NEB Class 11 · Past paper
The complete NEB Class 11 2069 exam paper for Physics, all 12 questions with solved model answers.
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Answer in brief any six questions. (a) How does the kinetic energy of a body change if its momentum is halved? (b) If the sun collapsed to a black hole, what effect on the orbit of the earth? (c) Can the direction of velocity of a body change when its acceleration is constant? (d) If A and B are non-zero vectors, is it possible for A x B and A . B both to be zero? (e) In hot-air ballooning why must the air be heated? (f) A man drops his briefcase in an elevator but it does not fall. What can be concluded? (g) An ice cube floats in a glass of water. When the ice melts, will the water level rise, fall or stay same?
(a) In terms of momentum, the kinetic energy of a body is $$KE = \frac{p^2}{2m}$$ so $KE \propto p^2$. If the momentum is halved, $p \to p/2$, then $$ \begin{aligned} KE' &= \frac{(p/2)^2}{2m} \ &= \frac{1}{4}\cdot\frac{p^2}{2m} \end{al...
Answer in brief any two questions. (a) Frozen water pipes often burst; will a mercury thermometer break if cooled below the freezing point of mercury? (b) If you add heat to an object, do you necessarily increase its temperature? (c) Why are the polar regions much cooler than equatorial regions despite periodic tilt toward the sun?
(a) Water is anomalous: it expands on freezing, so ice bursts pipes. Mercury, like most substances, contracts on solidifying, so on freezing it shrinks away from the glass. Hence a mercury thermometer will not burst when cooled below mer...
Answer in brief any one question. (a) Can a convex mirror ever form a real image? (b) A large object taken away from your eye appears smaller. Why?
(a) For a real object a convex mirror always gives a virtual, erect, diminished image (diverging mirror). It can form a real image only if the object is virtual, i.e. a converging beam is intercepted by the convex mirror before it forms ...
Answer in brief any one question. (a) Why is it dangerous to take shelter under a tree during lightning? (b) Can two electric lines of force ever intersect each other?
(a) A tree is tall and its moist, sappy interior conducts better than air, so it offers a preferred (low-resistance) path to ground: lightning tends to strike the tallest conductor. A person sheltering under it may be struck directly or ...
Answer any three questions. (a) State the parallelogram law of vector addition and derive the magnitude and direction of the resultant of two vectors inclined at angle theta. (b) What is a geostationary satellite? Obtain the total energy of a satellite orbiting the earth. (c) State the principle of conservation of linear momentum. How does Newton's third law lead to it? (d) Derive the energy stored in a stretched rod and define energy density.
(a) Parallelogram law: if two vectors are represented by the two adjacent sides of a parallelogram drawn from a point, their resultant is the diagonal through that point. For $\vec P$ and $\vec Q$ inclined at $\theta$, resolving the diag...
Answer any two questions. (a) Define linear and cubical expansivities of solids. Derive the variation in density with temperature. (b) Define adiabatic process. Show PV^gamma = constant. (c) Describe the working of a petrol engine with a P-V diagram.
(a) The linear expansivity $\alpha$ is the fractional increase in length per degree rise in temperature, and the cubical expansivity $\gamma$ is the fractional increase in volume per degree, with $\gamma = 3\alpha$.
The mass of a body is fixed, so its density is $\rho = m/V$. On heating through $\Delta\theta$ the volume becomes $V_2 = V_1(1 + \gamma,\Delta\theta)$, so the new density is
$$ \begin{aligned} \rho_2 &= \frac{m}{V_2} \ &= \frac{\rho_1}{1 + \gamma,\Delta\theta} \end{aligned} $$
and for small expansion this is approximately
$$\rho_2 \approx \rho_1(1 - \gamma,\Delta\theta)$$
showing that density decreases as temperature rises.
(b) An adiabatic process is one in which no heat enters or leaves the system ($dQ = 0$), as happens when a change is sudden or the system is well insulated.
From the first law, $dU + dW = 0$, so $nC_v,dT + P,dV = 0$. Using $PV = nRT$ and $R = C_p - C_v$ to eliminate $dT$,
$$C_v,V,dP + C_p,P,dV = 0$$
Dividing through by $C_v PV$,
$$\frac{dP}{P} + \gamma\frac{dV}{V} = 0$$
where $\gamma = C_p/C_v$. Integrating, $\ln P + \gamma\ln V$ is constant, which means
$$\boxed{PV^\gamma = \text{constant}}$$
(c) A petrol (Otto) engine works in four strokes, which trace out the following loop on the P-V (indicator) diagram:
The area enclosed by the loop equals the net work output per cycle, and the efficiency is
$$\eta = 1 - \frac{1}{r^{\gamma-1}}$$
with $r$ the compression ratio.
