MHT CET · Physics · Rotational Motion
Two spheres each of mass ' \(M\) ' and radius \(\frac{R}{2}\) are connected at the ends of massless rod of length ' \(2 \mathrm{R}\) '. What will be the moment of inertia of the system about an axis passing through centre of one of the spheres and perpendicular to the rod?
- A \(\frac{2}{3} \mathrm{MR}^2\)
- B \(\frac{5}{2} \mathrm{MR}^2\)
- C \(\frac{5}{21} \mathrm{MR}^2\)
- D \(\frac{21}{5} \mathrm{MR}^2\)
Answer & Solution
Correct Answer
(D) \(\frac{21}{5} \mathrm{MR}^2\)
Step-by-step Solution
Detailed explanation
From parallel axis theorem,
\(\mathrm{I}_{\mathrm{o}}=\mathrm{I}_{\mathrm{c}}+\mathrm{Mh}^2\)
Let the moment of inertia of sphere 1 be
\(\mathrm{I}_1=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2\)
and,
Let the moment of inertia of sphere 2 be
\(\mathrm{I}_2=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2\)
Moment of inertia of the rod \(\mathrm{I}_3=0\)
\(\therefore \quad\) Moment of inertia of the system,
\(\begin{aligned}
\mathrm{I} & =\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3 \\
\mathrm{I} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2+\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2 \\
& =\frac{4}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+4 \mathrm{MR}^2 \\
& =\frac{1}{5} \mathrm{MR}^2+4 \mathrm{MR}^2 \\
& =\frac{21}{5} \mathrm{MR}^2
\end{aligned}\)
\(\mathrm{I}_{\mathrm{o}}=\mathrm{I}_{\mathrm{c}}+\mathrm{Mh}^2\)
Let the moment of inertia of sphere 1 be
\(\mathrm{I}_1=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2\)
and,
Let the moment of inertia of sphere 2 be
\(\mathrm{I}_2=\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2\)
Moment of inertia of the rod \(\mathrm{I}_3=0\)
\(\therefore \quad\) Moment of inertia of the system,
\(\begin{aligned}
\mathrm{I} & =\mathrm{I}_1+\mathrm{I}_2+\mathrm{I}_3 \\
\mathrm{I} & =\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+\mathrm{M}(2 \mathrm{R})^2+\frac{2}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2 \\
& =\frac{4}{5} \mathrm{M}\left(\frac{\mathrm{R}}{2}\right)^2+4 \mathrm{MR}^2 \\
& =\frac{1}{5} \mathrm{MR}^2+4 \mathrm{MR}^2 \\
& =\frac{21}{5} \mathrm{MR}^2
\end{aligned}\)
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Physics
- In projectile motion two particles of masses \(\mathrm{m}_1\) and \(m_2\) have velocities \(\vec{V}_1\), and \(\vec{V}_2\) respectively at time \(\mathrm{t}=0\). Their velocities become \(\overrightarrow{\mathrm{V}_1{ }^{\prime}}\) and \(\overrightarrow{\mathrm{V}_2{ }^{\prime}}\) at time 2 t while still moving in air. The value of \(\left[\left(m_1 \overrightarrow{V_1{ }^{\prime}}+m_2 \overrightarrow{V_2{ }^{\prime}}\right)-\left(m_1 \overrightarrow{V_1}+m_2 \overrightarrow{V_2}\right)\right]\) is ( \(\mathrm{g}=\) acceleration due to gravity)MHT CET 2024 Medium
- Simple microscope is used to see the object first in blue light and then a red light. Due to the change from blue to red light, its magnifying powerMHT CET 2024 Easy
- A machine gun fires bullets of mass \(30 \mathrm{~g}\) with velocity of \(1000 \mathrm{~m} / \mathrm{s}\). The man holding the gun can exert a maximum force of \(300 \mathrm{~N}\) on it. How many bullets can he fire per second at most?MHT CET 2023 Easy
- A stationary object at \(4^{\circ} \mathrm{C}\) and weighing 3.5 kg falls from a height of 2000 m on snow mountain at \(0^{\circ} \mathrm{C}\). If the temperature of the object just before hitting the snow is \(0^{\circ} \mathrm{C}\) and the object comes to rest immediately then the quantity of ice that melts is
(Acceleration due to gravity \(=10 \mathrm{~m} / \mathrm{s}^2\), Latent heat of ice \(\left.=3.5 \times 10^5 \mathrm{~J} / \mathrm{kg}\right)\)MHT CET 2025 Medium - A child stands on a weighing machine inside a lift. When the lift is going down with acceleration \(\frac{g}{3}\), the machine shows a reading 20 N. When the lift goes upwards with acceleration \(\frac{g}{3}\), the reading would be ( \(\mathrm{g}=\) gravitational acceleration)MHT CET 2025 Medium
- If the apertaure of a telescope is decreased the resolving power willMHT CET 2010 Medium
More PYQs from MHT CET
- Which carbon atom of deoxy Ribose sugar in DNA does NOT contain the
bond?MHT CET 2018 Medium - Identify product 'C' in following reaction
\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \underset{\text { Pyridine }}{\stackrel{\mathrm{SOCl}_{2}}{\longrightarrow}} \mathrm{A} \stackrel{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}}{\longrightarrow} \mathrm{B} \underset{\Delta \text { , Pressure }}{\stackrel{\text { dil } \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}} \mathrm{C}\)MHT CET 2020 Medium - If , then K = ….MHT CET 2019 Medium
- In a game, 3 coins are tossed. A person is paid ₹ \(100\), if he gets all heads or all tails; and he is supposed to pay ₹ 40 , if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) isMHT CET 2024 Medium
- The value of \(\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}\) isMHT CET 2023 Medium
- \(\cos x \cdot \cos 7 x-\cos 5 x \cdot \cos 13 x=\)MHT CET 2020 Medium