Physics Guru

When a particle is thrown up vertically from the surface of the earth (Radius R), we have $v^2-u^2=2(-g)h$, $h << R$ (neglecting air, symbols have usual meanings). However, if $h << R$ condition is not met, prove that $v^2-u^2 > 2as$. Solution Using COME, $\frac {1}{2}mu^2 - \frac {GMm}{R} = \frac {1}{2} mv^2 - \frac {GMm}{R+h}$ $\Rightarrow v^2  - u^2 = 2GM ( \frac {1}{R+h} - \frac {1}{R} )$ $\Rightarrow v^2  - u^2 = 2gR^2 \frac {-h}{R(R+h)}=2gR \frac {-h}{R+h}$ $\Rightarrow v^2  - u^2 = 2(-g)h \frac {1}{1+\frac {h}{R}} > 2(-g)h$ $\Rightarrow v^2  - u^2 > 2as$; $a=-g$, $s=h$ Since $g$ falls with height, the speed falls gradually. So, it is obvious that $v^2$ will be more than what we have when $g$ is constant. The animation below shows two particles, launched at the exact same moment, with the exact same initial speed. Left particle: Constant g world Right particle: g decreases with height

Two sources of sound vibrate at the same wavelength \(\lambda\) and in phase. The setup is similar to the double slit experiment of light with one major variation, that is \(d = 2D\) (in YDSE, usually \(D \gg d\)). Will fringe width be $\frac{\lambda D}{d}$ or \(\frac{\lambda}{2}\) for \(y \ll D\)? Given that symbols have their usual meaning. Comprehensive Solution & Analysis At first glance, this question looks like a simple plug-and-chug problem. If you substitute \(d = 2D\) into the standard Young's Double-Slit Experiment (YDSE) fringe width formula \(w = \frac{\lambda D}{d}\), you get: \(w = \frac{\lambda D}{2D} = \frac{\lambda}{2}\) Algebraically, the options \(\lambda \cdot \frac{D}{d}\) and \(\frac{\lambda}{2}\) represent the exact same value. However, neither of them is correct for this setup. Here is why and how to solve it properly. 1. Why the Standard Formula Fails Here The standard formula \...

The Question Consider an object inside a pond full of water. At the critical angle C , the refracted ray just grazes the water-air interface. If the angle of incidence i > C , the light undergoes total internal reflection. Is the angle that the ray leaving the water-air interface makes with the upward normal a discontinuous function at i = C ? The Answer When the light grazes the water-air interface, the angle that the refracted ray makes with the upward normal is 90° . However, when the angle of incidence i is slightly greater than C , refraction ceases to occur and the light undergoes total internal reflection. The reflected ray then makes an angle of 180° − i with the upward normal. Thus, as i approaches C from below, the angle approaches 90°, whereas as i approaches C from above, it approaches 180° − C . Since these two limiting values are unequal, the angle that the ray leaving the water-air interface makes with the upw...

PHYSICS · RELATIVE MOTION The Stuntman, the Skis and the Rocket Launcher A fascinating problem in relative velocity Imagine a stuntman gliding effortlessly on skis through a frictionless region. He is carrying a rocket launcher and moving horizontally at a speed of v m/s. Suddenly, he fires a rocket with a muzzle speed of 4v . Here comes the interesting part: despite firing the rocket, the stuntman's velocity remains unchanged! THE CHALLENGE At what angle above the horizontal, measured relative to the ground, was the rocket launched? (Ignore air resistance.) Understanding the physics The key to solving this problem is to distinguish between the velocity of the rocket relative to the launcher and its velocity relative to the ground. Since the stuntman's velocity remains unchanged after firin...

Prove that, $[x] + \left[ {x + \frac{1}{n}} \right] + \left[ {x + \frac{2}{n}} \right] +$ ............... $+ \left[ {x + \frac{{n - 1}}{n}} \right] = [nx]$ Where, [ ] denotes greatest integer function. Solution Let, $x=I+f$ where I is the greatest integer part of $x$ and $f$ is the fractional part of $x$. We know that,  [{Integer + (.....)}] = Integer + [(.....)] $LHS = [I + f] + \left[ {I + f + \frac{1}{n}} \right] + \left[ {I + f + \frac{2}{n}} \right] +$ ............... $+ \left[ {I + f + \frac{{n - 1}}{n}} \right]$ $ = nI + [f] + \left[ {f + \frac{1}{n}} \right] + \left[ {f + \frac{2}{n}} \right] + $............... $+ \left[ {f + \frac{{n - 1}}{n}} \right]$ $ = nI + [f] + \left[ {f + \frac{1}{n}} \right] + \left[ {f + \frac{2}{n}} \right] + $.........$ + \left[ {f + \frac{p}{n}} \right] +$ ............$ + \left[ {f + \frac{{n - 1}}{n}} \right]$, where $p$ is min. value such that $\left[ {f + \frac{p}{n}} \right] = 1$ $ = nI + 0 + 0 + 0 + ........p$ times $+ \left[ {f + \frac{p}...

Evaluate, $I=\int\limits_{ - \infty }^\infty  {{e^{ - {x^2}}}dx}$ Solution $I=\int\limits_{ - \infty }^\infty  {{e^{ - {x^2}}}dx}=\int\limits_{ - \infty }^\infty  {{e^{ - {y^2}}}dy}$ ${I^2} = \left( {\int_{ - \infty }^\infty  {{e^{ - {x^2}}}} {\mkern 1mu} dx} \right)\left( {\int_{ - \infty }^\infty  {{e^{ - {y^2}}}} {\mkern 1mu} dy} \right) = \int_{ - \infty }^\infty  {\int_{ - \infty }^\infty  {{e^{ - ({x^2} + {y^2})}}} } {\mkern 1mu} dx{\mkern 1mu} dy$ Changing to polar co-ordinates, we have ${I^2} = \int_0^{2\pi } {\int_0^\infty  {{e^{ - {r^2}}}} } r{\mkern 1mu} dr{\mkern 1mu} d\theta  = \int_0^{2\pi } d \theta \int_0^\infty  r {e^{ - {r^2}}}{\mkern 1mu} dr$ [${x^2} + {y^2} = {r^2},dx{\mkern 1mu} dy = r{\mkern 1mu} dr{\mkern 1mu} d\theta $. The entire plane corresponds to $0\le r<\infty$ and $0\le\theta\le2\pi$.] Let $r^2=u$ ${I^2} = 2\pi \cdot\frac{1}{2}\int_0^\infty  {{e^{ - u}}} {\mkern 1mu} du = \pi \left[ { - {e^{ - u}}} ...

A satellite providing internet services revolves around Earth in a circular orbit at an altitude of $750km$. A ground station directly below the satellite transmits an electromagnetic signal to it and receives a response signal immediately after the satellite receives the original signal. Assume that the satellite is directly overhead during the communication, the signal travels at the speed of light, and processing delays and atmospheric effects are negligible. Calculate the minimum time taken by a signal to travel from the ground station to the satellite and back to the ground station. Compare this with a typical geostationary satellite explaining the justification behind using LEO in place of geostationary satellite taking into account the fact that TVs do operate well without LEOs. Solution Since the satellite is directly overhead, the signal travels vertically to the satellite and back to the ground station. $t_{LEO} = \frac {2\times 750 \times 10^3}{3.0 \times 10^8} = 5.0 ms$ A t...

Question: A balloon is in equilibrium in air while carrying a payload of mass $10m$. The mass of the balloon and the lifting gas is negligible compared with the payload. A small amount of gas is released from the balloon so that its volume decreases, while the density of the surrounding air remains constant. The payload is then reduced by releasing a mass $m$. What percentage decrease in the balloon's volume is required to maintain equilibrium after the payload is reduced? Solution Initially, the balloon is in equilibrium. Because the mass of the balloon and lifting gas is negligible, the buoyant force $B$ balances the weight of the initial payload:$$B = 10mg \quad \text{--- (1)}$$ When a mass $m$ is released from the payload, the remaining payload mass is $10m - m = 9m$. For the balloon to remain in equilibrium, the new buoyant force $B'$ must balance the weight of the remaining payload:$$B' = 9mg \quad \text{--- (2)}$$ Since the density of the surrounding air ($\rho$) rem...

Question: Assuming the existence of the compound shown, given the trivial name “Ravana,” what would its IUPAC name be? Answer: The intended IUPAC name is: 5-(2,3,4,5-tetramethylcyclopentyl)decane Decane: The parent chain contains 10 carbon atoms. 5-: The cyclopentyl substituent is attached to carbon 5 of the parent chain, numbered from the left in the given diagram. Cyclopentyl: The substituent is a five-membered carbon ring. 2,3,4,5-tetramethyl: The four methyl groups are attached to ring carbons 2, 3, 4, and 5, numbering the ring from its attachment carbon (carbon 1). Thus, the name describes a decane chain bearing a tetramethyl-substituted cyclopentyl group at carbon 5.

