Question

here is your nxt question ​

Answer 1

here is your answer in the attachment

Answer 2

pics post Nahi hora so I'm doing this by typing .....

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\underline{\underline{\huge\mathcal{\bf{\boxed{\huge\mathcal{~~QUESTION~~}}}}}}}$

is in attachment

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\underline{\underline{\huge\mathcal{\bf{\boxed{\huge\mathcal{!!ANSWER!!}}}}}}}$

$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\large\mathcal{\bf{\boxed{\large\mathcal{v=2.214\: m.s{}^{-1} \:and \:t=1.010\: sec}}}}}$

$\:\: \underline{\underline{\bf{\large\mathfrak{~~given~~}}}}$

mass=m=0.1 kg

length=l=2.45 m

g=9.8m/sec²

h=0.5 m

$\:\: \underline{\underline{\bf{\large\mathfrak{~~solution~~}}}}$

(1)

now tension at a point of height of h from the lower end of the rope is =mgh

now .... therefore the speed of the transverse wave .....

$v = \sqrt{ \frac{t}{m} } \\ = > v = \sqrt{ \frac{mgh}{m} } \\ = > v = \sqrt{gh} \\ = > v = \sqrt{9.8 \times 0.5} \\ = > v = 2.214 \: \: m.s {}^{ - 1}$

(2)

at lower end of rope ..tension t is =0

hence speed of transverse wave is zero

and at the upper end the velocity is ....

$= \sqrt{gl} \\ and \: acceleration \: is \: \\ = \frac{ \sqrt{gl} }{t} \\ therefore.... \\ \: l = \frac{1}{2} \times \frac{ \sqrt{gl} }{t} \times t {}^{2} \\ = > t = 2 \sqrt{ \frac{l}{g} } \\ = > t = 2 \sqrt{ \frac{2.5}{9.8} } \\ = > t = 1.010 \: \: sec$

$\huge\mathcal\green{\underline{hope\:\: this\:\: helps\:\: you}}$