Physics Guru

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}...