A person with normal eyesight can see visible light with wavelengths approximately in the range of $400 nm$ to $700 nm$ in air. A diver underwater, where the refractive index of water is $μ=\frac {4}{3}$, is shown a red warning light of wavelength $680 nm$ (in air) to alert him to a nearby danger. Will the diver be able to see the warning light as red? Justify your answer. [Note: A wavelength of $680 nm$ corresponds to red light in air, whereas $510 nm$ corresponds approximately to green light in air.] Answer The frequency $\nu = \frac {v}{\lambda}$ In water, the speed becomes $\frac {v}{\mu}$ and the wavelength becomes $\frac {\lambda}{\mu}$. So, the frequency does not change. The colour depends on the frequency and therefore, red appears as red. Although the wavelength in water is $λ_{water}=\frac {680}{4/3}=510 nm$, this does not mean that the light becomes green. The value $510 nm$ corresponds to green light when the wavelength is specified in air, not when it is the wavelength of ...

A block of mass $m$ is suspended from the ceiling of a stationary hanging elevator through an ideal spring of spring constant $k$. Suddenly, the cable of the elevator breaks and the elevator starts falling freely. What is the min. time the block will take to attain maximum speed with respect to the elevator? Solution Let us place the observer inside the elevator. Before the cable is cut, the spring is elongated. The moment the cable is cut and the elevator falls freely, the forces acting on the block are: $mg$ downwards Pseudo force $ma$ upwards (here $a = g$ as this is free fall, $a$ being the acceleration of the elevator during free fall) $kx$ upwards as the spring is elongated So, the net force upwards = $kx$ However, this is the condition for SHM. The time period of oscillation $T= 2\pi \sqrt \frac {m}{k}$ The speed is maximum at the mean position. The block crosses the mean position for the first time after time $t_{min}=\frac {T}{4}=\frac {\pi}{2} \sqrt {\frac {m}{k}}$

What minimum kinetic energy should be imparted to a projectile of mass 12 g, launched from point on the Moon's surface, so that it hits a target P, where OP=200m? The height of the target above the lunar surface is 50m. Take the acceleration due to gravity on the Moon to be one-sixth of that on Earth, with $g_{Earth} = 9.8 m /{s^2}$. Solution Using the typical symbols associated with projectile motion, we have, $x = u\cos \theta .t,\cos \theta  = \frac{x}{{ut}}$ and $y = u\sin \theta .t - \frac{1}{2}g{t^2},\sin \theta  = \frac{{y + \frac{1}{2}g{t^2}}}{{ut}}$ ${\sin ^2}\theta  + {\cos ^2}\theta  = 1 = \frac{{{x^2}}}{{{u^2}{t^2}}} + \frac{{{{\left( {y + \frac{1}{2}g{t^2}} \right)}^2}}}{{{u^2}{t^2}}}$ $\Rightarrow {u^2} = \frac{{{x^2}}}{{{t^2}}} + \frac{{{y^2} + yg{t^2} + \frac{1}{4}{g^2}{t^4}}}{{{t^2}}}$ $ \Rightarrow {u^2} = \frac{{{x^2}}}{{{t^2}}} + \frac{{{y^2}}}{{{t^2}}} + yg + \frac{1}{4}{g^2}{t^2}$ $ \Rightarrow {u^2} = \frac{{{r^2}}}{{{t^2}}} + \frac{{{g^2}{t^2}...

Question A fat man trying to touch his toes while keeping his legs straight is more inclined to fall in comparison to a lean man. True/False Answer True. A man with a prominent abdominal bulge may have his centre of mass displaced farther forward while bending to touch his toes with his legs straight. This can increase the perpendicular distance between the line of action of his weight and the ankle joints, thereby increasing the gravitational torque. Consequently, maintaining rotational equilibrium becomes more difficult, increasing the likelihood of falling forward.

Student: Normally there is a perception that, "speed kills" and that high speed is dangerous. But, planes fly at high speed and are generally considered safe. So, how do you address this paradox? Teacher: It all depends on the context. So, sometimes high speed can be dangerous but sometimes high speed is desired. In fact, sometimes high speed is essential. A very common case is that of circular motion. Consider an electron moving in a circular orbit in the Bohr model of the hydrogen atom. Electrostatic force, $\frac {k.e.e}{r^2}$ = centripetal force, $\frac {mv^2}{r}$ So, $v = e \sqrt \frac {k}{rm} = 1.6\times 10^{-19} \sqrt \frac {9\times 10^9}{0.53\times 10^{-10}\times 9.1\times 10^{-31}}$                                                        $= 1.6\times 3\times 10^{-19} \times 10^{24} \sqrt \frac {1000}{48.23}$         ...

