The angular momentum of a particle relative to a certain O varies L- -a- -b- t 2 where a- and b- are constant vectors with a- perpendicular to b- The torque t- relative to the point O acting on the particle when angle between - and L- is 45 o is-

The correct option is B τ⃗ =2b⃗√(| a || b |) [ Given * Angular momentum L⃗=a⃗+b⃗ t^2; Thurfor torque; ...

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Standard X

Physics

Torque

The angular m...

Question

The angular momentum of a particle relative to a certain O varies $\to L = \to a + \to b t^{2}$ where $\to a$ and $\to b$ are constant vectors with $\to a$ perpendicular to $\to b$ . The torque $\to t$ relative to the point O acting on the particle when angle between $\to τ$ and $\to L$ is $45^{o}$ is:

τ = 2 \to a \sqrt{\frac{| a |}{| b |}}

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\to τ = 2 \to b \sqrt{\frac{| a |}{| b |}}

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\to τ = 2 \to a \sqrt{\frac{| a |}{| b |}}

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Solution

The correct option is B $\to τ = 2 \to b \sqrt{\frac{| a |}{| b |}}$
$\begin{matrix} Given- * Angular momentum \to L = \to a + \to b t^{2} Thurfor torque \to τ = \frac{d}{d t} (\to L) = 2 \to b \cdot t \end{matrix}$
$\begin{matrix} Now at an instant when angle between torque vector and \to L vector is 45^{\circ} - \Rightarrow | \to t \cdot \to L | = | \to E | \cdot | \to L | \cdot cos 45^{\circ} \Rightarrow 2 | \to b |^{2} \cdot t^{3} = 2 | \to b | \cdot t \sqrt{| \to a |^{2} + | \to b |^{2} t^{4}} \cdot \frac{1}{\sqrt{2}} \Rightarrow 2 | \to b |^{2} t^{4} = | ¯ a |^{2} + | \to b |^{2} t^{4} \Rightarrow t = \sqrt{\frac{| ¯ a |}{| \to b |}} \Rightarrow \to τ = 2 \to b \cdot t = 2 \to b \sqrt{\frac{1 \to a ∣}{| ¯ b |}} \Rightarrow Hence, option (B) is correct. \end{matrix}$

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The angular momentum of a particle relative to a certain O varies →L=→a+→bt2 where →a and →b are constant vectors with →a perpendicular to →b. The torque →t relative to the point O acting on the particle when angle between →τ and →L is 45o is: