Physics Guru

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, $\tau =l.F=I_0 l^2B = I \alpha = \frac {5}{3} \mu l^3 \alpha$ $\therefore \alpha = \frac {3I_0B}{5\mu l}$