Question

Answerrrrrrrrr.......​

Answer 1

Answer:

(B)

Explanation:

PLS FIND SOLUTION.

Answer 2

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$\underline{\blacksquare\:\:\:\footnotesize{\red{\text{Figure:}}}}$

$\setlength{\unitlength}{1.5mm}\begin{picture}(30,20)\linethickness{0.08mm}\put(7,3.5){\line(1,0){22.5}}\put(30.5,-16){\line(0,1){18.5}}\put(25,0){\line(5,3){3.5}}\multiput(0,0)(1,0){26}{\line(-2,-3){1.6}}\linethickness{0.5mm}\put(0,0){\line(1,0){25}}\put(25,0){\line(0,-1){25}}\put(0,0){\dashbox{0.1}(7,7)[l]}\put(27,-23){\dashbox{0.1}(7,7)[l]}\put(29.5,2.5){\circle{2}}\linethickness{0.2mm}\put(7,6){\vector(1,0){22.5}}\put(20,6){\vector(-1,0){13}}\put(15,7){d}\put(2.5,3){ m_1 }\put(29.5,-20){ m_2 }\end{picture}$

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$\underline{\blacksquare\:\:\:\footnotesize{\red{\text{SolutioN:}}}}$

$\footnotesize{\bf{Let}\:,\text{ the acceleration of the arrangement is = a }}$

$\footnotesize{\text{and , the tension on the string is = T }}$

$\footnotesize{\text{Now , considering the horizontal motion of the block}}$

$\footnotesize{\text{of mass}\:m_1\:,}$

$\footnotesize{\red{\text{ T =}\:m_1a} \:----------(i)}$

$\footnotesize{\text{Now , considering the downward motion of the block}}$

$\footnotesize{\text{of mass}\:m_2\:,}$

$\footnotesize{\:\:\red{m_2g-T=m_2a}\:----------(ii)}$

$\footnotesize{\implies m_2g-m_1a=m_2a}$

$\footnotesize{\implies m_2g=m_1a+m_2a}$

$\footnotesize{\implies m_2g=a(m_1+m_2)}$

$\footnotesize{\implies a=\dfrac{m_2g}{(m_1+m_2)}}$

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$\footnotesize{\bf{\text{Now , from Law of motion ,}\:\:\:\:S=ut+\dfrac{1}{2}at^2}}$

$\footnotesize{\text{Here , u (initial velocity)=0 and S (distance)=d}}$

$\footnotesize{\therefore \:\:d=0\times t+\dfrac{1}{2}\big(\dfrac{m_2g}{m_1+m_2}\big)t^2}$

$\footnotesize{\implies d=\dfrac{m_2g}{2(m_1+m_2)}t^2}$

$\footnotesize{\implies t^2=\dfrac{2d(m_1+m_2)}{m_2g}}$

$\footnotesize{\implies\boxed{\red{\bf{t=\sqrt{\dfrac{2d(m_1+m_2)}{m_2g}}}}}\:\:\:\: option (B)}$

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