NEB Class 11 · Past paper
The complete NEB Class 11 2073 exam paper for Physics, all 12 questions with solved model answers.
Tap a question to open its answer.
Answer, in brief, any six questions: (a) Check dimensionally the correctness of Stoke's formula F = 6pinrv. (b) A woman in an elevator lets go of her briefcase but it does not fall to the floor. How is the elevator moving? (c) If there is a net force on a particle in uniform circular motion, why does its speed not change? (d) An astronaut in a ship orbiting Earth feels no gravity, but an astronaut on the Moon does. Why? (e) Does the centre of gravity of a solid body always lie within the material of the body? (f) If the polar ice melts, how will it affect the length of the day? (g) Small air bubbles rise slowly while big bubbles rise rapidly through a liquid. Why?
(a) Stokes' law is $F = 6\pi\eta r v$, and the numerical factor $6\pi$ is dimensionless. Using $[\eta] = \mathrm{ML^{-1}T^{-1}}$, $[r] = \mathrm{L}$ and $[v] = \mathrm{LT^{-1}}$, the right-hand side has dimensions $$ \begin{aligned} [6\p...
Answer, in brief, any two questions: (a) Groundnuts are fried along with sand. Why? (b) Dews are formed on a clear night but not on a cloudy night. Explain. (c) Explain the significance of the second law of thermodynamics.
(a) Sand grains are poor conductors that get very hot and, being finely divided, surround the nuts and distribute heat uniformly by direct contact. They act as a hot, evenly-heated medium so every nut is roasted uniformly without scorchi...
Answer, in brief, any one question: (a) A concave mirror is often used as an aid for applying cosmetics to the face. Why? (b) Sun glasses have curved surfaces but their power is zero. Why?
(a) When the face is held inside the focal length of a concave mirror (between pole and focus), the mirror forms a virtual, erect and magnified image. The enlarged upright image lets the user see fine detail of the face clearly, which is why concave (shaving/make-up) mirrors are used.
(b) The two surfaces of the sunglass lens are curved with the same radius of curvature (a meniscus of uniform thickness), so effectively $R_1 = R_2$. By the lens maker's formula,
$$ \begin{aligned} \frac{1}{f} &= (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \ &= 0 \end{aligned} $$
so the power is
$$ \begin{aligned} P &= \frac{1}{f} \ &= 0 \end{aligned} $$
The converging effect of one surface is exactly cancelled by the diverging effect of the other, so the net power is zero and objects are neither magnified nor reduced.
Answer, in brief, any one question: (a) What is potential gradient? How is it related to electric field intensity? (b) Can we give any desired charge to a capacitor? Explain.
(a) Potential gradient is the rate of change of electric potential with distance, $\dfrac{dV}{dx}$ (unit $\mathrm{V,m^{-1}}$). It is related to the electric field intensity by:
$$E = -\frac{dV}{dx}$$
The field points in the direction of decreasing potential, and its magnitude equals the potential gradient; the negative sign shows $\vec E$ is directed from high to low potential.
(b) No, not without limit. For a capacitor $Q = CV$, so raising the charge raises the voltage. Beyond a certain voltage the field between the plates exceeds the dielectric strength of the medium, the dielectric breaks down (sparks/leaks charge), and no more charge can be held. Thus the maximum charge is limited by the breakdown voltage:
$$Q_{max} = C V_{max}$$
Answer any three questions: (a) Define angle of repose. Show that the angle of repose and the angle of friction are equal. (b) What is escape velocity? Derive an expression for it on the Earth's surface. (c) Define moment of inertia. Obtain the moment of inertia of a thin uniform rod about an axis through the centre perpendicular to its length. (d) What is capillarity? Deduce a relation for the height of liquid rise in a capillary tube.
(a) Angle of repose: the maximum angle of an inclined plane at which a body just remains on the point of sliding down under its own weight.
On an incline of angle $\theta$, the body is in limiting equilibrium when the component of weight along the plane equals the limiting friction:
$$ \begin{aligned} mg\sin\theta &= \mu R \ &= \mu mg\cos\theta \end{aligned} $$
Cancelling $mg$ gives
$$\mu = \tan\theta$$
The angle of friction $\lambda$ is defined by $\mu = \tan\lambda$, so comparing the two, $\tan\theta = \tan\lambda$ and hence
$$\boxed{\theta = \lambda}$$
the angle of repose equals the angle of friction.
