JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 17)
A heavy ball of mass M is suspendeed from the ceiling of a car by a light string of mass m (m < < M). When the car is at rest, the speed of transverse waves in the string is 60 ms$$-$$1. When the car has acceleration a, the wave-speed increases to 60.5 ms$$-$$1. The value of a, in terms of gravitational acceleration g, is closest to :
$${g \over {30}}$$
$${g \over 5}$$
$${g \over 10}$$
$${g \over 20}$$
Explanation
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Resultant force on the ball of mass M when car is moving with a acceleration a is ,
Fnet = $$\sqrt {{{\left( {Mg} \right)}^2} + {{\left( {Ma} \right)}^2}} $$
= $$M\sqrt {{g^2} + {a^2}} $$
$$ \therefore $$ 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 = $${g \over 5}$$
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