HINTS & SOLUTIONS

Answer: Straight line with negative slope passing through the t-axis. Explanation: When a particle is thrown vertically upwards, its initial velocity $u$ is positive. Due to constant downward acceleration (gravity, $a = -g$), its velocity decreases linearly to zero at the highest point, and then becomes increasingly negative as it falls back down. The V-t graph is a straight line with a constant negative slope.

Answer: (b) Both A and R are true and R is not the correct explanation of A. Explanation: The size of the balloon increases because atmospheric pressure decreases with altitude, allowing the gas inside to expand (Boyle's Law). While the balloon material being stretchable is true, the primary physical reason for expansion is the drop in external pressure, not just the stretchability itself.

Answer: (a) Both A and R are true and R is the correct explanation of A. Explanation: An impulsive force is defined as a very large force acting for a very short time interval such that it produces a finite change in momentum (Impulse $J = F \times \Delta t = \Delta p$). The requirement to produce this finite momentum change in a minuscule time period is the exact reason the force must be extremely large.

Answer: (c) Zero Explanation: Work done is given by $W = F \cdot d$. While the cycle exerts a force of 200 N on the road (by Newton's Third Law), the road itself does not undergo any displacement ($d = 0$). Therefore, the work done *by* the cycle *on* the road is strictly zero.

Answer: (d) Total linear momentum Explanation: In all types of collisions (elastic, inelastic, perfectly inelastic) occurring in an isolated system, the total linear momentum is always conserved because there are no external unbalanced forces acting on the system.

Answer: (c) (A)-4, (B)-1, (C)-2, (D)-3 Explanation: Moment of Inertia ($I = mr^2$) $\to [ML^2T^0]$ (4). Torque ($\tau = r \times F$) $\to [ML^2T^{-2}]$ (1). Angular momentum ($L = mvr$) $\to [ML^2T^{-1}]$ (2). Radius of Gyration ($k$, a distance) $\to [M^0LT^0]$ (3).

Answer: (b) (A)-2, (B)-3, (C)-4, (D)-1 Explanation: Gravitational Intensity ($I = F/m$) $\to ms^{-2}$ or $Nkg^{-1}$ (2). Gravitational Potential ($V = W/m$) $\to Jkg^{-1}$ (3). Gravitational Potential Energy $\to$ Joule (J) (4). Gravitational constant ($G$) $\to Nm^2kg^{-2}$ (1).

Answer: (c) Longitudinal and Shear Explanation: When a helical spring is stretched by a load, there is an obvious macroscopic longitudinal increase in length. However, at a microscopic/material level, the wire of the spring is actually undergoing twisting or torsion. Therefore, it involves both longitudinal strain (macroscopically) and shear strain (microscopically in the wire).

Answer: (d) Energy Explanation: The First Law of Thermodynamics ($\Delta Q = \Delta U + \Delta W$) is essentially the principle of conservation of energy applied to a thermodynamic system.

Answer: (a) Isothermal process Explanation: Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to its volume, provided the temperature remains constant ($PV = \text{constant}$). A process occurring at constant temperature is called an isothermal process.

Answer: (b) Simple harmonic and time period is independent of the density of the liquid Explanation: The time period of an oscillating liquid column in a U-tube is given by $T = 2\pi\sqrt{\frac{h}{g}}$, where $h$ is the height of the liquid column in equilibrium. The density ($\rho$) of the liquid cancels out during the derivation, making the time period completely independent of it.

Answer: (a) Energy Explanation: A mechanical wave (like a longitudinal sound wave) propagates by transferring energy and momentum from one particle of the medium to the next through oscillations. However, the particles themselves only oscillate about their mean positions; there is no net bulk transport of matter.

Solution: Velocity-Time Graph: It is a graph plotted by taking time ($t$) along the x-axis and velocity ($v$) of a particle along the y-axis.
Uses: 1. The slope of the velocity-time graph gives the acceleration of the object. 2. The area under the velocity-time graph and the time axis gives the displacement covered by the object in a given time interval.

