# Physics 9702 Doubts | Help Page 90

__Question 462: [Waves > Stationary waves]__Speed of a transverse wave on a stretched string can be changed by adjusting tension of the string. A stationary wave pattern is set up on stretched string using an oscillator set at frequency of 650 Hz.

How must wave be changed to maintain the same stationary wave pattern if the applied frequency is increased to 750 Hz?

A Decrease the speed of the wave on the string.

B Decrease the wavelength of the wave on the string.

C Increase the speed of the wave on the string.

D Increase the wavelength of the wave on the string.

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q28*

__Solution 462:__**Answer: C.**

One complete wave is obtained (as
shown in the diagram). This corresponds to 1 wavelength.

The length of the string is
unchanged. To maintain the same wave pattern, that is, for only one complete
wave to be obtained again, the wavelength of the wave must remain constant.

Speed of wave, v = fλ

The wavelength λ needs to be kept
constant and the frequency is increased to 750Hz.

Wavelength λ = v / f

For λ
to have the same value, the speed of the wave, v, needs to be increased.

__Question 463: [Forces > Hooke’s law]__Spring of unextended length 40 mm is suspended from a fixed point. Load of 16 N is applied to the free end of the spring. This causes spring to extend so that its final length is five times its original length. The spring obeys Hooke’s Law.

What is the energy stored in the spring due to this extension?

A 1.3 J B 1.6 J C 2.6 J D 3.2 J

**Reference:**

*Past Exam Paper – June 2013 Paper 13 Q15*

__Solution 463:__**Answer: A.**

Original (unextended) length, L =
40mm

Load applied, F = 16N

Extension, e = 5(40) – 40 = 160mm

Energy stored in spring = ½ Fx = 0.5
(16) (160x10

^{-3}) = 1.28 = 1.3J

__Question 464: [Kinematics]__Two balls roll from rest down the same slope, but ball X reaches the bottom of the slope in half the time it takes Y. If ball X was accelerating at a ms

^{-2}, then Y's acceleration, in ms

^{-2}, was

A ¼ a B ½ a C 2 a D 4a

__Solution 464:__**Answer: A.**

Consider ball X.

Initial speed, u = 0. Acceleration = a. Time taken = t. Let the distance travelled = s

Equation for uniformly accelerated motion: s = ut + ½ at

^{2}

For ball X: s = 0 + 0.5at

^{2}= 0.5at

^{2}

Now, consider ball Y.

The distance travelled is the same (and equal to s). Initial speed, u = 0. Time taken = 2t. Let the acceleration = x.

Equation for uniformly accelerated motion: s = ut + ½ at

^{2}

For ball Y: s = 0 + 0.5x (2t)

^{2}= 2xt

^{2}

Since the distance s is the same for both cases, we can equate both equations.

2xt

^{2}= 0.5at

^{2}

Acceleration of ball Y, x = 0.5a / 2 = ¼ a

__Question 465: [Matter > Young modulus]__Lift is supported by two steel cables each of length 20 m.

Each of the cables consists of 100 parallel steel wires, each wire of cross-sectional area 3.2 × 10

^{–6}m

^{2}. Young modulus of steel is 2.1 × 10

^{11}N m

^{–2}.

Which distance does lift move downward when a man of mass 70 kg steps into it?

A 0.010 mm B 0.020 mm C 0.10 mm D 0.20 mm

**Reference:**

*Past Exam Paper – November 2013 Paper 13 Q22*

__Solution 465:__**Answer: C.**

Hooke’s law: F = ke

The lift is supported by 2 steel cables, each of which consists of 100 parallel steel wires. So, there is a total of 200 parallel steel wires in parallel.

For parallel spring,

Effective spring constant, k

_{eff}= k

_{1}+ k

_{2}+ k

_{3}+ …

First, we need to find the spring constant, k for 1 steel wire.

For 1 wire,

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

Hooke’s law: F = ke

Young modulus, E = (ke)L / Ae = kL / A

Spring constant, k = EA / L = (2.1 × 10

^{11}) (3.2 × 10

^{–6}) / 20 = 33600 Nm

^{-1}

Effective spring constant, k

_{eff}= 200k

Hooke’s law: F = k

_{eff }e

Mass of person = 70kg. Weight = mg = 700N (take g = 10 ms

^{-2})

Extension, e = F / k

_{eff}= 700 / (200 × 33600) = 0.00010m = 0.10 mm

__Question 466: [Work, Energy, Power]__Diagram shows lift system in which the elevator (mass m

_{1}) is partly counterbalanced by a heavy weight (mass m

_{2}).

At what rate does motor provide energy to the system when elevator is rising at a steady speed v? (g = acceleration of free fall)

A ½ m

_{1}v

^{2}B ½ (m

_{1}– m

_{2})v

^{2}C m

_{1}gv D (m

_{1}– m

_{2})gv

**Reference:**

*Past Exam Paper – November 2010 Paper 11 Q18*

__Solution 466:__**Answer: D.**

The rate at which the motor provides
energy is the power of the motor.

The system works as such: the mass m

_{2}works downwards under gravity, causing the elevator of mass m_{1}to move up. However, the motor also provides some energy in moving the elevator up.
The (gravitational) potential energy
of the masses are changing continuously and at a steady rate, since they are
moving at a steady speed.

GPE of elevator = m

_{1}gh_{1}where h_{1}is its height above the ground
Rate of change of GPE of elevator =
m

_{1}gh_{1}/ t = m_{1}gv
GPE of heavy weight = m

_{2}gh_{2}where h_{2}is its height above the ground
Rate of change of GPE of heavy weight
= m

_{2}gh_{2}/ t = m_{2}gv
Let the power of the motor be P.

m

_{1}gv = m_{2}gv + P
Power of motor = (m

_{1}– m_{2})gv
Choices A and B have units of
energy, not power. If B was correct, then there would have no need for the
heavy weight.

Salam,

ReplyDeleteI got another solution of solution 465,

I used (F/A)/(x/l) and got the write answer. Please respond to whether I can use that or else , will I get the wrong answer another time?

could you provide the complete calculations?

Deletethanks

E=(F/A) / (X/L)

ReplyDeleteF=70*9.81=686.7

Area = 2 * 100 * 3.2x10^-6 = 6.4 * 10^4

stress=686.7/6.4 * 10^4

length = 20

So I used this to solve

It's OK. However when they are not parallel, we cannot calculate the total area as you did.

Deleteok.. And all I wanted to say is thanks for all your support..This site realllllyyy helped me.. Jazakallah

ReplyDeleteNo problems. Alhamdulillah

Delete

ReplyDeletethanks

Great Site.

ReplyDeletein question 466, what do you mean by 'if B was correct,then there would have been no need for the heavy weight'?

ReplyDeleteB is definitely wrong as the system does not only involve KE, but also GPE.

DeleteAnd I seriously wish i had found this site sooner....its so helpful

ReplyDelete