Q.The average work done by a human heart while it beats once is 0.5 J. Calculate the power used by heart if it beats 72 times in a minute.

Step 1 — Choose the right principle

The question asks for the work done by the net force. The work–energy theorem states

This is the natural tool here: we are handed the speed as a function of position, which is precisely what we need for the endpoints. (You could find $F = ma = mv\frac{dv}{dx}$ and integrate $\int F\,dx$, but that is the same theorem done the long way.)

Step 2 — Write down the speed law

Squaring is more convenient than working with the speed itself, because kinetic energy needs $u^2$:

Step 3 — Evaluate at the two endpoints

At the initial point $x_i = 0$:

At the final point $x_f = 2\ \text{m}$:

$u_f^2 = 4(2)^3 = 4 \times 8 = 32~\text{m}^2\text{s}^{-2}$ …

Step 1 — Read the graph

The figure shows three curves against displacement $x$:

• $U(x)$ — an upward parabola, minimum (zero) at $x = 0$, maximum at $x = \pm A$.
• $K(x)$ — a downward parabola, maximum at $x = 0$, zero at $x = \pm A$.
• $E$ — a horizontal line (total energy is constant, as it must be for a conservative system).

The point $\pm x_0$ marked on the axis is where the two parabolas intersect. At an intersection the two functions are equal:

This is the entire physical content of the question.

Step 2 — The SHM energy expressions

For a particle of amplitude $A$ executing SHM under a restoring force $F = -kx$:

Potential energy (elastic PE stored at displacement $x$):

Total energy (constant — evaluate it at $x = A$, the extreme position, where the particle is momentarily at rest so $K = 0$ and all the energy is potential):

Kinetic energy (from conservation of mechanical energy):

These are exactly the two parabolas drawn.

Step 3 — Impose the crossing condition

Set $K(x_0) = U(x_0)$:

Cancel the common factor $\tfrac12 k$ (non-zero):

Taking the square root and keeping the magnitude:

Step 4 — The equivalent (and quicker) route

Because $K + U = E$ always, the condition $K = U$ means each is exactly half the total energy:

$U(x_0) = \frac{E}{2}$ …

Step 1 — Flux through the loop from the wire's field.

The wire's field at perpendicular distance $r$ is $B=\dfrac{\mu_0 I}{2\pi r}$. It is not uniform across the loop, so integrate: take a strip of width $dr$ and length $a$ at distance $r$,

Step 2 — Extract the mutual inductance.

By definition $\phi = M I$, so

$M$ depends only on the geometry — as it must.

Step 3 — Introduce the motion.

The loop moves away at constant speed, so $x=x_0+vt$ (increases linearly with $t$). Substituting,

Step 4 — Read off the shape of the graph.

• As $t$ increases, $\dfrac{a}{x_0+vt}$ decreases, so $M$ decreases monotonically. …

Step 1 — What "distance of closest approach" means

In Rutherford's experiment, an $\alpha$-particle (charge $+2e$) fired head-on at a nucleus (charge $+Ze$) is repelled by the Coulomb force. It decelerates, and at the point of closest approach it is momentarily at rest — all of its kinetic energy has been converted into electrostatic potential energy.

Step 2 — Apply conservation of energy

At launch (far away, PE $\approx0$): energy $=\tfrac12 mv^2$.

At closest approach (distance $d$, KE $=0$): energy $=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{d}$.

Equating:

Step 3 — Solve for $d$

Everything on the right except $v$ is fixed by the projectile and the target, so

The why is worth stating: the particle must "buy" its penetration with kinetic energy, and kinetic energy scales as $v^2$ — so it is $v^2$, not $v$, that sets how deep it gets.

Step 4 — Scale the velocity

Let the new velocity be $v'=2v$ and the new distance $d'$. Using the proportionality: …

Concept & Intuition

Power is the rate of doing work or transferring energy. Here, the gun does work by giving each bullet kinetic energy. The key is to find the total kinetic energy imparted to all bullets in one second, because “power” means energy per unit time. We don’t need force or acceleration — just the energy of the bullets leaving the gun.

Step-by-step solution

1. Find the number of bullets fired per second. The gun fires 120 bullets per minute.

2. Convert mass to kilograms.

   Each bullet has mass $100 \text{ g} = 0.1 \text{ kg}$.

   (Always use SI units: kg, m, s.)

3. Kinetic energy of one bullet.

Compute:

$120^2 = 14400$, so

4. Total kinetic energy per second (power). Since 2 bullets are fired each second,

Convert to kilowatts: $1440 \text{ W} = 1.44 \text{ kW}$.

5. Match with options. …

Step 1 — The kinetic-theory expression for pressure.

Pressure arises from molecular collisions with the walls. Kinetic theory gives

where $\bar E_k=\frac12 m\overline{v^2}$ is the average kinetic energy per molecule.

Step 2 — Impose constant volume.

With $N$ and $V$ fixed, the prefactor $\frac{2N}{3V}$ is a constant, so

(Equivalently, $\bar E_k = \frac32 k_B T$ and $P\propto T$ at constant $V$ — Gay-Lussac's law. The two statements are the same physics.)

