Which of the following is a square of a number ending with 3?
a) 139
b) 169
c) 89
d) 99

Answer

A, B, C

Explanation

Let M be the total mass of the earth. At any position x.
`:. (M_1)/(M) = (pxx(4)/(3)pix^3)/(pxx (4)/(3)pi r^3) xx (x^3)/(R^3)`
`rArr M_1 = (Mx^3)/(R^3)`
So, force on the particle is given by
`:. F_x = (GMm)/(x^2)`
` = (GMm)/(R^3)x ..... (1)`
So, acceleration of the mass 'M' at that position is given by,
`a_x = (GMm)/(R^3)`
`rArr (ax)/(x) = omega^2`
`=(GM)/(R^3) = (g)/(R ) (g = (GM)/(R^2))`
So, `T = 2pi sqrt((R )/(g))`
= time period of ocillation
(a) Now using velocity
= displacement equation
`upsilon = omega sqrt((A^2 - Y^2))`
[where, A = amplitude]
Given when,
`y = R, upsilon = sqrt(gR), w = sqrt((g)/(R ))`
`rArr sqrt(gR) = sqrt(((g)/(R )))/((A^2 - R^2))`
`rArr R^2 = A^2 - R^2`
[ Now, the phases of the particle at the point
P is grater then `(pi)/(2)` has less then `pi` and at
Q is grater then `pi` but less then `(3pi)/(2)` Let the times taken by the particle to reach the position P and Q be `t_1 and t_2` respectively, then using displacement equation].
y = r sin omega t`
We have
`R = sqrt2 R sin omega t`
`rArr omegat_1 = (3pi)/(4)`
We have, `R = sqrt2 R sin omega t_1`
`rArr omega t_2 = (5pi)/(4)`
`So, omega (t_2 - t_1) = (pi)/(2)`
`rArr t_2 - t_1 = (pi)/(2omega) = ((pi)/(2))/((R )/(g))`
Time taken by the particle to travel from P to Q is
`t_2 - t_1 = (pi)/(2) ((R )/(g))sec.`
(b) When the body is dropped from a height r, then applying conservation of energy, change in P.E. = gain in K.E.
`rArr (GMm)/(R ) - (GMm).(2R) = (1)/(2) m upsilon^2`
`rArr upsilon = sqrt(gR)`
Since the velocity is same at P, as in part (a) the body will taken same time to travel PQ.
(c ) When the body is projected vertically upward from P with a velocity. `(sqrt(gR)`its velocity will be zero at the highest point. The velocity of the body, when reaches P, again wil be
`upsilon = sqrt((gR))
Hece the body will take same time
`(pi)/(2) sqrt((R )/(g))` to travel P.Q.