Solution: Given: Initial velocity $u = 126 \text{ kmh}^{-1} = 126 \times \frac{5}{18} \text{ m/s} = 35 \text{ m/s}$. Final velocity $v = 0 \text{ m/s}$. Distance $s = 200 \text{ m}$. Using 3rd equation of motion: $v^2 - u^2 = 2as$. $0 - (35)^2 = 2 \times a \times 200 \implies -1225 = 400a \implies a = \frac{-1225}{400} = -3.06 \text{ m/s}^2$. Retardation is $3.06 \text{ m/s}^2$.
Using 1st equation of motion to find time: $v = u + at \implies 0 = 35 - 3.06t$. $t = \frac{35}{3.06} \approx 11.44 \text{ s}$. The car takes 11.44 seconds to stop.

Solution: According to Newton's second law, Force $F = \frac{\Delta p}{\Delta t}$ (rate of change of momentum). When a cricket player lowers his hands while catching a ball, he increases the time interval ($\Delta t$) over which the high velocity of the ball is reduced to zero. Since the change in momentum ($\Delta p$) is constant, increasing $\Delta t$ significantly decreases the impact force ($F$) exerted by the ball on his hands, preventing injury.

Solution: Angle of Friction ($\theta$): The angle which the resultant of limiting friction and normal reaction makes with the normal reaction. ($\tan\theta = \mu_s$).
Angle of Repose ($\alpha$): The minimum angle of an inclined plane with the horizontal such that a body placed on it just begins to slide down. Proof of Equality: For a body just sliding down an inclined plane of angle $\alpha$, the forces acting along the plane are $mg\sin\alpha$ (downward) and $f_s$ (limiting friction, upward). $f_s = mg\sin\alpha$ and Normal reaction $R = mg\cos\alpha$. Coefficient of friction $\mu_s = \frac{f_s}{R} = \frac{mg\sin\alpha}{mg\cos\alpha} = \tan\alpha$. Since $\tan\theta = \mu_s$ and $\tan\alpha = \mu_s$, it follows that $\tan\theta = \tan\alpha \implies \theta = \alpha$.

Solution: Pascal's Law: It states that pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. Proof: Consider an arbitrary triangular prism element of fluid in equilibrium inside a liquid. Let the areas of the faces be $A_1, A_2, A_3$ and the uniform pressures on them be $P_1, P_2, P_3$. The forces are $F_1=P_1 A_1$, $F_2=P_2 A_2$, $F_3=P_3 A_3$. Since the element is in equilibrium, balancing horizontal and vertical components (ignoring gravity for a very small element) yields $F_1 \sin\theta = F_2$ and $F_1 \cos\theta = F_3$. From geometry, $A_1 \sin\theta = A_2$ and $A_1 \cos\theta = A_3$. Dividing the force equations by area equations yields $P_1 = P_2 = P_3$. Hence, pressure is same in all directions. Solution (Alternative): Stoke's Law: It states that the backward viscous force ($F$) acting on a small spherical body falling through a viscous medium is directly proportional to the radius ($r$) of the body, its velocity ($v$), and the coefficient of viscosity ($\eta$) of the fluid.
Derivation by Dimensions: Let $F \propto \eta^a r^b v^c \implies F = k \eta^a r^b v^c$. Writing dimensions: $[MLT^{-2}] = [ML^{-1}T^{-1}]^a [L]^b [LT^{-1}]^c$. $[MLT^{-2}] = [M^a L^{-a+b+c} T^{-a-c}]$. Comparing powers of M, L, T: M: $a = 1$ T: $-a - c = -2 \implies -1 - c = -2 \implies c = 1$ L: $-a + b + c = 1 \implies -1 + b + 1 = 1 \implies b = 1$. Substituting values: $F = k \eta r v$. Experimentally, $k = 6\pi$. Thus, $F = 6\pi\eta r v$.

Solution: A perfectly rigid body does not undergo any change in its length, shape, or volume regardless of the amount of force (stress) applied to it. This means the strain produced is always zero. Since Modulus of Elasticity = $\frac{\text{Stress}}{\text{Strain}}$, and Strain = 0, both Young's modulus and Bulk modulus for a perfectly rigid body are Infinite ($\infty$).

