Question

Two bodies of masses 8kg and 32kg are placed at a distance 12m. A third body of mass 5kg is to be placed at such a point that the force acting on this body is zero. Find the position of that point.​

Answer 1

Explanation:

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$</p><p> </p><p></p><p>\begin{gathered}\dashrightarrow\sf\:\:(Diagonal)^2=(Length)^2+(Breadth)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=(BC)^2+(CD)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=(24\:cm)^2+(7\:cm)^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=576\:cm^2+49\:cm^2\\\\\\\dashrightarrow\sf\:\:(BD)^2=625\:cm^2\\\\\\\dashrightarrow\sf\:\:BD=\sqrt{625\:cm^2}\\\\\\\dashrightarrow\sf\:\:BD=\sqrt{25\:cm \times 25\:cm}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf BD=25\:cm}}\qquad\bigg\lgroup\bf Diagonal\bigg\rgroup\end{gathered} </p><p>⇢(Diagonal) </p><p>2</p><p> =(Length) </p><p>2</p><p> +(Breadth) </p><p>2</p><p> </p><p>⇢(BD) </p><p>2</p><p> =(BC) </p><p>2</p><p> +(CD) </p><p>2</p><p> </p><p>⇢(BD) </p><p>2</p><p> =(24cm) </p><p>2</p><p> +(7cm) </p><p>2</p><p> </p><p>⇢(BD) </p><p>2</p><p> =576cm </p><p>2</p><p> +49cm </p><p>2</p><p> </p><p>⇢(BD) </p><p>2</p><p> =625cm </p><p>2</p><p> </p><p>⇢BD= </p><p>625cm </p><p>2</p><p> </p><p> </p><p> </p><p>⇢BD= </p><p>25cm×25cm</p><p> </p><p> </p><p>⇢ </p><p>BD=25cm</p><p> </p><p> </p><p> </p><p> </p><p>⎩</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎧</p><p> </p><p> Diagonal </p><p>⎭</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎫</p><p> </p><p> </p><p> </p><p> </p><p></p><p>⠀</p><p></p><p>\therefore\:\underline{\textsf{Hence, Length of Diagonal is C) \textbf{25 cm}}}.∴ </p><p>Hence, Length of Diagonal is C) 25 cm</p><p> </p><p> .$