JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A thin circular disc is in the \(xy\) plane as shown in the figure. The ratio of its moment of inertia about \(z\) and \(z'\) axes will be
- A \(1:2\)
- B \(1:4\)
- C \(1:3\)
- D \(1:5\)
Answer & Solution
Correct Answer
(C) \(1:3\)
Step-by-step Solution
Detailed explanation
As we know, moment of inertia of a disc about an axis passing through \(C.G.\) and perpendicular to its plane, \({I_z}\, = \frac{{m{R^2}}}{2}\) Moment of inertia of a disc about a tangential axis perpendicular to its own plane, \({I_z}'\, = \frac{3}{{20}}m{R^2}\)…
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
- Two charges of \(-4 \ \mu \mathrm{C}\) and \(+4\ \mu \mathrm{C}\) are placed at the points \(A(1,0,4) \mathrm{m}\) and \(B(2,-1,5) \mathrm{m}\) located in an electric field \(\vec{E}=0.20 \hat{\mathrm{i}} \mathrm{V} / \mathrm{cm}\). The magnitude of the torque acting on the dipole is \(8 \sqrt{\alpha} \times 10^{-3} \mathrm{Nm}\), Where \(\alpha=\)___________JEE Mains 2024 Hard
- A particle having the same charge as of electron moves in a circular path of radius \(0.5
\,cm\) under the influence of a magnetic field of \(0.5\,T.\) If an electric field of \(100\,V/m\) makes it to move in a straight path, then the mass of the particle is (given charge of electron \(= 1.6 \times 10^{-19}\, C\) )JEE Mains 2019 Medium - A travelling microscope has \(20\) divisions per \(cm\) on the main scale while its Vernier scale has total \(50\) divisions and \(25\) Vernier scale divisions are equal to \(24\) main scale divisions, what is the least count of the travelling microscope \(..........\,cm\)JEE Mains 2022 Medium
- One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is \(4\) cm\(^2\). The gas is heated slowly to raise the temperature by \(1.2\,^\circ\)C during which the piston moves by \(25\) mm. The amount of heat supplied to the gas is ________ J. (Atmospheric pressure \(=100\) kPa, \(R=8.3\) J/mol·K) (Neglect mass of the piston)JEE Mains 2026 Hard
- The acceleration due to gravity at height \(h\) above the earth if \(h \ll R\) (radius of earth) is given byJEE Mains 2023 Medium
- A series \(LCR\) circuit is designed to resonate at an angular frequency \(\omega_{0}=10^{5} \,rad / s\). The circuit draws \(16\, W\) power from \(120\, V\) source at resonance. The value of resistance \('R'\) in the circuit is ...... \(\Omega\).JEE Mains 2021 Medium
More PYQs from JEE Mains
- A uniform circular disc of radius ' R ' and mass ' M ' is rotating about an axis perpendicular to its plane and passing through its centre. A small circular part of radius \(\mathrm{R} / 2\) is removed from the original disc as shown in the figure. Find the moment of inertia of the remaining part of the original disc about the axis as given above.
JEE Mains 2025 Medium - \( \text { If } S(x)=(1+x)+2(1+x)^2+3(1+x)^3+\ldots . \) \( +60(1+x)^{60}, x \neq 0 \text {, and }(60)^2 S(60)=a(b)^b+b\) where \(a, b N\), then \((a+b)\) equal to ...............JEE Mains 2024 Hard
- A bulb and a capacitor are connected in series across an AC supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb:JEE Mains 2024 Hard
- Let the normals at all the points on a given curve pass through a fixed point \((a, b) .\) If the curve passes through \((3,-3)\) and \((4,-2 \sqrt{2}),\) and given that \(a-2 \sqrt{2} b=3,\) then \(\left(a^{2}+b^{2}+a b\right)\) is equal to ..... .JEE Mains 2021 Hard
- Four identical solid spheres each of mass \('m'\) and radius \('a'\) are placed with their centres on the four corners of a square of side \('b'\). The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square isJEE Mains 2021 Hard
- If the solution \(y=y(x)\) of the differential equation \(\left(\mathrm{x}^4+2 \mathrm{x}^3+3 \mathrm{x}^2+2 \mathrm{x}+2\right) \mathrm{dy}-\left(2 \mathrm{x}^2+2 \mathrm{x}+3\right) \mathrm{dx}=0\) satisfies \(y(-1)=-\frac{\pi}{4}\), then \(y(0)\) is equal to :JEE Mains 2024 Medium