Solution: Joule is the SI unit of energy and erg is the CGS unit of energy. Dimensional formula of Energy = $[ML^2T^{-2}]$. Here $a=1, b=2, c=-2$. We know $n_2 = n_1 \left[\frac{M_1}{M_2}\right]^a \left[\frac{L_1}{L_2}\right]^b \left[\frac{T_1}{T_2}\right]^c$. Given $n_1 = 100$. $n_2 = 100 \left[\frac{1 \text{ kg}}{1 \text{ g}}\right]^1 \left[\frac{1 \text{ m}}{1 \text{ cm}}\right]^2 \left[\frac{1 \text{ s}}{1 \text{ s}}\right]^{-2}$ $n_2 = 100 \left[\frac{1000 \text{ g}}{1 \text{ g}}\right]^1 \left[\frac{100 \text{ cm}}{1 \text{ cm}}\right]^2 \left[1\right]^{-2}$ $n_2 = 100 \times 10^3 \times (10^2)^2 = 10^2 \times 10^3 \times 10^4 = 10^9$. Therefore, $100 \text{ Joules} = 10^9 \text{ ergs}$.

Solution: Force Constant ($k$): It is the restoring force set up in a spring per unit extension or compression. $k = F/x$.
Elastic Potential Energy Derivation: According to Hooke's Law, the restoring force $F$ is proportional to the displacement $x$: $F = -kx$. To stretch the spring by an additional small distance $dx$ against this restoring force, work done is $dW = -F dx = -(-kx) dx = kx dx$. Total work done to stretch the spring from $0$ to $x$ is: $W = \int_{0}^{x} kx dx = k \left[ \frac{x^2}{2} \right]_{0}^{x} = \frac{1}{2} kx^2$. This work is stored as elastic potential energy ($U$). Thus, $U = \frac{1}{2} kx^2$. Solution (Alternative): Conservation of Mechanical Energy for freely falling body: Let a body of mass $m$ fall from rest at height $h$ (Point A). At A: Velocity $v=0 \implies K.E. = 0$. Potential Energy $P.E. = mgh$. Total Energy $E_A = 0 + mgh = mgh$. At B (distance $x$ from A): Velocity $v^2 - 0 = 2gx \implies v^2 = 2gx$. $K.E. = \frac{1}{2}mv^2 = \frac{1}{2}m(2gx) = mgx$. Height from ground is $(h-x) \implies P.E. = mg(h-x)$. Total Energy $E_B = mgx + mg(h-x) = mgh$. At C (on ground): Velocity $V^2 - 0 = 2gh \implies V^2 = 2gh$. $K.E. = \frac{1}{2}mV^2 = mgh$. $P.E. = 0$. Total Energy $E_C = mgh + 0 = mgh$. Since $E_A = E_B = E_C = mgh$, total mechanical energy is conserved.

Solution: Angular Momentum ($L$): It is the moment of linear momentum of a rotating body about the axis of rotation. $\vec{L} = \vec{r} \times \vec{p}$.
Relation to Areal Velocity: Consider a particle of mass $m$ moving in a plane. Let its position vector change from $\vec{r}$ to $\vec{r} + d\vec{r}$ in time $dt$. The area $dA$ swept by the position vector in time $dt$ is the area of the triangle formed: $d\vec{A} = \frac{1}{2} (\vec{r} \times d\vec{r})$. Dividing by $dt$: $\frac{d\vec{A}}{dt} = \frac{1}{2} (\vec{r} \times \frac{d\vec{r}}{dt}) = \frac{1}{2} (\vec{r} \times \vec{v})$. Multiply and divide by mass $m$: $\frac{d\vec{A}}{dt} = \frac{1}{2m} (\vec{r} \times m\vec{v}) = \frac{1}{2m} (\vec{r} \times \vec{p})$. Since $\vec{r} \times \vec{p} = \vec{L}$ (Angular Momentum), we get: $\frac{dA}{dt} = \frac{L}{2m} \implies L = 2m \left(\frac{dA}{dt}\right)$. Thus, angular momentum is equal to twice the product of mass and areal velocity.