Answer any one question. (a) What is long sightedness? Discuss its causes and remedy. (b) Derive the lens maker's formula.
(a) Long sightedness, or hypermetropia, is a defect in which a person sees distant objects clearly but near objects appear blurred. The near point (the closest distance the eye can focus on) has shifted farther away than the normal 25 cm.
This happens because the eyeball is too short, so a sharp image would form behind the retina, or because the eye lens is too weak to converge the rays enough. Either way, rays from a near object are not brought to a focus on the retina.
The defect is corrected with a convex (converging) lens of suitable power placed in front of the eye. It bends the incoming rays inward a little, so the eye can now focus a near object on the retina and the near point is brought back to 25 cm.
(b) To derive the Lens maker's formula we consider refraction at the lens's two surfaces in turn. Take a thin lens of refractive index $\mu$ with surface radii $R_1$ and $R_2$.
At the first surface light travels from air into the glass, forming an intermediate image at distance $v_1$:
$$\frac{\mu}{v_1} - \frac{1}{u} = \frac{\mu - 1}{R_1}$$
At the second surface this intermediate image acts as the object as the light passes back into air:
$$\frac{1}{v} - \frac{\mu}{v_1} = \frac{1 - \mu}{R_2}$$
Adding the two equations eliminates $v_1$:
$$\frac{1}{v} - \frac{1}{u} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$
Since $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ for a lens, we obtain the Lens maker's formula:
$$\boxed{\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)}$$
Answer any one question. (a) State Gauss theorem and use it to find the electric field due to a plane charged conductor. (b) How can capacitors be connected to increase and decrease the effective capacitance? Find the expressions.
(a) Gauss's theorem states that the total flux through a closed surface equals $\dfrac{q{enc}}{\varepsilon0}$. For a plane charged conductor with surface charge density $\sigma$, the field just outside is normal to the surface. Take a sm...
Solve any three questions. (a) A man swims across a 600 m river at 4 km/h in still water; river flows 2 km/h. In what direction must he swim to reach the point exactly opposite, and how long? (b) A ballet dancer spins at 2.4 rev/s with moment of inertia I; folding arms gives 0.6I. Find new rate of spin. (c) A 145 g cricket ball at 14 m/s is hit back at 22 m/s; force acts 0.015 s. Find the average force. (d) Castor oil (viscosity 2.42 Ns/m^2, density 940 kg/m^3): find the terminal velocity of a steel ball of radius 2 mm.
(a) He must aim upstream so the current's push is exactly cancelled. With a swimming speed of $4\ \text{km/h}$ and a current of $2\ \text{km/h}$, the upstream component $v_{swim}\sin\theta$ has to equal the river speed, so
$$ \begin{aligned} \sin\theta &= \frac{v_{river}}{v_{swim}} \ &= \frac{2}{4} \ &= 0.5 \end{aligned} $$
which means he heads $\theta = 30^\circ$ upstream of the straight-across direction. Only the across component then carries him over, and that component is
$$ \begin{aligned} v_{swim}\cos\theta &= 4\cos30^\circ \ &= 3.464\ \text{km/h} \end{aligned} $$
Covering the $0.6\ \text{km}$ width at this speed takes
$$ \begin{aligned} t &= \frac{0.6}{3.464} \ &= 0.1732\ \text{h} \ &= 10.4\ \text{min} \end{aligned} $$
So he should swim $30^\circ$ upstream of straight across and reaches the opposite point in about $10.4$ minutes.
(b) As she folds her arms there is no external torque, so her angular momentum is conserved, $I_1\omega_1 = I_2\omega_2$. With $\omega_1 = 2.4\ \text{rev/s}$ and the moment of inertia dropping from $I$ to $0.6I$,
$$ \begin{aligned} \omega_2 &= \frac{I(2.4)}{0.6I} \ &= 4.0\ \text{rev/s} \end{aligned} $$
so she spins faster, at $4.0$ rev/s.
(c) Taking the incoming direction as positive, the $145\ \text{g}$ ball arrives at $u = 14\ \text{m/s}$ and leaves at $v = -22\ \text{m/s}$. Its change in momentum is
$$ \begin{aligned} \Delta p &= m(v - u) \ &= 0.145(-22 - 14) \ &= -5.22\ \text{kg m/s} \end{aligned} $$
The average force is this momentum change divided by the $0.015\ \text{s}$ contact time,
$$ \begin{aligned} F &= \frac{|\Delta p|}{t} \ &= \frac{5.22}{0.015} \ &= 348\ \text{N} \end{aligned} $$
so the bat exerts about $348\ \text{N}$, directed opposite to the ball's initial motion.