(b) Escape velocity is the minimum speed with which a body must be projected from a planet's surface to escape its gravitational field completely (reaching infinity with zero speed).
The kinetic energy given at the surface must equal the gain in potential energy in moving the mass $m$ from the surface (radius $R$) to infinity:
$$\tfrac{1}{2}mv_e^2 = \frac{GMm}{R}$$
Solving for the escape speed,
$$v_e = \sqrt{\frac{2GM}{R}}$$
Using $g = \dfrac{GM}{R^2}$, i.e. $GM = gR^2$, this simplifies to
$$v_e = \sqrt{2gR}$$
For Earth, with $g = 9.8\ \text{m/s}^2$ and $R = 6.4\times10^6\ \text{m}$, this gives $v_e \approx 11.2\ \text{km/s}$.
(c) Moment of inertia of a body about an axis is $I = \sum m_i r_i^2$, the sum of (mass $\times$ perpendicular-distance$^2$); it measures rotational inertia.
For a thin uniform rod of mass $M$, length $L$, linear density $\lambda = M/L$, axis through the centre $\perp$ to length, an element $dx$ at distance $x$ has mass $dm = \lambda,dx$:
$$I = \int_{-L/2}^{+L/2} x^2,\lambda,dx$$
Integrating,
$$I = \lambda\left[\frac{x^3}{3}\right]_{-L/2}^{L/2}$$
which evaluates to
$$ \begin{aligned} I &= \frac{\lambda}{3}\cdot\frac{2L^3}{8} \ &= \frac{\lambda L^3}{12} \end{aligned} $$
With $\lambda L = M$:
$$\boxed{I = \frac{ML^2}{12}}$$
(d) Capillarity is the rise or fall of a liquid in a fine (capillary) tube relative to the outside level, due to surface tension.
For a tube of radius $r$ in a liquid of surface tension $T$, density $\rho$, contact angle $\theta$, the upward pull of surface tension around the circumference balances the weight of the raised column of height $h$:
$$T\cos\theta \cdot 2\pi r = \pi r^2 h \rho g$$
Solving for the height of rise,
$$\boxed{h = \frac{2T\cos\theta}{r\rho g}}$$
Thus $h \propto \dfrac{1}{r}$: narrower tubes give greater rise.
Answer any two questions: (a) Define coefficients of real and apparent expansion of a liquid, and establish a relation between them. (b) Using the postulates of kinetic theory of gases, deduce an expression for the pressure of an ideal gas. (c) What is an isothermal process? Derive PV^gamma = constant [as stated].
(a) Coefficient of real expansion $\gamma_r$ = fractional increase in true volume of a liquid per degree rise in temperature. Coefficient of apparent expansion $\gamma_a$ = observed fractional increase per degree, measured relative to the (also expanding) containing vessel.
Because the vessel expands too, the observed (apparent) rise is less than the real rise by the vessel's expansion $\gamma_g$ (glass cubical expansivity):
$$\boxed{\gamma_r = \gamma_a + \gamma_g}$$
(b) Consider $N$ molecules each of mass $m$ in a cube of side $L$ (volume $V=L^3$). A molecule with $x$-velocity $u$ hits a wall and reverses, giving momentum change $2mu$; the time between successive hits on that wall is $2L/u$, so the force is $\dfrac{2mu}{2L/u} = \dfrac{mu^2}{L}$. Summing over all molecules and using $\overline{u^2}=\overline{v^2}/3$, the pressure is
$$ \begin{aligned} P &= \frac{F}{L^2} \ &= \frac{m}{L^3}\sum u_i^2 \ &= \frac{m}{V}\cdot\frac{N\overline{v^2}}{3} \end{aligned} $$
which can be written as
$$\boxed{P = \frac{1}{3}\frac{Nm}{V}\overline{v^2} = \frac{1}{3}\rho,\overline{v^2}}$$
where $\rho = Nm/V$ is the gas density and $\overline{v^2}$ the mean-square speed.
