# Physics 9702 Doubts | Help Page 164

__Question 818: [Work, Energy and Power]__
A body travelling with speed of 20 m
s

^{–1}has kinetic energy E_{k}.
If speed of the body is increased to
80 m s

^{–1}, what is its new kinetic energy?
A 4E

_{k}B 8E_{k}C 12E_{k}D 16E_{k}**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q18*

__Solution 818:__**Answer: D.**

Initial Kinetic energy = ½ mv

^{2}= ½ m(20)^{2}= E_{k}
Final Kinetic energy = ½ mv

^{2}= ½ m(80)^{2}
We need to write the final kinetic
energy in terms of E

_{k}which is itself equal to ½ m(20)^{2}.
Final Kinetic energy = ½ m (4×20)

^{2}= 16 × [½ m(20)^{2}] = 16 E_{k}

__Question 819: [Dynamics > Newton’s laws of motion]__
Water is pumped through hose-pipe at
a rate of 90 kg per minute. It emerges from the hose-pipe horizontally with
speed of 20 m s

^{–1}.
Which force is required from a
person holding the hose-pipe to prevent it moving backwards?

A 30 N B 270 N C
1800 N D 10 800 N

**Reference:**

*Past Exam Paper – November 2014 Paper 11 & 12 Q8*

__Solution 819:__**Answer: A.**

From Newton’s 2

^{nd}law,
Force = rate of change of momentum =
Δp / t

The emerging water has some momentum
and hence exerts a force on the person. From Newton’s 3

^{rd}law, the person should exert a force similar in magnitude but in the opposite direction to prevent the hose-pipe from moving backwards.
Force = rate of change of momentum =
Δp / t

In this case, the water is emerging
a constant speed, but the water is being pumped at a specific rate.

Δp / t = (Δm)v / t = (Δm / t) v

Δm / t = rate of water being pumped
= 90kg / min

1 min = 60s corresponds to 90kg of
water

1s corresponds to 90 / 60 = 1.5kg

Rate of water being pumped = Δm / t
= 1.5kg s

^{-1}
Force = Δp / t = (Δm / t) v = 1.5 × 20 = 30N

__Question 820: [Radioactivity]__
The first artificial radioactive
substance was made by bombarding aluminium,

^{27}_{13}Al, with α-particles. This produced an unstable isotope of phosphorus,^{30}_{15}P.
What was the by-product of this
reaction?

A an α-particle

B a β-particle

C a neutron

D a proton

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q40*

__Solution 820:__**Answer: C.**

In a reaction, the mass number and
proton number should be conserved.

^{27}

_{13}Al +

^{4}

_{2}α - - - >

^{30}

_{15}P +

^{Z}

_{A}X

27 + 4 = 30 + Z giving Z = 1

13 + 2 = 15 + A giving A = 0

^{1}

_{0}X is a neutron (

^{1}

_{0}n).

__Question 821: [Electric field]__
In diagram, the shaded area
represents a uniform electric field directed away from the observer (at
right-angles into the plane of the paper).

A horizontal beam of electrons
enters the field, travelling from left to right.

In which direction is this beam
deflected by the field?

A upwards (in the plane of the
paper)

B downwards (in the plane of the
paper)

C away from the observer

D towards the observer

**Reference:**

*Past Exam Paper – June 2010 Paper 11 Q29 & Paper 12 Q27 & Paper 13 Q26*

__Solution 821:__**Answer: D.**

Electric field is drawn from
positive to negative.

Since the electric field is said to
be into the plane of the paper, the negative is in the paper while positive is
out of the paper.

An electron, which is negatively
charged, is attracted to the positive. So the deflection of the electron is
towards observer (the field is directed away from the observer, so the position
of the positive charge is at the position of the observer).

__Question 822: [Current of Electricity]__
What describes the electric
potential difference between two points in wire that carries a current?

A the force required to move a unit
positive charge between the points

B the ratio of the energy dissipated
between the points to the current

C the ratio of the power dissipated
between the points to the current

D the ratio of the power dissipated
between the points to the charge moved

**Reference:**

*Past Exam Paper – November 2004 Paper 1 Q31 & June 2011 Paper 12 Q32*

__Solution 822:__**Answer: C.**

Power P dissipated = VI

p.d. V = P / I

The electric potential difference
between two points in wire that carries a current is described as the ratio of
the power dissipated between the points to the current. [C is correct]

Current and charge are different
quantities. [D is incorrect]

Power dissipated and energy dissipated
are different quantities. [B is incorrect]

__Question 823: [Pressure]__
The Mariana Trench in the Pacific
Ocean has a depth of about 10 km.

Assuming that sea water is
incompressible and has density of about 1020 kg m

^{–3}, what would be the approximate pressure at that depth?
A 10

^{5}Pa B 10^{6}Pa C 10^{7}Pa D 10^{8}Pa**Reference:**

*Past Exam Paper – June 2014 Paper 13 Q22*

__Solution 823:__**Answer: D.**

Pressure = hρg = 10000 ×1020 × 9.81 = 1×10

^{8 }Pa

__Question 824: [Matter > Deformation > Young modulus]__
A lift is supported by two steel
cables, each of length 10 m and diameter 0.5 cm.

The lift drops 1 mm when a man of
mass 80 kg steps into the lift.

What is the best estimate of value
of the Young modulus of the steel?

A 2 × 10

^{10}N m^{–2}
B 4
× 10

^{10}N m^{–2}
C 2 × 10

^{11}N m^{–2}
D 4 × 10

^{11}N m^{–2}**Reference:**

*Past Exam Paper – November 2012 Paper 13 Q25*

__Solution 824:__**Answer: C.**

Young modulus = Stress / Strain

Stress = Force F / Area A

Strain = extension e / Original
length L

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

The lift is supported by 2 wires
which are in parallel to each other. Both wires are made of steel, so they have
the same spring constant, k.

For wires in parallel,

Effective spring constant, k

_{eff}= k + k = 2k
Hooke’s law: F = k

_{eff}e
Weight of the person is
approximately = mg = 80 × 10 = 800N

Effective spring constant, k

_{eff}= F / e = 800 / 0.001 = 8.0×10^{5}Nm^{-1}
Effective spring constant, k

_{eff}= 2k = 8.0×10^{5}Nm^{-1}
Spring constant of 1 steel wire = 8.0×10

^{5}/ 2 = 4.0×10^{5}Nm^{-1}
For only 1 wire,

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

Spring constant, k = F / e = 4.0×10

^{5}Nm^{-1}
Young modulus = (F/e) (L/A) = kL / A
= (4.0×10

^{5}) × 10 / (π(0.005)^{2}/4) = 2 × 10^{11}N m^{–2}
OR Simply,

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

Since there are 2 wires, the total
cross-sectional area = 2[πd

^{2}/ 4] = πd^{2}/ 2
Young modulus = 800 × 10 / [(π(0.005)

^{2 }/ 2) × 0.001] = 2 × 10^{11}N m^{–2}
In solution 821 how can I know the uniform electric field is into the plane of the paper?

ReplyDeleteI suppose it would be mentioned in the question and indicated by a X.

DeleteAway from paper is indicated by a circle containing a dot

Solution 820: answer is C

ReplyDeleteWhere Z=1 and A=0, the particle is not a proton but it's a neutron.

corrected.

Deletethx