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4-1.Newton's Laws of Motion
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A ball is suspended on a thread from the ceiling of a car. The brakes are applied and the speed of car changes from $5\, m/sec$ to $\frac{5}{3}\, m/sec$ during the time interval of $3\,seconds$ . The angle that the thread will deviate from vertical in new equilibrium position :-
A$\theta = {\tan ^{ - 1}}\left( {\frac{1}{9}} \right)$
B$\theta = {\tan ^{ - 1}}\left( {\frac{8}{9}} \right)$
C$\theta = {\sin ^{ - 1}}\left( {\frac{1}{9}} \right)$
D$\theta = {\cos ^{ - 1}}\left( {\frac{1}{9}} \right)$
Solution
The situation is shown in figure.
When the car moves towards right with acceleration a the rope carrying the mass makes an angle $\theta$
Acceleration $a=\frac{1}{3}\left(5-\frac{5}{3}\right)=\frac{10}{9} \mathrm{m} / \mathrm{s}^{2}$
$\Rightarrow \tan \theta=\frac{\mathrm{ma}}{\mathrm{mg}}=\frac{\mathrm{a}}{\mathrm{g}} \Rightarrow \frac{10}{9 \times 10}=\frac{1}{9}$
When the car moves towards right with acceleration a the rope carrying the mass makes an angle $\theta$
Acceleration $a=\frac{1}{3}\left(5-\frac{5}{3}\right)=\frac{10}{9} \mathrm{m} / \mathrm{s}^{2}$
$\Rightarrow \tan \theta=\frac{\mathrm{ma}}{\mathrm{mg}}=\frac{\mathrm{a}}{\mathrm{g}} \Rightarrow \frac{10}{9 \times 10}=\frac{1}{9}$
Standard 11
Physics
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