(c) An isothermal process is one carried out at constant temperature ($T$ constant, so $\Delta U = 0$ for an ideal gas), performed slowly enough that heat exchange keeps the temperature fixed; it obeys Boyle's law, $PV = $ constant. (The relation $PV^\gamma = $ constant actually describes an adiabatic process, derived below.)
For an adiabatic change $dQ = 0$, so the first law gives $dU + P,dV = 0$, i.e. $nC_v,dT = -P,dV$. Differentiating $PV = nRT$ gives $P,dV + V,dP = nR,dT$. Eliminating $dT$ between these and using $R = C_p - C_v$,
$$C_v(P,dV + V,dP) = -R,P,dV$$
which tidies up to
$$ \begin{aligned} C_v V,dP &= -(C_v+R)P,dV \ &= -C_p P,dV \end{aligned} $$
Dividing through by $C_v PV$,
$$\frac{dP}{P} = -\gamma\frac{dV}{V}$$
where $\gamma = \dfrac{C_p}{C_v}$. Integrating gives $\ln P + \gamma\ln V = $ constant, that is
$$\boxed{PV^\gamma = \text{constant}}$$
Answer any one question: (a) What is chromatic aberration in a lens? Deduce the condition for achromatism of two thin lenses in contact. (b) Describe the structure and working of an astronomical telescope at normal adjustment with a ray diagram, and calculate its magnifying power.
(a) Chromatic aberration is the failure of a lens to bring light of all colours to the same focus: the refractive index, and hence the focal length, varies with colour, so violet focuses nearer than red and coloured fringes appear.
For a thin lens the power varies with colour through the dispersive power $\omega = \dfrac{df}{f}$. For two thin lenses of powers $P_1$ and $P_2$ (dispersive powers $\omega_1$ and $\omega_2$) in contact, the combined power is $P = P_1 + P_2$. Achromatism requires this combined power to be the same for red and violet, that is $dP = 0$, which gives
$$\omega_1 P_1 + \omega_2 P_2 = 0$$
or equivalently
$$\boxed{\frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0}$$
Since $\omega_1$ and $\omega_2$ are both positive, $f_1$ and $f_2$ must have opposite signs: one converging and one diverging lens of different materials, forming an achromatic doublet.
(b) Astronomical (refracting) telescope: it uses an objective lens of large focal length $f_o$ and large aperture together with an eyepiece of short focal length $f_e$, mounted coaxially with a separation $f_o + f_e$ at normal adjustment.
Parallel rays from a distant object form a real, inverted image at the objective's focus, which at normal adjustment coincides with the eyepiece's focus; the eyepiece then forms the final magnified virtual image at infinity, so the eye views it relaxed.
Ray diagram (described): parallel rays -> objective -> real inverted image $A'B'$ at the common focus -> eyepiece -> parallel rays emerging to the eye (image at infinity).
The magnifying power is the ratio of the angle subtended by the image to that subtended by the object,
$$\boxed{M = \frac{f_o}{f_e}}$$
and the tube length is $L = f_o + f_e$. A large $f_o$ and a small $f_e$ give high magnification.
Answer any one question: (a) What is electrostatic induction? Explain a method of charging a body positively by induction. (b) State and explain Gauss's theorem and use it to find the electric field due to a charged sphere at a point (i) outside and (ii) inside the sphere.
(a) Electrostatic induction is the redistribution of charge in a conductor caused by a nearby charged body, without contact.
To charge a body positively by induction using a negatively charged rod:
(b) Gauss's theorem: the total electric flux through any closed surface equals $1/\varepsilon_0$ times the net charge enclosed:
$$\oint \vec E\cdot d\vec A = \frac{q_{enc}}{\varepsilon_0}$$
Consider a sphere of radius $R$ carrying charge $q$ on its surface (a conducting shell).
(i) Outside ($r > R$): take a Gaussian sphere of radius $r$. By symmetry the field is radial and uniform over it, so
$$E,(4\pi r^2) = \frac{q}{\varepsilon_0}$$
which gives
$$\boxed{E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}}$$
that is, the sphere behaves as if all its charge were concentrated at its centre.
