NEB Class 11 · Past paper

Physics NEB Class 11 2070 question paper

The complete NEB Class 11 2070 exam paper for Physics, all 12 questions with solved model answers.

Past Papers2075 2073 2072 2070 2069 Model

Tap a question to open its answer.

1207012 marksHideAnswer
Answer, in brief, any six questions: (a) Angle between vectors A and B is theta. Find the magnitude and direction of A x B and A . B. (b) If y = a + bt + ct^2 (y distance, t time), find the dimension and unit of c. (c) In projectile motion, is there a point where acceleration is perpendicular to velocity? (d) A smooth stream of water narrows as it falls. Explain. (e) When four-legged animals walk they keep three legs on the ground. Why? (f) Why is soap solution a better cleansing agent than water? (g) A fan with blades takes longer to stop than without blades. Why?
(a) The cross product has magnitude $AB\sin\theta$ and points perpendicular to the plane of $\vec A$ and $\vec B$ (right-hand rule), $$\vec A\times\vec B = AB\sin\theta,\hat n$$ while the dot product is a scalar with no direction, $$\ve...
220704 marksHideAnswer
Answer, in brief, any two questions: (a) Does the cubical expansivity of a liquid depend on its original volume? (b) Animals curl into a ball when very cold. Why? (c) Define triple point and its significance.
(a) No. Cubical expansivity is a fractional change per degree: $$\gamma = \frac{\Delta V}{V,\Delta\theta}$$ It is a property of the material, independent of the original volume $V$. A larger volume expands more in absolute terms, but th...
320702 marksHideAnswer
Answer, in brief, any one question: (a) What is illuminance? On what factors does it depend? (b) Distinguish between real and virtual images.
(a) Illuminance is the luminous flux incident per unit area of a surface, $$E = \frac{\Phi}{A}$$ with SI unit the lux ($\mathrm{lm,m^{-2}}$). It depends on the luminous intensity of the source, the distance from the source (through the ...
420702 marksHideAnswer
Answer, in brief, any one question: (a) More charge can be stored on a metal if it is highly polished than when rough. Explain. (b) What factors determine the capacitance of a parallel plate capacitor?
(a) A rough surface has sharp points and edges where charge density and the surrounding field become very high (action of points), causing charge to leak away into the air by corona discharge. A highly polished surface has no such points...
5207012 marksHideAnswer
Answer any three questions: (a) Define work. Derive the work done by a variable force. (b) What is gravitational PE? Obtain an expression for it at distance r from Earth's centre. (c) Define terminal velocity and describe determining viscosity using Stokes' law. (d) Derive a relation between angular momentum and moment of inertia and hence define moment of inertia.
(a) Work is the product of force and the displacement in the direction of the force, $W = \vec F\cdot\vec s$. For a variable force, the work done over a small displacement $dx$ is $dW = F,dx$, so the total work is the integral

$$W = \int_{x_1}^{x_2} F,dx$$

which equals the area under the force-displacement graph.

(b) The gravitational potential energy of a body is the work done in bringing it from infinity to a point in the gravitational field. Bringing a mass $m$ from infinity to a distance $r$ from Earth's centre (mass $M$),

$$U = -\int_\infty^r \frac{GMm}{x^2}dx$$

Evaluating the integral,

$$\boxed{U = -\frac{GMm}{r}}$$

The energy is negative because gravity is attractive, and it rises toward zero as $r$ approaches infinity.

(c) Terminal velocity is the constant maximum velocity a body attains while falling through a viscous fluid, reached when its weight is balanced by the upthrust plus the viscous drag. At terminal velocity for a sphere of radius $a$ and density $\rho$ in a fluid of density $\sigma$ and viscosity $\eta$, the weight less upthrust equals the Stokes drag,

$$\tfrac{4}{3}\pi a^3(\rho-\sigma)g = 6\pi\eta a v_t$$

Solving for the terminal velocity,

$$v_t = \frac{2a^2(\rho-\sigma)g}{9\eta}$$

Measuring $v_t$ over a known distance for a ball of known $a$ and $\rho$ then gives the viscosity,

$$\boxed{\eta = \frac{2a^2(\rho-\sigma)g}{9v_t}}$$

(d) Summing the angular momentum of every particle, with $v_i = \omega r_i$,

$$ \begin{aligned} L &= \sum m_i v_i r_i \ &= \sum m_i(\omega r_i)r_i \ &= \left(\sum m_i r_i^2\right)\omega \end{aligned} $$

