Physics: Post your doubts here!

[Bishnu Dev]

Please answer me the iv question.... I tried to solve it but the answer that i got is twice the actual answers. i got Vb = 866 but the answer is 433... and For Va the answer is 250 and i got 500... Please tell me what's missing....

 

[student2]

see solution 120 at
http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-22.html

Cn u pls help with p5 2?

Edit: actually i jst want 2 knw hw 2 draw the graph fr mj 2011 51 pllss..

 

[Ogapatapata]

can someone explain to me why Y=2

 

[Thought blocker]

can someone explain to me why Y=2
View attachment 61980

I have shown neutron representation and made equation to solve for y.

 

[Ogapatapata]

Thanks i forgot add the 1 from the neutron.
God bless.

 

[nadeen64]

in paper 5, when i want to draw a hall probe in the diagram for question 1.. do i draw a rectangular block as the hall probe or something like this?

 

[Thought blocker]

in paper 5, when i want to draw a hall probe in the diagram for question 1.. do i draw a rectangular block as the hall probe or something like this? View attachment 61991

Yes. This will work.

 

[nadeen64]

can someone please show me how the diagram of question 1 for june 2010 p52 will be like?

 

[shahzaib ihsan]

Hello, i have doubt in the question of june 2016 physics varient 22 (aslevels) in Q1 (b) (v)
http://www.teachifyme.com/wp-content/uploads/2016/08/9702_s16_qp_22.pdf

Calculate the frictional force...

i can't understand the markscheme with the directions/components so if anyone can do a full detailed working which force/accelaration in which direction i will be grateful.

 

[shahzaib ihsan]

Hello, i have doubt in the question of june 2016 physics varient 22 (aslevels) in Q1 (b) (v)
http://www.teachifyme.com/wp-content/uploads/2016/08/9702_s16_qp_22.pdf

Calculate the frictional force...

i can't understand the markscheme with the directions/components so if anyone can do a full detailed working which force/accelaration in which direction i will be grateful.

Just found the answer:

-ma = mg - R

m= mass (95kg), R = resistive force(in this case frictional force), g= gravity (acceleration due to free fall {9.81})

so up the slope is negative direction, so resultant force is negative (ma) and r is also up the force so it is also negative. mg(weight) is pulling the object downwards, so it is in positive direction.

as (-ma) and (R) is acting up the slope, so (mg) responsible for down the slope must act down the slope, so down the slope component of g must be taken by sin20 x 9.81,
and then multiply the mass m (95kg) with component of g down the slope which is (sin20 x 9.81) to give the force acting down the slope.

so putting in values we get :

(-ma)- [m(sin20 x g) = (-R)

so we get a positive value of R after solving and then multiplying the equation with (-1)

 

[techgeek]

9702/02 Oct/Nov 2007, Q5:
If the width of the slit is increased, why are bright areas brighter, no change in dark areas and fewer fringes are seen?
Also if the separation of the slits is increased, there is no change in brightness. Can someone explain please?

 

[student2]

Copper braid provides ‘return’ for the signal in a coaxial cable
pls ny1 knws wht 'return' means?

 

[shahzaib ihsan]

9702/02 Oct/Nov 2007, Q5:
If the width of the slit is increased, why are bright areas brighter, no change in dark areas and fewer fringes are seen?
Also if the separation of the slits is increased, there is no change in brightness. Can someone explain please?

for the second question : if the separation of the slits is increased...
the question asks to state the appearance of fringes so you must include brightness of fringe and the fringe separation to score marks...
Brightness doesn't depend on slit separation, but on slit width. So to score two marks , If the slit separation is increased the brightness remains constant, but the separation of fringes will be smaller as d is inversely proportional to angle of diffraction.
now for the first question, it is of 3 marks. so only writing brightness and fringe separation will only score 2 marks. to score the third mark, you have to add another subject that is CONTRAST. now contrast is the difference between the bright pixels and dark pixels, in this case, bright and dark areas. As the width of EACH slit is increased, so both have same amount of light passing, so either full constructive or full destructive interference is happening. So THE DARK AREAS CAN"T BE MORE DARK AS there can't be more destructive interference than fill destructive interference. And as full destructive interference is happening, there can't be 'less 'destructive interference, so that dark can become less dark.And as for the bright becomes brighter, it is because now more light is passing through both slits, so the interference they produce is greater.
take an example: when both silts have original width, normal amount of light passes, lets say the max amplitude of both light waves is 2mm and minimum is -2mm. so when the interfere, the max amplitude is now 2+2=4, which is constructive interference and minimum is 2-2=0, which is destructive, which produces dark area. now lets say both slit width have been increased by double. now max amplitude of both waves is 4mm, and minimum is -4mm. Now when they both meet and constructive interference happens, the resultant wave has 4+4= 8mm which is constructive interference. now when they both meet and produce destructive interference, it is 4-4=0mm.
note that increasing the width of each slit by double, new constructive interference has twice the amplitude of original, hence it is twice brighter, but dark is still 0, so dark is not either more dark or less dark, but is constant, hence no change in dark area, but brighter area becomes brighter.
Now for the fewer fringes part:
you must know that when the wavelength of wave is equal to the slit width, maximum diffraction occurs. Now the original slits have greater width then the wavelength, so they produce less diffraction. But when the slit width is increased more, diffraction get more narrower i.e. very little diffraction takes place. Now fringes are actually the waves produced from interference of two separate waves. actually, waves from two slits diffract, and then both OVERLAP to produce interference pattern visible as fringes. when the slits have greater width, they diffract lesser, so overlapping is lesser, hence interference is lesser, so lesser number of fringes are produced. But you must know that increasing each slit width doesn't increase or decrease the angle of diffraction, but diffraction itself leading to low amount of overlapping and hence less interference patterns (fringes) producing.

