Physics Guru

A parallel-plate capacitor containing a polar dielectric with a dielectric constant $k$, connected to a battery with an e.m.f. $E$, is moved from room temperature ($25\text{ }^\circ\text{C}$) into a freezer. What happens to the electrostatic potential energy of the capacitor? (Assume thermal expansion of the plates is neglected and the battery's e.m.f. remains constant.) Answer At room temperature (25 °C), thermal energy causes molecules to vibrate and collide violently and randomly. This thermal chaos fights against the electric field from the battery, constantly knocking the molecular dipoles out of alignment. When capacitor is placed in a cool freezer, the thermal kinetic energy of the molecules drops significantly. With less thermal agitation disrupting them, the electric field becomes much more effective at lining up those molecular dipoles in an orderly fashion. Because more dipoles successfully align with the field, the material's internal polarization increases. A highe...

A dishonest carnival shop owner bends the barrel of an air rifle downward by $15^\circ$ near the middle, causing unsuspecting visitors to miss the balloon target. To compensate and successfully burst the balloon, at what approximate angle above the target should a clever visitor aim? (Neglect air resistance and effect of gravity considering short distance firing) Answer To understand how to compensate for the tampered rifle, we can break down the mechanics of the modification and the required adjustment: The Defect: Bending the rifle barrel downward by $15^\circ$ near its middle introduces a permanent angular offset to the muzzle. When the user aligns the sights of the rifle directly with the target ($0^\circ$), the altered trajectory of the barrel causes the projectile to exit at a $15^\circ$ downward angle relative to the line of sight, resulting in a miss below the balloon. The Compensation: To successfully hit the target, the visitor must counteract the $15^\circ$ downward deflecti...

Four persons are placed at the vertices of ABCD square frame of side a mounted on xy-plane having its center as the origin. The square frame is revolving with angular velocity $\vec \omega = \omega \hat k$. With what velocity vector should A throw a light ball with respect to himself when the square frame has all its sides parallel to x and y axis with point A lying in the second quadrant, so that the ball just falls downward as seen from the ground as if it were free falling? Solution The square frame is rotating anti-clockwise. The velocity of person A w.r.t. ground = $\frac {a\sqrt 2}{2}\omega (-cos 45^\circ \hat i - sin 45^\circ \hat j)$ = $-\frac {a\omega}{2} (\hat i + \hat j)$ The person should throw the ball at velocity = $+\frac {a\omega}{2} (\hat i + \hat j)$ This way, the initial velocity of ball would be 0 w.r.t. the ground and it will just fall downward like a free fall. Note that whether ABCD is labelled clockwise or anti-clockwise on square, it does not matter.

Do rockets on lunar missions, like Chandrayaan, take off at escape velocity? Answer Chandrayaan wants to reach the Moon, not simply escape Earth and fly off into deep space. So, the answer is no—not directly. Reaching Earth's escape velocity means having enough speed and energy to leave Earth on an unbound trajectory. But a lunar spacecraft isn't trying to escape Earth completely. It is trying to leave Earth orbit and enter a trajectory that intersects the Moon's orbit. The Moon itself orbits Earth, so even when Chandrayaan reaches the Moon, it is still within Earth's gravitational influence. Earth's gravity doesn't suddenly stop at the Moon. The spacecraft is still being pulled by Earth, while the Moon's gravity also acts on it. That's why a lunar mission is more accurately described as leaving Earth orbit, rather than escaping Earth's gravity altogether. Chandrayaan is first placed into Earth orbit. It then performs additional engine burns to raise...

