NEB Class 11 · Past paper
The official NEB Class 11 model questions for Physics, all 36 questions with solved model answers.
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Answer in brief, any six questions. (a) Does a dimensionally correct equation necessarily represent a correct physical relation? Justify your answer with one example. (b) A boatman wants to take his boat to a point just opposite on the other bank of the river. In which direction should he row the boat? Explain with the help of a suitable diagram. (c) If the net force acting on a body is zero, will the body necessarily remain in the rest position? Explain. (d) A body weighs more at the pole than at the equator, why? (e) Differentiate between conservative and non-conservative forces. (f) A solid ball and a hollow ball with the same mass and radius roll down a slope. Which one reaches the bottom first? (g) Machine parts are jammed in winter, why?
(a) No. Dimensional correctness is a necessary but not sufficient condition. It cannot detect dimensionless constants, nor errors in terms that share the same dimensions. Example: $s = ut + at^2$ is dimensionally correct (
Answer in brief, any two questions. (a) Why does a cycle tube burst in summer? (b) The triple point of water is chosen as a standard fixed point in modern thermometry, why? (c) Explain why the molar heat capacity at constant pressure is greater than that at constant volume.
(a) In summer the temperature rises. By the pressure law (Gay-Lussac), at fixed volume $\dfrac{P}{T} = \text{constant}$, so the pressure of the enclosed air rises with temperature. When the internal pressure exceeds the strength of the r...
Answer in brief, any one question. (a) Distinguish between chromatic aberration and spherical aberration. (b) The Sun is less bright in the morning and evening as compared to noon although its distance from the observer is almost the same. Why?
(a) Chromatic aberration is the failure of a lens to bring different colours to a common focus. Because the refractive index of the glass varies with wavelength, violet light bends more than red and focuses nearer the lens, so coloured fringes appear around the image. It is a defect of lenses and is corrected by combining a convex and a concave lens of different glasses into an achromatic doublet.
Spherical aberration, on the other hand, is the failure of a spherical lens or mirror to bring edge (marginal) rays and central (paraxial) rays to the same focus. Rays through the edge focus closer than those near the axis. This occurs even for a single colour, and it is reduced by using a parabolic surface or by using a stop to block the marginal rays.
(b) The reason lies in the path the sunlight takes through the atmosphere, not the distance to the Sun. At noon the Sun is almost overhead, so its light travels down through a short, nearly vertical thickness of air. In the morning and evening the Sun is near the horizon, so its rays enter at a slant and pass through a far greater thickness of atmosphere. Over this longer path much more light is scattered and absorbed by air molecules and dust (Rayleigh scattering strongly removing the shorter wavelengths), so less intensity reaches the observer and the Sun looks dimmer and redder, even though its distance is essentially unchanged.
Answer in brief, any one question. (a) Vehicles carrying inflammable materials usually have metallic ropes/chains touching the ground during motion. Why? (b) Mention the factors on which the capacitance of a parallel plate capacitor depends.
(a) As the vehicle moves, friction between the tyres, the body and the air generates static electric charge, which can accumulate to a high potential on the metal tank. A spark from this charge could ignite the inflammable vapour. The tr...
Answer any three questions. (a) State the triangle law of vector addition. Obtain an expression for the resultant of two vectors inclined at an angle. (b) What is banking of the road? Discuss the motion of a vehicle round a circular banked track. (c) Define angular momentum and moment of inertia. Establish a relation between them. (d) What is capillarity? Deduce an expression for the rise of a liquid in a capillary tube.
(a) Triangle law: if two vectors are represented in magnitude and direction by the two sides of a triangle taken in order, their resultant is represented by the third side taken in the opposite order. For two vectors $\vec P$ and $\vec Q$ inclined at angle $\theta$, the magnitude of the resultant is
$$R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$$
and its direction, as the angle $\alpha$ made with $\vec P$, is
$$\alpha = \tan^{-1}!\left(\frac{Q\sin\theta}{P + Q\cos\theta}\right)$$
(b) Banking means raising the outer edge of a curved road above the inner edge so the surface makes an angle $\theta$ with the horizontal. The horizontal component of the normal reaction then supplies the centripetal force, reducing the reliance on friction. For the ideal (frictionless) speed, the vertical and horizontal equations are
$$ \begin{aligned} N\cos\theta &= mg \ N\sin\theta &= \frac{mv^2}{r} \end{aligned} $$
Dividing the second by the first,
$$\tan\theta = \frac{v^2}{rg}$$
so the safe speed is
$$v = \sqrt{rg\tan\theta}$$
At this speed the vehicle rounds the bend with no tendency to skid.
