices of air and lens material. The height of image will be ________ cm.

  1. AOption (a)

    1

  2. BOption (b)

    0.5

  3. COption (c)

    1.2

  4. DOption (d)

    0.25

View answer & explanation

Correct answer

Answer: (a)A.

1

Why this answer is correct

Sol. μ2v−μ1u=μ2−μ1R \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}

1.54v−1(−40)=1.54−1−20 \frac{1.54}{v} - \frac{1}{(-40)} = \frac{1.54 - 1}{-20}

1.54v=−140−0.5420 \frac{1.54}{v} = -\frac{1}{40} - \frac{0.54}{20}

1.54v=−2.0840 \frac{1.54}{v} = \frac{-2.08}{40}

v=−29.61 cm v = -29.61 \text{ cm}

hiho=μ1vμ2u=1(−29.61)1.54(−40)=0.48 \frac{h_i}{h_o} = \frac{\mu_1 v}{\mu_2 u} = \frac{1(-29.61)}{1.54(-40)} = 0.48

hi=0.48×2=0.96≈1cm h_i = 0.48 \times 2 = 0.96 \approx 1 \text{cm}

Question 62

JEE Main 2026• Physics Section A

For a certain metal, when monochromatic light of wavelength λ \lambda is incident, the stopping potential for photoelectrons is 3V0 3V_0 . When the same metal is illuminated by light of wavelength 2λ 2\lambda , then the stopping potential becomes V0 V_0 . The threshold wavelength for photoelectric emission for the given metal is αλ \alpha\lambda . The value of α \alpha is ________.

  1. AOption (a)

    1

  2. BOption (b)

    4

  3. COption (c)

    2

  4. DOption (d)

    3

View answer & explanation

Correct answer

Answer: (b)B.

4

Why this answer is correct

Sol. hcλ−ϕ=3eV0…(i) \frac{hc}{\lambda} - \phi = 3eV_0 \dots (i)

hc2λ−ϕ=eV0…(ii) \frac{hc}{2\lambda} - \phi = eV_0 \dots (ii)

hcλ−ϕ=3[hc2λ−ϕ] \frac{hc}{\lambda} - \phi = 3 \left[ \frac{hc}{2\lambda} - \phi \right]

ϕ=hc4λ \phi = \frac{hc}{4\lambda}

hcλ0=hc4λ \frac{hc}{\lambda_0} = \frac{hc}{4\lambda}

λ0=4λ \lambda_0 = 4\lambda

α=4 \alpha = 4

Question 63

JEE Main 2026• Physics Section A

An electromagnetic wave travelling in x-direction is described by field equation Ey=300sin⁡ω(t−xc) E_y = 300 \sin \omega \left( t - \frac{x}{c} \right) . If the electron is restricted to move in y-direction only with speed of 1.5×106 m/s 1.5 \times 10^6 \text{ m/s} then ratio of maximum electric and magnetic forces acting on the electron is ________.

  1. AOption (a)

    200

  2. BOption (b)

    150

  3. COption (c)

    400

  4. DOption (d)

    300

View answer & explanation

Correct answer

Answer: (a)A.

200

Why this answer is correct

Sol. Ey=300sin⁡ω(t−xv) E_y = 300 \sin \omega \left( t - \frac{x}{v} \right)

FM=qvB…(i) F_M = qvB \dots (i)

Fe=qE…(ii) F_e = qE \dots (ii)

FeFm=qEqvB=EvB=cv=3×1081.5×106=200 \frac{F_e}{F_m} = \frac{qE}{qvB} = \frac{E}{vB} = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^6} = 200

Question 64

JEE Main 2026• Physics Section A

Angular momentum of an electron in a hydrogen atom is 3hπ \frac{3h}{\pi} , then the energy of the electron is ____ eV.

  1. AOption (a)
    • 1.51
  2. BOption (b)
    • 0.85
  3. COption (c)
    • 0.38
  4. DOption (d)
    • 0.28
View answer & explanation

Correct answer

Answer: (c)C.
  • 0.38

Why this answer is correct

Sol.  L = nh2π \frac{nh}{2\pi}

3hπ=nh2π \frac{3h}{\pi} = \frac{nh}{2\pi}

n = 6

E = −13.6n2 \frac{-13.6}{n^2} eV

= −13.662≈−0.38 \frac{-13.6}{6^2} \approx -0.38 eV

Question 65

JEE Main 2026• Physics Section A

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α×10−6 \alpha \times 10^{-6} J. The value of α \alpha is ______. (Take surface tension of liquid = 0.08 N/m)

  1. AOption (a)

    10

  2. BOption (b)

    7

  3. COption (c)

    8

  4. DOption (d)

    11

View answer & explanation

Correct answer

Answer: (b)B.