Solution: Escape Velocity ($v_e$): The minimum initial velocity required to project a body from the surface of a planet so that it completely overcomes the planet's gravitational pull and never returns.
Derivation: Potential energy of a mass $m$ on planet's surface is $U = -\frac{GMm}{R}$. Kinetic energy required to escape is $K = \frac{1}{2}mv_e^2$. To reach infinity where total energy is zero: $K + U = 0 \implies \frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0$. $\frac{1}{2}v_e^2 = \frac{GM}{R} \implies v_e = \sqrt{\frac{2GM}{R}}$. Since acceleration due to gravity $g = \frac{GM}{R^2} \implies GM = gR^2$. Substituting $GM$, we get $v_e = \sqrt{\frac{2(gR^2)}{R}} = \sqrt{2gR}$. Solution (Alternative): Gravitational Constant (G): It is numerically equal to the gravitational force of attraction between two unit masses placed at a unit distance apart. Gravity with Depth: Let $g$ be gravity on the surface of earth (radius R, density $\rho$). $g = \frac{GM}{R^2} = \frac{G (\frac{4}{3}\pi R^3 \rho)}{R^2} = \frac{4}{3}\pi G R \rho$. At depth $d$, the body is at distance $(R-d)$ from the center. Only the inner sphere of radius $(R-d)$ exerts force. $g_d = \frac{G M'}{(R-d)^2} = \frac{G (\frac{4}{3}\pi (R-d)^3 \rho)}{(R-d)^2} = \frac{4}{3}\pi G (R-d) \rho$. Dividing the equations: $\frac{g_d}{g} = \frac{R-d}{R} = 1 - \frac{d}{R}$. $g_d = g\left(1 - \frac{d}{R}\right)$. This shows $g$ decreases with depth. At the centre of the earth: Depth $d = R$. $g_{centre} = g\left(1 - \frac{R}{R}\right) = g(1 - 1) = 0$.

Solution: Given: Tension Force $F = 44500 \text{ N}$. Dimensions of rectangular cross-section = $15.2 \text{ mm} \times 19.1 \text{ mm}$. Area $A = (15.2 \times 10^{-3} \text{ m}) \times (19.1 \times 10^{-3} \text{ m}) = 290.32 \times 10^{-6} \text{ m}^2 = 2.9032 \times 10^{-4} \text{ m}^2$. Modulus of Elasticity (Shear) $G = 42 \times 10^9 \text{ N/m}^2$. We know, Modulus of Elasticity = $\frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\text{Strain}}$. Strain = $\frac{F}{A \times G} = \frac{44500}{(2.9032 \times 10^{-4}) \times (42 \times 10^9)}$. Strain = $\frac{44500}{121.93 \times 10^5} = \frac{44500}{12193000} \approx 3.65 \times 10^{-3}$.

Solution: Let 1 mole of an ideal gas be heated at constant volume. Since $\Delta V = 0$, work done $W = P\Delta V = 0$. By First Law of Thermodynamics, heat supplied $dQ_v = dU + 0 \implies C_v dT = dU$ (where $C_v$ is molar specific heat at constant volume). Now let the same gas be heated at constant pressure to achieve the same temperature rise $dT$. Heat supplied $dQ_p = dU + dW = dU + P dV$. Since $dQ_p = C_p dT$ (where $C_p$ is molar specific heat at constant pressure), we have: $C_p dT = C_v dT + P dV$. From Ideal Gas equation for 1 mole, $PV = RT$. Differentiating at constant pressure: $P dV = R dT$. Substituting this in the previous equation: $C_p dT = C_v dT + R dT$. Dividing by $dT$, we get $C_p = C_v + R \implies C_p - C_v = R$. (Mayer's Formula).