(c) Angular momentum $\vec L = \vec r \times \vec p$ is the moment of linear momentum about a point, and for a rigid body $L = I\omega$. Moment of inertia $I = \sum m_i r_i^2$ is the rotational inertia about an axis. To relate them, a single particle has
$$ \begin{aligned} L &= mvr \ &= m(r\omega)r \ &= (mr^2)\omega \ &= I\omega \end{aligned} $$
Summing over all particles of a rigid body,
$$ \begin{aligned} L &= \left(\sum m_i r_i^2\right)\omega \ &= I\omega \end{aligned} $$
(d) Capillarity is the rise or fall of a liquid in a fine tube due to surface tension. Balancing the weight of the raised column against the vertical pull of surface tension acting around the circumference $2\pi r$:
$$T\cos\theta ,(2\pi r) = (\pi r^2 h)\rho g$$
Solving for the height of rise,
$$h = \frac{2T\cos\theta}{r\rho g}$$
where $T$ is the surface tension, $\theta$ the contact angle, $r$ the tube radius and $\rho$ the liquid density. The rise is greater for a narrower tube.
Answer any two questions. (a) State the principle of calorimetry and use it to determine the specific latent heat of fusion of ice by the method of mixtures. (b) What is a black body? How is it realized in practice? Explain Stefan's law of black body radiation. (c) Describe a diesel engine and explain its working with the help of a PV diagram.
(a) Principle of calorimetry: when bodies at different temperatures are mixed and no heat escapes to the surroundings, the heat lost by the hotter body equals the heat gained by the colder one. To find the specific latent heat of fusion ...
Answer any one question. (a) What is lateral shift? Deduce an expression for it. How does the lateral shift change with the increase in the angle of incidence? (b) With the help of a labelled diagram, explain the construction and working of a compound microscope. Derive an expression for its magnifying power.
(a) Lateral shift is the perpendicular displacement between the incident ray (produced) and the emergent ray when light passes through a parallel-sided glass slab; the emergent ray is parallel to the incident ray but shifted sideways. If $t$ is the slab thickness, $i$ the angle of incidence and $r$ the angle of refraction, then $$\text{lateral shift } d = \frac{t\sin(i-r)}{\cos r}.$$ As the angle of incidence $i$ increases, $(i-r)$ increases and $\cos r$ decreases, so the lateral shift increases; it is maximum (equal to $t$) at grazing incidence ($i \to 90^\circ$).
(b) Compound microscope: a short-focus objective lens forms a real, inverted, magnified image of a nearby object placed just beyond its focus; this image lies within the focal length of the eyepiece, which acts as a simple magnifier to give a final virtual, inverted, highly magnified image at the least distance of distinct vision $D$. The magnifying power is the product of the two stages: $$ \begin{aligned} M &= m_o \times m_e \ &= \frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right)\approx \frac{L}{f_o}\cdot\frac{D}{f_e}, \end{aligned} $$ where $L$ is the tube length and $f_o, f_e$ are the objective and eyepiece focal lengths. Short $f_o$ and $f_e$ give high magnification.
Answer any one question. (a) Define electric field intensity and potential gradient. Obtain a relation between them. (b) State and explain Gauss's theorem and use it to find the electric field intensity due to a charged sphere (i) outside it and (ii) inside it.
(a) Electric field intensity $\vec E$ at a point is the force per unit positive test charge placed there, $\vec E = \vec F/q$ (units $\text{N C}^{-1}$ or $\text{V m}^{-1}$). Potential gradient is the rate of change of potential with dist...
Answer any three numerical questions. (a) Two masses 7 kg and 12 kg are connected at the two ends of a light inextensible string that passes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released. (b) A proposed communication satellite would revolve round the earth in a circular orbit in the equatorial plane at a height of 36000 km above the earth's surface. Find the period of revolution of the satellite in hours. (c) A steel wire of length 4.7 m and cross section 3x10^-5 m^2 stretches by the same amount as a copper wire of length 3.5 m and cross section 4x10^-5 m^2 under a given load. What is the ratio of the Young's modulus of steel to that of copper? (d) A body is vibrating with SHM of amplitude 15 cm and frequency 4 Hz. Calculate the velocity and acceleration when the displacement is 9 cm from the mean position.
(a) With the heavier mass falling, the unbalanced weight $(m_2-m_1)g$ drives the whole $m_1+m_2$ system, so the acceleration is
$$ \begin{aligned} a &= \frac{(m_2-m_1)g}{m_1+m_2} \ &= \frac{(12-7)(10)}{7+12} \ &= \frac{50}{19} \ &= 2.63\ \text{m s}^{-2} \end{aligned} $$
and the tension in the connecting string works out to
$$ \begin{aligned} T &= \frac{2m_1 m_2 g}{m_1+m_2} \ &= \frac{2(7)(12)(10)}{19} \ &= \frac{1680}{19} \ &= 88.4\ \text{N} \end{aligned} $$
So the masses accelerate at $2.63\ \text{m s}^{-2}$ and the string tension is $88.4\ \text{N}$.