(ii) Inside ($r < R$): a Gaussian surface here encloses no charge (all the charge sits on the outer surface), so
$$E,(4\pi r^2) = 0$$
and therefore
$$\boxed{E = 0}$$
The field inside a charged conducting sphere is zero.
Answer any three numerical questions: (a) A baseball leaves the bat at 37 m/s at 53 degrees. Find its position and velocity (magnitude and direction) after 2 s. (b) A 7 kg wagon on a frictionless surface has initial speed 4 m/s and is pushed 3 m by a 10 N force. Find final speed and acceleration. (c) A body of mass 200 g in SHM has amplitude 20 mm; maximum force 0.80 N. Find maximum velocity and period. (d) A steel wire density 8000 kg/m^3 weighs 20 g, length 2.5 m, extends 1 mm under 80 N. Find Young's modulus and energy stored.
(a) The ball leaves the bat at $u = 37\ \text{m/s}$ and $\theta = 53^\circ$; we want its position and velocity after $t = 2\ \text{s}$, taking $g = 9.8\ \text{m/s}^2$.
The initial velocity components are
$$ \begin{aligned} u_x &= 37\cos53^\circ \ &= 22.27\ \text{m/s} \ u_y &= 37\sin53^\circ \ &= 29.55\ \text{m/s} \end{aligned} $$
The horizontal motion is uniform, so after 2 s
$$ \begin{aligned} x &= u_x t \ &= 22.27\times2 \ &= 44.5\ \text{m} \end{aligned} $$
The vertical motion is under gravity, so
$$y = u_y t - \tfrac12 g t^2$$
Substituting the values,
$$ \begin{aligned} y &= 29.55(2) - \tfrac12(9.8)(4) \ &= 59.1 - 19.6 \ &= 39.5\ \text{m} \end{aligned} $$
For the velocity, the horizontal component is unchanged,
$$v_x = 22.27\ \text{m/s}$$
while the vertical component is
$$ \begin{aligned} v_y &= u_y - gt \ &= 29.55 - 19.6 \ &= 9.95\ \text{m/s} \end{aligned} $$
The speed is their resultant,
$$ \begin{aligned} v &= \sqrt{v_x^2+v_y^2} \ &= \sqrt{22.27^2+9.95^2} \ &= 24.4\ \text{m/s} \end{aligned} $$
directed at
$$ \begin{aligned} \alpha &= \tan^{-1}\frac{9.95}{22.27} \ &= 24.1^\circ \end{aligned} $$
above the horizontal. So after 2 s the ball is at $(44.5\ \text{m}, 39.5\ \text{m})$, moving at $24.4\ \text{m/s}$ at $24.1^\circ$ above the horizontal.
(b) The wagon has mass $m = 7\ \text{kg}$ and initial speed $u = 4\ \text{m/s}$, and is pushed $s = 3\ \text{m}$ by a force $F = 10\ \text{N}$ on a frictionless surface.
The acceleration comes straight from Newton's second law,
$$ \begin{aligned} a &= \frac{F}{m} \ &= \frac{10}{7} \ &= 1.43\ \text{m/s}^2 \end{aligned} $$
and the final speed follows from $v^2 = u^2 + 2as$:
$$ \begin{aligned} v^2 &= 16 + 2(1.43)(3) \ &= 16 + 8.57 \ &= 24.57 \end{aligned} $$
so
$$ \begin{aligned} v &= \sqrt{24.57} \ &= 4.96\ \text{m/s} \end{aligned} $$
So the acceleration is about $1.43\ \text{m/s}^2$ and the final speed about $4.96\ \text{m/s}$.
(c) Here $m = 200\ \text{g} = 0.200\ \text{kg}$, amplitude $A = 20\ \text{mm} = 0.020\ \text{m}$ and the maximum force is $F_{max} = 0.80\ \text{N}$.
