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STD 11 - 4.1 , newtons laws of motion — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 4.1 , newtons laws of motion4 Marks
MCQ
$A$ ball connected with string is released at an angle $45^o$ with the vertical as shown in figure. Then the acceleration of the box at this instant will be:

[Mass of the box is equal to mass of ball]

  • A
    $g/4$
  • ✓
    $g/3$
  • C
    $g/2$
  • D
    none of these
✓

Answer

Correct option: B.
$g/3$
b
Let acceleration of box at this is $'a'$(towards right). $F B D$ of ball $w.r.t.$ box

Net force in $O P$ ditection is zero

$\therefore T+\frac{m a}{\sqrt{2}}=\frac{m g}{\sqrt{2}}$

$F B D$ of ball $w.r.t.$ ground

$T \cos 45^{\circ}=m a$

or $T=\sqrt{2} m a$

Substituting in Eq. $( i ),$ we get

$a=\frac{g}{3}$

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