Question

refer to the attachment for the Question...

class 11.. physcis​

Answer 1

The correct answer is option : (b) 0.64 x 10⁻¹ N.m

Given: Diameter of solid cylinder (D) = 4.9 cm

           twisting couple per unit twist (T/θ) = 0.1 N.m

Inner and outer diameter of hollow cylinder D₁ & D₂ are 8cm & 10cm respectively.

• Also they are of same material.

From torsion equation we have:

   T/J  =  τ/R  =  Gθ/L

T - Torque or twisting moment

J - polar moment of inertia

τ - shear stress

R - Radius of the shaft

G - Modulus of rigidity

θ - Angle of twist

L - length of shaft

Polar moment of inertia of solid cylinder is given by,

$J_{1} = \frac{\pi D^4}{32} = \frac{\pi *9.8^4}{32} \ cm^4$

polar moment of inertia of hollow cylinder is also given by ,

$J_{2} = \frac{\pi *(D_{1} ^4 - D_{2}^4) }{32} = \frac{\pi *(10^4-8^4)}{32} \ cm^4$

from torsion equation,

 T/θ = GJ/L , for same material & similar type of cylinder

⇒ T/θ ∝ J

⇒ (T/θ)₁ / (T/θ)₂ = J₁ / J₂

⇒ 0.1 / (T/θ)₂ = [π 9.8⁴ / 32] / [π (10⁴-8⁴) / 32]

∴(T/ θ)₂ = 0.064 N.m or 0.64 x 10⁻¹ N.m

HOPE THIS HELPS YOU !!    : )

Answer 2

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

Here the Concept of Torsional Rigidity (C) and Shear Modulus (G) has been used. We know that, twisting a couple per unit twist is the torque required to deform a cylinder for a unit rotation for a unit length when there is a external torque acting on it. Firstly we shall apply the values in the formula of Torsional Rigidity to find the equation of first case. Then we shall do it for second solid. After doing this, we shall apply values and find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Torsional\;Rigidity,\;C\;=\;\bf{\dfrac{\pi \eta\:r^{4}}{2l}}}}$

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★ Solution :-

Given,

» Twisting couple of solid cylinder = C₁ = 0.1 N/m

» Radius of the solid cylinder = R = 4.9 cm = 4.9 × 10⁻² m

» Outer radius of hollow cylinder = r₂ = 5 cm = 5 × 10⁻² m

» Inner radius of hollow cylinder = r₁ = 4 cm = 4 × 10⁻² m

Since both solids are made of same material. Then,

• Coefficient of rigidity = η
• Length of solid = 2l
• Torsional Rigidity of Couple of Hollow Cylinder = C₂

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~ For Torsional Rigidity of Solid Cylinder ::

$\\\;\;\sf{:\rightarrow\;\;Torsional\;Rigidity,\;C\;=\;\bf{\dfrac{\pi \eta\:r^{4}}{2l}}}$

$\\\;\;\sf{:\Rightarrow\;\;C_{1}\;=\;\bf{\dfrac{\pi \eta\:R^{4}}{2l}}}$

$\\\;\;\bf{:\Rightarrow\;\;C_{1}\;=\;\bf{\dfrac{\pi \eta\:R^{4}}{2l}\;=\;0.1\;\;N\:m^{-1}}}$

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~ For Torsional Rigidity of Hollow Cylinder ::

$\\\;\;\sf{:\rightarrow\;\;Torsional\;Rigidity,\;C\;=\;\bf{\dfrac{\pi \eta\:r^{4}}{2l}}}$

For hollow cylinders, this formula is given as,

$\\\;\;\bf{:\Rightarrow\;\;C_{2}\;=\;\bf{\dfrac{\pi \eta\:(r_{2} ^{4}\;-\;r_{1} ^{4})}{2l}}}$

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~ For the value of C₂ ::

For finding the value of C₂, we must divide the C₂ by C₁. Then,

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{C_{1}}\;=\;\bf{\dfrac{\bigg(\dfrac{\pi \eta\:(r_{2} ^{4}\;-\;r_{1} ^{4})}{2l}\bigg)}{\bigg(\dfrac{\pi \eta\:R^{4}}{2l}\bigg)}}}$

We already know that value of C₁ that is 0.1 . By applying that, we get,

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{\bigg(\dfrac{\pi \eta\:(r_{2} ^{4}\;-\;r_{1} ^{4})}{2l}\bigg)}{\bigg(\dfrac{\pi \eta\:R^{4}}{2l}\bigg)}}}$

Now cancelling π and η (since they are same for both solids) , we get,

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{\bigg(\dfrac{(r_{2} ^{4}\;-\;r_{1} ^{4})}{2l}\bigg)}{\bigg(\dfrac{R^{4}}{2l}\bigg)}}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{(r_{2} ^{4}\;-\;r_{1} ^{4})\;\times\;2l}{R^{4}\;\times\;2l}}}$

Cancelling 2l, we get,

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{(r_{2} ^{4}\;-\;r_{1} ^{4})}{R^{4}}}}$

Now let's apply the values of the variables,

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{(5\;\times\;10^{-2})^{4}\;-\;(4\;\times\;10^{-2})^{4}}{(4.9\;\times\;10^{-2})^{4}}}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{(5)^{4}\;\times\;10^{2}\;-\;(4)^{4}\;\times\;10^{2}}{(4.9)^{4}\;\times\;10^{2}}}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{10^{2}\;\times\;[(5)^{4}\;-\;(4)^{4}]}{10^{2}\;\times\;(4.9)^{4}}}}$

Cancelling, 10⁻² , we get,

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{5^{4}\;-\;4^{4}}{4.9^{4}}}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{625\;-\;256}{576.5}}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{\dfrac{369}{576.5}}}$

$\\\;\sf{\Longrightarrow\;\;\dfrac{C_{2}}{0.1}\;=\;\bf{0.64}}$

$\\\;\sf{\Longrightarrow\;\;C_{2}\;=\;\bf{0.64\;\times\;0.1}}$

$\\\;\bf{\Longrightarrow\;\;C_{2}\;=\;\bf{0.64\;\times\;10^{-1}\;\;N\:m^{-1}}}$

S. I. unit of C = N/m

So , option b.) 0.64 × 10⁻¹ N/m is correct.

$\\\;\underline{\boxed{\tt{Hence,\;\;required\;\;twisting\;\;couple\;\;=\;\bf{0.64\;\times\;10^{-1}\;N\:m^{-1}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;C\;=\;\dfrac{G}{\theta}}$

$\\\;\sf{\leadsto\;\;\eta\;=\;\dfrac{Force\;\times\;Distance}{Area\;\times\;Velocity}}$

$\\\;\sf{\leadsto\;\;Angular\;Velocity\;=\;\dfrac{Angle}{Time}}$

$\\\;\sf{\leadsto\;\;Angular\;Acceleration\;=\;\dfrac{Angular\;Velocity}{Time}}$

$\\\;\sf{\leadsto\;\;Moment\;of\;Inertia\;=\;Mass\;\times\;(Distance)^{2}}$