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Physics: Post your doubts here!

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Muskan Achhpilia said:
Can anyone help with the following MCQs
View attachment 46182
Answer A

View attachment 46183
Answer B

View attachment 46184
Answer C

Thank you so much!
Click to expand...
Well I solved all of them and found that one of your answer is wrong. That is of kinetic energy.
So Here you go :
15)
So Clock wise moment = Anti Clock wise moment
Therefore 300 * 0.4 = 200 * 0.8
--> 40 Anticlockwise direction.
Alternative :
By taking the torque clockwise at the centre:-
=>300 x 0.4 - 200 x 0.4 = 40 Nm (clockwise)
=>Thus to balance it required torque is 40 Nm (anticlockwise)
=>(B)

13)
Already answer. Here in some more detail,
The ratio of the radius of the pulley to the arm is 0.2m/0.6m. Multiplying that by 900N gives you 300N. There are two force vectors, so 150N needs to be applied to each arm.
In other words,
To lift the weight, the cord needs to exert at least 900 N of force.
This means the disc needs to exert at least (900 N * 0.20 m) of torque.
This means the two forces on the lever needs to each exert at least (1/2 * 900 N * 0.20 m) of torque.
This means each force on the lever needs to be at least (1/2 * 900 N * 0.20 m / 0.60 m).
-> 150N

10)
This is the question where ms might not be correct or you haven't looked properly,
X = 1/2 x m x (-v)^2
Y = 1/2 x 2m x v^2 = m x v^2
Ratio of X/Y :
1/2 x m x v^2
--------------- = 1 / 2
m x v^2

Hope you got all of them. :)
 
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A

akshay 999

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Nov 88 p1 Num 8 and the num 11




11 Water waves of wavelength 4 m are produced by two generators, S1 and S2, as shown. Each generator, when operated by itself, produces waves which have an amplitude A at P, which is 3m from S1 and 5 m from S2.

upload_2014-8-7_9-41-54.png




When the generators are operated in phase, what is the amplitude of oscillation at P?



A 0 B ½ A C A D 2A
 
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akshay 999 said:
Nov 88 p1 Num 8 and the num 11




11 Water waves of wavelength 4 m are produced by two generators, S1 and S2, as shown. Each generator, when operated by itself, produces waves which have an amplitude A at P, which is 3m from S1 and 5 m from S2.

View attachment 46358




When the generators are operated in phase, what is the amplitude of oscillation at P?



A 0 B ½ A C A D 2A
Click to expand...
I dont get the question.
Just fill in this :¬
Year :
Question numbers :
Session :
The detail you provide me, was like this :¬
Year : 2088 :/
Question numbers : 8 n 11
Session : Winter.
Yes? Then what is in the year :eek: 2088? how?
 
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akshay 999 said:
Month: November
Year :1988
paper : 1
Question number : 8

and by the way can u solve the number 11 that i have post
Click to expand...
I never solved 19th century paper. Please get me a link.
Bud, my iPhone is not with me nor my Mac. I have my dabba phone through which I cant see the pic you posted, its like corrupted.
 
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Plz someone can help me out wiz zis questions

Question 1 : An organ pipe of effective length 0.6 m is closed at one end. Given that the speed of sound in air is 300 m/s . What is the two lowest resonant frequencies ???

Question 2 : A suspension bridge is to be built across a valley where it is known that the wind can gust at 5 second intervals. It is estimated that the speed of transverse wave along the span of the bridge would be 400 m/s. The danger of the resonant motions in the bridge at its fundamental frequency would be greatest if the span had a length of ???

Question 3 : A taut wire is clamped at two points 1.0 m apart. It is plucked near one end. Which are the three longest wavelengths present on the vibrating wire??

Question 4 : A string fixed at both ends and of length L is plucked at its midpoint and emits its fundamental note of frequency f1. When the string is plucked at a different point , the first overtone frequency f2 is also produced. Find f2/f1 and v , where v is the speed of transverse waves in the string???
 
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akshay 999 said:
Plz someone can help me out wiz zis questions

Question 1 : An organ pipe of effective length 0.6 m is closed at one end. Given that the speed of sound in air is 300 m/s . What is the two lowest resonant frequencies ???

Question 2 : A suspension bridge is to be built across a valley where it is known that the wind can gust at 5 second intervals. It is estimated that the speed of transverse wave along the span of the bridge would be 400 m/s. The danger of the resonant motions in the bridge at its fundamental frequency would be greatest if the span had a length of ???

Question 3 : A taut wire is clamped at two points 1.0 m apart. It is plucked near one end. Which are the three longest wavelengths present on the vibrating wire??