 

[Bishnu Dev]

i am a bit confused about the DEFINE type question in paper 2. Sometimes, we've to write formula and sometimes qualitative definition. How do i know whether the question is asking for definition or formula?

 

[shahzaib ihsan]

i am a bit confused about the DEFINE type question in paper 2. Sometimes, we've to write formula and sometimes qualitative definition. How do i know whether the question is asking for definition or formula?

hmm , i get confused too, but the thing i have noticed there are 3 key words: define, state and explain. state and explain you have to write qualitatively but in define there can be some options. Like if they say define work done, then you can write forcex distance travelled in direction of force. but when they say define pd, you cant write work done/charge but 'energy converted from electrical to other forms per unit charge'. i guess this is because work done /charge can also be applied to emf, so you have to write qualitatively..

 

[techgeek]

for the second question : if the separation of the slits is increased...
the question asks to state the appearance of fringes so you must include brightness of fringe and the fringe separation to score marks...
Brightness doesn't depend on slit separation, but on slit width. So to score two marks , If the slit separation is increased the brightness remains constant, but the separation of fringes will be smaller as d is inversely proportional to angle of diffraction.
now for the first question, it is of 3 marks. so only writing brightness and fringe separation will only score 2 marks. to score the third mark, you have to add another subject that is CONTRAST. now contrast is the difference between the bright pixels and dark pixels, in this case, bright and dark areas. As the width of EACH slit is increased, so both have same amount of light passing, so either full constructive or full destructive interference is happening. So THE DARK AREAS CAN"T BE MORE DARK AS there can't be more destructive interference than fill destructive interference. And as full destructive interference is happening, there can't be 'less 'destructive interference, so that dark can become less dark.And as for the bright becomes brighter, it is because now more light is passing through both slits, so the interference they produce is greater.
take an example: when both silts have original width, normal amount of light passes, lets say the max amplitude of both light waves is 2mm and minimum is -2mm. so when the interfere, the max amplitude is now 2+2=4, which is constructive interference and minimum is 2-2=0, which is destructive, which produces dark area. now lets say both slit width have been increased by double. now max amplitude of both waves is 4mm, and minimum is -4mm. Now when they both meet and constructive interference happens, the resultant wave has 4+4= 8mm which is constructive interference. now when they both meet and produce destructive interference, it is 4-4=0mm.
note that increasing the width of each slit by double, new constructive interference has twice the amplitude of original, hence it is twice brighter, but dark is still 0, so dark is not either more dark or less dark, but is constant, hence no change in dark area, but brighter area becomes brighter.
Now for the fewer fringes part:
you must know that when the wavelength of wave is equal to the slit width, maximum diffraction occurs. Now the original slits have greater width then the wavelength, so they produce less diffraction. But when the slit width is increased more, diffraction get more narrower i.e. very little diffraction takes place. Now fringes are actually the waves produced from interference of two separate waves. actually, waves from two slits diffract, and then both OVERLAP to produce interference pattern visible as fringes. when the slits have greater width, they diffract lesser, so overlapping is lesser, hence interference is lesser, so lesser number of fringes are produced. But you must know that increasing each slit width doesn't increase or decrease the angle of diffraction, but diffraction itself leading to low amount of overlapping and hence less interference patterns (fringes) producing.

that was great
Thanks

 

[Albeit]

When resistance of thermister is 2200 please explain which led will glow?

 

[student2]

fr specific latent heat of vaporisation, is it
energy supplied + gain in heat energy = mLf?
cn any1 confirm??

 

[Bishnu Dev]

hmm , i get confused too, but the thing i have noticed there are 3 key words: define, state and explain. state and explain you have to write qualitatively but in define there can be some options. Like if they say define work done, then you can write forcex distance travelled in direction of force. but when they say define pd, you cant write work done/charge but 'energy converted from electrical to other forms per unit charge'. i guess this is because work done /charge can also be applied to emf, so you have to write qualitatively..

hey bro.... can you explain me about the change in momentum stuff.... Please correct me if i am wrong here. Change in momentum = Final Momentum - Initial Momentum, but mark scheme says Initial Momentum - Final momentum... Actually i am confused about the SIGN in the final answer or during the working...

 

[Bishnu Dev]

b (iii)

why isn't the mass in equilibrium at t=0? Since the weight + force applied to pull the mass = tension in the spring? the resultant force is zero.