(b) The satellite orbits at radius $r = R + h = 6.4\times10^6 + 3.6\times10^7 = 4.24\times10^7\ \text{m}$, and with $GM = 6.67\times10^{-11}\times6\times10^{24} = 4.00\times10^{14}$ the period is
$$ \begin{aligned} T &= 2\pi\sqrt{\frac{r^3}{GM}} \ &= 2\pi\sqrt{\frac{(4.24\times10^7)^3}{4.00\times10^{14}}} \end{aligned} $$
Working out the ratio inside the root,
$$ \begin{aligned} T &= 2\pi\sqrt{\frac{7.62\times10^{22}}{4.00\times10^{14}}} \ &= 2\pi\sqrt{1.905\times10^{8}} \end{aligned} $$
which gives
$$ \begin{aligned} T &= 2\pi(1.380\times10^4) \ &= 8.67\times10^4\ \text{s} \ &= \frac{8.67\times10^4}{3600} \ &= 24.1\ \text{h} \end{aligned} $$
The period is about $24$ hours, which is exactly why such a satellite stays fixed over one point (geostationary).
(c) Both wires carry the same load and stretch by the same amount, and since $Y = \dfrac{FL}{A,\Delta L}$ with $F$ and $\Delta L$ common to both, the ratio reduces to the geometry alone:
$$ \begin{aligned} \frac{Y_s}{Y_c} &= \frac{L_s/A_s}{L_c/A_c} \ &= \frac{L_s A_c}{L_c A_s} \ &= \frac{(4.7)(4\times10^{-5})}{(3.5)(3\times10^{-5})} \ &= \frac{18.8}{10.5} \ &= 1.79 \end{aligned} $$
So $Y_s : Y_c = 1.79 : 1$.
(d) Here $A = 0.15\ \text{m}$, $f = 4\ \text{Hz}$ so $\omega = 2\pi f = 25.13\ \text{rad s}^{-1}$, and we want the motion at $y = 0.09\ \text{m}$ from the mean position. The speed there is
$$ \begin{aligned} v &= \omega\sqrt{A^2-y^2} \ &= 25.13\sqrt{0.15^2-0.09^2} \ &= 25.13\sqrt{0.0144} \ &= 3.02\ \text{m s}^{-1} \end{aligned} $$
and the acceleration, which always points back toward the mean position, is
$$ \begin{aligned} a &= \omega^2 y \ &= (25.13)^2(0.09) \ &= 56.9\ \text{m s}^{-2} \end{aligned} $$
So at $9\ \text{cm}$ displacement the velocity is $3.02\ \text{m s}^{-1}$ and the acceleration is $56.9\ \text{m s}^{-2}$ directed toward the centre.
Answer any two numerical questions. (a) A steel wire 10 m long and 5 mm in diameter is fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10 degrees C. (b) Calculate the rms speed of thermal agitation of the molecules of helium in a vessel at 30 degrees C. Density of helium at STP is 0.1785 kg/m^3. (c) A quantity of dry air at 27 degrees C is compressed (i) isothermally and (ii) adiabatically to one third of its volume. Find the change in temperature in each case.
(a) A steel wire of length $L = 10\ \text{m}$ and diameter $d = 5\times10^{-3}\ \text{m}$ is cooled by $\Delta T = 10\ \text{K}$, with $Y = 2\times10^{11}\ \text{N m}^{-2}$ and $\alpha = 1.2\times10^{-5}\ \text{K}^{-1}$. Its cross-sectional area is
$$ \begin{aligned} A &= \frac{\pi d^2}{4} \ &= \frac{\pi(5\times10^{-3})^2}{4} \ &= 1.963\times10^{-5}\ \text{m}^2 \end{aligned} $$
Because the fixed supports stop the wire from contracting, a tensile stress $Y\alpha\Delta T$ develops, so the increase in tension is
$$ \begin{aligned} F &= Y A \alpha \Delta T \ &= (2\times10^{11})(1.963\times10^{-5})(1.2\times10^{-5})(10) \ &= 471\ \text{N} \end{aligned} $$
The tension therefore increases by about $471\ \text{N}$.
(b) For helium we use $v_{rms} = \sqrt{\dfrac{3P_0}{\rho_0}}$ at STP and then scale it as $\sqrt{T}$. At STP $P_0 = 1.013\times10^5\ \text{Pa}$, $\rho_0 = 0.1785\ \text{kg m}^{-3}$ and $T_0 = 273\ \text{K}$, while here $T = 30^\circ\text{C} = 303\ \text{K}$. The rms speed at STP is
$$ \begin{aligned} v_{rms}(T_0) &= \sqrt{\frac{3(1.013\times10^5)}{0.1785}} \ &= \sqrt{1.703\times10^6} \ &= 1305\ \text{m s}^{-1} \end{aligned} $$
Scaling to $303\ \text{K}$,
$$ \begin{aligned} v_{rms}(303) &= 1305\sqrt{\frac{303}{273}} \ &= 1305(1.0535) \ &= 1375\ \text{m s}^{-1} \end{aligned} $$
So $v_{rms} \approx 1.38\times10^3\ \text{m s}^{-1}$.
(c) Here $T_1 = 27^\circ\text{C} = 300\ \text{K}$, the gas is compressed to $V_2 = V_1/3$, and $\gamma = 1.4$.
(i) In the isothermal case the temperature is held constant, so
$$\Delta T = 0$$
(ii) In the adiabatic case $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$, which gives
$$ \begin{aligned} T_2 &= T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} \ &= 300(3)^{0.4} \ &= 300(1.552) \ &= 465.5\ \text{K} \end{aligned} $$
so the temperature change is
$$ \begin{aligned} \Delta T &= 465.5 - 300 \ &= +165.5\ \text{K} \end{aligned} $$
Thus there is no change isothermally, while adiabatically the temperature rises by about $165\ \text{K}$.
The image obtained with a converging lens is erect and three times the length of the object. The focal length is 20 cm. Calculate the object and image distances.
An erect, magnified image from a converging lens must be virtual, and that fixes the sign of the magnification, which we then feed into the lens formula. The lens is converging, so $f = +20\ \text{cm}$. Because the image is erect and thr...
A capacitor of capacitance 6 microfarad is charged to a potential of 150 V. Its potential falls to 90 V when another uncharged capacitor is connected in parallel to it. Find the capacitance of the second capacitor.
A capacitor $C1=6\ \mu\text{F}$ is charged to $V1=150\ \text{V}$, and its potential falls to a common value $V=90\ \text{V}$ when an initially uncharged capacitor $C2$ is connected in parallel. Charge is conserved when the two are joined...
What is accuracy of measurement? Write the dimensional formula of force, gravitational constant, work and specific heat capacity.
Accuracy of a measurement is the closeness of the measured value to the true (accepted) value of the quantity. A measurement is accurate when the systematic error is small, so the result agrees with the standard value.
Dimensional formulae: $$ \begin{aligned} \text{Force } (=ma) &: [MLT^{-2}]\ \text{Gravitational constant } G;\left(F=\tfrac{Gm_1m_2}{r^2}\right) &: [M^{-1}L^{3}T^{-2}]\ \text{Work } (=Fd) &: [ML^{2}T^{-2}]\ \text{Specific heat capacity } \left(Q=mc,\Delta\theta\right) &: [L^{2}T^{-2}K^{-1}] \end{aligned} $$
State and prove the triangle law of vector addition.
Triangle law: if two vectors are represented in magnitude and direction by two sides of a triangle taken in the same order, their resultant is represented by the third side taken in the opposite order. Proof. Let
A projectile is fired with a velocity of 320 m/s at an angle of 30 degrees to the horizontal. Find the time of flight and horizontal range.
The projectile is fired with $u = 320\ \text{m s}^{-1}$ at $\theta = 30^\circ$, taking $g = 10\ \text{m s}^{-2}$. For the time of flight we use $$T = \frac{2u\sin\theta}{g}$$ Substituting the values, $$ \begin{aligned} T &= \frac{2(320)(...
Define angular velocity and derive a relation between angular velocity and linear velocity.
Angular velocity $\omega$ is the rate of change of angular displacement, $\omega = \dfrac{d\theta}{dt}$ (units $\text{rad s}^{-1}$).
For a particle moving on a circle of radius $r$, an angular displacement $d\theta$ corresponds to an arc length $ds = r,d\theta$. Dividing both sides by $dt$,
$$\frac{ds}{dt} = r\frac{d\theta}{dt}$$
and since $ds/dt = v$ and $d\theta/dt = \omega$, this gives
$$v = r\omega$$
In vector form $\vec v = \vec\omega \times \vec r$, so the linear velocity is directly proportional to the angular velocity and to the radius, and is directed along the tangent to the circle.
What is energy crisis? Explain various sources of energy found in Nepal.
Energy crisis is the shortage (or unreliable supply) of usable energy to meet demand, arising from depletion of conventional fuels, rising consumption and inadequate generation and distribution.
Sources of energy in Nepal:
Emphasis on renewable hydropower and solar is the main strategy to ease Nepal's energy crisis.
State and prove the conservation of linear momentum.
Principle: in the absence of a net external force, the total linear momentum of a system remains constant. To prove it, consider two bodies of masses $m1$ and $m2$ with initial velocities $u1$ and $u2$ that collide and then move off with...
What is Archimedes's principle? How can you verify this law in the laboratory?
Archimedes's principle: when a body is wholly or partly immersed in a fluid, it experiences an upward buoyant force (upthrust) equal to the weight of the fluid displaced, $$U = V\rho g,$$ where $V$ is the immersed volume, $\rho$ the fluid density.
Verification: weigh a solid in air with a spring balance, giving $W_1$. Fully immerse it in water held in a can fitted with an overflow spout, and record the apparent weight $W_2$; the displaced water overflows into a pre-weighed beaker. The apparent loss in weight is $W_1 - W_2$. Weigh the collected overflow water; its weight is found to equal $W_1 - W_2$. Since the loss of weight equals the weight of displaced water, Archimedes's principle is verified.
