IIT JEE , AIEEE Forum - Physics , Modern Physics

ayshwarya

2 4

binding energy per nucleon 4 1H & 2 He r 1.1 Me V & 7MeV respectively den d energy released when 2 1H2 nulei fuse 2 form a 2He4 is ans is -2.8 MeV

D frequencyof revolution in nth orbit of Bohr model of an atom is fn den d variation of log(fn/f1) wid log (n) can b qualitatively represented as

(draw d graph 4 it plz giv d explanation 4 it)

d ball of mass m falls vertically from a height h & collides wid a block of equal mass m which is moving horizontally on a horizontal surface , velocity of d block at d instant of collision is u. coefficient of friction btw d block & d horizontal surface is

=0.2 & d coefficient of restitution is 0.5 den d velocity of d block decreases by

a) 0 b)0.3 rt(2gh) c) 2 rt(2gh) d) u

2 identical positiv charges r fixed on d Y-axis at equal distances from d origin O a negativly charged particle starts on d -ve X-axis at a large distance from O moves along d X-axis passes thru O & goes far away from origin 2wards +ve X-axis amgnitude of d acceleration a due 2 +ve charges is plotted as a function of its +ve coordinate x as

(draw d graph btw 'a' on Y-axis & x on X -axis)

ayshwarya

rates assured plz answer

amulye

plz ans dees quesn any1 rates assured

ayshwarya

hey guys dont anybody hav guts to do dis sums areeeeeee kya yaaron

amulye

oyy jaldi bolo yaaron paints milega plz mey aapse request kar rahin hun mey aur kis basha mein samjaoon koi bhaiyya bolo na

zorro

pls check data for first question...I think its incorrect....

and in question 3 none option is matching..

I m getting 0.1(2gh)^1/2..

amulye

na yaar how do u get it can u explain ur medod plz

ayshwarya

mahmmi jaldi bolo mein aur wait nahin kar sakti hoon mein janna chahti hoon is ki ans

ayshwarya

plz tell me ive typed it 2months back but even im nt getting d soln

allamraju

1)B.E of hydrogen=2(1.1)=2.2Mev,that of helium=4(7)=28Mev.So,the energy released during the reaction is change in B.E=28-2(2.2)=23.6Mev[I feel this is right though i got a diff. ans]

allamraju

2)mvr=nh/2pi $\alpha$ n $\Rightarrow$ r²w $\alpha$ n,since r_n is proportional to n²

$\Rightarrow$ w is proportional to 1/n³ and so is frequency $\Rightarrow$ log(f_n/f₁)=log(1/n³)=-3logn

So,the graph is a straight line passing thro' origin and slope -3.

Am I correct?

amulye

thanx 4 helping me in dis case all of dem was busy but u helped in dis cas even i 2 got d same ans in 1st sum but d ans is given diff

amulye

i want more ans

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1)B.E of hydrogen=2(1.1)=2.2Mev,that of helium=4(7)=28Mev.So,the energy released during the reaction is change in B.E=28-2(2.2)=23.6Mev[I feel this is right though i got a diff. ans]

2)mvr=nh/2pinr2wn,since rn is proportional to n2

w is proportional to 1/n3 and so is frequencylog(fn/f1)=log(1/n3)=-3logn

So,the graph is a straight line passing thro' origin and slope -3.

2)mvr=nh/2pi $\alpha$ n $\Rightarrow$ r²w $\alpha$ n,since r_n is proportional to n²

$\Rightarrow$ w is proportional to 1/n³ and so is frequency $\Rightarrow$ log(f_n/f₁)=log(1/n³)=-3logn