Recognising the bracket as the moment of inertia $I = \sum m_i r_i^2$,

$$\boxed{L = I\omega}$$

The moment of inertia is thus the rotational analogue of mass: $I = L/\omega$, the angular momentum per unit angular velocity, which measures a body's resistance to change in its rotational motion.
620708 marksHideAnswer
Answer any two questions: (a) Define coefficient of real and apparent expansion of a liquid and derive a relation between them. (b) What is a perfect gas? Prove the average KE of a gas molecule is proportional to absolute temperature. (c) Derive the work done during adiabatic expansion of an ideal gas. Does internal energy change?
(a) The coefficient of real expansion $\gammar$ is the fractional increase in the true volume of a liquid per degree rise in temperature, while the coefficient of apparent expansion $\gammaa$ is the fractional increase we actually observ...
720704 marksHideAnswer
Answer any one question: (a) What is lateral shift? Derive its value. How does it change with angle of incidence? (b) What is chromatic aberration? Show that for a lens it equals the product of dispersive power and focal length of mean light.
(a) The lateral shift is the perpendicular displacement between the incident ray (produced straight through) and the emergent ray when light passes through a parallel-sided glass slab. The emergent ray comes out parallel to the incident ...
820704 marksHideAnswer
Answer any one question: (a) What is electrostatic induction? How can you charge a body positively by induction? (b) State and explain Gauss's law and use it to find the electric field due to a line charge.
(a) Electrostatic induction is the redistribution of a conductor's charge caused by a nearby charged body, without any contact. To charge a body positively using a negative rod: bring the rod near the conductor, so the near face becomes ...
9207012 marksNumericalHideAnswer
Answer any three numerical questions: (a) A 6 kg box is pushed at constant speed 0.35 m/s with mu_k = 0.12; find F. Then find F for constant acceleration 0.18 m/s^2. (b) A string breaks at 25 N; a 500 g mass on a 1 m string is rotated in a horizontal circle. Find the greatest rpm without breaking. (c) Work done in stretching a steel wire 100 cm long, area 0.03 cm^2, load 100 N. (d) A body in SHM: amplitude 15 cm, frequency 4 Hz. Find maximum acceleration and velocity.
(a) While the $6\ \text{kg}$ box moves at constant speed the applied force only has to overcome kinetic friction, so

$$ \begin{aligned} F &= \mu_k mg \ &= 0.12\times6\times9.8 \ &= 7.06\ \text{N} \end{aligned} $$

To give it an acceleration of $0.18\ \text{m/s}^2$ instead, the force must also supply $ma$ on top of friction:

$$ \begin{aligned} F &= ma + \mu_k mg \ &= 6(0.18) + 7.06 \ &= 8.14\ \text{N} \end{aligned} $$

So $F = 7.06\ \text{N}$ keeps it moving steadily, and $F = 8.14\ \text{N}$ accelerates it.

(b) The string tension supplies the centripetal force, $T = m\omega^2 r$, and it snaps once $T$ reaches $25\ \text{N}$. With $m = 0.5\ \text{kg}$ and $r = 1\ \text{m}$ the greatest angular speed follows from

$$ \begin{aligned} \omega^2 &= \frac{T}{mr} \ &= \frac{25}{0.5\times1} \ &= 50 \end{aligned} $$

so $\omega = 7.07\ \text{rad/s}$. Converting to revolutions per minute,

$$ \begin{aligned} n &= \frac{\omega}{2\pi} \ &= \frac{7.07}{6.283} \ &= 1.125\ \text{rev/s} \ &= 67.5\ \text{rev/min} \end{aligned} $$

The wire can therefore be spun at most about $67.5$ times per minute.

(c) The $100\ \text{N}$ load first stretches the wire ($L = 1\ \text{m}$, $A = 3\times10^{-6}\ \text{m}^2$, $Y = 2\times10^{11}\ \text{N/m}^2$) by

$$ \begin{aligned} x &= \frac{FL}{AY} \ &= \frac{100\times1}{3\times10^{-6}\times2\times10^{11}} \ &= 1.67\times10^{-4}\ \text{m} \end{aligned} $$

Because the tension builds up gradually from zero, the work stored is $\tfrac12 Fx$:

$$ \begin{aligned} W &= \tfrac12(100)(1.67\times10^{-4}) \ &= 8.3\times10^{-3}\ \text{J} \end{aligned} $$

so the work done is about $8.3\times10^{-3}\ \text{J}$ (roughly $8.3\ \text{mJ}$).