Step 3 — Why the others fail.

• (A) Average potential energy — in an ideal gas the molecules are assumed to have no intermolecular potential energy, so this is identically zero and cannot determine $P$. …

Step 1 — What emf means.

The emf $\varepsilon$ of a cell is defined as the work done per unit charge by the source in driving charge around the circuit:

Step 2 — The charge delivered in time $t$.

A steady current $I$ transports

of charge through the circuit in time $t$.

Step 3 — The total work done by the battery.

Equivalently, the power delivered by the source is $P = \varepsilon I$, so $W = Pt = \varepsilon I t$.

Step 4 — Where that energy goes (why the other options are wrong).

Using $\varepsilon = I(R+r)$:

$W = \varepsilon I t = I^{2}Rt + I^{2}rt.$ …

When a charge $\Delta q$ moves through a potential difference $V$, the work done by the electric field is $\Delta W = V \Delta q$. Since current $I = \Delta q / \Delta t$, we have $\Delta W = I V \Delta t$. This is the energy supplied by the battery.

The key question is: where does this energy go? In a normal conductor (like a metal wire), the charge carriers (electrons) do not accelerate indefinitely. They collide repeatedly with the lattice ions, transferring their gained kinetic energy to the lattice as heat. So the energy does not stay as kinetic energy of the charges — it becomes thermal energy of the conductor.

Let’s examine each statement carefully.

1. Statement (I): "the kinetic energy of the charge increases by $IV\Delta t$."

   If the charges kept gaining kinetic energy, they would accelerate without limit. In a steady current, the drift velocity is constant — the energy gained from the field is lost in collisions. So the kinetic energy of the charge carriers does not increase by $IV\Delta t$. This statement is false.

2. Statement (II): "the electric potential energy of the charge decreases by $IV\Delta t$."

   As a charge moves from higher to lower potential, its electric potential energy decreases by exactly $V \Delta q = I V \Delta t$. This is the very definition of potential difference. This statement is true.

3. Statement (III): "the thermal energy of the conductor increases by $IV\Delta t$." …

The key idea comes from Bohr's model of the hydrogen atom. In this model, the electron moves in circular orbits around the nucleus under the influence of the Coulomb force. The period of revolution is simply the time taken to complete one full circle: $T = \frac{2\pi r}{v}$, where $r$ is the radius of the orbit and $v$ is the orbital speed. To find how $T$ depends on the principal quantum number $n$, we need to know how both $r$ and $v$ scale with $n$.

From Bohr's postulates, two results are central:

1. The radius of the $n^{th}$ orbit is proportional to $n^2$: $r_n \propto n^2$.
2. The velocity of the electron in the $n^{th}$ orbit is inversely proportional to $n$: $v_n \propto \frac{1}{n}$.

These come from quantising angular momentum ($mvr = \frac{nh}{2\pi}$) and equating the Coulomb force to the centripetal force. Let's combine them step by step.

1. Write the expression for the period. The time for one revolution is the circumference divided by speed:

2. Substitute the proportionalities. Since $r_n \propto n^2$ and $v_n \propto \frac{1}{n}$, we get:

3. Check the constants. If you want the exact expression, it is: $T_n = \frac{4 \varepsilon_0^2 h^3 n^3}{m e^4},$ …

Step 1 — The force on the charge.

In a uniform field $\vec E$, a charge $q$ experiences

which is constant in magnitude and direction. Released from rest, the particle accelerates along $\vec E$ (for $q>0$), so its displacement $y$ is parallel to the force.

Step 2 — Work done by that force.

For a constant force parallel to the displacement,

Step 3 — Work–energy theorem.

The particle is released from rest, so $K_i=0$, and the electric force is the only force doing work. Hence

Step 4 — Cross-check with kinematics.

$a=\frac{qE}{m},\qquad v^{2}=u^{2}+2ay=0+2\left(\frac{qE}{m}\right)y=\frac{2qEy}{m},$ …

Step 1 — The physics.

When a charge $q$ is accelerated through a potential difference $V$, the electric field does work

on it. If the charge starts from rest and no other force acts, the work–energy theorem says all of this work becomes kinetic energy:

Step 2 — Solve for $v$.

Step 3 — Substitute the given values ($m = 1\ \mathrm{kg}$, $q = 2\ \mathrm{C}$, $V = 1\ \mathrm{V}$; all already in SI units, so no conversion is needed):

$v = \sqrt{\frac{2 \times (2\ \mathrm{C}) \times (1\ \mathrm{V})}{1\ \mathrm{kg}}} = \sqrt{4} = 2\ \mathrm{m\,s^{-1}}$ …

Step 1 — The half-life law.

Radioactive decay is exponential, $N=N_0e^{-\lambda t}$, which is most conveniently written in terms of the half-life $t_{1/2}$:

where $n$ is the number of half-lives elapsed. (Mass is proportional to the number of undecayed nuclei, so the same formula applies to the mass.)

Step 2 — Number of half-lives.

Step 3 — Remaining mass.

Tracing it half-life by half-life makes it transparent: …