JEE 2016 :: Advanced 2016 :: Physics 2016

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016

A transparent slab of thickness d has a refractive index n (z)that increases with z. Here, z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices n₁ and n₂ (> n₂ ) as shown in the figure. A ray of light is incident with angle θ_i , from medium 1 and emerges in medium 2 with refraction angle θ_f with a lateral displacement l.

Which of the following statement(s) is (are) true

Answer:

Explanation:

From Snell's law,

$n \sin \theta= constant$

$n_{1} \sin \theta_{i}= n_{2} \sin \theta_{f} $

Further, I will depend on n₁ and n(z). But it will be independent of n₂

A length-scale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (k_B) the absolute temperature (T), the number per unit volume (n ) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression (s) for / is (are) dimensionally correct?

Answer:

Explanation:

$[n]=[L^{-3}]; [q]=[AT]$

$[\epsilon]=[M^{-1}L^{-3}A^{2}T^{4}]$

$[T]=[L]$

$[l]=[L]$

$[k_{B}]=[M^{1}L^{2}T^{-2}K^{-1}]$

(a)  RHS

$\sqrt{\frac{[L^{-3}A^{2}T^{2}]}{[M^{-1}L^{-3}T^{4}A^{2}][M^{1}L^{2}T^{-2}K^{-1}][K]}}$

$\sqrt{\frac{[L^{-3}A^{2}T^{2}]}{[L^{-1}A^{2}T^{2}]}}$

$\sqrt{[L^{-2}]}$=$[L^{-1}]$   Wrong

(b) RHS

$\sqrt{\frac{[M^{-1}L^{-3}T^{4}A^{2}][M^{1}L^{2}T^{-2}K^{-1}][K]}{[L^{-3}A^{2}T^{2}]}}$

$\sqrt{\frac{[L^{-1}A^{2}T^{2}]}{[L^{-3}A^{2}T^{2}]}}$=[L]  Correct

(c) RHS

$\sqrt{\frac{[A^{2}T^{2}]}{[M^{-1}L^{-3}T^{4}A^{2}][L^{-2}][M^{1}L^{2}T^{-2}K^{-1}][K]}}$

$\sqrt{[L^{3}]}$ Wrong

(d) RHS

$\sqrt{\frac{[A^{2}T^{2}]}{[M^{-1}L^{-3}T^{4}A^{2}][L^{-1}][M^{1}L^{2}T^{-2}K^{-1}]}}$

$\sqrt{\frac{[A^{2}T^{2}]}{[L^{-2}A^{2}T^{2}]}}$= [L]   Correct

A conducting loop in the shape of a right-angled isosceles triangle of height 10 cm is kept such that the 90⁰ vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in the counterclockwise direction and increased at a constant rate of 1OAs^-1  which of the following statement is (are) true?

Answer:

Explanation:

By reciprocity theorem of mutual induction, it can be assumed that current in the infinite wire is varying at l0A"/s and EMF is induced in the triangular loop.

Flux of magnetic field through the triangle loop, if current in infinite wire is Q, can be calculated as follows:

$d\phi=\frac{\mu_{0}i}{2\pi y}.2ydy$

$d\phi=\frac{\mu_{0}i}{\pi}dy\Rightarrow \phi=\frac{\mu_{0}i}{\pi}\left(\frac{1}{\sqrt{2}}\right)$

$EMF=\mid \frac{\text{d}\phi}{\text{d}t}= \frac{\mu_{0}}{\pi}\left(\frac{1}{\sqrt{2}}\right).\frac{\text{d}i}{\text{dt}}$

$=\frac{\mu_{0}}{\pi}(10cm)\left(10\frac{A}{s}\right)=\frac{\mu_{0}}{\pi}$ volt

If we assume the current in the wire towards right then as the flux in the loop increases we know that the induced current in the wire is counter clockwise. Hence, the current in the wire is towards right. Field due to triangular loop at the location of infinite wire is into the paper. Hence, the force on the infinite wire is away from the loop. By cylindrical symmetry about infinite wire, rotation of triangular loop will not cause any additional EMF.

The position vector r of particle of mass m is given by the following equation $r(t)=\alpha t^{3}\hat{i}+\beta t^{2}\hat{j}$ where , $\alpha= \frac{10}{3}ms^{-3}$, $\beta=5 ms^{-2}$ and m=0.1 kg .At t = 1s ,which of the following statement is (are ) true about the particle?

Answer:

Explanation:

 $r=\alpha t^{3}\hat{i}+\beta t^{2}\hat{j}$

$v=\frac{\text{d}r}{\text{d}t}=3\alpha t^{2}\hat{i}+2\beta t\hat{j}$

$a=\frac{\text{d}^{2}}{\text{d}t^{2}}=6\alpha t\hat{i}+2\beta \hat{j}$

At t=1 s,

(a) $v=3\times \frac{10}{3}\times 1\hat{i}+2\times 5\times 1\hat{j}$

$=(10\hat{i}+10\hat{j})m/s$

(b) $\hat{L}=\hat{r}\times \hat{p}$

$\left(\frac{10}{3}\times 1\hat{i}+5\times 1\hat{j}\right)\times 0.1(10\hat{i}+10\hat{j})$

$=\left(-\frac{5}{3}\hat{k}\right)Nms$

(c) $F=ma$

$=m\times \left(6\times \frac{10}{3}\times 1\hat{i}+2\times 5\hat{j}\right)$

$=(2\hat{i}+\hat{j})N$

A plano-convex lens is made of material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is (are) true?

Answer:

Explanation:

Case 1

Using lens formula

$\frac{1}{60}+\frac{1}{30}=\frac{1}{f}$

$\Rightarrow \frac{1}{f}=\frac{1}{60}+\frac{2}{60}$

$\Rightarrow f=20 cm $

Further $\frac{1}{f_{1}}=(n-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)$

$f_{1}=\frac{R}{n-1}=+20 cm $

Case 2

Using mirror formula

$\frac{1}{10}-\frac{1}{30}=\frac{1}{f_{2}}$

$\frac{3}{30}-\frac{1}{30}=\frac{1}{f_{2}}=\frac{2}{30}$

$f_{2}=15=\frac{R}{2} \Rightarrow R=30$

$\frac{R}{n-1}=+20 cm = \frac{30}{n-1}$

$\Rightarrow 2n-2 = 3$

$\Rightarrow f_{1}=+20 cm$

Refractive index of lens is 2.5. Radius of curvature of convex surface is 30 cm. Faint image is erect and virtual. Focal length of lens is 20 cm .

• Advanced 2016 - Physics 2016
• Advanced 2016 - Chemistry 2016
• Advanced 2016 - Mathematics 2016