Question 4 : A string fixed at both ends and of length L is plucked at its midpoint and emits its fundamental note of frequency f1. When the string is plucked at a different point , the first overtone frequency f2 is also produced. Find f2/f1 and v , where v is the speed of transverse waves in the string???
Click to expand...
What problem you are facing in this? Look I am online from my dabba phone as I said, so will reply slower.
2)
5s intervals is a frequency f = 1/T = 1/5 = 0.2Hz.

Since v = fλ, the wavelength for transverse wave on the bridge is given by
400 = 0.2λ
λ = 2000m

At the fundametal frequency, the bridge's length iequals λ/2 (node to node), so the danger is greatest at 2000/2 = 1000m

4)
Rememeber the distance between 2 nodes is λ/2. See link (http://www.solitaryroad.com/c1031/ole.gif)

The fundamental (f₁) is sometimes called the 1st harmonic.
If the string is length L, then L= λ₁/2. So λ₁ = 2L.

The first overtone (f₂) is is sometimes called the 2nd harmonic.
If the string is length L, then L= 2(λ₂/2) = λ₂
_________________________________

v = λ₁f₁ = (2L)f₁ = 2Lf₁

v = λ₂f₂ = Lf₂

Therfore
2Lf₁ = Lf₂
f₂/f₁ = 2
 
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aliciaa said:
Hi, I just obtained my AS Results for physics,I got a 72 as I didn't have time to finish quite a number of questions. Is it possible for me to still get an A for my A level? Or should I resit my AS?
Click to expand...
some one from my school scored a C in AS and an A* in A levels of the same subject you just need to give your best
 
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Assalam alikum I have some doubts in As physics paper 1 (VARIENT1) I would be really gratetful if anyone helps me with them
M/J/05 Q 14 24 36
O/N/05 Q 30
M/J/06 Q 9
O/N/06 Q 21
M/J/07 Q 20
O/N/07 Q 11
M/J/08 Q 14 16
O/N/08 Q 32
M/J/09 Q 18
O/N/12 Q 10 29
 
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XP member said:
Assalam alikum I have some doubts in As physics paper 1 (VARIENT1) I would be really gratetful if anyone helps me with them
M/J/05 Q 14 24 36
O/N/05 Q 30
M/J/06 Q 9
O/N/06 Q 21
M/J/07 Q 20
O/N/07 Q 11
M/J/08 Q 14 16
O/N/08 Q 32
M/J/09 Q 18
O/N/12 Q 10 29
Click to expand...
s05 :¬
14)
untitled-png.44408


24)
This concerns values you need to keep in memory unless they are on the data sheets, i'm not sure. Either ways, visible light falls in the range of
400 nm< λ < 700 nm. We can take any value in this range to represent out wavelength. Suppose we take 600 nm.
Then, the number of wavelengths in 1 meter = 1 / (600 * 10^-9) = 1.6667 * 10^6. So our answer is closest to B. Therefore, that's our answer.

36)
V=IR

total rsistance=450
voltage=6
V=IR
I=>6/450=0.013

VOLTAGE IN 180 resistor
V=IR
V=0.013*180=2.4 volts

Or simply do proportionaly= 180/450 * 6
=2.4

w05 :¬
30)
Electric field lines go from + to -, so the +ve particle will move down towards the -ve side.

s06 :¬
9)
when the mass is at the lowest point, its kinetic energy = 0.
hence its velocity is also = 0 m/s
that means, either B or D.
since the first part of the graph shows the "UPWARD" velocity of the mass, hence the ans is D

w06 :¬
21)
17.5 *10^6 = 830*g*h + 1000*g*h
17.5 * 10^6 = 8142.3x + 9810(2000 - x)
17.5 *10^6 = 8142.3x + 19 620 000 - 9810x
-2 120 000= -1667.7x
1271.2 = x
~ 1270 m (3 s.f.)

s07 :¬
20)
X is in tension as it is opposing the weight of the horizontal bar and the foce W...Y is aso in tension as it is opposing W..Z is in compression as it is between a force from X and a force of Y and W

w07:¬
11)
q11-png.44357


s08 :¬
14)
moment = force x distance
clockwise moment = anticlockwise moment
W x a + F x h = W x 2a

16)
Potential energy decrease.
There is no work done from bringing the positive charge vertically.

w08 :¬
32)
All the wires are connected in parallel.
So 1/R = (1/10)X6 + 1/100
Therefore, 1/R = 61/100
R = 100/61 = 1.6 ohms.

s09:¬
18)
pressure = hdg. But here h is 2h since on one side liquid rises by h and another side liquid falls by h.

w12_11 :¬
10)
earth has atmosphere...air present so deviated path
on moon there is no air resistance so perfect semicircle...

29)
Very simple concept. Read your text book.
 
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