Solution: Postulates of Kinetic Theory of Gases: 1. A gas consists of a very large number of identical, tiny spherical particles called molecules. 2. The molecules are in constant, random, rapid motion, colliding with each other and the walls of the container. 3. All collisions are perfectly elastic (no loss of kinetic energy). 4. The size of the molecules is negligible compared to the average distance between them. 5. There are no intermolecular forces of attraction or repulsion between the molecules except during collisions.
Kinetic Interpretation of Temperature: From kinetic theory, pressure $P = \frac{1}{3}\frac{M}{V} v_{rms}^2$. $PV = \frac{1}{3} M v_{rms}^2$. Multiply and divide by 2: $PV = \frac{2}{3} \left(\frac{1}{2} M v_{rms}^2\right) = \frac{2}{3} E$ (where $E$ is total translational kinetic energy). From ideal gas equation for 1 mole, $PV = RT$. Therefore, $\frac{2}{3} E = RT \implies E = \frac{3}{2} RT$. This shows that the average translational kinetic energy of a gas is directly proportional to its absolute temperature ($E \propto T$).

Solution: Simple Harmonic Motion (SHM): It is a special type of periodic motion in which the restoring force acting on the particle is directly proportional to its displacement from the mean position and is always directed towards the mean position ($F = -kx$).
Velocity Derivation: Displacement of a particle in SHM is $y = A \sin(\omega t)$ (starting from mean position). Velocity $v = \frac{dy}{dt} = \frac{d}{dt}(A \sin\omega t) = A\omega \cos\omega t$. Since $\cos\omega t = \sqrt{1 - \sin^2\omega t}$ and $\sin\omega t = y/A$: $v = A\omega \sqrt{1 - (y/A)^2} = \omega \sqrt{A^2 - y^2}$.
Acceleration Derivation: Acceleration $a = \frac{dv}{dt} = \frac{d}{dt}(A\omega \cos\omega t) = -A\omega^2 \sin\omega t$. Since $y = A\sin\omega t$, we get $a = -\omega^2 y$.

Solution: Triangle Law of Vector Addition: If two vectors acting simultaneously on a body are represented in magnitude and direction by two sides of a triangle taken in order, then their resultant is completely represented by the third closing side of the triangle taken in the opposite order.
Analytical Treatment: Let vectors $\vec{P}$ and $\vec{Q}$ act at angle $\theta$. Represent them by sides OA and AB of triangle OAB. Resultant $\vec{R}$ is OB. Draw perpendicular BM on extended OA. In right $\Delta ABM$: $BM = Q\sin\theta$ and $AM = Q\cos\theta$. In right $\Delta OMB$: $OB^2 = OM^2 + BM^2 = (OA + AM)^2 + BM^2$. $R^2 = (P + Q\cos\theta)^2 + (Q\sin\theta)^2 = P^2 + 2PQ\cos\theta + Q^2\cos^2\theta + Q^2\sin^2\theta$. $R^2 = P^2 + Q^2(\cos^2\theta + \sin^2\theta) + 2PQ\cos\theta = P^2 + Q^2 + 2PQ\cos\theta$. Magnitude: $R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$. Direction ($\beta$ with $\vec{P}$): $\tan\beta = \frac{BM}{OM} = \frac{Q\sin\theta}{P + Q\cos\theta}$. Solution (Alternative): Projectile: Any object thrown into space with some initial velocity and then allowed to move freely strictly under the influence of gravity alone is called a projectile.
Parabolic Path: Let a projectile be fired with velocity $u$ at angle $\theta$. Horizontal component $u_x = u\cos\theta$ (constant, as $a_x=0$). Vertical component $u_y = u\sin\theta$ ($a_y = -g$). After time $t$, horizontal distance $x = (u\cos\theta)t \implies t = \frac{x}{u\cos\theta}$. Vertical distance $y = u_y t - \frac{1}{2}gt^2 = (u\sin\theta)t - \frac{1}{2}gt^2$. Substitute $t$: $y = u\sin\theta \left(\frac{x}{u\cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2$. $y = (\tan\theta)x - \left(\frac{g}{2u^2\cos^2\theta}\right)x^2$. This is a quadratic equation ($y = ax - bx^2$), proving the trajectory is a parabola.
Maximum Height ($H$): At max height, final vertical velocity $v_y = 0$. Using $v_y^2 - u_y^2 = 2a_y y \implies 0 - (u\sin\theta)^2 = -2gH \implies H = \frac{u^2\sin^2\theta}{2g}$.
Horizontal Range ($R$): Distance covered during Time of Flight ($T = \frac{2u\sin\theta}{g}$). $R = u_x \times T = (u\cos\theta) \times \left(\frac{2u\sin\theta}{g}\right) = \frac{u^2 (2\sin\theta\cos\theta)}{g} = \frac{u^2\sin(2\theta)}{g}$.