(d) The steel ball ($\rho_s = 7800\ \text{kg/m}^3$, radius $r = 2\times10^{-3}\ \text{m}$) settles at its terminal velocity in the castor oil ($\eta = 2.42\ \text{Ns/m}^2$, $\rho_l = 940\ \text{kg/m}^3$), given by Stokes' law:
$$v = \frac{2}{9}\frac{r^2(\rho_s - \rho_l)g}{\eta}$$
Substituting the values,
$$v = \frac{2}{9}\frac{(2\times10^{-3})^2(7800-940)(9.8)}{2.42}$$
which simplifies to
$$ \begin{aligned} v &= \frac{2}{9}\times\frac{0.2689}{2.42} \ &= 0.0247\ \text{m/s} \end{aligned} $$
So the terminal velocity is about $0.0247\ \text{m/s}$ (roughly $2.5\ \text{cm/s}$).
Solve any two questions. (a) Air at 273 K, 1.01x10^5 N/m^2 has 2.7x10^25 molecules/m^3. How many per m^3 at 223 K and 1.33x10^-4 N/m^2? (b) 10 g steam at 100 C is passed into 100 g water and 10 g ice at 0 C. Find the resulting temperature. (c) A 50 cm^3 glass vessel filled with mercury is heated 20 C to 60 C. What volume of mercury overflows?
(a) Air at $T_1 = 273\ \text{K}$ and $P_1 = 1.01\times10^{5}\ \text{N/m}^2$ has $n_1 = 2.7\times10^{25}$ molecules per m³, and we want the number density at $T_2 = 223\ \text{K}$ and $P_2 = 1.33\times10^{-4}\ \text{N/m}^2$. Since the number density is $n = \dfrac{P}{kT}$, the two states are related by $n_2 = n_1\dfrac{P_2}{P_1}\dfrac{T_1}{T_2}$,
$$n_2 = 2.7\times10^{25}\times\frac{1.33\times10^{-4}}{1.01\times10^{5}}\times\frac{273}{223}$$
Evaluating the factors,
$$ \begin{aligned} n_2 &= 2.7\times10^{25}\times(1.317\times10^{-9})\times(1.224) \ &= 4.35\times10^{16}\ \text{molecules/m}^3 \end{aligned} $$
So $n_2 \approx 4.4\times10^{16}$ molecules per m³.
(b) Here $10\ \text{g}$ of steam at $100^\circ$C is passed into $100\ \text{g}$ of water and $10\ \text{g}$ of ice at $0^\circ$C (work in calories with $L_{steam}=540$, $L_{ice}=80$, $c_w=1$). The heat released as the steam condenses is
$$10\times540 = 5400\ \text{cal}$$
and the heat needed to melt the ice is
$$10\times80 = 800\ \text{cal}$$
The $110\ \text{g}$ of water (the original 100 g plus the 10 g of melted ice) then warms from $0$ to $T$, while the $10\ \text{g}$ of condensed steam cools from $100$ to $T$. Balancing the heat,
$$ \begin{aligned} 5400 + 10(100 - T) &= 800 + 110T \ 6400 - 10T &= 800 + 110T \ 5600 &= 120T \ T &= 46.7^\circ\text{C} \end{aligned} $$
Since $0 < T < 100$, all the steam condenses and all the ice melts, and the final temperature is about $46.7^\circ$C.
(c) A $50\ \text{cm}^3$ glass vessel full of mercury is heated from $20^\circ$C to $60^\circ$C, so $\Delta\theta = 40$. The overflow equals the apparent expansion of the mercury relative to the glass, whose coefficient is
$$ \begin{aligned} \gamma_{app} &= \gamma_{Hg} - \gamma_{glass} \ &= 1.82\times10^{-4} - 0.27\times10^{-4} \ &= 1.55\times10^{-4}\ \text{K}^{-1} \end{aligned} $$
so the volume that overflows is
$$ \begin{aligned} \Delta V &= V\gamma_{app}\Delta\theta \ &= 50\times1.55\times10^{-4}\times40 \ &= 0.31\ \text{cm}^3 \end{aligned} $$
About 0.31 cm³ of mercury overflows.
An object is imaged by a lens on a screen 30 cm to the right of the lens. When the lens is moved 4 cm right, the screen must be moved 4 cm left to refocus. Determine the focal length of the lens.