(d) For the SHM the angular frequency is $\omega = 2\pi f = 25.13\ \text{rad/s}$, with amplitude $A = 0.15\ \text{m}$. The acceleration is greatest at the extreme position:

$$ \begin{aligned} a_{max} &= \omega^2 A \ &= (25.13)^2(0.15) \ &= 94.7\ \text{m/s}^2 \end{aligned} $$

and the velocity is greatest at the mean position:

$$ \begin{aligned} v_{max} &= \omega A \ &= 25.13\times0.15 \ &= 3.77\ \text{m/s} \end{aligned} $$

So $a_{max} \approx 94.7\ \text{m/s}^2$ and $v_{max} \approx 3.77\ \text{m/s}$.
1020708 marksNumericalHideAnswer
Answer any two numerical questions: (a) A 300 g copper calorimeter has 500 g water at 15 C; a 560 g aluminium ball at 100 C is dropped, temperature rises to 22 C. Find specific heat of aluminium. (b) A rod length 20 cm, area 3.14 cm^2, one end 100 C, other in ice 0 C; 25 g ice melts in 5 min. Find thermal conductivity. (c) A petrol engine consumes 25 kg petrol/hr, calorific value 11.4x10^6 cal/kg, power 99.75 kW. Find efficiency.
(a) An aluminium ball of mass $m = 560\ \text{g} = 0.560\ \text{kg}$ at $100^\circ$C is dropped into a copper calorimeter ($300\ \text{g} = 0.300\ \text{kg}$) holding water ($500\ \text{g} = 0.500\ \text{kg}$) at $15^\circ$C, and the mix...
1120704 marksNumericalHideAnswer
A refracting telescope has an objective of focal length 1 m and an eyepiece of focal length 2 cm. A real image of the sun, 10 cm in diameter, is formed on a screen 24 cm from the eyepiece. What angle does the sun subtend at the objective?
Two lenses act in series here: the objective forms a small real image of the sun, and the eyepiece then projects that image, enlarged, onto the screen. By working back from the screen we can find the size of the small image, and from its size the angle the sun subtends.

We are given the objective focal length $f_o = 1\ \text{m}$, the eyepiece focal length $f_e = 2\ \text{cm}$, and the final real image of the sun, $H = 10\ \text{cm}$ tall, formed at $v = 24\ \text{cm}$ from the eyepiece. The objective forms its image at its focal plane, where the image height is $h_i = f_o,\alpha$ and $\alpha$ is the angle the sun subtends at the objective. That image is the object for the eyepiece, so once we know $h_i$ we can find $\alpha$.

For the eyepiece we use $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ with $f_e = 2\ \text{cm}$ and $v = 24\ \text{cm}$:

$$ \begin{aligned} \frac{1}{u} &= \frac{1}{2} - \frac{1}{24} \ &= \frac{11}{24} \end{aligned} $$

so $u = 2.18\ \text{cm}$, and the eyepiece magnification is

$$ \begin{aligned} m &= \frac{v}{u} \ &= \frac{24}{2.18} \ &= 11 \end{aligned} $$

The $10\ \text{cm}$ screen image is $m$ times the eyepiece's object, so the intermediate image formed by the objective was

$$ \begin{aligned} h_i &= \frac{H}{m} \ &= \frac{10}{11} \ &= 0.909\ \text{cm} \ &= 9.09\times10^{-3}\ \text{m} \end{aligned} $$

Using $h_i = f_o,\alpha$, the angle the sun subtends at the objective is

$$ \begin{aligned} \alpha &= \frac{h_i}{f_o} \ &= \frac{9.09\times10^{-3}}{1} \ &= 9.09\times10^{-3}\ \text{rad} \ &= 0.52^\circ \end{aligned} $$

Therefore the sun subtends about $9.1\times10^{-3}$ rad (about $0.52^\circ$) at the objective.
1220703 marksNumericalHideAnswer
A parallel plate air capacitor of capacitance 245x10^-12 F has a charge of 0.148 uC on each plate. Find the potential difference and electric field intensity if the plate separation is 5 mm.
The capacitor has $C = 245\times10^{-12}\ \text{F}$ with a charge $Q = 0.148\ \mu\text{C} = 0.148\times10^{-6}\ \text{C}$ on each plate, and the plate separation is $d = 5\ \text{mm} = 5\times10^{-3}\ \text{m}$. The potential difference ...