Write the differences between transverse waves and longitudinal waves.
Transverse wave: the particles of the medium vibrate perpendicular to the direction of wave propagation; it advances as crests and troughs; can be polarised; needs a medium that supports shear (solids, or a surface). Example: waves on a string, light (in the electromagnetic sense).
Longitudinal wave: the particles vibrate parallel to the direction of propagation; it advances as compressions and rarefactions; cannot be polarised; propagates through solids, liquids and gases. Example: sound in air.
| Feature | Transverse | Longitudinal |
|---|---|---|
| Particle vibration | $\perp$ to propagation | parallel to propagation |
| Form | crests and troughs | compressions and rarefactions |
| Polarisation | possible | not possible |
| Media | mainly solids/surfaces | solids, liquids, gases |
Write down Newton's formula for the velocity of sound and explain Laplace's correction.
Newton's formula: Newton assumed that sound propagates through a gas isothermally, so the relevant elasticity is the isothermal bulk modulus, which equals the pressure $P$. $$ \begin{aligned} v &= \sqrt{\frac{E{iso}}{\rho}} \ &= \sqrt{...
Describe pitch, loudness and quality of a musical sound.
Define radius of curvature and focal length of a spherical mirror. Establish a relation between them.
Radius of curvature $R$ is the radius of the sphere of which the mirror surface forms a part (distance from pole to centre of curvature $C$). Focal length $f$ is the distance from the pole $P$ to the principal focus $F$, where paraxial rays parallel to the axis converge (or appear to diverge).
Relation: consider a paraxial ray parallel to the axis striking the mirror at $M$ and reflected through $F$. By the law of reflection the angle at $C$ (angle of incidence) equals angle $MFP$, and for paraxial rays $FM \approx FP$ and $CM \approx CP = R$. Geometry gives $CF = FP$, so $F$ bisects $CP$: $$f = \frac{R}{2}.$$ The focal length of a spherical mirror is half its radius of curvature.
Determine the position and nature of the image formed by a concave lens of 30 cm focal length when an object is placed 20 cm from it.
We put the given distances into the lens formula, taking care with the signs, and then use the magnification to describe the image. A concave (diverging) lens has $f = -30\ \text{cm}$, and the real object gives $u = -20\ \text{cm}$. Rear...
Describe the working of an astronomical telescope and obtain an expression for its magnifying power.
An astronomical (refracting) telescope has a large, long-focus objective lens and a short-focus eyepiece. Parallel rays from a distant object are brought to a real, inverted image at the focus of the objective. This image lies at (or jus...
What are the elements of the earth's magnetic field? Explain them briefly.
The earth's magnetic field at a place is completely specified by three magnetic elements: - Declination ($\theta$): the angle in the horizontal plane between the geographic (true) north and the magnetic north (the direction of the horizo...
State and explain Faraday's law of electromagnetic induction. Derive an expression for the emf induced in an ac generator.
Faraday's law: whenever the magnetic flux linked with a circuit changes, an emf is induced whose magnitude equals the rate of change of flux linkage, and by Lenz's law it opposes the change. $$\varepsilon = -N\frac{d\Phi}{dt}$$ AC genera...
Distinguish between heat and temperature. Write a relation between Celsius and Fahrenheit scales.
Heat is a form of energy that flows from a hotter body to a colder one because of their temperature difference. It depends on the mass, nature and temperature change of the body, and is measured in joules. Temperature is the degree of ho...
Explain conductors and insulators with examples.
Conductors are materials that allow electric charge (free electrons) to move through them easily, because they have a large number of loosely bound free electrons. They have very low resistivity. Examples: copper, silver, aluminium, and ...
What are cathode rays? Write some properties of cathode rays.
Cathode rays are streams of fast-moving electrons emitted from the cathode of a discharge tube when a high potential difference is applied across the electrodes at very low gas pressure. Properties: - They travel in straight lines and ca...
Describe nuclear fusion and nuclear fission with examples.
Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei of comparable mass (plus a few neutrons) with the release of energy, usually triggered by neutron capture. The mass defect is released as energy ($E = \Delta m,c^2$). Example: $$^{235}{92}\text{U} + {}^{1}{0}n \to {}^{141}{56}\text{Ba} + {}^{92}{36}\text{Kr} + 3,{}^{1}_{0}n + \text{energy}.$$
Nuclear fusion is the combining of two light nuclei into a heavier nucleus, again with release of energy, requiring very high temperature and pressure to overcome Coulomb repulsion. It powers the Sun. Example: $$^{2}{1}\text{H} + {}^{3}{1}\text{H} \to {}^{4}{2}\text{He} + {}^{1}{0}n + 17.6\ \text{MeV}.$$ Per nucleon, fusion of light nuclei releases even more energy than fission of heavy nuclei.
Explain the death of stars.
A star lives by fusing hydrogen into helium in its core; the outward radiation pressure balances gravity. When the core hydrogen is exhausted, the balance is lost and the star's fate depends on its mass: - The star first swells into a re...
Define acceleration due to gravity. Derive an expression for the variation of 'g' with altitude and depth. What is free fall?
Acceleration due to gravity $g$ is the acceleration produced in a freely falling body by the Earth's gravitational pull. At the surface it is $g = \dfrac{GM}{R^2}$.
Variation with altitude $h$ (above the surface): at a height $h$ the distance from the centre becomes $R+h$, so
$$ \begin{aligned} g_h &= \frac{GM}{(R+h)^2} \ &= g\left(1+\frac{h}{R}\right)^{-2} \end{aligned} $$
For $h\ll R$ this is approximately
$$g_h \approx g\left(1 - \frac{2h}{R}\right)$$
so $g$ decreases with height.
Variation with depth $d$ (below the surface): assuming uniform density $\rho$, only the sphere of radius $(R-d)$ contributes,
$$g_d = \frac{G\left(\tfrac{4}{3}\pi (R-d)^3\rho\right)}{(R-d)^2}$$
which simplifies to
$$ \begin{aligned} g_d &= \frac{4}{3}\pi G\rho (R-d) \ &= g\left(1 - \frac{d}{R}\right) \end{aligned} $$
so $g$ also decreases with depth, falling to zero at the centre.
Free fall is the motion of a body under the action of gravity alone, with no other force such as air resistance acting. Every freely falling body has the same acceleration $g$, whatever its mass.
What is a simple pendulum? Show that the motion of a simple pendulum is simple harmonic and calculate its time period. The length of a second's pendulum at a place is 102 cm. What will be the acceleration due to gravity at that place?
A simple pendulum is a heavy point mass (bob) suspended by a light, inextensible string from a rigid support, free to swing in a vertical plane.
To show its motion is simple harmonic, take a small angular displacement $\theta$. The restoring force is the tangential component of gravity, and for small angles $\sin\theta \approx \theta$:
$$F = -mg\sin\theta \approx -mg\theta$$
Writing the arc displacement as $x = L\theta$, this becomes
$$F = -\frac{mg}{L}x$$
so the acceleration is
$$ \begin{aligned} a &= -\frac{g}{L}x \ &= -\omega^2 x \end{aligned} $$
which is exactly the SHM condition with $\omega = \sqrt{g/L}$. The motion is therefore simple harmonic, and its time period is
$$ \begin{aligned} T &= \frac{2\pi}{\omega} \ &= 2\pi\sqrt{\frac{L}{g}} \end{aligned} $$
For the numerical part, a second's pendulum has period $T = 2\ \text{s}$ and here the length is $L = 1.02\ \text{m}$. Squaring the period formula and rearranging for $g$ gives
$$g = \frac{4\pi^2 L}{T^2}$$
Substituting the values,
$$ \begin{aligned} g &= \frac{4\pi^2 (1.02)}{2^2} \ &= \pi^2(1.02) \ &= 9.87(1.02) \ &= 10.07\ \text{m s}^{-2} \end{aligned} $$
So the acceleration due to gravity at that place is about $10.07\ \text{m s}^{-2}$.
Define the coefficients of linear expansion and cubical expansion. Establish a relation between them. How much heat is required to heat 10 litres of water from 15 degrees C to 55 degrees C? Specific heat capacity of water is 4200 J/kg K.
The coefficient of linear expansion $\alpha$ is the fractional increase in length per unit rise in temperature,
$$\alpha = \frac{\Delta L}{L,\Delta\theta}$$
and the coefficient of cubical (volume) expansion $\gamma$ is the fractional increase in volume per unit rise in temperature,
$$\gamma = \frac{\Delta V}{V,\Delta\theta}$$
To relate them, take a cube of side $L$, so $V = L^3$. When heated, each side becomes $L(1+\alpha\Delta\theta)$, so the new volume is
$$V' = L^3(1+\alpha\Delta\theta)^3 \approx L^3(1 + 3\alpha\Delta\theta)$$
neglecting higher powers of the small quantity $\alpha\Delta\theta$. Comparing this with $V' = V(1+\gamma\Delta\theta)$ gives
$$\gamma = 3\alpha$$
For the numerical part, 10 litres of water has mass $m = 10\ \text{kg}$ (density $1000\ \text{kg m}^{-3}$), with $c = 4200\ \text{J kg}^{-1}\text{K}^{-1}$ and a temperature rise $\Delta\theta = 55 - 15 = 40\ \text{K}$. The heat required is
$$ \begin{aligned} Q &= mc,\Delta\theta \ &= 10 \times 4200 \times 40 \ &= 1.68\times10^{6}\ \text{J} \end{aligned} $$
So the heat required is $1.68\ \text{MJ}$.
Describe the important constituents of the universe. Explain the working of a radio telescope to study astronomical objects.
Constituents of the universe: the observable universe is made up of
Radio telescope: many astronomical objects emit radio waves. A large parabolic dish collects these weak signals and reflects them to a focus, where an antenna (the feed) converts them into a tiny electrical signal. This is amplified by a low-noise receiver, filtered, and recorded and processed by a computer to build maps or spectra of the source. Because radio waves pass through dust clouds and can be observed day or night in most weather, radio telescopes reveal objects (pulsars, quasars, cold gas clouds) that are hidden from optical telescopes, and several dishes can be combined (interferometry) for high resolution.
(a) No. Dimensional correctness is a necessary but not sufficient condition. It cannot detect dimensionless constants, nor errors in terms that share the same dimensions. Example: is dimensionally correct (
(a) In summer the temperature rises. By the pressure law (Gay-Lussac), at fixed volume , so the pressure of the enclosed air rises with temperature. When the internal pressure exceeds the strength of the r...
(a) Triangle law: if two vectors are represented in magnitude and direction by the two sides of a triangle taken in order, their resultant is represented by the third side taken in the opposite order. For two vectors and inclined at angle , the magnitude of the resultant is
and its direction, as the angle made with , is
(b) Banking means raising the outer edge of a curved road above the inner edge so the surface makes an angle with the horizontal. The horizontal component of the normal reaction then supplies the centripetal force, reducing the reliance on friction. For the ideal (frictionless) speed, the vertical and horizontal equations are
Dividing the second by the first,
so the safe speed is
At this speed the vehicle rounds the bend with no tendency to skid.
(c) Angular momentum is the moment of linear momentum about a point, and for a rigid body . Moment of inertia is the rotational inertia about an axis. To relate them, a single particle has
Summing over all particles of a rigid body,
(d) Capillarity is the rise or fall of a liquid in a fine tube due to surface tension. Balancing the weight of the raised column against the vertical pull of surface tension acting around the circumference :
Solving for the height of rise,
where is the surface tension, the contact angle, the tube radius and the liquid density. The rise is greater for a narrower tube.
(a) Lateral shift is the perpendicular displacement between the incident ray (produced) and the emergent ray when light passes through a parallel-sided glass slab; the emergent ray is parallel to the incident ray but shifted sideways. If is the slab thickness, the angle of incidence and the angle of refraction, then As the angle of incidence increases, increases and decreases, so the lateral shift increases; it is maximum (equal to ) at grazing incidence ().
(b) Compound microscope: a short-focus objective lens forms a real, inverted, magnified image of a nearby object placed just beyond its focus; this image lies within the focal length of the eyepiece, which acts as a simple magnifier to give a final virtual, inverted, highly magnified image at the least distance of distinct vision . The magnifying power is the product of the two stages:
where is the tube length and are the objective and eyepiece focal lengths. Short and give high magnification.
(a) Electric field intensity at a point is the force per unit positive test charge placed there, (units or ). Potential gradient is the rate of change of potential with dist...
(a) With the heavier mass falling, the unbalanced weight drives the whole system, so the acceleration is
and the tension in the connecting string works out to
So the masses accelerate at and the string tension is .
(b) The satellite orbits at radius , and with the period is
Working out the ratio inside the root,
which gives
The period is about hours, which is exactly why such a satellite stays fixed over one point (geostationary).
(c) Both wires carry the same load and stretch by the same amount, and since with and common to both, the ratio reduces to the geometry alone:
So .
(d) Here , so , and we want the motion at from the mean position. The speed there is
and the acceleration, which always points back toward the mean position, is
So at displacement the velocity is and the acceleration is directed toward the centre.
(a) A steel wire of length and diameter is cooled by , with and . Its cross-sectional area is
Because the fixed supports stop the wire from contracting, a tensile stress develops, so the increase in tension is
The tension therefore increases by about .
(b) For helium we use at STP and then scale it as . At STP , and , while here . The rms speed at STP is
Scaling to ,
So .
(c) Here , the gas is compressed to , and .
(i) In the isothermal case the temperature is held constant, so
(ii) In the adiabatic case , which gives
so the temperature change is
Thus there is no change isothermally, while adiabatically the temperature rises by about .
An erect, magnified image from a converging lens must be virtual, and that fixes the sign of the magnification, which we then feed into the lens formula. The lens is converging, so . Because the image is erect and thr...
A capacitor is charged to , and its potential falls to a common value when an initially uncharged capacitor is connected in parallel. Charge is conserved when the two are joined...
Accuracy of a measurement is the closeness of the measured value to the true (accepted) value of the quantity. A measurement is accurate when the systematic error is small, so the result agrees with the standard value.
Dimensional formulae:
The projectile is fired with at , taking . For the time of flight we use Substituting the values, $$ \begin{aligned} T &= \frac{2(320)(...
