Physics NEB Class 11 2075 question paper

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1. 1207512 marksHideAnswer

   Answer in brief, any six questions. (a) A and B are two non-zero vectors. If |A x B| = A . B, what is the angle between A and B? (b) How does the shape of the surface of mercury look in a capillary tube dipped in it? Explain with a figure and proper justification. (c) Which one gives the feeling of heaviness: a kilogram of cotton or a kilogram of lead? Why? (d) Both 'the work done by a force' and 'the torque produced by a force' are products of force and the position vector. How can one make the difference between the two? (e) A stone tied at one end of a string is whirled in a vertical circle. Where will the tensions in the string be maximum and minimum? Show them in a diagram. (f) The position of a certain object in SHM is given as x = 0.05 cos(290t + 2.5), where x is in metre and t in second. What are the amplitude, period and initial phase angle for this motion? (g) In a printed document the unit of universal gravitational constant is given as N m Kg^-2. Check its correctness from dimensional analysis.

   (a) The magnitudes of the two products are $|\vec A\times\vec B| = AB\sin\theta$ and $\vec A\cdot\vec B = AB\cos\theta$. Setting them equal,

   $$AB\sin\theta = AB\cos\theta$$

   which gives $\tan\theta = 1$, so $\theta = 45^\circ$.

   (b) Mercury does not wet glass and its cohesive force exceeds its adhesion to glass, so it is depressed in the tube and its surface (meniscus) is convex upward, bulging up in the middle. The contact angle is obtuse ($>90^\circ$). $;\cap$ (convex meniscus, level lower than outside).

   (c) Both have the same mass (1 kg) and hence the same weight, so neither is truly heavier. However, cotton occupies a much larger volume, so it experiences a greater upthrust from the air; its apparent weight is slightly less, and the lead feels heavier.

   (d) Work is the scalar (dot) product $W = \vec F\cdot\vec s$, which gives a scalar. Torque is the vector (cross) product $\vec\tau = \vec r\times\vec F$, which gives a vector perpendicular to the plane of $\vec r$ and $\vec F$. So work uses $\cos\theta$ (the component along the displacement) while torque uses $\sin\theta$ (the perpendicular component).

   (e) The tension is maximum at the lowest point, where $T = \tfrac{mv^2}{r}+mg$, and minimum at the highest point, where $T = \tfrac{mv^2}{r}-mg$, because gravity adds to the tension at the bottom but opposes the required centripetal force at the top.

   (f) Comparing with $x = A\cos(\omega t + \phi)$: the amplitude is $A = 0.05\ \text{m}$ and the angular frequency is $\omega = 290\ \text{rad s}^{-1}$, so the period is $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{290} = 0.0217\ \text{s}$ and the initial phase angle is $\phi = 2.5\ \text{rad}$.

   (g) From $F = \dfrac{Gm_1m_2}{r^2}$ we have $G = \dfrac{Fr^2}{m_1m_2}$, whose unit is $\dfrac{\text{N}\cdot\text{m}^2}{\text{kg}^2} = \text{N m}^2,\text{kg}^{-2}$. The printed unit $\text{N m kg}^{-2}$ has $\text{m}^1$ instead of $\text{m}^2$, so it is incorrect.

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2. 220754 marksHideAnswer

   Answer in brief, any two questions. (a) How can the Kelvin scale, in principle, be designed by an experiment based on the ideal gas law? (b) What is the physical meaning of emissivity? Write its unit. (c) Point out the differences between saturated and unsaturated vapour pressures with examples.

   (a) For a fixed mass of an ideal gas at constant volume, the pressure is proportional to the absolute temperature: $P \propto T$. Measuring the pressure of a gas at the triple point of water (assigned $273.16\ \text{K}$) and at an unknow...

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3. 320752 marksHideAnswer

   Answer in brief, any one question. (a) A ray of light originating in air falls perpendicularly on the shorter face of a right-angled isosceles (90-45-45) triangular glass prism of refractive index 1.5. Find the angle of deviation in this case, in a diagram. (b) What do lumen and lux signify? Explain.

   (a) The ray enters normally through one of the shorter faces, so it passes in without bending, and then strikes the hypotenuse from inside at $45^\circ$. The critical angle for the glass is

   $$ \begin{aligned} C &= \sin^{-1}!\left(\frac{1}{\mu}\right) \ &= \sin^{-1}!\left(\frac{1}{1.5}\right) \ &= 41.8^\circ \end{aligned} $$

   Since $45^\circ > C$, the ray undergoes total internal reflection at the hypotenuse and turns through $90^\circ$, emerging normally from the other face. The angle of deviation is therefore $90^\circ$ (the prism acts as a right-angle reflector).

   (b) Lumen is the SI unit of luminous flux, the total amount of visible light energy emitted per second by a source (weighted for the eye's response). Lux is the SI unit of illuminance, the luminous flux falling per unit area of a surface, with $1\ \text{lux} = 1\ \text{lumen m}^{-2}$. So the lumen measures the light output of a source, while the lux measures how brightly a surface is lit.

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4. 420752 marksHideAnswer

   Answer in brief, any one question. (a) What is the physical significance of relative permittivity of a material placed between two plates of a capacitor? (b) How can a body be charged with positive electricity by the method of induction? Explain.

   (a) The relative permittivity (dielectric constant) $\varepsilon_r$ tells us how many times the capacitance increases when the material fills the space between the plates compared with vacuum, so that $C = \varepsilon_r C_0$. Physically it measures how strongly the dielectric becomes polarised, which weakens the field between the plates and so lets the capacitor store more charge at the same voltage.

   (b) To charge a body positively by induction, bring a negatively charged rod near (but not touching) the body, so the near face acquires induced positive charge and the far face negative. While the rod is held in place, earth the body momentarily by touching it, and electrons from the far (negative) face flow away to earth. Remove the earth connection first, then remove the rod. The body is left with a net positive charge, opposite in sign to the inducing rod.

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5. 5207512 marksHideAnswer

   Answer any three questions. (a) Explain the difference between conservative and non-conservative forces. Also state and prove the law of conservation of linear momentum. (b) Explain Hooke's law with a necessary figure and derive an expression for the energy density in a stretched wire. (c) What is the physical meaning of moment of inertia of a rigid body? Also derive its expression for a thin uniform rod about an axis passing through one end and perpendicular to its length. (d) What is the difference between the conceptual design of a conical pendulum and that of a simple pendulum? Derive relations for the time period and frequency of a conical pendulum.

   (a) A conservative force does work that is independent of the path taken (zero over a closed loop), and this work can be stored as potential energy; gravity and the spring force are examples. A non-conservative force does path-dependent work and dissipates energy, friction being the usual example.

   For the conservation of linear momentum, take two bodies that collide. By Newton's third law the mutual forces are equal and opposite and act for the same time $t$, so their impulses cancel when added, giving

   $$m_1 v_1 + m_2 v_2 = m_1 u_1 + m_2 u_2$$

   That is, the total momentum is unchanged when no external force acts.

   (b) Hooke's law states that within the elastic limit stress is proportional to strain,

   $$\text{stress} = Y\times\text{strain}$$

   so the graph of load against extension is a straight line. When a wire is stretched the force grows from $0$ to $F$, so the work done is $W = \tfrac{1}{2}F e$. Dividing this by the volume $V = AL$ gives the energy density,

   $$ \begin{aligned} u &= \frac{W}{AL} \ &= \frac{\tfrac{1}{2}Fe}{AL} \end{aligned} $$

   which rearranges to

   $$ \begin{aligned} u &= \frac{1}{2}\cdot\frac{F}{A}\cdot\frac{e}{L} \ &= \frac{1}{2},\text{stress}\times\text{strain} \end{aligned} $$

   (c) The moment of inertia measures a body's resistance to a change in its rotational motion, $I = \sum m_i r_i^2$, and it depends on both the mass and how that mass is distributed about the axis. For a thin rod of mass $M$ and length $L$ about an axis through one end and perpendicular to its length, take the mass per unit length $\lambda = M/L$; an element $dx$ at distance $x$ contributes $dI = (\lambda,dx)x^2$. Integrating over the rod,

   $$ \begin{aligned} I &= \int_0^L \lambda x^2,dx \ &= \lambda\frac{L^3}{3} \end{aligned} $$

   and substituting $\lambda = M/L$,

   $$ \begin{aligned} I &= \frac{M}{L}\cdot\frac{L^3}{3} \ &= \frac{ML^2}{3} \end{aligned} $$

   (d) In a conical pendulum the bob moves in a horizontal circle, so the string sweeps out a cone at a fixed angle $\theta$, whereas a simple pendulum swings to and fro in a vertical plane. For a conical pendulum of string length $L$ and semi-vertical angle $\theta$, the vertical component of the tension balances the weight,

   $$T\cos\theta = mg$$

   while the horizontal component provides the centripetal force (with $r = L\sin\theta$),

   $$ \begin{aligned} T\sin\theta &= \frac{mv^2}{r} \ &= m\omega^2 r \end{aligned} $$

   Dividing the second relation by the first gives $\tan\theta = \dfrac{\omega^2 L\sin\theta}{g}$, so

   $$\omega = \sqrt{\frac{g}{L\cos\theta}}$$

   The time period and frequency then follow as

   $$ \begin{aligned} \mathcal{T} &= 2\pi\sqrt{\frac{L\cos\theta}{g}} \ f &= \frac{1}{2\pi}\sqrt{\frac{g}{L\cos\theta}} \end{aligned} $$

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6. 620758 marksHideAnswer

   Answer any two questions. (a) On what factors does the thermal conductivity of a solid depend? Describe Searle's experimental method to determine the thermal conductivity of a solid. (b) Define real and apparent expansivities of a liquid. Derive the relation between them. (c) Describe the working principle of a petrol engine with the help of its PV diagram.

   (a) The thermal conductivity $K$ is a property of the material itself, but the rate of heat flow through a bar also depends on its cross-sectional area $A$ and the temperature gradient $\dfrac{d\theta}{dx}$, through $\dfrac{dQ}{dt} = -KA\dfrac{d\theta}{dx}$. In Searle's method a thick, well-lagged metal bar has steam heating one end and cooling water flowing past the other. Two thermometers a distance $x$ apart read $\theta_1$ and $\theta_2$, while the cooling water (mass $m$ collected in time $t$, warming from $\theta_3$ to $\theta_4$) carries away exactly the heat conducted along the bar. At the steady state these two heat rates are equal:

   $$KA\frac{\theta_1-\theta_2}{x} = mc,\frac{\theta_4-\theta_3}{t}$$

   Solving for the conductivity,

   $$K = \frac{mc(\theta_4-\theta_3)x}{A(\theta_1-\theta_2)t}$$

   (b) The real expansivity $\gamma_r$ is the true fractional increase in a liquid's volume per kelvin, whereas the apparent expansivity $\gamma_a$ is the increase we observe relative to the vessel, which is itself expanding. If the vessel has cubical expansivity $\gamma_g$, the true expansion is the observed expansion plus that of the container:

   $$\gamma_r = \gamma_a + \gamma_g$$

   (c) A petrol engine runs on the four-stroke Otto cycle. A petrol-air mixture is first drawn in (intake), then compressed adiabatically (compression stroke); a spark ignites it so the pressure jumps at almost constant volume (heat addition); the hot gas expands adiabatically and does useful work (power stroke); and finally, after the pressure drops at constant volume (heat rejection), the burnt gas is pushed out (exhaust). On the PV diagram this appears as two adiabatic curves (compression and expansion) joined by two constant-volume lines, and the area enclosed by the loop equals the net work delivered per cycle.

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7. 720754 marksHideAnswer

   Answer any one question. (a) What are the defects of vision known to you? How are myopic and hypermetropic defects removed? Explain with necessary theory. (b) Derive the expression for the focal length in Lensmaker's formula.

   (a) The common defects of vision are myopia (short sight), hypermetropia (long sight), presbyopia and astigmatism.

   In myopia, distant objects focus in front of the retina, so the far point lies closer than infinity. It is corrected with a diverging (concave) lens whose focal length equals the far-point distance, giving a power

   $$P = -\frac{1}{x}$$

   where $x$ is the far point.

   In hypermetropia, near objects would focus behind the retina, so a converging (convex) lens is used to form a virtual image at the near point $N$ of an object placed at 25 cm, giving

   $$\frac{1}{f} = \frac{1}{0.25} - \frac{1}{N}$$

   (b) The Lensmaker's formula comes from applying refraction at the lens's two spherical surfaces (radii $R_1$ and $R_2$, refractive index $n$) in turn. At the first surface, light travels from air into glass, giving

   $$\frac{n}{v_1} - \frac{1}{u} = \frac{n-1}{R_1}$$

   At the second surface, the first image acts as the object as light passes back into air, giving

   $$\frac{1}{v} - \frac{n}{v_1} = \frac{1-n}{R_2}$$

   Adding the two equations eliminates $v_1$,

   $$\frac{1}{v} - \frac{1}{u} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

   Since $\dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{f}$ for a thin lens, this becomes

   $$\boxed{\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$

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8. 820754 marksHideAnswer

   Answer any one question. (a) Three capacitors are connected in series with a cell. The same three capacitors are connected in parallel with the same cell. Which combination is larger in magnitude? Also find an expression for their value in series combination. (b) State Gauss's law of electrostatics and use it to find the electric field intensity due to a plane charged conductor.

   (a) For the parallel combination the total capacitance is $C_p = C_1 + C_2 + C_3$, whereas for the series combination

   $$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

   which is always smaller than the smallest capacitor. So the parallel combination has the larger capacitance. For the series expression, the same charge $Q$ sits on each capacitor and the voltages add,

   $$ \begin{aligned} V &= V_1 + V_2 + V_3 \ &= \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \end{aligned} $$

   Since $V = Q/C_s$, dividing through by $Q$ gives

   $$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$

   (b) Gauss's law: the flux through a closed surface is

   $$\oint\vec E\cdot d\vec A = \frac{q_{enc}}{\varepsilon_0}$$

   For a plane charged conductor with surface charge density $\sigma$, take a cylindrical Gaussian pillbox with one flat face (area $A$) just outside the conductor and the other inside, where $E = 0$. Flux passes only through the outer face, so

   $$E,A = \frac{\sigma A}{\varepsilon_0}$$

   which gives

   $$E = \frac{\sigma}{\varepsilon_0}$$

   directed normally outward from the conductor's surface.

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9. 9207512 marksNumericalHideAnswer

   Solve any three questions. (a) An artificial satellite revolves round the earth in 2.5 hours in a circular orbit. Find the height of the satellite above the earth, assuming the earth to be a sphere of radius 6370 km. (b) Two forces of 1.5 N and 2.0 N act vertically at the two ends of a metre scale. Where and in what direction should a force be applied so that the scale remains horizontally stable? (c) A water reservoir tank of capacity 250 m^3 is situated at a height of 20 m above the water level. What is the power of an electric motor used to fill the tank in 3 hours if the efficiency of the motor is 70%? (d) A body is projected upwards making an angle theta with the horizontal at a velocity of 300 m/s. Find the value of theta so that the horizontal range is maximum. Hence find its range and time of flight.

   (a) The satellite's period is $T = 2.5\ \text{h} = 9000\ \text{s}$, the Earth's radius is $R = 6.37\times10^6\ \text{m}$, and with $g = 10\ \text{m s}^{-2}$ we have $GM = gR^2 = 4.058\times10^{14}$. From $T = 2\pi\sqrt{r^3/GM}$, the orbital radius satisfies

   $$ \begin{aligned} r^3 &= \frac{GM,T^2}{4\pi^2} \ &= \frac{(4.058\times10^{14})(9000)^2}{4\pi^2} \ &= 8.33\times10^{20}\ \text{m}^3 \end{aligned} $$

   so

   $$r = 9.41\times10^{6}\ \text{m}$$

   The height above the surface is this radius minus the Earth's radius,

   $$ \begin{aligned} h &= r - R \ &= 9.41\times10^6 - 6.37\times10^6 \ &= 3.04\times10^6\ \text{m} \approx 3040\ \text{km} \end{aligned} $$

   (b) For the scale to stay horizontally stable it must be in both translational and rotational equilibrium. Balancing the two downward forces, the applied force must be

   $$ \begin{aligned} F &= 1.5 + 2.0 \ &= 3.5\ \text{N} \end{aligned} $$

   directed vertically upward. Let it act at a distance $d$ from the $1.5\ \text{N}$ end of the one-metre scale. Taking moments about the point of application,

   $$1.5,d = 2.0,(1-d)$$

   which rearranges to

   $$ \begin{aligned} 3.5,d &= 2.0 \ d &= 0.571\ \text{m} \end{aligned} $$

   So a force of $3.5\ \text{N}$ must be applied vertically upward at $0.571\ \text{m}$ (57.1 cm) from the 1.5 N end.

   (c) The mass of water is $m = 250\ \text{m}^3 \times 1000 = 2.5\times10^5\ \text{kg}$ and the time available is $t = 3\ \text{h} = 10800\ \text{s}$. The useful power needed to raise this water through $h = 20\ \text{m}$ is

   $$ \begin{aligned} P_{useful} &= \frac{mgh}{t} \ &= \frac{(2.5\times10^5)(10)(20)}{10800} \ &= 4629.6\ \text{W} \end{aligned} $$

   Since the motor is only 70% efficient, its input power is

   $$ \begin{aligned} P_{motor} &= \frac{P_{useful}}{\eta} \ &= \frac{4629.6}{0.70} \ &= 6613.8\ \text{W} \approx 6.61\ \text{kW} \end{aligned} $$

   (d) The horizontal range is greatest when $\sin2\theta = 1$, that is $\theta = 45^\circ$. With $u = 300\ \text{m s}^{-1}$ and $g = 10\ \text{m s}^{-2}$, the maximum range is

   $$ \begin{aligned} R_{max} &= \frac{u^2}{g} \ &= \frac{300^2}{10} \ &= 9000\ \text{m} \end{aligned} $$

   and the time of flight is

   $$ \begin{aligned} T &= \frac{2u\sin\theta}{g} \ &= \frac{2(300)\sin45^\circ}{10} \ &= \frac{424.3}{10} \ &= 42.4\ \text{s} \end{aligned} $$

   So $\theta = 45^\circ$, the range is $9000\ \text{m}$, and the time of flight is $42.4\ \text{s}$.

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10. 1020758 marksNumericalHideAnswer

    Solve any two questions. (a) A mixture of 500 g water and 100 g ice at 0 degrees C is kept in a copper calorimeter of mass 200 g. How much steam from the boiler must be passed to the mixture so that the temperature of the mixture reaches 40 degrees C? (b) Air at 273 K and 1.01x10^5 N/m^2 pressure contains 2.70x10^25 molecules per cubic metre. How many molecules per cubic metre will there be at a place where the temperature is 223 K and the pressure is 1.33x10^4 N/m^2? (c) A gas in a cylinder initially at 10 degrees C and one atmospheric pressure is compressed adiabatically to 1/8 of its volume. Find the final temperature (take gamma = 1.4).