Keep going

Important questions

Chapter notes Full syllabus

Other papers

2075 paper 2073 paper 2072 paper 2069 paper Model questions

NEB Class 11 · Past paper

Physics NEB Class 11 2070 question paper

The complete NEB Class 11 2070 exam paper for Physics, all 12 questions with solved model answers.

Past Papers2075 2073 2072 2070 2069 Model

Tap a question to open its answer.

1207012 marksHideAnswer
Answer, in brief, any six questions: (a) Angle between vectors A and B is theta. Find the magnitude and direction of A x B and A . B. (b) If y = a + bt + ct^2 (y distance, t time), find the dimension and unit of c. (c) In projectile motion, is there a point where acceleration is perpendicular to velocity? (d) A smooth stream of water narrows as it falls. Explain. (e) When four-legged animals walk they keep three legs on the ground. Why? (f) Why is soap solution a better cleansing agent than water? (g) A fan with blades takes longer to stop than without blades. Why?
(a) The cross product has magnitude $AB\sin\theta$ and points perpendicular to the plane of $\vec A$ and $\vec B$ (right-hand rule), $\vec A\times\vec B = AB\sin\theta\,\hat n$ while the dot product is a scalar with no direction, $$\ve...
(a) The cross product has magnitude $AB\sin\theta$ and points perpendicular to the plane of $\vec A$ and $\vec B$ (right-hand rule), $$\vec A\times\vec B = AB\sin\theta,\hat n$$ while the dot product is a scalar with no direction, $$\ve...
220704 marksHideAnswer
Answer, in brief, any two questions: (a) Does the cubical expansivity of a liquid depend on its original volume? (b) Animals curl into a ball when very cold. Why? (c) Define triple point and its significance.
(a) No. Cubical expansivity is a fractional change per degree: $\gamma = \frac{\Delta V}{V\,\Delta\theta}$ It is a property of the material, independent of the original volume $V$ . A larger volume expands more in absolute terms, but th...
(a) No. Cubical expansivity is a fractional change per degree: $$\gamma = \frac{\Delta V}{V,\Delta\theta}$$ It is a property of the material, independent of the original volume $V$. A larger volume expands more in absolute terms, but th...
320702 marksHideAnswer
Answer, in brief, any one question: (a) What is illuminance? On what factors does it depend? (b) Distinguish between real and virtual images.
(a) Illuminance is the luminous flux incident per unit area of a surface, $E = \frac{\Phi}{A}$ with SI unit the lux ( $\mathrm{lm\,m^{-2}}$ ). It depends on the luminous intensity of the source, the distance from the source (through the ...
(a) Illuminance is the luminous flux incident per unit area of a surface, $$E = \frac{\Phi}{A}$$ with SI unit the lux ($\mathrm{lm,m^{-2}}$). It depends on the luminous intensity of the source, the distance from the source (through the ...
420702 marksHideAnswer
Answer, in brief, any one question: (a) More charge can be stored on a metal if it is highly polished than when rough. Explain. (b) What factors determine the capacitance of a parallel plate capacitor?
(a) A rough surface has sharp points and edges where charge density and the surrounding field become very high (action of points), causing charge to leak away into the air by corona discharge. A highly polished surface has no such points...
5207012 marksHideAnswer
Answer any three questions: (a) Define work. Derive the work done by a variable force. (b) What is gravitational PE? Obtain an expression for it at distance r from Earth's centre. (c) Define terminal velocity and describe determining viscosity using Stokes' law. (d) Derive a relation between angular momentum and moment of inertia and hence define moment of inertia.
(a) Work is the product of force and the displacement in the direction of the force, $W = \vec F\cdot\vec s$ . For a variable force, the work done over a small displacement $dx$ is $dW = F\,dx$ , so the total work is the integral

$W = \int_{x_1}^{x_2} F\,dx$

which equals the area under the force-displacement graph.

(b) The gravitational potential energy of a body is the work done in bringing it from infinity to a point in the gravitational field. Bringing a mass $m$ from infinity to a distance $r$ from Earth's centre (mass $M$ ),

$U = -\int_\infty^r \frac{GMm}{x^2}dx$

Evaluating the integral,

$\boxed{U = -\frac{GMm}{r}}$

The energy is negative because gravity is attractive, and it rises toward zero as $r$ approaches infinity.

