QJEEM19Phe885: JEE Main 2019 Physics - Class 11 JEEM19Phe885 Solution: A heavy ball of mass M is suspendeed from the ceiling of a car…

Resultant force on the ball of mass M when car is moving with a acceleration a is ,

F_net = $\sqrt {{{\left( {Mg} \right)}^2} + {{\left( {Ma} \right)}^2}}$

   = $M\sqrt {{g^2} + {a^2}}$

$\therefore$    T = M $\sqrt {{g^2} + {a^2}}$

We know,

Velocity, V = $\sqrt {{T \over \mu }}$

When Car is at rest then,

   60 = $\sqrt {{{Mg} \over \mu }}$    . . . . (1)

and when is moving then

   60.5 = $\sqrt {{{M\sqrt {{g^2} + {a^2}} } \over \mu }}$    . . . . (2)

By dividing (2) by (1) we get,

${{60.5} \over {60}} = \sqrt {{{\sqrt {{g^2} + {a^2}} } \over g}}$

$\Rightarrow$     $\left( {1 + {{0.5} \over {60}}} \right)$ = ${\left( {{{{g^2} + {a^2}} \over {{g^2}}}} \right)^{{1 \over 4}}}$

$\Rightarrow$     ${{{g^2} + {a^2}} \over {{g^2}}}$ = ${\left( {1 + {{0.5} \over {60}}} \right)^4}$

$\Rightarrow$     ${{{g^2} + {a^2}} \over {{g^2}}}$ = 1 + 4 $\times$ ${{{0.5} \over {60}}}$ [Using Binomial approximation]

$\Rightarrow$     ${{{g^2} + {a^2}} \over {{g^2}}}$ = 1 + ${1 \over {30}}$

$\Rightarrow$    1 + ${{{a^2}} \over {{g^2}}}$ = 1 + ${1 \over {30}}$

$\Rightarrow$     ${a \over g}$ = ${1 \over {\sqrt {30} }}$

$\Rightarrow$    a = ${g \over {\sqrt {30} }}$

$\therefore$    Closest answer, a = ${g \over 5}$