$I = \int {\frac{{1 + x}}{{{e^{ - x}}{{(1 + x{e^x})}^n}}}dx = ?} ,n \ne 1$ Solution Let, $1 + x{e^x} = t$ $(x{e^x} + {e^x})dx = dt$ $ \Rightarrow {e^x}(1 + x)dx = dt$ Now, $I = \int {\frac{{{e^x}(1 + x)dx}}{{{{(1 + x{e^x})}^n}}}}  = \int {\frac{{dt}}{{{t^n}}}}  = \frac{{{t^{ - n + 1}}}}{{ - n + 1}} + C$ $\therefore I = \frac{{{{(1 + x{e^x})}^{1 - n}}}}{{1 - n}} + C$

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 $= \frac {q}{4\epsilon_0}$

Find a if ${S_n} = {1^4} + {2^4} + .......... + {n^4} = a{n^5} + b{n^4} + c{n^3} + d{n^2} + en + f$, where n is a natural number and a, b, c, d, e, f are constants. Solution We have, ${S_{n - 1}} = {1^4} + {2^4} + .......... + {(n - 1)^4}$ $\therefore {S_n} - {S_{n - 1}} = {n^4}$ $\therefore a\{ {n^5} - {(n - 1)^5}\}  + b\{ {n^4} - {(n - 1)^4}\}  + c\{ {n^3} - {(n - 1)^3}\}  + ..... = {n^4}$ In LHS, $n^4$ can only come from $a\{ {n^5} - {(n - 1)^5}\} $ as other terms have lower power on n. Using binomial expansion of ${(n - 1)^5}$ and equating the coefficient of $n^4$, $a.5 = 1$ $ \Rightarrow a = \frac{1}{5}$

There are two identical right-triangular ramps. Their vertical faces are parallel and facing each other, separated by a horizontal distance of 5 m, and their upper ends are at the same height. The inclined surface of each ramp makes an angle of 15° with the horizontal. A small toy car approaches the first ramp along a horizontal road and moves onto the inclined surface with constant speed. What is the minimum speed with which the car should enter the ramp so that it clears the gap and lands safely on the inclined surface of the second ramp? Neglect air resistance. Take $𝑔=10 m/s^2$. Solution The car leaves the first ramp with speed u at $15^\circ$. For the minimum speed, its projectile range must be 5 m. $R=\frac {u^2sin 2\theta}{g}$ $\therefore 5=\frac {u^2sin30}{10}$ $\Rightarrow u = 10m/s$

A negative point charge -q is brough near an isolated metal cube. (A) The cube becomes positively charged (B) The cube becomes negatively charged (C) The interior of the cube remains charge free and the surface gets nonuniform charge distribution (D) Protons inside the metal cube move towards -q making the interior of the cube negatively charged and the surface close to -q positively charged Answer The charge does not stay inside the metal cube. The charge has to reside on the surface. Moreover, the free electrons move farther away from -q so there is nonuniform surface charge distribution. Hence, (C)

At t = 0, three particles A, B and C are situated at the vertices of an equilateral triangle ABC of side 1.5d. Each particle moves with constant speed v such that A always has its velocity along AB, B along BC and C along CA. At what time will the particles meet? Solution Velocity of A along AB = $v$ Velocity of B along BA = $v cos 60^\circ = 0.5v$ Relative velocity of B towards A = $v+0.5v=1.5v$ Displacement covered by B to reach A with respect to A = 1.5d time = $\frac {d_{rel}}{v_{rel}}=\frac {1.5d}{1.5v}=\frac {d}{v}$

Why are small steps advisable while walking on wet path? Answer Let F be the force that the person exerts on the ground. The ground exerts equal and opposite force F on the person. Let $\theta$ be the angle that the force F exerted on the person by the ground makes with the vertical. The force F that ground exerts on the person can be resolved into two components: $F cos \theta = N$ (normal reaction) $F sin \theta = Friction \leq \mu. N$ So, $F sin \theta \leq \mu. F cos \theta $ So, $tan \theta \leq \mu $ or $(tan \theta)_{max} = \mu $ The coefficient of friction $\mu $ decreases when the surface is wet. So, the maximum permissible $\theta $ also decreases. Hence, the legs should be kept more nearly vertical, which is achieved by taking smaller steps.

