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51.

Determine the binding energy of satellite of mass `1000 kg` revolving in a circular orbit around the Earth when it is close to the surface of Earth. Hence find kinetic energy and potential energy of the satellite. [Mass of Earth `=6xx10^(24)kg`, radius of Earth `=6400km`, gravitational constant `G=6.67xx10^(-11)Nm^(2)//kg^(2)`]

Answer» Given:
Mass of satellite, `m=1000kg`
Mass of earth, `M_(e)=6xx10^(24)kg`
Radius of earth, `R_(e)=6400km`
`=6.4xx10^(6)` metre
Gravitational constant, `G=6.67xx10^(-11)Nm^(2)//kg^(2)`
Binding energy of satellite`=?`
Kinetic energy and potential energy of satellite `=?`
Binding energy of satellite close to the surface of earth is given by
Binding energy `=+-(1)/(2)*(GM_(e)m)/(R_(e))`
`=(1)/(2)xx(6.67xx10^(-11)xx6xx10^(24)xx1000)/(6.4xx10^(6))`
`=3.1265xx10^(10)J`
Kinetic energy, `K.E.=(1)/(2)*(GM_(e)m)/(R_(e))`
`K.E.=3.1265xx10^(10)J`
Potential energy, `P.E.=-(GM_(e)m)/(R_(e))`
`:.P.E.=-2K.E.`
`P.E.=-6.2530xx10^(10)J`
Discussion
52.

Write a summary of the above extract with the help a of outline given below and suggest a suitable title Winter - chilika lake - migratory birds - secoind largest lake - conductive atmosphere- artifical mounds - variety of birds - effects fof pesticides / chemicals - harmony

Answer» With the onset of winter the nalatian bird sancturary inside the chilika lake is agog with the presence of thousands of migratory birds from faraway places like siberia iraq iran and cnetrl europe according to wildlife officers the mirgatin of birds to the second largest lake in tehe workd has increased due to conducive atomosphere prevailing there the birds have taken shelter in the artifical mounds inside the sancturary and variety of birds have been spottded pesticides and chemicals have resutled in loss of natural habitat birds and animlas have a realationship and their harmonous presence is benefical for he continuity of a sanctuary
Discussion
53.

Show that `int(1)/(x^(2)sqrt(a^(2)+x^(2)))dx=(-1)/(a^(2))(sqrt(a^(2)+x^(2)))/(x)+c`

Answer» `I=int(1)/(x^(2)sqrt(a^(2)+x^(2)))dx`
Let `x=atantheta`
`dx=asec^(2)thetad theta`
`I=int(asec^(2)thetad theta)/(a^(2)tan^(2)thetasqrt(a^(2)+a^(2)tan^(2)theta))`
`I=int(sec^(2)theta)/(a^(2)tan^(2)thetasqrt(1+tan^(2)theta))d theta`
`=(1)/(a^(2))int(sec^(2)theta)/(tan^(2)thetasectheta)d theta`
`=(1)/(a^(2))int(1xxcos^(2)theta)/(costhetaxxsin^(2)theta)d theta`
`=(1)/(a^(2))int(costheta)/(sin^(2)theta)d theta`
Put `sintheta=t`
`rArrcosthetad theta=dt`
`I=(1)/(a^(2))int(dt)/(t^(2))`
`=(-1)/(a^(2)t)+c`
`=-(1)/(a^(2)sintheta)+c`
`=-(cosectheta)/(a^(2))+c`
`=-(sqrt(1+cot^(2)theta))/(a^(2))+c`
`=-(sqrt(1+(a^(2))/(x^(2))))/(a^(2))+c`
`=-(sqrt(x^(2)+a^(2)))/(a^(2)x)+c`
Discussion
54.

Find the expected value ,variance and standard deviation of random variable X whose probability mass function (p.m.f.) is given below.