Angular velocity is the rate of change of angular displacement, (units ).
For a particle moving on a circle of radius , an angular displacement corresponds to an arc length . Dividing both sides by ,
and since and , this gives
In vector form , so the linear velocity is directly proportional to the angular velocity and to the radius, and is directed along the tangent to the circle.
Principle: in the absence of a net external force, the total linear momentum of a system remains constant. To prove it, consider two bodies of masses and with initial velocities and that collide and then move off with...
Archimedes's principle: when a body is wholly or partly immersed in a fluid, it experiences an upward buoyant force (upthrust) equal to the weight of the fluid displaced, where is the immersed volume, the fluid density.
Verification: weigh a solid in air with a spring balance, giving . Fully immerse it in water held in a can fitted with an overflow spout, and record the apparent weight ; the displaced water overflows into a pre-weighed beaker. The apparent loss in weight is . Weigh the collected overflow water; its weight is found to equal . Since the loss of weight equals the weight of displaced water, Archimedes's principle is verified.
Transverse wave: the particles of the medium vibrate perpendicular to the direction of wave propagation; it advances as crests and troughs; can be polarised; needs a medium that supports shear (solids, or a surface). Example: waves on a string, light (in the electromagnetic sense).
Longitudinal wave: the particles vibrate parallel to the direction of propagation; it advances as compressions and rarefactions; cannot be polarised; propagates through solids, liquids and gases. Example: sound in air.
| Feature | Transverse | Longitudinal |
|---|---|---|
| Particle vibration | to propagation | parallel to propagation |
| Form | crests and troughs | compressions and rarefactions |
| Polarisation | possible | not possible |
| Media | mainly solids/surfaces | solids, liquids, gases |
Newton's formula: Newton assumed that sound propagates through a gas isothermally, so the relevant elasticity is the isothermal bulk modulus, which equals the pressure . $$ \begin{aligned} v &= \sqrt{\frac{E{iso}}{\rho}} \ &= \sqrt{...
Radius of curvature is the radius of the sphere of which the mirror surface forms a part (distance from pole to centre of curvature ). Focal length is the distance from the pole to the principal focus , where paraxial rays parallel to the axis converge (or appear to diverge).
Relation: consider a paraxial ray parallel to the axis striking the mirror at and reflected through . By the law of reflection the angle at (angle of incidence) equals angle , and for paraxial rays and . Geometry gives , so bisects : The focal length of a spherical mirror is half its radius of curvature.
We put the given distances into the lens formula, taking care with the signs, and then use the magnification to describe the image. A concave (diverging) lens has , and the real object gives . Rear...
The earth's magnetic field at a place is completely specified by three magnetic elements: - Declination (): the angle in the horizontal plane between the geographic (true) north and the magnetic north (the direction of the horizo...
Faraday's law: whenever the magnetic flux linked with a circuit changes, an emf is induced whose magnitude equals the rate of change of flux linkage, and by Lenz's law it opposes the change. AC genera...
Nuclear fission is the splitting of a heavy nucleus into two lighter nuclei of comparable mass (plus a few neutrons) with the release of energy, usually triggered by neutron capture. The mass defect is released as energy (). Example:
Nuclear fusion is the combining of two light nuclei into a heavier nucleus, again with release of energy, requiring very high temperature and pressure to overcome Coulomb repulsion. It powers the Sun. Example: Per nucleon, fusion of light nuclei releases even more energy than fission of heavy nuclei.
Acceleration due to gravity is the acceleration produced in a freely falling body by the Earth's gravitational pull. At the surface it is .
Variation with altitude (above the surface): at a height the distance from the centre becomes , so
For this is approximately
so decreases with height.
Variation with depth (below the surface): assuming uniform density , only the sphere of radius contributes,
which simplifies to
so also decreases with depth, falling to zero at the centre.
Free fall is the motion of a body under the action of gravity alone, with no other force such as air resistance acting. Every freely falling body has the same acceleration , whatever its mass.
A simple pendulum is a heavy point mass (bob) suspended by a light, inextensible string from a rigid support, free to swing in a vertical plane.
To show its motion is simple harmonic, take a small angular displacement . The restoring force is the tangential component of gravity, and for small angles :
Writing the arc displacement as , this becomes
so the acceleration is
which is exactly the SHM condition with . The motion is therefore simple harmonic, and its time period is
For the numerical part, a second's pendulum has period and here the length is . Squaring the period formula and rearranging for gives
Substituting the values,
So the acceleration due to gravity at that place is about .
The coefficient of linear expansion is the fractional increase in length per unit rise in temperature,
and the coefficient of cubical (volume) expansion is the fractional increase in volume per unit rise in temperature,
To relate them, take a cube of side , so . When heated, each side becomes , so the new volume is
neglecting higher powers of the small quantity . Comparing this with gives
For the numerical part, 10 litres of water has mass (density ), with and a temperature rise . The heat required is
So the heat required is .