7

Why this answer is correct

Sol.  V1=V2 V_1 = V_2

43πR3=512(43πr3) \frac{4}{3} \pi R^3 = 512 \left( \frac{4}{3} \pi r^3 \right)

R=(512)1/3r R = (512)^{1/3} r

1 = 8r

r = 18 \frac{1}{8} mm

ΔSE=SE2−SE1 \Delta SE = SE_2 - SE_1

= 512[T×4πr2]−[T×4πR2] 512 \left[ T \times 4\pi r^2 \right] - \left[ T \times 4\pi R^2 \right]

= 4πT[512r2−R2] 4\pi T [512r^2 - R^2]

= 4π(0.08)[512(18)2−12]×10−6≈7×10−6 4\pi (0.08) \left[ 512 \left( \frac{1}{8} \right)^2 - 1^2 \right] \times 10^{-6} \approx 7 \times 10^{-6}

α=7 \alpha = 7

Question 66

JEE Main 2026• Physics Section A

The dimensional formula of 12ϵ0E2 \frac{1}{2} \epsilon_0 E^2 (ϵ0 \epsilon_0 = permittivity of vacuum and E = electric field) is MaLbTc M^a L^b T^c . The value of 2a - b + c = ________.

  1. AOption (a)

    0

  2. BOption (b)

    1

  3. COption (c)

    -1

  4. DOption (d)

    2

View answer & explanation

Correct answer

Answer: (b)B.

1

Why this answer is correct

12ϵ0E2 \frac{1}{2} \epsilon_0 E^2 = Energy density

∴[12ϵ0E2]=ML2T−2L3=ML−1T−2 \therefore \left[ \frac{1}{2} \epsilon_0 E^2 \right] = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}

∴ \therefore a = 1, b = -1, c = -2

∴ \therefore 2a - b + c = 2 + 1 - 2 = 1

Question 67

JEE Main 2026• Physics Section A

The diameter of a wire measured by a screw gauge of least count 0.001 cm is 0.08 cm. The length measured by a scale of least count 0.1 cm is 150 cm. When a weight of 100 N is applied to the wire, the extension in length is 0.5 cm, measured by a micrometer of least count 0.001 cm. The error in the measured Young's modulus is α×109N/m2 \alpha \times 10^9 N/m^2 . The value of α \alpha is ________.

(Ignore the contribution of the load to Young's modulus error calculation)

  1. AOption (a)

    1.3

  2. BOption (b)

    1.65

  3. COption (c)

    0.13

  4. DOption (d)

    0.25

View answer & explanation

Correct answer

Answer: (b)B.

1.65

Why this answer is correct

Y = FℓAΔℓ \frac{F\ell}{A\Delta\ell}

∴ΔYY=ΔFF+Δℓℓ+2Δdd+Δ(Δℓ)Δℓ \therefore \frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta \ell}{\ell} + 2\frac{\Delta d}{d} + \frac{\Delta(\Delta\ell)}{\Delta\ell}

[A=πd24⇒ΔAA=2Δdd] \left[ A = \pi \frac{d^2}{4} \Rightarrow \frac{\Delta A}{A} = \frac{2\Delta d}{d} \right]

Since F is known exactly ⇒ΔF=0 \Rightarrow \Delta F = 0

ΔYY=0.1150+2×0.0010.08+0.0010.5=0.0277 \frac{\Delta Y}{Y} = \frac{0.1}{150} + \frac{2 \times 0.001}{0.08} + \frac{0.001}{0.5} = 0.0277

∴ΔY=0.0277Y \therefore \Delta Y = 0.0277 Y

Y = 100×1.5π4(0.08100)2×0.5×10−2=5.97×1010N/m2 \frac{100 \times 1.5}{\frac{\pi}{4} \left( \frac{0.08}{100} \right)^2 \times 0.5 \times 10^{-2}} = 5.97 \times 10^{10} N/m^2

∴ΔY=0.0277×5.97×1010N/m2=1.65×109N/m2 \therefore \Delta Y = 0.0277 \times 5.97 \times 10^{10} N/m^2 = 1.65 \times 10^9 N/m^2

Question 68

JEE Main 2026• Physics Section A

The velocity of a particle is given as

v⃗=−xi^+2yj^−zk^m/s \vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k} m/s . The magnitude of acceleration at point (1, 2, 4) is ________ m/s2 m/s^2 .

  1. AOption (a)

    6 \sqrt{6}

  2. BOption (b)

    9

  3. COption (c)

    33 \sqrt{33}

  4. DOption (d)

    0

View answer & explanation

Correct answer

Answer: (b)B.

