# Physics 9702 Doubts | Help Page 38

__Question 234: [Work, Energy, Power]__The potential energy of a body when it is at point P a distance x from a reference point O is given by V = kx

^{2}, where k is a constant. What is the force acting on the body when it is at P?

A 2kx in the direction OP

B kx in the direction OP

C zero

D kx in the direction PO

E 2kx in the direction PO

**Reference:**

*Past Exam Paper – J84 / II / 2*

__Solution 234:__**Answer: E.**

**Force F = - dV / dx**

where V is the potential energy and x is the distance from the reference point O.

V = kx

^{2}

dV / dx = 2kx

Force F = - dV / dx = - 2kx

The potential energy V is from point O to point P (in the direction OP). The negative sign for the force indicates that it is in the opposite direction from that of V. So, the direction of the force is in the direction PO.

__Question 235: [Measurements > Uncertainties]__Student finds density of a liquid by measuring its mass and its volume. Following is a summary of his measurements.

mass of empty beaker = (20 ± 1) g

mass of beaker + liquid = (70 ± 1) g

volume of liquid = (10.0 ± 0.6) cm

He correctly calculates the density of liquid as 5.0 g cm^{3}^{–3}.

What is uncertainty in this value?

A 0.3 g cm

^{–3}B 0.5 g cm

^{–3}C 0.6 g cm

^{–3}D 2.6 g cm

^{–3}

**Reference:**

*Past Exam Paper – June 2010 Paper 12 Q3*

__Solution 235:__**Answer: B.**

Mass, m of the liquid = 70 – 20 = 50g

Uncertainty in mass of liquid = ± (1 + 1) = ± 2g

So, mass of liquid = 50 ± 2 g

Volume, V of liquid = (10.0 ± 0.6) cm

^{3}

**Density, ρ = m / V**

**Δρ / ρ = Δm/m + ΔV/V**

Δρ = ρ [Δm/m + ΔV/V] = 5 [(2/50) + (0.6/10)] = 0.5gcm

^{-3}

__Question 236: [Work, Energy, Power]__A constant force is applied to a body which is initially stationary but free to move in the direction of the force. Assuming that the effects of friction are negligible, which of the following graphs best represents the variation of P, the power supplied, with time t?

**Reference:**

*Past Exam Paper – N89 / I / 4*

__Solution 236:__**Answer: E.**

Power = Work done / time = Force x Velocity = Fv

Applying a force to a body causes the body to accelerate. The velocity of an accelerating body increases with time. [So, the power (= Fv) increases with time] [A and B are incorrect]

Since the force is constant, the acceleration is also constant. So, the increase in velocity is constant with time. Thus, (since the power = Fv and F is constant,) the increase in power is constant with time.

The change in power P with respect to time t is represented by the gradient of the P-t graph. So, the graph should have a constant gradient. [C and D are incorrect]

__Question 237: [Electric field > Potential]__**(a)**Define electric potential at a point

**(b)**A charged particle is accelerated from rest in vacuum through potential difference V. Show that final speed v of particle is given by expression

v = √(2Vq / m)

where (q/m) is ration of charge to
mass (the specific charge) of the particle.

**(c)**A particle with specific charge +9.58 × 10

^{7}C kg

^{–1}is moving in vacuum towards fixed metal sphere, as illustrated in Fig.

Initial speed of particle is 2.5 ×
10

^{5}m s^{–1}when it is a long distance from sphere.
Sphere is positively charged and has
potential of +470 V.

Use expression in (b) to determine
whether particle will reach surface of the sphere.

**Reference:**

*Past Exam Paper – June 2013 Paper 41 & 43 Q4*

__Solution 237:__**(a)**

The electric potential at a point is
the work done moving unit positive charge from infinity (to the point).

**(b)**

{As the particle moves in
the vacuum its electric potential energy changes.}

From the conservation of energy,

(Gain in) kinetic energy of particle
= change (loss) in potential energy of particle

½ mv

^{2}= qV leading to speed v = √(2Vq / m)**(c)**

{This question can be
answered in different methods, all of which consisting of calculating a
quantity and compare it with the corresponding data given in the question. The
question gives data for specific charge (q/m), speed (v) and potential (V).}

Note that the equation
v = √(2Vq / m) gives the final speed of a particle of mass m and
charge q being accelerated from rest in a vacuum through a potential difference
V.

