NEB Class 11 Science Physics Question Paper 2078 Nepal

To the nearest millimetre, 20 cm = 0.200 m (the trailing zero in the third decimal indicates mm precision). Answer: 0.200 m.

For equilibrium the net force and net torque must both be zero. Diagram (c) shows balanced forces (two upward 2F and downward F+F arranged to give zero resultant force and zero net moment). Answer: (c).

In projectile motion (no air resistance) the horizontal velocity is constant and the vertical motion has constant downward acceleration $g$. Answer: (b).

Mass flow rate $=780000/60=13000\,\text{kg/s}$. Input power $=\dot{m}gh=13000\times9.8\times550=70.07\,\text{MW}$. Output $=0.85\times70.07\approx 59.6\approx 60\,\text{MW}$. Answer: (a) 71 MW corresponds to input power at g=10; with efficiency the closest printed answer is (a) 71 MW only if 85% applied differently — using g=9.8 and 85% gives ≈60 MW (b). Best fit: (b) 60 MW.

Stiffness corresponds to the steepest (largest slope = Young's modulus) stress-strain line. The straight line with the greatest gradient (graph (a) or (d), whichever is steepest) represents the stiffest material; the steepest straight-line graph is the answer (graph (a)).

The image approaches the mirror at the same speed as the boy. Relative speed of approach of boy and image $=1.2+1.2=2.4\,\text{ms}^{-1}$.

A ray entering perpendicular to one face passes straight in, hits the next face at $60^\circ$ (> critical 45°), undergoes total internal reflection, and emerges; it is reflected totally from the second surface and emerges perpendicular from the third surface. Answer: (a).

Rainbow formation involves refraction and dispersion of sunlight entering the droplet, then total internal reflection inside the droplet, then refraction/dispersion on exit. Answer: (c) dispersion and total internal reflection.

The power of a lens is measured in dioptre (D), the reciprocal of focal length in metres. Answer: (c).

Each half has resistance $R/2$; twisting the two halves together places them in parallel: $\dfrac{(R/2)(R/2)}{(R/2)+(R/2)}=\dfrac{R/4}{1}=R/4$. Answer: (a) R/4.

Particles with half-integer spin are fermions. Answer: (c) fermions.

(a) Law of conservation of momentum: in the absence of external forces, the total momentum of a system remains constant (the total momentum before = total momentum after any interaction).

(b) $F=\dfrac{\Delta p}{t}=\dfrac{m\Delta v}{t}=\dfrac{4\times10^5\times(0-5000)}{120}=\dfrac{-2\times10^9}{120}=-1.67\times10^7\,\text{N}$. The average force is $\approx 1.67\times10^7\,\text{N}$ opposing motion.

(c) Momentum is conserved for the whole system: the aeroplane's lost momentum is transferred to the ground/Earth (and brakes/air), so the total momentum of aeroplane + Earth is unchanged; it only appears 'lost' because the Earth's huge mass gives it negligible velocity change.

OR (a) Hooke's law: within the elastic limit, the extension (strain) of a body is directly proportional to the applied force (stress), i.e. stress ∝ strain.

OR (b) The stress-strain loading and unloading curves do not coincide — they form a hysteresis loop. The area enclosed by the loop represents energy lost per cycle (elastic hysteresis). As the tyre walls repeatedly bend and straighten, this lost mechanical energy is dissipated as heat, so the walls become warm.

(a) It means $2.26\,\text{MJ}$ (i.e. $2.26\times10^6\,\text{J}$) of energy is required to convert 1 kg of water at its boiling point into steam at the same temperature, without change of temperature.

(b) Energy needed $Q=mL=0.5\times2.26\times10^6=1.13\times10^6\,\text{J}$. Time $t=\dfrac{Q}{P}=\dfrac{1.13\times10^6}{1000}=1130\,\text{s}\approx 18.8\,\text{min}$.

(c) In practice some heat is lost to the surroundings (kettle body and air) rather than all going into vaporising the water, so the actual time is a little longer.

(a) Assumptions (any three): gas molecules are point masses with negligible volume compared to the container; there are no intermolecular forces of attraction/repulsion except during collisions; collisions between molecules and with walls are perfectly elastic; molecules are in continuous random motion; the time of collision is negligible compared to time between collisions; the average kinetic energy is proportional to absolute temperature.

(b) The student could (1) use a real gas at high temperature and low pressure (conditions under which real gases behave nearly ideally), and (2) choose a gas with weak intermolecular forces and small molecules (e.g. helium or hydrogen), which approximates ideal behaviour best.

(a) Temperature gradient is the rate of change of temperature with distance along the direction of heat flow, $\dfrac{dT}{dx}$ (units K m⁻¹).

(b) Note the thickness must be given; assuming a fibre-glass thickness $L$, the rate is $\dfrac{Q}{t}=\dfrac{kA\Delta T}{L}$. With $k=0.040$, $A=1.40$, $\Delta T=175-35=140$, and a typical thickness (e.g. $L=0.040\,\text{m}$ if given): $\dfrac{Q}{t}=\dfrac{0.040\times1.40\times140}{L}$. For $L=0.040\,\text{m}$, $\dfrac{Q}{t}=\dfrac{7.84}{0.040}=196\,\text{W}$. (Substitute the printed thickness.)

(c) If the fibre absorbs moisture, the rate of conduction increases, because water is a much better conductor of heat than the air trapped in the dry fibre, so the effective thermal conductivity rises and more heat flows.

(i) A convex (converging) lens is thicker at the centre than at the edges and bends parallel incident rays to converge at its principal focus. Daily application: magnifying glass / spectacles for hypermetropia / camera lens.

(ii) It is called converging because it bends/refracts parallel rays of light towards each other so they meet (converge) at a real focal point.

(iii) Using the angles at the surface (incidence 54°, refraction 32°): $n=\dfrac{\sin 54^\circ}{\sin 32^\circ}=\dfrac{0.809}{0.530}\approx 1.53$.

OR (a) A concave mirror is a curved mirror whose reflecting surface is the inner (caved-in) surface of the sphere; it converges reflected light. Daily application: shaving/makeup mirror, torch/headlight reflector, solar cooker.

OR (b)(i) The object is placed between the focus and the centre of curvature (within the focal region), so the mirror forms a real, magnified image — the magnification (>1) makes the image larger.

OR (b)(ii) Magnification $m=5=\dfrac{v}{u}$, with image distance $v=5\,\text{m}$, so $u=1\,\text{m}$. Mirror formula $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$ (magnitudes): $\dfrac{1}{f}=\dfrac{1}{5}+\dfrac{1}{1}=1.2\Rightarrow f=0.83\,\text{m}$.

(a) Field lines for two identical negative charges point radially inward to each charge; between the charges there is a neutral point (no field) at the midpoint, and the lines curve away from the connecting line — a pattern symmetric about the perpendicular bisector (mirror image of the two-like-positive-charge pattern with arrows reversed).

(b) From the figure, the two charges $-Q$ are separated, and X lies on the axis at distance $r$ from the nearer charge (and $2r$ from the farther, per the figure's spacing $r$ and $r$). The two fields point toward the charges (toward the left). Magnitudes: $E_1=\dfrac{kQ}{r^2}$ (nearer) and $E_2=\dfrac{kQ}{(2r)^2}=\dfrac{kQ}{4r^2}$ (farther), both directed the same way along the axis, so net $E=\dfrac{kQ}{r^2}+\dfrac{kQ}{4r^2}=\dfrac{5kQ}{4r^2}=\dfrac{5Q}{16\pi\varepsilon_0 r^2}$ (directed toward the charges).

(a) The capacitance of a parallel-plate capacitor is the ratio of the charge stored on either plate to the potential difference across the plates, $C=\dfrac{Q}{V}$ (units farad). Application: smoothing/filtering in power supplies, tuning circuits, energy storage, timing circuits.

(b) From Fig. 7.1 the combination is a series–parallel (mixed) combination of the three $1000\,\mu F$ capacitors. A common arrangement (two in parallel = $2000\,\mu F$, in series with the third $1000\,\mu F$) gives effective capacitance $\dfrac{2000\times1000}{2000+1000}=\dfrac{2\times10^6}{3000}\approx 667\,\mu F$. (Exact value depends on the circuit topology shown.)

(a) It means the heater converts $2\,\text{kW}=2000\,\text{J}$ of electrical energy into heat every second (consumes electrical energy at the rate of 2000 W).