The maximum restoring force in SHM is $F_{max} = m\omega^2 A$, so
$$ \begin{aligned} \omega^2 &= \frac{F_{max}}{mA} \ &= \frac{0.80}{0.200\times0.020} \ &= 200 \end{aligned} $$
giving
$$\omega = 14.14\ \text{rad/s}$$
The maximum velocity is then
$$ \begin{aligned} v_{max} &= \omega A \ &= 14.14\times0.020 \ &= 0.283\ \text{m/s} \end{aligned} $$
and the period is
$$ \begin{aligned} T &= \frac{2\pi}{\omega} \ &= \frac{2\pi}{14.14} \ &= 0.444\ \text{s} \end{aligned} $$
So $v_{max} \approx 0.283\ \text{m/s}$ and $T \approx 0.44\ \text{s}$.
(d) The steel wire has density $\rho = 8000\ \text{kg/m}^3$, mass $0.020\ \text{kg}$, length $L = 2.5\ \text{m}$, and extends $\Delta L = 10^{-3}\ \text{m}$ under a load $F = 80\ \text{N}$.
Its volume follows from the mass and density,
$$ \begin{aligned} V &= \frac{m}{\rho} \ &= \frac{0.020}{8000} \ &= 2.5\times10^{-6}\ \text{m}^3 \end{aligned} $$
and dividing by the length gives the cross-sectional area,
$$ \begin{aligned} A &= \frac{V}{L} \ &= \frac{2.5\times10^{-6}}{2.5} \ &= 1.0\times10^{-6}\ \text{m}^2 \end{aligned} $$
Young's modulus is stress over strain,
$$Y = \frac{F L}{A,\Delta L}$$
Substituting the values,
$$ \begin{aligned} Y &= \frac{80\times2.5}{(1.0\times10^{-6})(10^{-3})} \ &= \frac{200}{10^{-9}} \ &= 2.0\times10^{11}\ \text{N/m}^2 \end{aligned} $$
and the elastic energy stored is
$$ \begin{aligned} W &= \tfrac12 F,\Delta L \ &= \tfrac12(80)(10^{-3}) \ &= 0.04\ \text{J} \end{aligned} $$
So $Y = 2.0\times10^{11}\ \text{N/m}^2$ and the energy stored is $0.04\ \text{J}$.
Answer any two numerical questions: (a) A 200 g metal block at 150 C is dropped into a copper calorimeter (270 g) with 150 cm^3 water at 27 C; final temperature 40 C. Find specific heat of metal. (b) A pot with 8.5 mm steel bottom, area 0.15 m^2, water at 100 C, 390 g evaporated every 3 min. Find temperature of the lower surface. (c) A Carnot engine has 25% efficiency with sink at 9 C. By how much must the source temperature rise to raise efficiency to 70%?
(a) A metal block of mass $m_1 = 200\ \text{g} = 0.200\ \text{kg}$ at $150^\circ$C is dropped into a copper calorimeter of mass $m_c = 270\ \text{g} = 0.270\ \text{kg}$ holding $150\ \text{cm}^3 = 0.150\ \text{kg}$ of water at $27^\circ$C, and the mixture settles at $40^\circ$C (take $c_w = 4200$ and $c_{Cu} = 400\ \text{J kg}^{-1}\text{K}^{-1}$). By calorimetry, the heat lost by the metal equals the heat gained by the water and the calorimeter,
$$m_1 c_1(150-40) = m_w c_w(40-27) + m_c c_{Cu}(40-27)$$
Putting in the numbers,
$$ \begin{aligned} 0.200,c_1(110) &= 0.150(4200)(13) + 0.270(400)(13) \ 22,c_1 &= 8190 + 1404 \ &= 9594 \ c_1 &= 436\ \text{J kg}^{-1}\text{K}^{-1} \end{aligned} $$
The specific heat of the metal is about $436\ \text{J kg}^{-1}\text{K}^{-1}$.