Answer» Mean , `E(X)=sumXP(X)`
`=1xx(1)/(5)+2xx(2)/(5)+3xx(2)/(5)`
`=(1)/(5)+(4)/(5)+(6)/(5)=(11)/(5)`
Variance `(X)=sumX^(2)P(X)-[E(X)]^(2)`
`=1^(2)xx(1)/(5)+2^(2)xx(2)/(5)+3^(2)xx(2)/(5)-((11)/(5))^(2)`
`=(1)/(5)+(8)/(5)+(18)/(5)-(121)/(25)=(27)/(5)-(121)/(25)`
`=(135-121)/(25)=(14)/(25)`
S.D `sqrt("Variance")`
`=sqrt((14)/(25))=(3*74)/(5)`
`0*748`
Discussion
55.

A rectangle has an area of 50 `"cm"^(2)`. Find its dimensions for least perimeter.

Answer» Let the length of rectangle be x cm and breadth of rectangle be y cm .
Given, Area = `50cm^(2)`
`thereforexy=50` . . . (i)
Perimeter of rectangle, P =2 (x+y)
`=2(x+(50)/(x))` [ From (i)]
`=(2(x^(2)+50))/(x)`
Now, `(dP)/(dx)=2(1-(50)/(x^(2)))`
Put, `(dP)/(dx)=0`,
`rArr2(1-(50)/(x^(2)))=0`
`rArr1-(50)/(x^(2))=0`
`rArrx^(2)=50`
`rArrx=5sqrt(2)m`
Put the value of x in equation (i),
`5sqrt(2)y=50`
`rArry=(50)/(5sqrt(2))`
`rArry=5sqrt(2)`
Again, `(d^(2)P)/(dx^(2))=(100)/(x^(3))=` Positive
`therefore` P is minimum.
`therefore` Dimensions when its perimeter is the least are `5sqrt(2)` m and `5sqrt(2)` m. lt
Discussion
56.

Solve the different equation : `x+y(dy)/(dx)=sec(x^(2)+y^(2))`. Also find the particular solution if x=y=0.

Answer» The given differential equation is,
`x+y(dy)/(dx)=sec(x^(2)+y^(2))` . . . (i)
Let `x^(2)+y^(2)=t`
`2x+2y(dy)/(dx)=(dt)/(dx)`
`2(x+y(dy)/(dx))=(dt)/(dx)`
`x+y(dy)/(dx)=(1)/(2)(dt)/(dx)`
From equation (i) , `(1)/(2)(dt)/(dx)=sect`
`int(dt)/(sect)=int2dx`
`intcostdt=2x+c`
`sint=2x+c`
`sin(x^(2)+y^(2))=2x+c`
When x =0, y=0, we have
`sin0=0+c` c=0
`therefore` The particular solution is given by,
`sin(x^(2)+y^(2))=2x`
Discussion
57.

If f(x)`=x^(2)+a`, for `xge0`, f(x)`=2sqrt(x^(2)+1)+b,"for"xlt0andf((1)/(2))=2`, is continuous at x = 0 , find a and b .

Answer» Given, `f(x)={{:(x^(2)+a",","for"xge0),(2sqrt(x^(2)+1)+b",","for"xlt0):}`
Now , `f((1)/(2))=((1)/(2))^(2)+a`
According to question,
`(1)/(4)+a=2`
`a=2-(1)/(4)`
`a=(7)/(4)`
f is continuous at x =0.
R.H.L.
`lim_(xto0^(+))f(x)=lim_(xto0^(+))(x^(2)+a)`
`=0^(2)+a`
=a
L.H.L.
`lim_(xto0^(-))f(x)=lim_(xto0^(-))2sqrt(x^(2)+1)+b`
=2 +b
According to question,
R.H.L. = L.H.L.
a=2+b
But `a=(7)/(4)`
`(7)/(4)=2+b`
`(7)/(4)-2=b`
`b=-(1)/(4)`
Discussion
58.

Each of the total fice questions in amultiple choice examination has four choices, only one of which is correct.A student is attempting to guess the amswer . The renadom varible X is the number of question answerred correctly. What is the probability that the student will giveat least one correct answer ?

Answer» Let p = probability that the given answer is correct. Q= probability that the given answer is wrong. `p=(1)/(4),q=1-(1)/(4)=(3)/(4)` P
Probability that the student will given wrong answers of all questions.
`=(3)/(4)xx(3)/(4)xx(3)/(4)xx(3)/(4)xx(3)/(4)`
`=(243)/(1024)`
Probability that the student will given at least one correct answer
`=1-(243)/(1024)`
Discussion