Physics Guru

Case (I) A video image of a man, carried by light, enters a spherical planet made of a transparent material with refractive index 𝜇 > 1. The light emerges from the diametrically opposite side of the planet, where it is viewed. The diameter of the planet is 𝑑. Case (II) The same video image is viewed after the light has travelled a distance d through vacuum. In which case will the man appear younger to the observer? Assume that the man continues to age normally while the light is travelling. Answer The key idea is that we see an object as it was when the light left it. The light slows down in case (I) so that the observer sees the image from an earlier time. Image from an earlier time means younger man. Mathematically, speed of light in case (I), $v=\frac {c}{𝜇}$ Time taken to cross diameter, $t= \frac {𝜇d}{c}$ In case (II), time taken to cover the same distance d in vacuum, $t'=\frac {d}{c}$ Since, 𝜇 > 1, t > t' So, in Case (I), the light takes longer to reach the...

If $a\sin x = b\sin \left( {x + \frac{{2\pi }}{3}} \right) = c\sin \left( {x + \frac{{4\pi }}{3}} \right) \neq 0$, prove that $(a+b+c)^2=a^2+b^2+c^2$. Solution Dividing by $abc$ ($\because a,b,c \neq 0$), $\frac{{\sin x}}{{bc}} = \frac{{\sin \left( {x + \frac{{2\pi }}{3}} \right)}}{{ca}} = \frac{{\sin \left( {x + \frac{{4\pi }}{3}} \right)}}{{ab}} = \frac{{\sin x + \sin \left( {x + \frac{{2\pi }}{3}} \right) + \sin \left( {x + \frac{{4\pi }}{3}} \right)}}{{bc + ca + ab}}$ $\therefore (ab + bc + ca)\sin x = bc\left\{ {\sin x + \underbrace {\sin \left( {x + \frac{{2\pi }}{3}} \right) + \sin \left( {x + \frac{{4\pi }}{3}} \right)}_{}} \right\}$ $ \Rightarrow (ab + bc + ca)\sin x = bc\left\{ {\sin x + 2\sin \left( {x + \pi } \right)\cos \left( {\frac{\pi }{3}} \right)} \right\}$ $ \Rightarrow (ab + bc + ca)\sin x = bc\left\{ {\sin x - 2\sin x.\frac{1}{2}} \right\} = 0$ $\therefore ab + bc + ca = 0 \because \sin x \ne 0$ Now, ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca) = {a^2...

A small drone is launched from a helicopter flying at a height of 720 m and moving horizontally at a constant velocity of 180 km/h. The velocity of the drone is 90 km/h with respect to the helicopter having same direction as that of helicopter and remains constant until it loses power after flying for 4 minutes. How far away from its launching point will the drone land? [Ignore air resistance. Take $g=10 m/s^2$.] Solution Velocity of drone w.r.t. the ground when it is launched=180+90=270 kmph Horizontal distance covered in 4 minutes = 270 kmph . 4 min. = 18 km Time to fall t after it loses power can be obtained from the equation, $720 m = \frac {1}{2} gt^2$ $144 = t^2$ t = 12 sec Horizontal distance covered after it loses power = 270 kmph. 12 sec = 0.9 km Total horizontal distance = 18 km + 0.9 km = 18.9 km

Find the area bounded by the function $y=\frac {1+sgn(sinx)}{2}$ with x-axis from 0 to $2\pi$ when sgn represents signum or sign function. Solution For $0 < x < \pi$, sinx > 0, so sgn(sinx)=1 giving y=1. For $\pi < x < 2\pi$, sinx < 0, so sgn(sinx)=-1 giving y=0. So, y is a square wave. Area bounded with x-axis = area of rectangle + 0 = $\pi.1 + 0 = \pi$ sq. unit

At t = 0, four particles A, B, C and D are situated at the vertices of a square ABCD of side d. Each particle moves with constant speed v such that A always has its velocity along AB, B along BC, C along CD and D along DA. At what time will the particles meet? Solution Let us consider particles A and B. The relative velocity of B w.r.t. A can be obtained from the right triangle formed by the two velocity vectors: $v_{rel}=\sqrt 2 .v$ The component of $v_{rel}$ along BA = $v_{rel} cos45^\circ $ = v This component will decrease the distance d. So, $t=\frac {d}{v}$

A small drone launched at an angle of $45^\circ$ with horizontal moves with constant velocity of $\sqrt {200}$ m/s. Its power shuts down just 4 second after the launch. Find the horizontal range of the drone. [Ignore air, $g=10 m/s^2$] Solution Motion with constant velocity $d = ucos\theta.t = 40m$ $h = usin\theta.t = 40m$ Projectile motion $y = -40 = usin\theta .t - \frac {1}{2} g t^2 = 10.t-\frac {1}{2}.10.t^2=10t-5t^2$ $\therefore 5t^2-10t-40=0$ $\Rightarrow t^2-2t-8=0$ $\Rightarrow (t-4)(t+2) = 0$, t = 4 sec $R=ucos\theta .t=10.4 = 40 m$ Range = d+R = 40+40 m = 80 m

A charge q is placed at a distance of $\frac {a}{\sqrt {24}}$ above the centre of a horizontal, equilateral triangular surface of edge a. Find the flux of the electric field through the equilateral triangular surface. Solution Let us evaluate if we can have regular tetrahedron as the closed Gaussian surface. The altitude of tetrahedron having side a is given by $H = a \sqrt {\frac {2}{3}}$ Distance of centroid from any face = $\frac {H}{4} = \frac {a}{\sqrt {24}}$ So, we can imagine a regular tetrahedron as the closed Gaussian surface with q placed at the centroid. Total flux $= \frac {q}{\epsilon_0} = 4 \times $ flux though one surface So, flux through one triangular surface $= \frac {q}{4\epsilon_0}$