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QJEEM24Phf71f: JEE Main 2024 Physics - Class 11 JEEM24Phf71f Solution: Ratio of radius of gyration of a hollow sphere to that of a…

Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis ABA B as shown in figure is 8/x\sqrt{8 / x} . The value of xx is :

JEE Main 2024 (Online) 5th April Morning Shift Physics - Rotational Motion Question 15 English

Multiple Choice Options

  • A 34
  • B 51
  • C 67
  • D 17

Step-by-Step Verified Solution

For hollow sphere

amp;I1=23mR2=(23R)2mamp;∴K1=23R\begin{aligned}& I_1=\frac{2}{3} m R^2=\left(\sqrt{\frac{2}{3}} R\right)^2 m \\& \therefore \quad K_1=\sqrt{\frac{2}{3}} R\end{aligned}

For solid cylinder

amp;I2=14mR2+m3(4R)2amp;=(6712R)2mamp;∴K2=6712Ramp;⇒K1K2=23×1267amp;=867\begin{aligned}& I_2=\frac{1}{4} m R^2+\frac{m}{3}(4 R)^2 \\& =\left(\sqrt{\frac{67}{12}} R\right)^2 m \\& \therefore \quad K_2=\sqrt{\frac{67}{12}} R \\& \Rightarrow \frac{K_1}{K_2}=\sqrt{\frac{2}{3} \times \frac{12}{67}} \\& =\sqrt{\frac{8}{67}}\end{aligned}

⇒x=67\Rightarrow \quad x=67

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