Question

A car is moving in a circular horizontal track of radius 10 m
with a constant speed of 10 m/s. A bob is suspended from
the roof of the car by a light wire of length 1.0 m. The angle
made by the wire with the vertical is [NEET Kar. 2013]
(a) 0° (b) π/3
(c) π/6
(d)π/4

Answer 1

$\purple{\underline{\underline{\pink{\boxed{\boxed{\red{\star{\sf Question:}}}}}}}}$

$\rm{A \ car \ is \ moving \ in \ a \ circular \ horizontal}$

$\rm{track \ of \ radius \ 10 \ m \ with \ a \ constant}$

$\rm{ speed \ 10 ms^-1 \ bob \ is \ suspended \ from}$

$\rm{the \ roof \ of \ the \ car \ by \ a \ light \ wire \ of}$

$\rm{length \ 1.0 m. \ The \ angle \ made \ by \ the \ wire }$

$\rm{with \ the \ vertical \ is}$

$\rm{ [NEET \ Kar. \ 2013]}$

$\rm{(a) 0°}$

$\rm{(b) π/3}$

$\rm{(c) π/6}$

$\rm{(d)π/4}$

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$\purple{\underline{\underline{\orange{\boxed{\boxed{\green{\star{\sf Answer:}}}}}}}}$

$\rm{\orange{\underline{\purple{The \ angle \ made \ by \ the \ wire \ with }}}}$

$\rm{\orange{\underline{\purple{the \ vertical \ is }}}}$

$\rm{\orange{\boxed{\purple{\star{(d)π/4}}}}}$

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$\sf\large\underline\pink{Given:}$

$\rm{Radius \ of \ the \ curved \ path \ : \ 10m}$

$\rm{Speed = 10 ms^-1}$

$\rm{Length \ of \ wire \ = \ 1.0m}$

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$\sf\large\underline\blue{To \ Find}$

$\rm{Angle \ made \ by \ wire \ with \ vertical}$

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$\purple{\underline{\underline{\blue{\boxed{\boxed{\pink{\star{\sf Solution:}}}}}}}}$

$\rm{We \ are \ given }$

$\rm{Radius \ of \ the \ curved \ path \ : \ 10m}$

$\rm\implies{r=10m}$

$\rm{Speed (v) = 10 ms^-1}$

$\rm{Length \ of \ wire \ = \ 1.0m}$

$\rm{Let \ \theta \ be \ the \ angel \ made \ by \ wire}$

$\rm{with \ vertical}$

$\rm{From \ the \ figure \ (check\ the \ attachment)}$

$\rm{We \ know ,}$

$\rm{T sin\theta \ = \ \frac{mv^{2}}{r} \ ......[centripetal \ force] \ ....(1)}$

$\rm{T cos\theta \ = \ mg \ ........[ force] \ ....(2)}$

$\rm{From \ equation \ (1) \ and \ (2)}$

$\rm{\frac{T sin\theta}{T cos\theta} \ = \ \frac{mv^{2}}{r} \times \frac{1}{mg}}$

$\rm\therefore{\green{\boxed{\orange{tan\theta \ = \ \frac{v^2}{rg}}}}}$

$\rm\implies{tan\theta \ = \ \frac{10^2}{10 \times 10}}$

$\rm\implies{tan\theta \ = \ \frac{10^2}{10 \times 10}}$

$\rm\implies{tan\theta \ = \ 1}$

$\rm\implies{\theta \ = \ tan^{-1}}$

$\rm\implies{\theta \ = \ \frac{\pi}{4} \ ......... \ (as \ tan(45°) =1)}$

$\rm{\orange{\underline{\purple{The \ angle \ made \ by \ the \ wire \ with }}}}$

$\rm{\orange{\underline{\purple{the \ vertical \ is }}}}$

$\rm{\orange{\boxed{\purple{\star{(d)π/4}}}}}$

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$\rm\red{Hope \ it \ helps \ :))}$