An `alpha` particle of `K.E. 10^(-12)J` exhibits back scattering from a gold nucleus Z=79. What can be the maximum possible radius of the gold nucleus?
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Prove that `tan alpha+2 tan 2 alpha+2^(2)tan^(2)alpha+..............+2^(n-1)tan2^(n-1)alpha+2^(@)cot2^(@)alpha=cotalpha`
We know `tan theta=cot theta-2 cot 2 theta`. Putting `theta=alpha,2alpha,2^(2)alpha,....................` in (i), we get `tan alpha=(cot alpha-2 cot 2alpha)` `2(tan2alpha)=2(cot 2 alpha-2 cot 2 ^(2)alpha)` `2^(2)(tan2^(2)alpha)=2^(2)(cot2^(2)alpha-2cot2^(3)alpha)` `2^(n-1)(tan2^(n-1)alpha)=2^(n-1)(cot2^(n-1)alpha-2cos2^(n)alpha)` Adding, `tan alpha+2tan...
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If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal to
Correct Answer - A We have, `A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]` `=A(-alpha, -beta)`
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If `alpha, beta, gamma` are three real numbers and `A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),
Correct Answer - A::B::D `A=[(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)][(cos alpha,cos beta,cos gamma),(sin alpha,sin beta,sin gamma),(0,0,0)]` Clearly, A is symmetric and `|A|=0`, hence, singular and not invertiable. Also, `A A^(T)...
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If `f(x)=x^(11)+x^9-x^7+x^3+1` and `f(sin^(-1)(sin8)=alpha,alpha` is constant, then `f(tan^(-1)(t a n8)` is equal to `alpha` (b) `alpha-2` (c) `alpha+
Correct Answer - D `f(x) + f(-x) = 2` Now `(sin^(-1) (sin 8)) = 3pi - 8 = y` And `(tan^(-1) (tan 8)) = (8 - 3pi) = -y` Hence, `f(y)...
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The value of `(alpha^3)/2cos e c^2(1/2tan^(-1)alpha/beta)+(beta^3)/2sec^2(1/2tan^(-1)(beta/alpha))i se q u a lto` `(alpha+beta)(alpha^2+beta^2)` (b) `
Correct Answer - C `(alpha^(3))/(2) cosec^(2) ((1)/(2) tan^(-1). (alpha)/(beta)) + (beta^(3))/(2) sec^(2) ((1)/(2) tan^(-1).(beta)/(alpha))` `= alpha^(3) (1)/(1 - cos(tan^(-1) ((alpha)/(beta)))) + beta^(3) (1)/(1 + cos (tan^(-1).(beta)/(alpha)))` `= alpha^(3) (1)/(1 -cos (cos^(-1)...
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(a) Why is an alpha particle, rather than neutrons and protons, emitted from an unstable nucleus? (b) A nucleus `.^(A)X` alpha decays to produce Y. Th
Correct Answer - (b) A=228
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If `2sec^2 alpha- sec^4 alpha-2 cosec^2 alpha+cosec^4 alpha=15/4` then `tan alpha=`
Correct Answer - A::D `2sec^2alpha-sec^4alpha-2cosec^2alpha+cosec^4alpha=15/4` `or 2(sec^2alpha-cosec^2alpha)+(cosec^2alpha+sec^2alpha)xx(cosec^2alpha-sec^2alpha-sec^2alpha)=15/4` `or (cosec^2alpha-sec^2alpha)[cosec^2alpha+sec^2alpha-2]=15/4` `or 4(cot^2alpha-tan^2alpha)[cosec^2alpha+sec^2alpha)=15` `or4(cot^4alpha-tan^4alpha)=15` `or 4(1-tan^8alpha)=15tan^4alpha` `or 4tan^8alpha+15tan^4alpha` `or 4tan^8alpha+154tan^4alpha-4=0` `or (4tan^4alpha-1)(tan^4alpha+4)=0` `or tan^4alpha=1/4ot tan^4alpha=-4` `or tan^2alpha=pm1/2` `or tan^2alpha=+1/2` `or tanalpha=pm1/sqrt2`
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Let `alpha,beta` be two real numbers satisfying the following relations `alpha^2+beta^2=5, 3(alpha^5+beta^5)=11(alpha^3+beta^3)1.` Possible value of `
Correct Answer - A `alpha^(2)+beta^(2)=5` `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` `(alpha^(5)+beta^(5))/(alpha^(3)+beta^(3))=(11)/(3)` ` :. ((alpha^(3)+beta^(3))(alpha^(2)+beta^(2))-(alpha^(2)beta^(2)(alpha+beta)))/(alpha^(3)+beta^(3))=(11)/(3)` `:. alpha^(2)+beta^(2)-(alpha^(2)beta^(2)(alpha+beta))/((alpha+beta)(alpha^(2)+beta^(2)-alphabeta))=(11)/(3)` ` :. 5-(alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` ltbrlt `:. (25-5alphabeta-alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` Let `alphabeta=t` `(25-5t-t^(2))/(5-t)=(11)/(3)` `75-15t-3t^(2)=55-11 t` `75-15t-3t^(2)-55+11t=0` `-3t^(2)-4t+20=0` `(t-2)(3t+10)=0` ` :. t=2` or `(-10)/(3)` So `alpha...
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There are two sets of parallel lines, their equations being `x cos alpha+y sin alpha=p` and `x sin alpha- y cos alpha=p` , `p=1,2,3,….n` and `alpha in
Correct Answer - C `(c )` `"^(n)C_(2)*^(n)C_(2)=225impliesn=6`
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An alpha particle of velocity `1.6xx10^(7) ms^(-1)` approaches a gold nucleus (Z=79). Calculate the distance of the closest approach. Mass of an alpha
Correct Answer - `4.3 xx 10^(-14)m` , the size of the nucleus cannot be greater than `4.3 xx 10^(-14)m`.
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