The lens has a single focal length, so the two arrangements (before and after the lens is moved) must give the same $f$. Equating them lets us find the object distance and hence $f$. At first the screen (image) is $v1 = 30\ \text{cm}$ to...
Two charges +1x10^-6 C and -4x10^-6 C are separated by 2 m. Determine the position of the null point.
The charges are $+1\times10^{-6}\ \text{C}$ and $-4\times10^{-6}\ \text{C}$, separated by $2\ \text{m}$. For two unlike charges the null point (where $\vec E = 0$) lies outside the pair, on the side of the smaller charge. Let it be a dis...
(a) In terms of momentum, the kinetic energy of a body is so . If the momentum is halved, , then $$ \begin{aligned} KE' &= \frac{(p/2)^2}{2m} \ &= \frac{1}{4}\cdot\frac{p^2}{2m} \end{al...
(a) Parallelogram law: if two vectors are represented by the two adjacent sides of a parallelogram drawn from a point, their resultant is the diagonal through that point. For and inclined at , resolving the diag...
(a) The linear expansivity is the fractional increase in length per degree rise in temperature, and the cubical expansivity is the fractional increase in volume per degree, with .
The mass of a body is fixed, so its density is . On heating through the volume becomes , so the new density is
and for small expansion this is approximately
showing that density decreases as temperature rises.
(b) An adiabatic process is one in which no heat enters or leaves the system (), as happens when a change is sudden or the system is well insulated.
From the first law, , so . Using and to eliminate ,
Dividing through by ,
where . Integrating, is constant, which means
(c) A petrol (Otto) engine works in four strokes, which trace out the following loop on the P-V (indicator) diagram:
The area enclosed by the loop equals the net work output per cycle, and the efficiency is
with the compression ratio.
(a) Long sightedness, or hypermetropia, is a defect in which a person sees distant objects clearly but near objects appear blurred. The near point (the closest distance the eye can focus on) has shifted farther away than the normal 25 cm.
This happens because the eyeball is too short, so a sharp image would form behind the retina, or because the eye lens is too weak to converge the rays enough. Either way, rays from a near object are not brought to a focus on the retina.
The defect is corrected with a convex (converging) lens of suitable power placed in front of the eye. It bends the incoming rays inward a little, so the eye can now focus a near object on the retina and the near point is brought back to 25 cm.
(b) To derive the Lens maker's formula we consider refraction at the lens's two surfaces in turn. Take a thin lens of refractive index with surface radii and .
At the first surface light travels from air into the glass, forming an intermediate image at distance :
At the second surface this intermediate image acts as the object as the light passes back into air:
Adding the two equations eliminates :
Since for a lens, we obtain the Lens maker's formula:
(a) Gauss's theorem states that the total flux through a closed surface equals . For a plane charged conductor with surface charge density , the field just outside is normal to the surface. Take a sm...
(a) He must aim upstream so the current's push is exactly cancelled. With a swimming speed of and a current of , the upstream component has to equal the river speed, so
which means he heads upstream of the straight-across direction. Only the across component then carries him over, and that component is
Covering the width at this speed takes
So he should swim upstream of straight across and reaches the opposite point in about minutes.
(b) As she folds her arms there is no external torque, so her angular momentum is conserved, . With and the moment of inertia dropping from to ,
so she spins faster, at rev/s.
(c) Taking the incoming direction as positive, the ball arrives at and leaves at . Its change in momentum is
The average force is this momentum change divided by the contact time,
so the bat exerts about , directed opposite to the ball's initial motion.
(d) The steel ball (, radius ) settles at its terminal velocity in the castor oil (, ), given by Stokes' law:
Substituting the values,
which simplifies to
So the terminal velocity is about (roughly ).
(a) Air at and has molecules per m³, and we want the number density at and . Since the number density is , the two states are related by ,
Evaluating the factors,
So molecules per m³.
(b) Here of steam at C is passed into of water and of ice at C (work in calories with , , ). The heat released as the steam condenses is
and the heat needed to melt the ice is
The of water (the original 100 g plus the 10 g of melted ice) then warms from to , while the of condensed steam cools from to . Balancing the heat,
Since , all the steam condenses and all the ice melts, and the final temperature is about C.
(c) A glass vessel full of mercury is heated from C to C, so . The overflow equals the apparent expansion of the mercury relative to the glass, whose coefficient is
so the volume that overflows is
About 0.31 cm³ of mercury overflows.
The lens has a single focal length, so the two arrangements (before and after the lens is moved) must give the same . Equating them lets us find the object distance and hence . At first the screen (image) is to...
The charges are and , separated by . For two unlike charges the null point (where ) lies outside the pair, on the side of the smaller charge. Let it be a dis...