Question

refer to the attachment for the question..

class 11... physics​

Answer 1

The correct answer is option : (c) 3

Given : inital angular velocity (ω)= 0 ;

 θ₁ , θ₂ are angular displacements in 1st and 2nd seconds.

 Let ∝ be the uniform angular accelration

For unifromly accelerated bodies we have,

θ = ωt + 1/2 ∝ t²

θ₁ = 0 + 1/2 ∝ 1²

⇒ θ₁ = ∝/2 rad

also,

θ₂ = (0 + 1/2 ∝2² ) - (0 + 1/2 ∝1²)

⇒ θ₂ = 2∝  - ∝/2

⇒ θ₂ = 3∝/2 rad

so, θ₂ / θ₁ = (3∝/2) /  (∝/2)

∴  θ₂ / θ₁ = 3

HOPE THIS HELPS YOU !!   : )

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Equations of Rotational Motion has been used. We are given the Angular Displacement done by the wheel whose value is in radians. So first we can calculate values of both the displacements and then find their ratio.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\theta\;=\;\bf{\omega_{0}\:t\;+\;\dfrac{1}{2}\:\alpha\;t^{2}}}}$

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★ Solution :-

Given,

» Initial angular velocity of wheel = ω₀ = 0

» Time taken to rotate through θ₁ = 1 sec

» Time taken to rotate through θ₁ + θ₂ = 1 + 1 = 2 sec

» Angular Acceleration = α

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~ For the value of θ₁ ::

By applying the values in the formula, we get,

$\\\;\;\sf{:\rightarrow\;\;\theta_{1}\;=\;\bf{\omega_{0}\:t\;+\;\dfrac{1}{2}\:\alpha\;t^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;\theta_{1}\;=\;\bf{0\:\times\:t\;+\;\dfrac{1}{2}\:\alpha\;(1)^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;\theta_{1}\;=\;\bf{0\;+\;\dfrac{1}{2}\:\alpha\;1}}$

$\\\;\;\underline{\underline{\bf{:\rightarrow\;\;\theta_{1}\;=\;\bf{\dfrac{1}{2}\:\alpha}}}}$

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~ For the value of θ₂ ::

Firstly we must calculate the value of θ₁ and θ₂, which is given as,

$\\\;\;\sf{:\rightarrow\;\;\theta_{1}\;+\;\theta_{2}\;=\;\bf{\omega_{0}\:t\;+\;\dfrac{1}{2}\:\alpha\;t^{2}}}$

$\\\;\;\sf{:\rightarrow\;\;\theta_{1}\;+\;\theta_{2}\;=\;\bf{0\:\times\:t\;+\;\dfrac{1}{2}\:\alpha\;(2)^{2}}}$

$\\\;\;\bf{:\rightarrow\;\;\theta_{1}\;+\;\theta_{2}\;=\;\bf{\dfrac{1}{2}\:\alpha\;4}}$

Now by applying the value of θ₁ here, we get,

$\\\;\;\bf{:\rightarrow\;\;\dfrac{1}{2}\:\alpha\;+\;\theta_{2}\;=\;\bf{\dfrac{1}{2}\:\alpha\;4}}$

$\\\;\;\bf{:\rightarrow\;\;\theta_{2}\;=\;\bf{\bigg(\dfrac{1}{2}\:\alpha\;4\bigg)\;-\;\dfrac{1}{2}\:\alpha}}$

$\\\;\;\bf{:\rightarrow\;\;\theta_{2}\;=\;\bf{\bigg(\dfrac{4\alpha\;-\;\alpha}{2}\bigg)}}$

$\\\;\;\underline{\underline{\bf{:\rightarrow\;\;\theta_{2}\;=\;\bf{\dfrac{3}{2}\:\alpha}}}}$

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~ For the ratio θ₂ : θ₁ ::

$\\\;\;\sf{:\mapsto\;\;\dfrac{\theta_{2}}{\theta_{1}}\;=\;\bf{\dfrac{\bigg(\dfrac{3}{2}\:\alpha\bigg)}{\bigg(\dfrac{1}{2}\:\alpha\bigg)}}}$

Cancelling α , we get,

$\\\;\;\sf{:\mapsto\;\;\dfrac{\theta_{2}}{\theta_{1}}\;=\;\bf{\dfrac{\bigg(\dfrac{3}{2}\bigg)}{\bigg(\dfrac{1}{2}\bigg)}}}$

$\\\;\;\sf{:\mapsto\;\;\dfrac{\theta_{2}}{\theta_{1}}\;=\;\bf{\dfrac{3\;\times\;2}{1\;\times\;3}}}$

Cancelling 2, we get,

$\\\;\;\sf{:\mapsto\;\;\dfrac{\theta_{2}}{\theta_{1}}\;=\;\bf{\dfrac{3}{1}}}$

$\\\;\;\bf{:\mapsto\;\;\theta_{2}\;:\;\theta_{1}\;=\;\bf{3\;:\;1}}$

So, option c.) 3 is correct.

$\\\;\underline{\boxed{\tt{The\;\;ratio\;\;of\;\;\dfrac{\theta_{2}}{\theta_{1}}\;=\;\bf{\dfrac{3}{1}\;=\;3}}}}$

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★ More formulas to know :-

$\\\;\sf{\leadsto\;\;\omega\;=\;\omega_{0}\;+\;\alpha t}$

$\\\;\sf{\leadsto\;\;\omega^{2}\;=\;\omega_{0} ^{2}\;+\;2\alpha \theta}$

$\\\;\sf{\leadsto\;\;K\;=\;\sqrt{\dfrac{I}{M}}}$

$\\\;\sf{\leadsto\;\;\Delta W\;=\;\tau \Delta \theta}$

$\\\;\sf{\leadsto\;\;\tau\;=\;I\;\times\;\alpha}$