(b) A pot with a steel base of thickness $d = 8.5\ \text{mm} = 8.5\times10^{-3}\ \text{m}$ and area $A = 0.15\ \text{m}^2$ holds water at $100^\circ$C, and $390\ \text{g}$ evaporates every $3\ \text{min} = 180\ \text{s}$ (take $L_v = 2.26\times10^6\ \text{J/kg}$ and $k_{steel}\approx 50\ \text{W m}^{-1}\text{K}^{-1}$). The rate of heat needed to boil off the water is
$$ \begin{aligned} \frac{Q}{t} &= \frac{m L_v}{t} \ &= \frac{0.390\times2.26\times10^6}{180} \ &= 4897\ \text{W} \end{aligned} $$
This heat is conducted through the base, $\dfrac{Q}{t} = \dfrac{kA(\theta - 100)}{d}$, so the temperature rise across the steel is
$$ \begin{aligned} \theta - 100 &= \frac{(Q/t),d}{kA} \ &= \frac{4897\times8.5\times10^{-3}}{50\times0.15} \ \theta - 100 &= \frac{41.62}{7.5} \ &= 5.55 \ \theta &= 105.6^\circ\text{C} \end{aligned} $$
The lower surface is at about $105.6^\circ$C (using $k_{steel}=50$; the result scales with $k$).
(c) A Carnot engine is 25% efficient with its sink at $9^\circ\text{C} = 282\ \text{K}$, and we want the source temperature that would raise the efficiency to 70%. From $\eta = 1 - T_2/T_1$, the initial source temperature is
$$ \begin{aligned} T_1 &= \frac{T_2}{1-\eta_1} \ &= \frac{282}{0.75} \ &= 376\ \text{K} \end{aligned} $$
For an efficiency of 0.70 the source would need to be at
$$ \begin{aligned} T_1' &= \frac{282}{1-0.70} \ &= \frac{282}{0.30} \ &= 940\ \text{K} \end{aligned} $$
so the required rise is
$$ \begin{aligned} \Delta T &= 940 - 376 \ &= 564\ \text{K} \end{aligned} $$
The source temperature must be raised by $564\ \text{K}$ (that is, by $564^\circ$C).
A ray of light is refracted through a prism of angle 60 degrees. Find the angle of incidence so that the emergent ray just grazes the second face. Refractive index = 1.45.
The phrase "just grazes the second face" means the ray strikes that face at exactly the critical angle, so total internal reflection is on the verge of happening and $r_2 = C$. We first find that critical angle, then work back through the prism to the first face.
For a prism of refractive index $\mu = 1.45$, the critical angle is
$$ \begin{aligned} \sin C &= \frac{1}{\mu} \ &= \frac{1}{1.45} \ &= 0.6897 \end{aligned} $$
so $C = 43.6^\circ$. The two refraction angles inside a prism satisfy $r_1 + r_2 = A$, and with $A = 60^\circ$ this gives
$$ \begin{aligned} r_1 &= 60^\circ - 43.6^\circ \ &= 16.4^\circ \end{aligned} $$
Applying Snell's law at the first face,
$$ \begin{aligned} \sin i &= \mu\sin r_1 \ &= 1.45\times\sin16.4^\circ \ &= 1.45\times0.2823 \ &= 0.4093 \end{aligned} $$
so the required angle of incidence is $i = \sin^{-1}(0.4093) = 24.2^\circ$.
Two capacitors of 4 uF and 12 uF are connected in series across a 200 V battery. Find the charge and potential difference across each.
In a series combination the two capacitors share the same charge, so we first find the equivalent capacitance, then the common charge, and finally the voltage across each.
With $C_1 = 4\ \mu\text{F}$ and $C_2 = 12\ \mu\text{F}$ in series, the equivalent capacitance comes from
$$ \begin{aligned} \frac{1}{C_s} &= \frac{1}{4} + \frac{1}{12} \ &= \frac{3+1}{12} \ &= \frac{4}{12} \end{aligned} $$
so $C_s = 3\ \mu\text{F}$. Across the $200\ \text{V}$ battery this stores a charge that is the same on each capacitor,
$$ \begin{aligned} Q &= C_s V \ &= 3\times10^{-6}\times200 \ &= 6\times10^{-4}\ \text{C} \ &= 600\ \mu\text{C} \end{aligned} $$
The potential difference across each is then this charge divided by its own capacitance. For the first,
$$ \begin{aligned} V_1 &= \frac{Q}{C_1} \ &= \frac{600\ \mu\text{C}}{4\ \mu\text{F}} \ &= 150\ \text{V} \end{aligned} $$
and for the second,
$$ \begin{aligned} V_2 &= \frac{Q}{C_2} \ &= \frac{600\ \mu\text{C}}{12\ \mu\text{F}} \ &= 50\ \text{V} \end{aligned} $$
As a check these add up to $150 + 50 = 200\ \text{V}$, the battery voltage. So each capacitor holds $600\ \mu\text{C}$, with $V_1 = 150\ \text{V}$ and $V_2 = 50\ \text{V}$.