A projectile is projected from the origin with a speed of $20m{s^{ - 1}}$ at an angle of ${63^\circ }$ above the horizontal. After a delay of 1s, a drone is launched from the same point with constant velocity. The motion of both the projectile and the drone takes place in the vertical xz-plane. If the drone strikes the projectile 2s after the projectile is launched, find the velocity of the drone. Ignore air resistance. [$g = 10m{s^{ - 2}}$, $\sin {63^\circ } = \frac{4}{5}$ and $\cos {63^\circ } = \frac{3}{5}$] Solution For projectile motion, $x = 20\cos {63^\circ }.2 = 24m$ $z = 20\sin {63^\circ }.2 - \frac{1}{2}{.10.2^2} = 32 - 20 = 12m$ Position vector of the projectile when the drone strikes it,  $\vec r = 24\hat i + 12\hat k$ m Velocity of drone = $\frac{{\vec r}}{1} = 24\hat i + 12\hat k$ $m{s^{ - 1}}$

Find the minimum value of the function $f(x, y)$ for $x, y > 0:$ $f(x,y)=\sqrt{x^{2}+1}+\sqrt{\frac{y^{2}}{x^{2}}+4}+\sqrt{\frac{9}{y^{2}}+9}$ Solution Let, $a = x,b = \frac{y}{x},c = \frac{3}{y}$ Then $a,b,c > 0$ and $abc = x.\frac{y}{x}.\frac{3}{y} = 3$ So $f = \sqrt {{a^2} + 1}  + \sqrt {{b^2} + 4}  + \sqrt {{c^2} + 9} $ Now, Minkowski inequality (in its 2-dimensional form) for ${a_i},{b_i} \ge 0$, $\sum\limits_{i = 1}^n {\sqrt {a_i^2 + b_i^2} }  \ge \sqrt {{{\left( {\sum\limits_{i = 1}^n {{a_i}} } \right)}^2} + {{\left( {\sum\limits_{i = 1}^n {{b_i}} } \right)}^2}} $ $\therefore f = \sqrt {{a^2} + {1^2}}  + \sqrt {{b^2} + {2^2}}  + \sqrt {{c^2} + {3^2}}  \ge \sqrt {{{(a + b + c)}^2} + {{(1 + 2 + 3)}^2}} $ $\therefore f \ge \sqrt {{{(a + b + c)}^2} + 36} $ Now, $A.M. \ge G.M$ or $\frac{{a + b + c}}{3} \ge {(abc)^{1/3}}$ $ \Rightarrow a + b + c \ge {3^{4/3}}$ $\therefore f \ge \sqrt {{3^{8/3}} + 36} $ $ \Rightarrow f \ge \sqrt {{{9.9}^{1/3}} + 9.4} ...

What should be the minimum value of amplitude so that the particle speed can match or exceed the wave speed in sinusoidal progressive wave $y = A sin (ωt − \frac {2\pi}{\lambda}x)$? (A) Particle speed can never exceed wave speed (B) $A_{min} = \lambda$ (C) $A_{min} = \frac {\lambda}{2\pi}$ (D) Particles are not present as the wave travels in vacuum Solution Particle speed = $|\frac {\partial y}{\partial t}| = Aω |cos (ωt − \frac {2\pi}{\lambda}x)|$ So, particle speed is less than or equal to Aω. Wave speed = $\nu.\lambda = \frac {ω}{2\pi}.\lambda$ If particle speed were to match or exceed wave speed, $Aω |cos (ωt − \frac {2\pi}{\lambda} x)| \geq \frac {ω}{2\pi}.\lambda$ $\therefore A \geq \frac{{\frac{\lambda }{{2\pi }}}}{{\left| {cos\left( {\omega t - \frac{{2\pi }}{\lambda }x} \right)} \right|}}$ For A to be min. cosine should be maximum = 1 $A_{min} \geq \frac {\lambda}{2\pi}$ When $A_{min} = \frac {\lambda}{2\pi}$, particle speed can match wave speed periodically and when $A_{min} ...