(c) Terminal velocity is the constant maximum velocity a body attains while falling through a viscous fluid, reached when its weight is balanced by the upthrust plus the viscous drag. At terminal velocity for a sphere of radius $a$ and density $\rho$ in a fluid of density $\sigma$ and viscosity $\eta$ , the weight less upthrust equals the Stokes drag,

$\tfrac{4}{3}\pi a^3(\rho-\sigma)g = 6\pi\eta a v_t$

Solving for the terminal velocity,

$v_t = \frac{2a^2(\rho-\sigma)g}{9\eta}$

Measuring $v_t$ over a known distance for a ball of known $a$ and $\rho$ then gives the viscosity,

$\boxed{\eta = \frac{2a^2(\rho-\sigma)g}{9v_t}}$

(d) Summing the angular momentum of every particle, with $v_i = \omega r_i$ ,
$\begin{aligned} L &= \sum m_i v_i r_i \\ &= \sum m_i(\omega r_i)r_i \\ &= \left(\sum m_i r_i^2\right)\omega \end{aligned}$
Recognising the bracket as the moment of inertia $I = \sum m_i r_i^2$ ,

$\boxed{L = I\omega}$

The moment of inertia is thus the rotational analogue of mass: $I = L/\omega$ , the angular momentum per unit angular velocity, which measures a body's resistance to change in its rotational motion.
(a) Work is the product of force and the displacement in the direction of the force, $W = \vec F\cdot\vec s$. For a variable force, the work done over a small displacement $dx$ is $dW = F,dx$, so the total work is the integral

$$W = \int_{x_1}^{x_2} F,dx$$

which equals the area under the force-displacement graph.

(b) The gravitational potential energy of a body is the work done in bringing it from infinity to a point in the gravitational field. Bringing a mass $m$ from infinity to a distance $r$ from Earth's centre (mass $M$),

$$U = -\int_\infty^r \frac{GMm}{x^2}dx$$

Evaluating the integral,

$$\boxed{U = -\frac{GMm}{r}}$$

The energy is negative because gravity is attractive, and it rises toward zero as $r$ approaches infinity.

(c) Terminal velocity is the constant maximum velocity a body attains while falling through a viscous fluid, reached when its weight is balanced by the upthrust plus the viscous drag. At terminal velocity for a sphere of radius $a$ and density $\rho$ in a fluid of density $\sigma$ and viscosity $\eta$, the weight less upthrust equals the Stokes drag,

$$\tfrac{4}{3}\pi a^3(\rho-\sigma)g = 6\pi\eta a v_t$$

Solving for the terminal velocity,

$$v_t = \frac{2a^2(\rho-\sigma)g}{9\eta}$$

Measuring $v_t$ over a known distance for a ball of known $a$ and $\rho$ then gives the viscosity,

$$\boxed{\eta = \frac{2a^2(\rho-\sigma)g}{9v_t}}$$

(d) Summing the angular momentum of every particle, with $v_i = \omega r_i$,

$$ \begin{aligned} L &= \sum m_i v_i r_i \ &= \sum m_i(\omega r_i)r_i \ &= \left(\sum m_i r_i^2\right)\omega \end{aligned} $$