Consider an experiment involving a supercar having virtually no upper limit on its capacity to run at high speeds fitted with wheels whose rotational inertia can be neglected. Let the coefficient of friction between the tyres of the car and the ground to be $0.25$. Find the minimum time $t$ it will take to race $500 m$ starting from rest on a horizontal stretch.  At the end of $500 m$ stretch there is incline plane at angle $\theta$ very close to $15^\circ$ ($sin \theta = 0.25$). The engine of car is turned of at the end of $500 m$ stretch and car is allowed to move freely. After how much time will the car come to a halt? [$g=10 m/s^2$] Solution For horizontal motion, we have maximum frictional force $= \mu N = \mu mg = ma$ So, maximum $a = \mu g = 0.25  \times 10 = 2.5 m/s^2$ $s=ut+\frac {1}{2}at^2$ So, $500=\frac {1}{2}\times 2.5 \times t^2$ $\Rightarrow t = 20s$ For motion in incline plane, Deceleration $= gsin\theta $ Deceleration $=10\times 0.25 = 2.5 m/s^2$ Since deceler...

One edge of a square conducting loop made of uniform wire is mounted on a vertical non-conducting light pole. The light pole is free to rotate about the vertical axis passing through it. A uniform magnetic field $B$ points towards the right side such that loop's plane is parallel to the field as shown in the figure. The current $I_0$ is switched on at $t=0$. Find the instantaneous angular acceleration of the loop at $t=0$. The mass per unit length of the uniform wire is $\mu$ and length of one edge is $l$. Solution Force $F$ on the right edge $F=I_0lB$. This force will create torque which will rotate the loop. Torque $\tau =l.F=I_0 l^2B$ We have $\tau = I\alpha$ $I=\mu l.l^2+\mu l.\frac {l^2}{3}+\mu l.\frac {l^2}{3}$ $\therefore I=\frac {5}{3} \mu l^3 $ Now, $\tau =l.F=I_0 l^2B = I \alpha = \frac {5}{3} \mu l^3 \alpha$ $\therefore \alpha = \frac {3I_0B}{5\mu l}$

During an excavation in the southern part of India, a bundle of old leaves bearing inscriptions was discovered. During carbon dating, a sample of the leaves was found to have a C-14 activity of 14.40 disintegrations per gram per minute. A fresh, identical sample of the leaves has a C-14 activity of 15.12 disintegrations per gram per minute. Determine the age of the old leaves. The half-life of C-14 is 5730 years. Solution We have, $t = 3.32{t_{1/2}}{\log _{10}}\frac{{{{[A]}_0}}}{{[A]}}$ $\Rightarrow t = 3.32 \times 5730 \times {\log _{10}}\frac{{15.12}}{{14.40}}$ $ \Rightarrow t = 19023.6 \times {\log _{10}}1.05$ $ \Rightarrow t \approx 19023.6 \times 0.02119 \approx 403$ years

A small projectile is launched with velocity $u$ in $xz$ vertical plane in a gravity-free large hall from origin $(0, 0, 0)$ at an angle of $60^\circ$ from $x-axis$. At what position on $x-axis$ should another projectile be launched at the same instant with identical speed $u$ in the $xz$ plane at an angle of $120^\circ$ from $x-axis$ so that the two projectiles collide after time $t$? (ignore air resistance) Answer:  As can be seen in the figure, we have an equilateral triangle.  The distance between the two, $OA = OP = AP = ut$.  So, the position of A $\equiv (ut, 0, 0)$.

A small decorative 50 mW LED lamp is connected to a 3.0 V DC button cell. The circuit is left open, with a 2.0 cm gap between two exposed copper terminals inside a terracotta diya. The LED is mounted at the top of the diya, while the button cell is hidden at the bottom. Person X proposes pouring ordinary tap water into the diya so that the water bridges the gap between the two terminals. He says that distilled water is generally considered safe but is of no use here. Person Y says tap water causes a short circuit. So, pouring it would burn the circuit. Who is correct? Answer Usually, it is considered that water and electricity do not play well together, and water is a no-no as far as electricity is concerned. This perception has a lot to do with high voltages, including domestic supply voltages, and the dangers posed by water during the rainy season or in bathrooms. In this problem, however, we are dealing with a low-voltage, small-current situation. Tap water is not a short circuit in...

In the circuit shown, the battery is connected across AB. Will the drift velocity of electrons be maximum in the smallest resistance $R$ as there would be least resistance for the electrons to move in $R$? (all resistors are made of wires with identical cross sectional area and identical material) Answer The given circuit is actually balanced Wheatstone bridge as $\frac {4R}{2R} = \frac {8R}{4R}$ which means no current flows through the resistance $R$. So, even though the resistance $R$ is smallest, the drift velocity is not going to be maximum in $R$. In fact the drift velocity would be zero in $R$.  In the top we would have $2R+4R=6R$ and in the bottom we would have $4R+8R=12R$. $6R$ and $12R$ would be in parallel, so the higher current would flow through $6R$. So, the current would be maximum in $2R$ and $4R$ both.  Now, $I=nAev_d$ So, drift velocity is same for resistive wires having identical current, cross-sectional area and identical material. Hence, the drift velocity ...