9

Why this answer is correct

v⃗=−xi^+2yj^−zk^ \vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}

a⃗=dv⃗dt=−dxdti^+2dydtj^−dzdtk^ \vec{a} = \frac{d\vec{v}}{dt} = -\frac{dx}{dt}\hat{i} + 2\frac{dy}{dt}\hat{j} - \frac{dz}{dt}\hat{k}

= −(−x)i^+2(2y)j^−(−z)k^=xi^+4yj^+zk^ -(-x)\hat{i} + 2(2y)\hat{j} - (-z)\hat{k} = x\hat{i} + 4y\hat{j} + z\hat{k}

∴a⃗ \therefore \vec{a} at (1,2,4) = i^+8j^+4k^ \hat{i} + 8\hat{j} + 4\hat{k}

∴∣a⃗∣=9m/s2 \therefore |\vec{a}| = 9 m/s^2

Question 69

JEE Main 2026• Physics Section A

The position of an object having mass 0.1 kg as a

function of time t is given as r⃗=(10t2i^+5t3j^)m \vec{r} = (10t^2\hat{i} + 5t^3\hat{j}) m .

At t = 1 s, which of the following statements are

correct?

A. The linear momentum p⃗=(2i^+1.5j^)kg.m/s \vec{p} = (2\hat{i} + 1.5\hat{j}) kg.m/s .

B. The force acting on the object F⃗=(2i^+3j^)N \vec{F} = (2\hat{i} + 3\hat{j}) N .

C. The angular momentum of the object about its

origin L⃗=15k^Js \vec{L} = 15\hat{k} Js .

D. The torque acting on the object about its origin

τ⃗=20k^Nm \vec{\tau} = 20\hat{k} Nm .

Choose the correct answer from the option given

below.

  1. AOption (a)

    A, B and C only

  2. BOption (b)

    B, C and D only

  3. COption (c)

    A, C and D only

  4. DOption (d)

    A, B and D only

View answer & explanation

Correct answer

Answer: (d)D.

A, B and D only

Why this answer is correct

Given r⃗=10t2i^+5t3j^⇒r⃗t=1=10i^+5j^ \vec{r} = 10t^2\hat{i} + 5t^3\hat{j} \Rightarrow \vec{r}_{t=1} = 10\hat{i} + 5\hat{j}

v⃗=dr⃗dt=20ti^+15t2j^⇒v⃗t=1sec=20i^+15j^ \vec{v} = \frac{d\vec{r}}{dt} = 20t\hat{i} + 15t^2\hat{j} \Rightarrow \vec{v}_{t=1sec} = 20\hat{i} + 15\hat{j}

a⃗=d2r⃗dt2=20i^+30tj^⇒a⃗t=1sec=20i^+30j^ \vec{a} = \frac{d^2\vec{r}}{dt^2} = 20\hat{i} + 30t\hat{j} \Rightarrow \vec{a}_{t=1sec} = 20\hat{i} + 30\hat{j}

∴P⃗=mv⃗=2i^+1.5j^ \therefore \vec{P} = m\vec{v} = 2\hat{i} + 1.5\hat{j}

F⃗=ma⃗=2i^+3j^ \vec{F} = m\vec{a} = 2\hat{i} + 3\hat{j}

τ⃗=r⃗×F⃗=20k^ \vec{\tau} = \vec{r} \times \vec{F} = 20\hat{k}

L⃗=r⃗×P⃗=5k^ \vec{L} = \vec{r} \times \vec{P} = 5\hat{k}

Question 70

JEE Main 2026• Physics Section A

A planet (P1) (P_1) is moving around the star of mass 2M in the orbit of radius R. Another planet (P2) (P_2) is moving around another star of mass 4M in a orbit

of radius 2R. Ratio of time periods of revolution of

P2 P_2 and P1 P_1 is ________.

  1. AOption (a)

    12 \frac{1}{2}

  2. BOption (b)

    2

  3. COption (c)

    4

  4. DOption (d)

    14 \frac{1}{4}

View answer & explanation

Correct answer

Answer: (b)B.

2

Why this answer is correct

From Kepler's law,

T2=4π2GMsr3 T^2 = \frac{4\pi^2}{GM_s}r^3

∴TP1=2πG(2M)R3/2 \therefore T_{P_1} = \frac{2\pi}{\sqrt{G(2M)}} R^{3/2}

TP2=2πG(4M)(2R)3/2 T_{P_2} = \frac{2\pi}{\sqrt{G(4M)}} (2R)^{3/2}

∴TP2TP1=(2)3/24=21×22×22=2 \therefore \frac{T_{P_2}}{T_{P_1}} = \frac{(2)^{3/2}}{\sqrt{4}} = \frac{\sqrt{2}}{1} \times \frac{2\sqrt{2} \times \sqrt{2}}{2} = 2

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