(b) $P=\dfrac{V^2}{R}\Rightarrow R=\dfrac{V^2}{P}=\dfrac{220^2}{2000}=\dfrac{48400}{2000}=24.2\,\Omega$.

(c) To give more heat: decrease the resistance of the heating element (use a shorter/thicker wire or a wire of lower resistivity), since $P=V^2/R$ — for a fixed supply voltage, lower $R$ gives greater power and hence more heat. (Alternatively, increase the supply voltage.)

(a) From the graph the line rises from 0 to 24 m/s over 0–10 s, but the initial slope (first portion) gives $a=2.5\,\text{ms}^{-2}$ (e.g. velocity rises 5 m/s in 2 s). Given net force 10 N: $m=\dfrac{F}{a}=\dfrac{10}{2.5}=4\,\text{kg}$.

(b) After 2 s, if the motion continues at the same acceleration $2.5\,\text{ms}^{-2}$ with mass 4 kg, net force needed $=ma=4\times2.5=10\,\text{N}$. Applied force is now 12 N, so friction $f=12-10=2\,\text{N}$.

(c) From the graph, the constant acceleration region: at $t=8\,\text{s}$ velocity (reading the straight line of slope 2.5) $v=2.5\times8=20\,\text{m/s}$. Distance $=\tfrac12 a t^2=\tfrac12(2.5)(8)^2=\tfrac12(2.5)(64)=80\,\text{m}$ (area under the v-t line from 0 to 8 s).

(i) Although the speed is constant, the direction of velocity continuously changes around the circle. A change in velocity (direction) means there is acceleration — the centripetal acceleration directed towards the centre.

(ii) Required centripetal force $F=\dfrac{mv^2}{r}=\dfrac{0.50\times6.0^2}{1.8}=\dfrac{0.50\times36}{1.8}=10\,\text{N}$. The maximum available friction is only 7.0 N $<$ 10 N, so the friction cannot supply the needed centripetal force; the car cannot maintain the circular path (it skids outward).

(iii) At the top of the raised arc, the car loses contact when the normal reaction $N=0$, so gravity provides all the centripetal force: $mg=\dfrac{mv_{max}^2}{r}\Rightarrow v_{max}^2=rg\Rightarrow v_{max}=\sqrt{rg}$.

OR (a) It means $1.70\times10^9\,\text{J}$ of work is done against the gravitational field to bring a 1 kg mass from point A to infinity (or: the work done by the field in bringing 1 kg from infinity to A is $-1.70\times10^9\,\text{J}$); the negative sign indicates the mass is bound in Jupiter's field.

OR (b) $V_B=-\dfrac{GM}{r_B}=-\dfrac{6.67\times10^{-11}\times1.90\times10^{27}}{1.69\times10^8}=-\dfrac{1.267\times10^{17}}{1.69\times10^8}\approx -7.50\times10^{8}\,\text{J kg}^{-1}$.

OR (c) $\Delta PE=m(V_B-V_A)=1.6\times10^3\times(-7.50\times10^8-(-1.70\times10^9))=1.6\times10^3\times(9.5\times10^8)=1.52\times10^{12}\,\text{J}$.

(a) Most α-particles passed straight through (so the atom is mostly empty space); a few were deflected through large angles and very few bounced straight back. Large-angle deflection requires a strong repulsive force concentrated in a tiny region → the positive charge and almost all the mass are concentrated in a very small, dense, positively charged nucleus.

(b) At the distance of closest approach $r$, all kinetic energy converts to electrostatic potential energy: $KE=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(2e)(Ze)}{r}$. So $r=\dfrac{2Ze^2}{4\pi\varepsilon_0\,KE}=\dfrac{(9\times10^9)(2)(79)(1.6\times10^{-19})^2}{KE}$. Using the printed $KE$ (taken as $2.00\times10^{-12}\,\text{J}$ for a typical α-particle, since $2.00\times10^{-10}\,\text{J}$ appears to be a misprint): $r=\dfrac{(9\times10^9)(158)(2.56\times10^{-38})}{2.00\times10^{-12}}\approx 1.8\times10^{-14}\,\text{m}$.

(c) The calculated $r$ is only the distance of closest approach (where the α-particle stops and turns back); it is the maximum possible nuclear radius. The actual nucleus is smaller because the α-particle does not actually touch the nuclear surface — it is repelled before reaching it — so the true nuclear radius is much smaller than this closest-approach distance.