(a) Stokes' law is , and the numerical factor is dimensionless. Using , and , the right-hand side has dimensions $$ \begin{aligned} [6\p...
(a) When the face is held inside the focal length of a concave mirror (between pole and focus), the mirror forms a virtual, erect and magnified image. The enlarged upright image lets the user see fine detail of the face clearly, which is why concave (shaving/make-up) mirrors are used.
(b) The two surfaces of the sunglass lens are curved with the same radius of curvature (a meniscus of uniform thickness), so effectively . By the lens maker's formula,
so the power is
The converging effect of one surface is exactly cancelled by the diverging effect of the other, so the net power is zero and objects are neither magnified nor reduced.
(a) Potential gradient is the rate of change of electric potential with distance, (unit ). It is related to the electric field intensity by:
The field points in the direction of decreasing potential, and its magnitude equals the potential gradient; the negative sign shows is directed from high to low potential.
(b) No, not without limit. For a capacitor , so raising the charge raises the voltage. Beyond a certain voltage the field between the plates exceeds the dielectric strength of the medium, the dielectric breaks down (sparks/leaks charge), and no more charge can be held. Thus the maximum charge is limited by the breakdown voltage:
(a) Angle of repose: the maximum angle of an inclined plane at which a body just remains on the point of sliding down under its own weight.
On an incline of angle , the body is in limiting equilibrium when the component of weight along the plane equals the limiting friction:
Cancelling gives
The angle of friction is defined by , so comparing the two, and hence
the angle of repose equals the angle of friction.
(b) Escape velocity is the minimum speed with which a body must be projected from a planet's surface to escape its gravitational field completely (reaching infinity with zero speed).
The kinetic energy given at the surface must equal the gain in potential energy in moving the mass from the surface (radius ) to infinity:
Solving for the escape speed,
Using , i.e. , this simplifies to
For Earth, with and , this gives .
(c) Moment of inertia of a body about an axis is , the sum of (mass perpendicular-distance); it measures rotational inertia.
For a thin uniform rod of mass , length , linear density , axis through the centre to length, an element at distance has mass :
Integrating,
which evaluates to
With :
(d) Capillarity is the rise or fall of a liquid in a fine (capillary) tube relative to the outside level, due to surface tension.
For a tube of radius in a liquid of surface tension , density , contact angle , the upward pull of surface tension around the circumference balances the weight of the raised column of height :
Solving for the height of rise,
Thus : narrower tubes give greater rise.
(a) Coefficient of real expansion = fractional increase in true volume of a liquid per degree rise in temperature. Coefficient of apparent expansion = observed fractional increase per degree, measured relative to the (also expanding) containing vessel.
Because the vessel expands too, the observed (apparent) rise is less than the real rise by the vessel's expansion (glass cubical expansivity):
(b) Consider molecules each of mass in a cube of side (volume ). A molecule with -velocity hits a wall and reverses, giving momentum change ; the time between successive hits on that wall is , so the force is . Summing over all molecules and using , the pressure is
which can be written as
where is the gas density and the mean-square speed.
(c) An isothermal process is one carried out at constant temperature ( constant, so for an ideal gas), performed slowly enough that heat exchange keeps the temperature fixed; it obeys Boyle's law, constant. (The relation constant actually describes an adiabatic process, derived below.)
For an adiabatic change , so the first law gives , i.e. . Differentiating gives . Eliminating between these and using ,
which tidies up to
Dividing through by ,
where . Integrating gives constant, that is
(a) Chromatic aberration is the failure of a lens to bring light of all colours to the same focus: the refractive index, and hence the focal length, varies with colour, so violet focuses nearer than red and coloured fringes appear.
