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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

The Speed Paradox

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, $v = e \sqrt \frac {k}{rm} = 1.6\times 10^{-19} \sqrt \frac {9\times 10^9}{0.53\times 10^{-10}\times 9.1\times 10^{-31}}$                                                        $= 1.6\times 3\times 10^{-19} \times 10^{24} \sqrt \frac {1000}{48.23}$         ...

The Supercar Experiment

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...

Rotating Current Loop

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

Dating the Ancient Leaves

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

Motion in a Gravity-Free Hall

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 $\equiv (ut, 0, 0)$.

Water as Diya Oil

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...

Drift Velocity in a Resistive Network

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 ...