{Both the particle and the
sphere are positively charged. So, they would repel each other. So, work should
be done (energy is required) to bring the particle towards the sphere. This would cause the
particle to lose its kinetic energy (this energy is taken from the kinetic energy of the particle) – its speed decreases.
So, if the kinetic energy becomes zero (its speed becomes zero)

__before__the particle reaches the surface of the metal sphere, then the particle will not reach the surface of the sphere.}**EITHER**

{In this method, we
calculate the

__maximum__potential V (remember that the potential V is reducing the speed of the particle) of the sphere such that the speed of the particle (of initial velocity 2.5x10^{5}ms^{-1}) is just zero at the surface of the sphere (that is it reaches the sphere). So, if the given value of V is greater, then the speed of the particle becomes zero before reaching the surface and if V is smaller, then the particle will still have some speed at the surface – it reaches the surface of the sphere. From the equation for the velocity, shown in (b)}
(2.5 × 10

^{5})^{2}= 2 × V × 9.58 × 10^{7}
Potential V = 330 V

This (the
maximum potential possible for which the particle can reach the surface)
is less than 470 V and so ‘no’ (the particle does not
reach the surface)

**OR**

{In this method, we
calculate the minimum initial speed that the particle needs to have in order to
reach the surface of a metal sphere at a potential of 470 V. This minimum speed corresponds to the minimum
kinetic energy required for the particle to reach the surface. The higher the
initial speed, the higher is the initial kinetic energy. So, if the value given
for the velocity in the question (2.5x10

^{5}ms^{-1}) is less than the calculated value, it means that the particle does not have enough kinetic energy to reach the surface. If the velocity given is higher than that calculated, then the particle reaches the surface. From the equation for the velocity, shown in (b)}
v = (2 × 470 × 9.58 × 10

^{7})^{0.5}
Speed v = 3.0 × 10

^{5}m s^{–1}
This (minimum
speed required for the particle to reach the surface) is greater than
2.5 × 10

^{5}m s^{–1}and so ‘no’ (the particle does not reach the surface)**OR**

{How does the specific
charge relate to the velocity and the potential. As explained above, the
velocity of the particle relates to its kinetic energy. Since both the particle
and the metal sphere are positively charged (the metal sphere has a positive
potential), they tend to repel each other, preventing the particle from
reaching the surface if it does not have enough kinetic energy. The specific
charge is the ratio of charge to mass. That is, the greater the specific
charge, the greater will its charge (compared to its mass) be. So, the greater
the specific charge, the greater is the repulsion. So, we need to find the
maximum specific charge of a particle to reach the surface in the situation
described in the question.}

{In this method, we
calculate the maximum charge-to-mass ratio (specific charge) of a positively-charged
particle, with initial velocity 2.5x105ms

^{-1}, to reach the surface of a metal sphere at a potential of 470V. If the specific charge given in the question (9.58x10^{7}Ckg^{-1}) is greater than the calculated value, then the particle will not reach the surface. If the specific charge given is less than the calculated value, then the particle reaches the surface of the metal sphere. From the equation for the velocity, shown in (b)}
(2.5 × 10

^{5})^{2}= 2 × 470 × (q/m)
Specific charge, (q/m) = 6.6 × 10

^{7}C kg^{–1}
This (maximum
value of specific charge for the particle to reach the surface) is less
than 9.58 × 10

^{7}C kg^{–1}and so ‘no’ (the particle does not reach the surface)

__Question 238: [Matter]__**(a)**Define density

**(b)**Liquid of density ρ fills container to a depth h, as illustrated in Fig.

Container has vertical sides and
base of area A.

(i) State, in terms of A, h and ρ,
the mass of liquid in container.

(ii) Hence derive an expression for
pressure p exerted by liquid on the base of the container. Explain your
working.

**(c)**Density of liquid water is 1.0 g cm

^{–3}. Density of water vapour at atmospheric pressure is approximately 1/1600 gcm

^{-3}.

Determine ratio

(i) volume of water vapour to volume
of equal mass of liquid water

(ii) mean separation of molecules in
water vapour to mean separation of molecules in liquid water

**(d)**State evidence for

(i) molecules in solids and liquids
having approximately same separation

(ii) strong rigid forces between
molecules in solids.

**Reference:**

*Past Exam Paper – June 2007 Paper 2 Q3*

__Solution 238:__**(a)**Density is defined as the mass per unit volume

**(b)**

(i) Mass of liquid in container = Ahρ

(ii)

Pressure p = Force / Area

Force (on the base) {= weight of
liquid} = Ahρg

Pressure p {= Ahρg / A} = hρg

**(c)**

(i)

{Density = mass / volume.
Volume = mass / density. For comparison, the same mass should be used in both
states. Let the mass be 1 in both cases.

Volume of liquid water = 1
/ 1 = 1cm

^{3}
Volume of water vapour = 1
/ 1600 cm

^{3}
Ratio of vol of water
vapour to volume of liquid water = (1/1600) / 1 = 1600}

Ratio = 1600 or 1600:1

(ii)

{Ratio of volumes = 1600.
This is in 3 dimensions. So, the linear separation whici is in 1 dimension is
given by the cube-root of the ratio of volumes. [Units of volume: m

^{3}. Unit of separation: m]}
Ratio = [1600]

^{1/3}= 11.7**(d)**

(i) The density of solids are
liquids are (about) equal.

(ii) The forces being strong result
a fixed volume. The forces being rigid cause the solid to retain its shape / do
not flow / little deformation.