For a thin lens the power varies with colour through the dispersive power . For two thin lenses of powers and (dispersive powers and ) in contact, the combined power is . Achromatism requires this combined power to be the same for red and violet, that is , which gives
or equivalently
Since and are both positive, and must have opposite signs: one converging and one diverging lens of different materials, forming an achromatic doublet.
(b) Astronomical (refracting) telescope: it uses an objective lens of large focal length and large aperture together with an eyepiece of short focal length , mounted coaxially with a separation at normal adjustment.
Parallel rays from a distant object form a real, inverted image at the objective's focus, which at normal adjustment coincides with the eyepiece's focus; the eyepiece then forms the final magnified virtual image at infinity, so the eye views it relaxed.
Ray diagram (described): parallel rays -> objective -> real inverted image at the common focus -> eyepiece -> parallel rays emerging to the eye (image at infinity).
The magnifying power is the ratio of the angle subtended by the image to that subtended by the object,
and the tube length is . A large and a small give high magnification.
(a) Electrostatic induction is the redistribution of charge in a conductor caused by a nearby charged body, without contact.
To charge a body positively by induction using a negatively charged rod:
(b) Gauss's theorem: the total electric flux through any closed surface equals times the net charge enclosed:
Consider a sphere of radius carrying charge on its surface (a conducting shell).
(i) Outside (): take a Gaussian sphere of radius . By symmetry the field is radial and uniform over it, so
which gives
that is, the sphere behaves as if all its charge were concentrated at its centre.
(ii) Inside (): a Gaussian surface here encloses no charge (all the charge sits on the outer surface), so
and therefore
The field inside a charged conducting sphere is zero.
(a) The ball leaves the bat at and ; we want its position and velocity after , taking .
The initial velocity components are
The horizontal motion is uniform, so after 2 s
The vertical motion is under gravity, so
Substituting the values,
For the velocity, the horizontal component is unchanged,
while the vertical component is
The speed is their resultant,
directed at
above the horizontal. So after 2 s the ball is at , moving at at above the horizontal.
(b) The wagon has mass and initial speed , and is pushed by a force on a frictionless surface.
The acceleration comes straight from Newton's second law,
and the final speed follows from :
so
So the acceleration is about and the final speed about .
(c) Here , amplitude and the maximum force is .
The maximum restoring force in SHM is , so
giving
The maximum velocity is then
and the period is
So and .
(d) The steel wire has density , mass , length , and extends under a load .
Its volume follows from the mass and density,
and dividing by the length gives the cross-sectional area,
Young's modulus is stress over strain,
Substituting the values,
and the elastic energy stored is
So and the energy stored is .
(a) A metal block of mass at C is dropped into a copper calorimeter of mass holding of water at C, and the mixture settles at C (take and ). By calorimetry, the heat lost by the metal equals the heat gained by the water and the calorimeter,
Putting in the numbers,
The specific heat of the metal is about .
(b) A pot with a steel base of thickness and area holds water at C, and evaporates every (take and ). The rate of heat needed to boil off the water is
This heat is conducted through the base, , so the temperature rise across the steel is
The lower surface is at about C (using ; the result scales with ).
(c) A Carnot engine is 25% efficient with its sink at , and we want the source temperature that would raise the efficiency to 70%. From , the initial source temperature is
For an efficiency of 0.70 the source would need to be at
so the required rise is
The source temperature must be raised by (that is, by C).
The phrase "just grazes the second face" means the ray strikes that face at exactly the critical angle, so total internal reflection is on the verge of happening and . We first find that critical angle, then work back through the prism to the first face.
For a prism of refractive index , the critical angle is
so . The two refraction angles inside a prism satisfy , and with this gives
Applying Snell's law at the first face,
so the required angle of incidence is .
In a series combination the two capacitors share the same charge, so we first find the equivalent capacitance, then the common charge, and finally the voltage across each.
With and in series, the equivalent capacitance comes from
so . Across the battery this stores a charge that is the same on each capacitor,
The potential difference across each is then this charge divided by its own capacitance. For the first,
and for the second,
As a check these add up to , the battery voltage. So each capacitor holds , with and .