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Physics: Post your doubts here!

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Assalamoalaikum wr wb!

I need some help with the significant figures..plz..

I'm confused about it. What we kearnt is it shud be having the same no. of sig. fig as in the data...but ms doesnt seem to follow this everytime :/
 
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Please post the answers.

(a) As the pole is in equilibrium,
C.W moments = A.C.W moments
mg*(1.5) = Tsin30*(2.5)
25*9.8*1.5 = Tsin30*(2.5)
T = 294 N

(b) Let the required force be 'R'
As the pole is in equilibrium,
Sum of upward forces = Sum of downward forces
Tcos30 = mg + R
294*cos30 = 25*9.8 + R
254.6 = 245 + R
R = 9.6 N

Hope this is right.
 
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Assalamoalaikum wr wb!

I need some help with the significant figures..plz..

I'm confused about it. What we kearnt is it shud be having the same no. of sig. fig as in the data...but ms doesnt seem to follow this everytime :/
mmm well i think u should follow the sig fig in the question it self mostly its given ! if they r given in 2 sgf then use that etc... !
 
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Assalamoalaikum wr wb!

I need some help with the significant figures..plz..

I'm confused about it. What we kearnt is it shud be having the same no. of sig. fig as in the data...but ms doesnt seem to follow this everytime :/
I felt exactly the same about the significant figures and the mark scheme. However I realised that always give the same number of or one better s.f than the question. Trust me, you wont get it wrong.
 

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Waalaikumassalam wr wb!
Across P and Q is a variable resistor connected....and the resistance of it can be varied from zero to 50 ...

Now in series I remains the same...I=V/R

So when the resistance across the variable resistor is zero...the voltage is ofcourse going to be zero.

And when it is 50

I = V/R = 9 / 60

V across P and Q = I x R = 9/60 x 50 = 7.5 V

Hence the answer is B.

Note: In this case the units will get cancelled out in the end for resistance...and it's an MCQ so I haven't converted the kilo ohms to ohms.. Otherwise u are supposed to do that conversion first!
 
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State which combination, of any. Of monochromatic light and metal surface could give raise to photo-electric emission. Give a quantitative explanation of your answer.
 
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View attachment 4315 I'll be obliged if sum1 can solve this for me.
I will say T=250/sin30
1378bf52.jpg
 
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help me in mcq no 9 and 35 of may june 2002 paper 1..... explain in detail.
 

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kindly explain q.10 worksheet 9 C part! how is the acceleration of the proton zero at right angles to the field? plz help

Why not at right angles to the field? Imagine the proton traveling between the two parallel plates in the figure above this question. There will be a downward force on it and you calculate it from the electric field strength as F/Q so answer is charge on proton X 2.40 × 10–6 V m–1.
 
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help me in mcq no 9 and 35 of may june 2002 paper 1..... explain in detail.
Q9:
As it is an elastic collision,
Total K.E after collision = Total K.E before collision
= (1/2)m(v)2 + (1/2)m(-v)2
= m(v)2
So, B is correct.

And in Q 35, just test all the combinations to find out the correct one.
 
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