Recognising the bracket as the moment of inertia $I = \sum m_i r_i^2$,

$$\boxed{L = I\omega}$$

The moment of inertia is thus the rotational analogue of mass: $I = L/\omega$, the angular momentum per unit angular velocity, which measures a body's resistance to change in its rotational motion.
620708 marksHideAnswer
Answer any two questions: (a) Define coefficient of real and apparent expansion of a liquid and derive a relation between them. (b) What is a perfect gas? Prove the average KE of a gas molecule is proportional to absolute temperature. (c) Derive the work done during adiabatic expansion of an ideal gas. Does internal energy change?
(a) The coefficient of real expansion $\gammar$ is the fractional increase in the true volume of a liquid per degree rise in temperature, while the coefficient of apparent expansion $\gammaa$ is the fractional increase we actually observ...
(a) The coefficient of real expansion $\gammar$ is the fractional increase in the true volume of a liquid per degree rise in temperature, while the coefficient of apparent expansion $\gammaa$ is the fractional increase we actually observ...
720704 marksHideAnswer
Answer any one question: (a) What is lateral shift? Derive its value. How does it change with angle of incidence? (b) What is chromatic aberration? Show that for a lens it equals the product of dispersive power and focal length of mean light.
(a) The lateral shift is the perpendicular displacement between the incident ray (produced straight through) and the emergent ray when light passes through a parallel-sided glass slab. The emergent ray comes out parallel to the incident ...
820704 marksHideAnswer
Answer any one question: (a) What is electrostatic induction? How can you charge a body positively by induction? (b) State and explain Gauss's law and use it to find the electric field due to a line charge.
(a) Electrostatic induction is the redistribution of a conductor's charge caused by a nearby charged body, without any contact. To charge a body positively using a negative rod: bring the rod near the conductor, so the near face becomes ...
9207012 marksNumericalHideAnswer
Answer any three numerical questions: (a) A 6 kg box is pushed at constant speed 0.35 m/s with mu_k = 0.12; find F. Then find F for constant acceleration 0.18 m/s^2. (b) A string breaks at 25 N; a 500 g mass on a 1 m string is rotated in a horizontal circle. Find the greatest rpm without breaking. (c) Work done in stretching a steel wire 100 cm long, area 0.03 cm^2, load 100 N. (d) A body in SHM: amplitude 15 cm, frequency 4 Hz. Find maximum acceleration and velocity.
(a) While the $6\ \text{kg}$ box moves at constant speed the applied force only has to overcome kinetic friction, so
$\begin{aligned} F &= \mu_k mg \\ &= 0.12\times6\times9.8 \\ &= 7.06\ \text{N} \end{aligned}$
To give it an acceleration of $0.18\ \text{m/s}^2$ instead, the force must also supply $ma$ on top of friction:
$\begin{aligned} F &= ma + \mu_k mg \\ &= 6(0.18) + 7.06 \\ &= 8.14\ \text{N} \end{aligned}$
So $F = 7.06\ \text{N}$ keeps it moving steadily, and $F = 8.14\ \text{N}$ accelerates it.

(b) The string tension supplies the centripetal force, $T = m\omega^2 r$ , and it snaps once $T$ reaches $25\ \text{N}$ . With $m = 0.5\ \text{kg}$ and $r = 1\ \text{m}$ the greatest angular speed follows from
$\begin{aligned} \omega^2 &= \frac{T}{mr} \\ &= \frac{25}{0.5\times1} \\ &= 50 \end{aligned}$
so $\omega = 7.07\ \text{rad/s}$ . Converting to revolutions per minute,
$\begin{aligned} n &= \frac{\omega}{2\pi} \\ &= \frac{7.07}{6.283} \\ &= 1.125\ \text{rev/s} \\ &= 67.5\ \text{rev/min} \end{aligned}$
The wire can therefore be spun at most about $67.5$ times per minute.

(c) The $100\ \text{N}$ load first stretches the wire ( $L = 1\ \text{m}$ , $A = 3\times10^{-6}\ \text{m}^2$ , $Y = 2\times10^{11}\ \text{N/m}^2$ ) by
$\begin{aligned} x &= \frac{FL}{AY} \\ &= \frac{100\times1}{3\times10^{-6}\times2\times10^{11}} \\ &= 1.67\times10^{-4}\ \text{m} \end{aligned}$
Because the tension builds up gradually from zero, the work stored is $\tfrac12 Fx$ :
$\begin{aligned} W &= \tfrac12(100)(1.67\times10^{-4}) \\ &= 8.3\times10^{-3}\ \text{J} \end{aligned}$
so the work done is about $8.3\times10^{-3}\ \text{J}$ (roughly $8.3\ \text{mJ}$ ).

(d) For the SHM the angular frequency is $\omega = 2\pi f = 25.13\ \text{rad/s}$ , with amplitude $A = 0.15\ \text{m}$ . The acceleration is greatest at the extreme position:
$\begin{aligned} a_{max} &= \omega^2 A \\ &= (25.13)^2(0.15) \\ &= 94.7\ \text{m/s}^2 \end{aligned}$
and the velocity is greatest at the mean position:
$\begin{aligned} v_{max} &= \omega A \\ &= 25.13\times0.15 \\ &= 3.77\ \text{m/s} \end{aligned}$
So $a_{max} \approx 94.7\ \text{m/s}^2$ and $v_{max} \approx 3.77\ \text{m/s}$ .
(a) While the $6\ \text{kg}$ box moves at constant speed the applied force only has to overcome kinetic friction, so

$$ \begin{aligned} F &= \mu_k mg \ &= 0.12\times6\times9.8 \ &= 7.06\ \text{N} \end{aligned} $$

To give it an acceleration of $0.18\ \text{m/s}^2$ instead, the force must also supply $ma$ on top of friction:

$$ \begin{aligned} F &= ma + \mu_k mg \ &= 6(0.18) + 7.06 \ &= 8.14\ \text{N} \end{aligned} $$

So $F = 7.06\ \text{N}$ keeps it moving steadily, and $F = 8.14\ \text{N}$ accelerates it.

