Question

answer this qsn.....???​

Answer 1

ANSWER :-

$\implies \bf \: (b) \: \dfrac{F}{2m} \dfrac{x}{ \sqrt{a ^{2} - x ^{2} } }$

TO FIND :-

$\implies \bf \: The \: \: acceleration \: .$

SOLUTION :-

➵ Let the tension in each string be T. Hence the net force on the 'P'. should be zero, (m = 0).

$\to \bf \: F \: = m \times a \: \: \: \: \: \: \: \: \: \: \: \: \big\{ \because \: m \: = 0 \big \} \\ \\ \to \boxed{ \red{ \bf \: F = 0}}$

$\therefore \bf \: \: F \: = 2 \: T \cos \: \{ 90 \: - \theta \} \\ \\ \to \bf \: F \: = 2 \: T \: \sin \theta \\ \\ \to \boxed{ \red{ \bf \: T \: = \dfrac{F}{2 \: \sin \theta } }}$

➵ Now, come in mass (m).

$\to \bf \: \cos \theta \: = \dfrac{x}{a} \\ \\ \to \bf \: \sec \theta \: = \dfrac{a}{x} \\ \\ \because \bf \: \sec ^{2}\theta \: - \tan ^{2} \theta \: = 1 \\ \\ \to \bf \: \tan \theta \: = \sqrt{ \sec ^{2} \theta - 1 } \\ \\ \to \bf \: \tan \theta \: = \sqrt{ \dfrac{a ^{2} }{x ^{2} } - 1} \\ \\ \to \boxed{ \red{ \bf \: \dfrac{1}{ \tan \theta } = \dfrac{x}{ \sqrt{a ^{2} - x ^{2} } } }}$

$\to \bf \: Now \: Acceleration = \dfrac{T \: \cos \theta}{m} \\ \\ \to \bf \: \dfrac{F}{2 \: \sin \theta} \: . \dfrac{cos \theta}{m} \\ \\ \to \bf \dfrac{F}{2 \: \tan \theta \: .m } \\ \\ \to \boxed{ \red{\bf \: \dfrac{F}{2m} \: \dfrac{x}{ \sqrt{a ^{2} - x ^{2} } } }}$

Hence Option (b) is correct.

Answer 2

Answer:

daani answer b aah ne ? ....

Explanation:

answer cheppu