# Physics 9702 Doubts | Help Page 92

__Question 472: [Stationary waves]__
A string fixed at both ends and of
length L is plucked at its midpoint and emits its fundamental note of f

_{1}. When the string is plucked at a different point, the first overtone frequency f_{2}is also produced. Which one of the following correctly gives both f_{2 }/ f_{1}and v, where v is the speed of transverse waves in the string?
f

_{2 }/ f_{1 }v
A 2 f

_{1}L
B 0.5 f

_{1}L
C 2 f

_{2}L
D 0.5 f

_{2}L
E 2 f

_{2}L / 2**Reference:**

*Past Exam Paper – J81 / II / 13*

__Solution 472:__**Answer: C.**

The higher overtone frequencies are
always integral multiple of the fundamental frequency, that is, the overtone
frequencies can be 2f

_{1}, 3f_{1}, 4f_{1}, ….
The first overtone frequency f

_{2}= 2f_{1}
So, the ration f

_{2}/ f_{1}= 2f_{1}/ f_{1}= 2 [B and D are incorrect]
Consider the fundamental frequency.

Since the string is fixed at both
ends, nodes will be found at these ends. The fundamental frequency consist of
only these nodes. It can be seen that half a wave will be formed at this
frequency. That is, when the frequency = f

_{1}, the length of the string, L = λ / 2.
Wavelength λ = 2L

Speed v = f λ = f

_{1}(2L) = 2f_{1}L
Since f

_{2}/ f_{1}= 2,
Fundamental frequency f

_{1}= f_{2}/ 2
Speed v = 2 (f

_{2}/2) L = f_{2}L
At the first overtone frequency, a
complete wave in formed in the string.

__Question 473: [Dynamics > Momentum]__
Isolated system consists of two
bodies on which no external forces act. The two bodies collide with each other
and stick together on impact.

Which row correctly compares total
kinetic energy and total momentum of the bodies before and after the collision?

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q11*

__Solution 473:__**Answer: B.**

For any isolated system, the law of
conservation of momentum applies, that is, the total momentum before collision
is equal to the total momentum after collision. [A
and C are incorrect]

Kinetic energy = ½ mv

^{2}
Let the speeds of the two bodies
before collision be v

_{1}and v_{2}. Let the speed of the 2 bodies after collision be v_{f}.
From the conservation of linear
momentum,

mv

_{1}+ mv_{2}= (2m)v_{f}
Speed v

_{f}= (v_{1}+ v_{2}) / 2
Total kinetic energy before
collision = ½ mv

_{1}^{2}+ ½ mv_{2}^{2}
Total kinetic energy after collision
= ½ (2m) [(v

_{1}+ v_{2}) / 2]^{2}
Therefore, the kinetic energy after
collision is different from that before collision.

__Question 474: [Current of Electricity > Potential difference]__
Six resistors, each of resistance 5
Ω, are connected to 2 V cell of negligible internal resistance.

What is potential difference between
terminals X and Y?

A 2 / 3 V B 8 / 9 V C
4 / 3 V D 2 V

**Reference:**

*Past Exam Paper – November 2002 Paper 1 Q36*

__Solution 474:__**Answer: A.**

To obtain the potential difference
between terminals X and Y, we first need to know the values of the potential at
X and Y.

The positive terminal of the cell
can be taken to be at a potential of 2V and the negative terminal to be at a potential
of 0V.

It can be seen that the circuit
consists of 2 loops, with a junction at the top and one at the bottom. Since there
are no components connected between the upper junction and the positive
terminal of the cell, it can be concluded that the upper junction at the same
potential of 2V. Similarly, the lower junction is at a potential of 0V.

Thus, in any loop, going from up and
down, the potential decreases across each components until it is zero at the
lower junction. Current flows from the positive terminal of the cell to the
negative one. So, following the direction of current flow, the potential
decreases after going across a component.

From Kirchhoff’s law, the sum of
p.d. in any loop is equal to the e.m.f in the circuit.

Consider the first loop (where
terminal X is connected).

From the potential divider equation,

p.d. across the upper resistor = [5
/ (5+5+5)] x 2 = 2/3 V

Note that this is the potential
DIFFERENCE across the upper resistor. It is also the p.d. across each resistor
since they all have the same resistance.

Since the upper junction is at a
potential of 2V,

Potential at terminal X = 2 – 2/3 =
4/3 V

Consider the second loop (where
terminal Y is connected).

From the upper junction to terminal
Y, there are 2 resistors and the p.d. across each is 2/3V.

Potential at terminal Y = 2 – 2 (2/3)
= 2/3 V

Potential difference between X and Y
= 4/3 – 2/3 = 2/3 V

__Question 475: [Measurement > Uncertainty]__
Steel wire is stretched in an
experiment to determine Young modulus for steel.

Uncertainties in the measurements
are given below.

measurement
uncertainty

load
on wire ±2%

length
of wire ±0.2%

diameter
of wire ±1.5%

extension
±1%

What is the percentage uncertainty
in Young modulus?

A 1.3% B 1.8% C
4.7% D 6.2%

**Reference:**

*Past Exam Paper – November 2014 Paper 11 & 12 Q4*

__Solution 475:__**Answer: D.**

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

where F is the load on the wire, L
is the length of wire, A is the cross-sectional area of the wire and e is the
extension

Cross-sectional area of wire = πd

^{2}/ 4
where d is the diameter of the wire

Young modulus, E = 4FL / πd

^{2}e
(ΔE / E) x 100% = [(ΔF / F) + (ΔL / L) + 2(Δd / d) + (Δe / e)] x 100%

Percentage uncertainty, (ΔE / E) x 100% = 2 + 0.2 + 2(1.5) + 1 = 6.2%

.

__Question 476: [Work done]__
Graph is a force-extension graph for
a wire that is being stretched.

How much work needs to be done by
tensile force, to two significant figures, to cause extension of 7.0 mm?

A 0.088 J B 0.12 J C
0.53 J D 120 J

**Reference:**

*Past Exam Paper – June 2012 Paper 12 Q24*

__Solution 476:__**Answer: B.**

The work done by the tensile force
is given by the area under the force-extension graph from extension = 0mm to
7.0mm.

Since the work done is given in
joules, we need to convert the unit of extension from mm to m.

The area can be approximated by the
area of a triangle formed by the x-axis, point (0, 0) and point (3.5, 21) +
area of a trapezium formed as shown.

Area of triangle = ½ (3.5x10

^{-3}) (21) = 0.03675J
At extension = 7mm, force = 25N.

Area of trapezium = ½ (sum of
parallel sides) (height)

Area of trapezium = ½ (21 + 25) ([7.0
– 3.5]x10

^{-3}) = 0.0805J
Total work done = 0.03675 + 0.0805 =
0.11725 = 0.12J

__Question 477: [Pressure]__
The Mariana Trench in 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 – November 2010 Paper 11 Q19*

__Solution 477:__**Answer: D.**

Depth h = 10km = 10 000 m

Density ρ = 1020
kg m

^{–3}
Assume that the acceleration of free
fall, g = 10 ms

^{-2}.
Pressure = hρg = (10 000) (1020) (10) = 1.02x10

^{8}Pa
Can you please uplaod the worked examples of the difficult mcq qustions june 2015 please

ReplyDeleteState which question, which variant.

DeletePlease can you upload the solutions for june mcq 2015 and as well as for paper 2 and paper 4

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