(b) The string tension supplies the centripetal force, $T = m\omega^2 r$, and it snaps once $T$ reaches $25\ \text{N}$. With $m = 0.5\ \text{kg}$ and $r = 1\ \text{m}$ the greatest angular speed follows from

$$ \begin{aligned} \omega^2 &= \frac{T}{mr} \ &= \frac{25}{0.5\times1} \ &= 50 \end{aligned} $$

so $\omega = 7.07\ \text{rad/s}$. Converting to revolutions per minute,

$$ \begin{aligned} n &= \frac{\omega}{2\pi} \ &= \frac{7.07}{6.283} \ &= 1.125\ \text{rev/s} \ &= 67.5\ \text{rev/min} \end{aligned} $$

The wire can therefore be spun at most about $67.5$ times per minute.

(c) The $100\ \text{N}$ load first stretches the wire ($L = 1\ \text{m}$, $A = 3\times10^{-6}\ \text{m}^2$, $Y = 2\times10^{11}\ \text{N/m}^2$) by

$$ \begin{aligned} x &= \frac{FL}{AY} \ &= \frac{100\times1}{3\times10^{-6}\times2\times10^{11}} \ &= 1.67\times10^{-4}\ \text{m} \end{aligned} $$

Because the tension builds up gradually from zero, the work stored is $\tfrac12 Fx$:

$$ \begin{aligned} W &= \tfrac12(100)(1.67\times10^{-4}) \ &= 8.3\times10^{-3}\ \text{J} \end{aligned} $$

so the work done is about $8.3\times10^{-3}\ \text{J}$ (roughly $8.3\ \text{mJ}$).

(d) For the SHM the angular frequency is $\omega = 2\pi f = 25.13\ \text{rad/s}$, with amplitude $A = 0.15\ \text{m}$. The acceleration is greatest at the extreme position:

$$ \begin{aligned} a_{max} &= \omega^2 A \ &= (25.13)^2(0.15) \ &= 94.7\ \text{m/s}^2 \end{aligned} $$

and the velocity is greatest at the mean position:

$$ \begin{aligned} v_{max} &= \omega A \ &= 25.13\times0.15 \ &= 3.77\ \text{m/s} \end{aligned} $$

So $a_{max} \approx 94.7\ \text{m/s}^2$ and $v_{max} \approx 3.77\ \text{m/s}$.
1020708 marksNumericalHideAnswer
Answer any two numerical questions: (a) A 300 g copper calorimeter has 500 g water at 15 C; a 560 g aluminium ball at 100 C is dropped, temperature rises to 22 C. Find specific heat of aluminium. (b) A rod length 20 cm, area 3.14 cm^2, one end 100 C, other in ice 0 C; 25 g ice melts in 5 min. Find thermal conductivity. (c) A petrol engine consumes 25 kg petrol/hr, calorific value 11.4x10^6 cal/kg, power 99.75 kW. Find efficiency.
(a) An aluminium ball of mass $m = 560\ \text{g} = 0.560\ \text{kg}$ at $100^\circ$ C is dropped into a copper calorimeter ( $300\ \text{g} = 0.300\ \text{kg}$ ) holding water ( $500\ \text{g} = 0.500\ \text{kg}$ ) at $15^\circ$ C, and the mix...
(a) An aluminium ball of mass $m = 560\ \text{g} = 0.560\ \text{kg}$ at $100^\circ$C is dropped into a copper calorimeter ($300\ \text{g} = 0.300\ \text{kg}$) holding water ($500\ \text{g} = 0.500\ \text{kg}$) at $15^\circ$C, and the mix...
1120704 marksNumericalHideAnswer
A refracting telescope has an objective of focal length 1 m and an eyepiece of focal length 2 cm. A real image of the sun, 10 cm in diameter, is formed on a screen 24 cm from the eyepiece. What angle does the sun subtend at the objective?
Two lenses act in series here: the objective forms a small real image of the sun, and the eyepiece then projects that image, enlarged, onto the screen. By working back from the screen we can find the size of the small image, and from its size the angle the sun subtends.

We are given the objective focal length $f_o = 1\ \text{m}$ , the eyepiece focal length $f_e = 2\ \text{cm}$ , and the final real image of the sun, $H = 10\ \text{cm}$ tall, formed at $v = 24\ \text{cm}$ from the eyepiece. The objective forms its image at its focal plane, where the image height is $h_i = f_o\,\alpha$ and $\alpha$ is the angle the sun subtends at the objective. That image is the object for the eyepiece, so once we know $h_i$ we can find $\alpha$ .

