JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2022
- Easy
- MCQ
The terminal velocity \(\left( v _{ t }\right)\) of the spherical rain drop depends on the radius ( \(r\) ) of the spherical rain drop as
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(r^{2}\)
Step-by-step Solution
Detailed explanation
\(v _{ t }=\frac{2}{9} \frac{ gr ^{2}\left(\rho_{ p }-\rho_{1}\right)}{\eta} ; \quad v _{ t } \propto r ^{2}\)
More questions from Physics
- When a particle executes simple Harmonic motion, the nature of graph of velocity as function of displacement will be.JEE Mains 2022 Medium
- The current in a coil of self inductance \(2.0 \,H\) is increasing according to \(I =2 \sin \left( t ^{2}\right) A\). The amount of energy spent during the period when current changes from \(0\) to \(2\, A\) is......\(J.\)JEE Mains 2022 Medium
- A ball of mass \(0.5 \; kg\) is dropped from the height of \(10 \; m\). The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is \(\dots \; m\). (Use \(g =10 \; m / s ^{2}\)).JEE Mains 2022 Medium
- Find the mutual inductance in the arrangement, when a small circular loop of wire of radius ' \(R\) ' is placed inside a large square loop of wire of side \(L\) \(( L \gg R )\). The loops are coplanar and their centres coincide :
JEE Mains 2023 Medium - A small particle of mass \(m\) moves in such a way that its potential energy \(U=\frac{1}{2} m \omega^2 r ^2\) where \(\omega\) is constant and \(r\) is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of \(n^{\text {th }}\) orbit will be proportional to.JEE Mains 2023 Medium
- A vessel contains \(0.15\) m\(^3\) of a gas at pressure \(8\) bar and temperature \(140°\)C with \(c_p = 3R\) and \(c_v = 2R\). It is expanded adiabatically till pressure falls to \(1\) bar. The work done during this process is _______ kJ. (\(R\) is gas constant)JEE Mains 2026 Hard
More PYQs from JEE Mains
- The position vector of the centre of mass \(\vec r\, cm\) of an asymmetric uniform bar of negligible area of cross-section as shown in figure is
JEE Mains 2019 Medium - The position of the centre of mass of a uniform semi-circular wire of radius \('R'\) placed in \(x-y\) plane with its centre at the origin and the line joining its ends as \(x\)-axis is given by \(\left(0, \frac{x R}{\pi}\right)\). Then, the value of \(|x|\) is ...... .JEE Mains 2021 Medium