A heavy ball of mass M is suspended 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

Correct answer: C. (g)/(5) For standing car, v = (T)/(μ) = (Mg)/(μ) For accelerating car, υ′ = ( Mg ) ^ 2 + ( Ma ) ^ 2 μ ⇒ (60.5)/(60) = ( Mg ) ^ 2 + ( Ma ) ^ 2 Mg ⇒ (1 + (1)/(120)) ^ 2 = ( Mg ) ^ 2 + ( Ma ) ^ 2 Mg ⇒ 1 + (2)/(120) = 1 + a ^ 2 g ^ 2 (2)/(120) = (1)/(2) a ^ 2 g ^ 2 ∴ a = g (4)/(120) = g 30 (g)/(5)

JEE Main 20199 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Morning Shift
Subject: Physics
Chapter: Gravitation

JEE Main 2019 · 9 January · Morning Shift · Physics · Gravitation

A heavy ball of mass M is suspended from the ceiling of a car by a...

A heavy ball of mass

M

is suspended from the ceiling of a car by a light string of mass

m\left(m ≪ M\right) .

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 closed to A heavy ball of mass M is suspended from the ceiling of a car by a light string of mass m(m ≪ M) . When the car is at

Options

A $\dfrac{g}{10}$
B $\dfrac{g}{20}$
C $\dfrac{g}{5}$
D $\dfrac{g}{30}$

Correct answer

C. $\dfrac{g}{5}$

Step-by-step solution

For standing car,

v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{Mg}{\mu}}

For accelerating car,

υ′ = \sqrt{\dfrac{\sqrt{{( Mg )}^{2} + {( Ma )}^{2}}}{\mu}}

⇒ \dfrac{60.5}{60} = \sqrt{\dfrac{\sqrt{{( Mg )}^{2} + {( Ma )}^{2}}}{Mg}}

⇒ {\left(1 + \dfrac{1}{120}\right)}^{2} = \dfrac{\sqrt{{( Mg )}^{2} + {( Ma )}^{2}}}{Mg}

⇒ 1 + \dfrac{2}{120} = \sqrt{1 + \dfrac{{a}^{2}}{{g}^{2}}}

\approx \dfrac{2}{120} = \dfrac{1}{2}\dfrac{{a}^{2}}{{g}^{2}}

∴ a = g\sqrt{\dfrac{4}{120}} = \dfrac{g}{\sqrt{30}} \approx \dfrac{g}{5}

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