Rotational Motion

JEE Main level, numerical answer, medium

11 of 280

Question 11 of 280

A uniform solid cylinder of length LL and radius RR has moment of inertia about its axis equal to I1I_{1}. A small co-centric cylinder of length L/2L / 2 and radius R/3R / 3 carved from this cylinder has moment of inertia about its axis equals to I2I_{2}. The ratio I1/I2I_{1} / I_{2} is ____\_\_\_\_.

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Answer: 162

For solid cylinder: I=12mR2I = \frac{1}{2}mR^2
Large cylinder: mass m1=ρπR2Lm_1 = \rho \pi R^2 L, so I1=12ρπR2L×R2=ρπR4L2I_1 = \frac{1}{2}\rho \pi R^2 L \times R^2 = \frac{\rho \pi R^4 L}{2}
Small cylinder (carved out): length L/2, radius R/3, mass m2=ρπ(R/3)2(L/2)=ρπR2L18m_2 = \rho \pi(R/3)^2(L/2) = \frac{\rho \pi R^2 L}{18}
I2=12×ρπR2L18×(R/3)2=12×ρπR2L18×R29=ρπR4L324I_2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times (R/3)^2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times \frac{R^2}{9} = \frac{\rho \pi R^4 L}{324}
I1I2=ρπR4L/2ρπR4L/324=3242=162\frac{I_1}{I_2} = \frac{\rho \pi R^4 L/2}{\rho \pi R^4 L/324} = \frac{324}{2} = 162

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