Please solve the questions given in the photograph from FIITJEE AITS.

just after first collision let velociy of bob is v then

applying momentam conservation

initial momantam of system(P_i) = (m/2) (u)

 u = (2gL)^1/2 so P_i = m(2gL)^1/2/2

final momemtam (P_f) = mv

 mv = m(2gL)^1/2 /2                          (from momentam conservation)

   v = (gL/2)^1/2             ................1

KE of bob is mv²/2

KE = mgL/4      ...............2

when bob is at max height then its KE is converted into PE

 mgL/4 = mgH_max

   L/4 = H_max     

from floor height of bob is L+H_max = L/4+L = 5L/4                                  (Q15 => option c)

Approved

let tension in string just after collision is T then

T - mg = mv²/r (centripital force)                                         { tension is in upward direction , weight is in downward direction }

T = mg + mv²/L

v = (gL/2)^1/2  so

T = 3mg/2               (ans14 =>c )

just after second collision , let velocity of bob is v₁ & tension in string is T ...

after collision this system becomes conical pendulam so verticle component of tension is balanced by weight &

centripital force is provided by horizontal component of tension ...

let string makes an angle of @ with verticle then

Tcos@ = weight = 2mg       ............1      (weight of bob + particle)

Tsin@ = 2mv₁² /r              .............2        (r is radius of its path)

r = Lsin@         .........3

from 1,2 & 3

v₁² = {gLsin@tan@}                  ..................4

time period of this system = T = 2pir/v₁

                                           = 2pi(Lsin@)/{gLsin@tan@}^1/2           ........................5

now , @ can be calculated by using trignometry ... height of this cone = L - L/4 = 3L/4

                                                                      slant height = L

              cos@ = 3/4

so , time perion will be T = 2pi(3L/4g)^1/2

approve if u like my ans