__Question 239: [Matter > Young modulus]__Two wires each have length 1.8 m and diameter 1.2 mm. One wire has Young modulus 1.1 x 10

^{2}Pa and the other has 2.2 x 10

^{11}Pa. One end of each wire is attached to same fixed point and the other end of each is attached to same load of 75 N so that each has same extension. Calculate the extension of the wires.

**Reference:**

*???*

__Solution 239:__Young modulus, E = Stress / Strain

Stress = Force, F / (cross-sectional) Area, A

Strain = extension / original length = e / L

Young modulus, E = (F/A) / (e/L) = FL / Ae

Extension, e = FL / AE

For both wires:

Original length, L = 1.8m

Diameter, d = 1.2x10

^{-3}m

Cross-sectional area, A = πd

^{2}/4 = 1.131x10

^{-6}m

^{2}

Force = 75N

The springs are in parallel and are attached to the same load of 75N.

Consider 2 springs, of spring constant k

_{1}and k

_{2}, attached to a load in parallel. The effective ‘spring constant’ of the combination is given by k

_{eff}= k

_{1}+ k

_{2}

From Hooke’s law: Force, F = ke where k is the spring constant.

So, extension, e = F / k = FL / AE

1 / k = L / AE giving the spring constant k = AE / L

For wire of Young modulus E = 1.1x10

^{11}Pa,

Spring constant, k

_{1}= A (1.1x10

^{11}) / L

For wire of Young modulus E = 2.2x10

^{11}Pa,

Spring constant, k

_{2}= A (2.2x10

^{11}) / L

Effective spring constant, k

_{eff}= k

_{1}+ k

_{2}= A (1.1x10

^{11}+ 2.2x10

^{11}) / L

Hooke’s law (for the combination): F = k

_{eff }e

Extension, e = F / keff = [75 x 1.8] / [(1.131x10

^{-6}) (1.1x10

^{11}+ 2.2x10

^{11})]

Extension, e = 3.617x10

^{-4}m

__Question 240: [Waves > Double Slit experiment]__**(a)**Fig shows variation with time t of displacement y of wave W as it passes point P. Wave has intensity I.

Second wave X of same frequency as
wave W also passes point P. Wave has intensity ½I. Phase difference between the
2 waves is 60°. On Fig, sketch variation with time t of displacement y of wave
X.

**(b)**In double-slit interference experiment using light of wavelength 540 nm, separation of slits is 0.700 mm. Fringes are viewed on screen at distance of 2.75 m from double slit, as illustrated.

Calculate separation of
fringes observed on screen.

**(c)**State effect, if any, on appearance of fringes observed on screen when following changes are made, separately, to double-slit arrangement in (b):

(i) Width of each slit increased but
separation remains constant:

(ii) Separation of slits increased:

**Reference:**

*Past Exam Paper – November 2007 Paper 2 Q5*

__Solution 240:__**(a)**

The amplitude should be between 6.5
squares and 7.5 squares on 3 peaks

{Intensity, I is
proportional to (amplitude, A)

^{2}. So, A^{2}prop to I. For half I, amplitude prop to (½)^{0.5}I = 0.71I. For wave W, intensity I is represented by 10 squares. So, X should be represented by about 7 squares}
Correct phase

{Phase diff of 180

^{o}= half wavelength which is represented by 15 squares on the graph. So, a phase diff of 60^{o}= 15 x (60/180) = 5 squares. So, graph of X should be lead or lag by amount of 5 squares}**(b)**

λ = ax / D

540x10

^{-9}= (0.700x10^{-3})x / 2.75
Separation of fringes, x = 2.12mm

**(c)**

(i)

The separation of the fringes
remains the same

+ Choose any 2:

The bright areas are brighter

No change occur to the dark areas

Fewer fringes are observed

(ii)

Separation of slits increased:

The separation of the fringes is
smaller

No change occurs in the brightness

I need detailed solutions for the following please:

ReplyDelete02/M/J/07 Q.3(c)

02/O/N/07 Q.5(a)

02/O/N/08 Q.6(b)

21/M/J/09 Q.5

22/M/J/09 Q.2(b)

21/O/N/09 Q.2(b), Q.5(c)

21/M/J/10 Q.2(b)(ii),Q.3(b)(iii)

22/M/J/10 Q.4(b)(ii),(c),Q.5(a)(ii)

22/O/N/10 Q.2(b)

21/M/J/11 Q.1(b)

22/M/J/11 Q.2(b)(ii)

11/M/J/12 Q.12

11/M/J/13 Q.12

12/M/J/13 Q.9

13/M/J/13 Q.11

J2007 P2 Q3 is explained above. The others will be explained gradually (not all at once). Check again regularly.

DeleteN200& P2 Q5 is also explained above

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The June 2009 Paper 2s are already solved at

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http://physics-ref.blogspot.com/2014/11/9702-june-2009-paper-22-worked.html

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N2009 P2 Q2 is explained at

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N2009 P2 Q5 is explained at

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Links for June 2010 P21, P22 and November 2010 P22 are already available. Check at the respective pages.

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