For the eyepiece we use $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ with $f_e = 2\ \text{cm}$ and $v = 24\ \text{cm}$ :
$\begin{aligned} \frac{1}{u} &= \frac{1}{2} - \frac{1}{24} \\ &= \frac{11}{24} \end{aligned}$
so $u = 2.18\ \text{cm}$ , and the eyepiece magnification is
$\begin{aligned} m &= \frac{v}{u} \\ &= \frac{24}{2.18} \\ &= 11 \end{aligned}$
The $10\ \text{cm}$ screen image is $m$ times the eyepiece's object, so the intermediate image formed by the objective was
$\begin{aligned} h_i &= \frac{H}{m} \\ &= \frac{10}{11} \\ &= 0.909\ \text{cm} \\ &= 9.09\times10^{-3}\ \text{m} \end{aligned}$
Using $h_i = f_o\,\alpha$ , the angle the sun subtends at the objective is
$\begin{aligned} \alpha &= \frac{h_i}{f_o} \\ &= \frac{9.09\times10^{-3}}{1} \\ &= 9.09\times10^{-3}\ \text{rad} \\ &= 0.52^\circ \end{aligned}$
Therefore the sun subtends about $9.1\times10^{-3}$ rad (about $0.52^\circ$ ) at the objective.
Two lenses act in series here: the objective forms a small real image of the sun, and the eyepiece then projects that image, enlarged, onto the screen. By working back from the screen we can find the size of the small image, and from its size the angle the sun subtends.

We are given the objective focal length $f_o = 1\ \text{m}$, the eyepiece focal length $f_e = 2\ \text{cm}$, and the final real image of the sun, $H = 10\ \text{cm}$ tall, formed at $v = 24\ \text{cm}$ from the eyepiece. The objective forms its image at its focal plane, where the image height is $h_i = f_o,\alpha$ and $\alpha$ is the angle the sun subtends at the objective. That image is the object for the eyepiece, so once we know $h_i$ we can find $\alpha$.

For the eyepiece we use $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ with $f_e = 2\ \text{cm}$ and $v = 24\ \text{cm}$:

$$ \begin{aligned} \frac{1}{u} &= \frac{1}{2} - \frac{1}{24} \ &= \frac{11}{24} \end{aligned} $$

so $u = 2.18\ \text{cm}$, and the eyepiece magnification is

$$ \begin{aligned} m &= \frac{v}{u} \ &= \frac{24}{2.18} \ &= 11 \end{aligned} $$

The $10\ \text{cm}$ screen image is $m$ times the eyepiece's object, so the intermediate image formed by the objective was

$$ \begin{aligned} h_i &= \frac{H}{m} \ &= \frac{10}{11} \ &= 0.909\ \text{cm} \ &= 9.09\times10^{-3}\ \text{m} \end{aligned} $$

Using $h_i = f_o,\alpha$, the angle the sun subtends at the objective is

$$ \begin{aligned} \alpha &= \frac{h_i}{f_o} \ &= \frac{9.09\times10^{-3}}{1} \ &= 9.09\times10^{-3}\ \text{rad} \ &= 0.52^\circ \end{aligned} $$

Therefore the sun subtends about $9.1\times10^{-3}$ rad (about $0.52^\circ$) at the objective.
1220703 marksNumericalHideAnswer
A parallel plate air capacitor of capacitance 245x10^-12 F has a charge of 0.148 uC on each plate. Find the potential difference and electric field intensity if the plate separation is 5 mm.
The capacitor has $C = 245\times10^{-12}\ \text{F}$ with a charge $Q = 0.148\ \mu\text{C} = 0.148\times10^{-6}\ \text{C}$ on each plate, and the plate separation is $d = 5\ \text{mm} = 5\times10^{-3}\ \text{m}$ . The potential difference ...
The capacitor has $C = 245\times10^{-12}\ \text{F}$ with a charge $Q = 0.148\ \mu\text{C} = 0.148\times10^{-6}\ \text{C}$ on each plate, and the plate separation is $d = 5\ \text{mm} = 5\times10^{-3}\ \text{m}$. The potential difference ...

Keep going

Important questions

Chapter notes Full syllabus

Other papers

2075 paper 2073 paper 2072 paper 2069 paper Model questions