Solution: (a) Centripetal Force: The external force required to make a body move in a circular path with uniform speed. It always acts along the radius directed towards the center of the circular path. Example: Tension in a string whirling a stone, gravitational force on the moon. Expression: $F = \frac{mv^2}{r}$ or $F = mr\omega^2$.

(b) Friction is a necessary evil: It is an evil because it causes wear and tear of machinery parts, dissipates energy as heat reducing efficiency, and opposes the desired relative motion. It is necessary because without friction, we could not walk without slipping, vehicles could not brake or turn safely, and we couldn't grip objects, write, or light a match. Since it possesses both indispensable benefits and unavoidable drawbacks, it is termed a necessary evil. Solution (Alternative): (a) Time to stop: Given $F = -50 \text{ N}$ (retarding), $m = 20 \text{ kg}$, $u = 15 \text{ m/s}$, $v = 0$. Acceleration $a = F/m = -50/20 = -2.5 \text{ m/s}^2$. Using $v = u + at \implies 0 = 15 - 2.5t \implies 2.5t = 15 \implies t = 6 \text{ seconds}$.

(b) Newton's Second Law: The rate of change of momentum of a body is directly proportional to the applied unbalanced external force and takes place in the direction in which the force acts. Proof: Force $F \propto \frac{dp}{dt}$. Let $p = mv$. $F \propto \frac{d(mv)}{dt}$. Since mass $m$ is constant for non-relativistic speeds, $F \propto m \frac{dv}{dt}$. We know acceleration $a = \frac{dv}{dt}$. So, $F \propto ma \implies F = kma$. By defining 1 unit of force as the force that produces 1 unit acceleration in 1 unit mass, $k=1$. Thus, $\vec{F} = m\vec{a}$.

Solution: (a) Modes of vibration in Closed Organ Pipe: A closed organ pipe is closed at one end. A node must form at the closed end (displacement zero) and an antinode at the open end. First mode (Fundamental): Length $L = \lambda_1/4 \implies \lambda_1 = 4L$. Frequency $\nu_1 = v/\lambda_1 = \frac{v}{4L}$. Second mode (First Overtone): Contains one node and antinode inside. $L = 3\lambda_2/4 \implies \lambda_2 = 4L/3$. Frequency $\nu_2 = \frac{v}{\lambda_2} = \frac{3v}{4L} = 3\nu_1$. (3rd harmonic). Third mode (Second Overtone): $L = 5\lambda_3/4 \implies \lambda_3 = 4L/5$. Frequency $\nu_3 = \frac{5v}{4L} = 5\nu_1$. (5th harmonic). The ratio of frequencies of harmonics produced is $\nu_1 : \nu_2 : \nu_3 ... = 1 : 3 : 5 : 7$. Only odd harmonics are present.

(b) Elastic Potential Energy of Spring: Let a spring of force constant $k$ be stretched by a distance $x$. According to Hooke's law, restoring force $F = -kx$. To stretch it further by $dx$ against this force, the external work done is $dW = -F dx = -(-kx)dx = kx dx$. Total work done from $x=0$ to $x=x_0$ is $W = \int_{0}^{x_0} kx dx = k \left[\frac{x^2}{2}\right]_0^{x_0} = \frac{1}{2}kx_0^2$. This work is stored as the elastic potential energy $U$. Hence, $U = \frac{1}{2}kx^2$.