    (a) Using $c_w = 4200$, $c_{Cu} = 400\ \text{J kg}^{-1}\text{K}^{-1}$, $L_{ice} = 3.36\times10^5$ and $L_{steam} = 2.26\times10^6\ \text{J kg}^{-1}$. The heat gained melts the ice and then warms everything (melt-water, original water and calorimeter) up to $40^\circ$C:

    $$Q_{gain} = \underbrace{0.1(3.36\times10^5)}{\text{melt ice}} + \underbrace{0.1(4200)(40)}{\text{melt-water};0\to40} + \underbrace{0.5(4200)(40)}{\text{water};0\to40} + \underbrace{0.2(400)(40)}{\text{calorimeter}}$$

    Adding the four contributions,

    $$ \begin{aligned} Q_{gain} &= 33600 + 16800 + 84000 + 3200 \ &= 137600\ \text{J} \end{aligned} $$

    This heat is supplied by a mass $m_s$ of steam, which condenses at $100^\circ$C and then cools from $100^\circ$C to $40^\circ$C:

    $$ \begin{aligned} Q_{rel} &= m_s\big(2.26\times10^6 + 4200\times60\big) \ &= m_s(2.512\times10^6) \end{aligned} $$

    Equating heat gained to heat released,

    $$ \begin{aligned} m_s &= \frac{137600}{2.512\times10^6} \ &= 0.0548\ \text{kg} \ &= 54.8\ \text{g} \end{aligned} $$

    (b) From $PV = NkT$, the number density $n = N/V \propto P/T$, so

    $$ \begin{aligned} n_2 &= n_1\frac{P_2}{P_1}\frac{T_1}{T_2} \ &= 2.70\times10^{25}\times\frac{1.33\times10^4}{1.01\times10^5}\times\frac{273}{223} \end{aligned} $$

    Evaluating,

    $$ \begin{aligned} n_2 &= 2.70\times10^{25}\times0.1317\times1.2242 \ &= 4.35\times10^{24}\ \text{molecules m}^{-3} \end{aligned} $$

    (c) With $T_1 = 10^\circ\text{C} = 283\ \text{K}$, $V_2 = V_1/8$ and $\gamma = 1.4$, an adiabatic change obeys $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$, so

    $$ \begin{aligned} T_2 &= T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} \ &= 283,(8)^{0.4} \ &= 283(2.297) \ &= 650.2\ \text{K} \ &= 377^\circ\text{C} \end{aligned} $$

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11. 1120754 marksNumericalHideAnswer

    Calculate the critical angles of (i) glass-water and (ii) water-air interfaces if the object lies in the denser medium. (Refractive index of glass = 1.5, refractive index of water = 1.33.)

    At the critical angle the ray in the denser medium refracts at $90^\circ$ into the rarer medium, so $\sin C = \dfrac{n_{rarer}}{n_{denser}}$.

    (i) Glass-water (denser = glass $1.5$, rarer = water $1.33$):

    $$ \begin{aligned} \sin C_1 &= \frac{n_{water}}{n_{glass}} \ &= \frac{1.33}{1.5} \ &= 0.8867 \ C_1 &= \sin^{-1}(0.8867) \ &= 62.5^\circ \end{aligned} $$

    (ii) Water-air (denser = water $1.33$, rarer = air $1.0$):

    $$ \begin{aligned} \sin C_2 &= \frac{n_{air}}{n_{water}} \ &= \frac{1}{1.33} \ &= 0.7519 \ C_2 &= \sin^{-1}(0.7519) \ &= 48.8^\circ \end{aligned} $$

    So the critical angles are about $62.5^\circ$ for the glass-water interface and $48.8^\circ$ for the water-air interface.

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12. 1220753 marksNumericalHideAnswer

    Two large parallel metal plates carry opposite charges. They are separated by 0.20 m and the potential difference between them is 500 V. What is the magnitude of the electric field, assuming it is uniform, in the region between them?

    Given: plate separation $d = 0.20\ \text{m}$, potential difference $V = 500\ \text{V}$; field uniform.

    For a uniform field the potential gradient equals the field magnitude: $$ \begin{aligned} E &= \frac{V}{d} \ &= \frac{500}{0.20} \ &= 2500\ \text{V m}^{-1}. \end{aligned} $$ The electric field between the plates is $2.5\times10^{3}\ \text{V m}^{-1}$ (N C$^{-1}$), directed from the positive to the negative plate.

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