Small water droplets of radius 0.01 mm are formed in the upper atmosphere and falling with a terminal velocity of 10 cm/s. Due to condensation, if 8 such droplets are coalesced and formed a larger drop, the new terminal velocity will be ………..cm/s.
Correct Answer: 40
Approach Solution - 1
To solve the problem of determining the new terminal velocity of the larger droplet formed by the coalescence of 8 smaller droplets, we can use the physics of terminal velocity and volume conservation. First, consider that the terminal velocity \( v \) of a droplet can be expressed as \( v \propto r^2 \), where \( r \) is the radius of the droplet. When droplets coalesce, their combined volume equals the volume of the larger drop. Given the radius \( r_0 = 0.01 \) mm for each of the smaller droplets and their terminal velocity \( v_0 = 10 \) cm/s, we can calculate the new radius of the larger droplet. The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \). Therefore, the total initial volume of 8 small droplets is \( 8 \times \frac{4}{3} \pi r_0^3 \) and the volume of the larger drop is \( \frac{4}{3} \pi R^3 \). Equating these, we get:
\( R^3 = 8r_0^3 \)
Solving for \( R \), we have:
\( R = 8^{1/3}r_0 = 2r_0 \)
Substituting \( r_0 = 0.01 \) mm, we find:
\( R = 2 \times 0.01 \text{ mm} = 0.02 \text{ mm} \)
Next, calculate the terminal velocity of the larger droplet, using \( v \propto r^2 \). The new terminal velocity \( V \) is given by:
\( \frac{V}{v_0} = \left(\frac{R}{r_0}\right)^2 \)
Substitute the values:
\( \frac{V}{10} = \left(\frac{0.02}{0.01}\right)^2 = 2^2 = 4 \)
Therefore, \( V = 4 \times 10 = 40 \text{ cm/s} \)
The calculated terminal velocity of the larger droplet is indeed 40 cm/s, which fits perfectly within the given range of 40 to 40 cm/s.
Approach Solution -2
The terminal velocity (\(V_t\)) of a droplet is given by:
\[ V_t =\( \frac{2R^2 (\rho - \sigma) g}{9\eta}\), \]
where \(R\) is the radius of the droplet.
When 8 small droplets coalesce:
Mass of larger drop \(M \propto 8m\), \quad \(M = \(\frac{4}{3}\)\pi R^3 (\rho)\).
The radius of the larger drop is:
\[ R_{\text{large}}^3 = 8R_{\text{small}}^3 \implies R_{\text{large}} = 2R_{\text{small}}. \]
Since terminal velocity (\(V_t\)) is proportional to \(R^2\):
\[ V_t \propto R^2 \implies V_t^{\text{large}} = 4V_t^{\text{small}}. \]
For the smaller droplet:
\[ V_t^{\text{small}} = 10 \, \text{cm/s}. \]
Thus:
\[ V_t^{\text{large}} = 4 \cdot 10 = 40 \, \text{cm/s}. \]
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Fluid Mechanics Questions
- In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 ms–1 and 65 ms–1 respectively. If the wing area is 2 m2 the lift of the wing is _______ N.
(Given density of air = 1.2 kg m-3) - A plane is in level flight at constant speed and each of its two wings has an area of 40 m2 . If the speed of the air is 180 km/h over the lower wing surface and 252 km/h over the upper wing surface, the mass of the plane is ________kg. (Take air density to be 1 kg m–3 and g = 10 ms–2
- Given below are two statements:
Statement I: When the speed of liquid is zero everywhere, the pressure difference at any two points depends on the equation $$ P_1 - P_2 = \rho g (h_2 - h_1). $$ Statement II: In the ventury tube shown, $$ 2gh = v_1^2 - v_2^2. $$
In the light of the above statements, choose the most appropriate answer from the options given below. - Given below are two statements :
Statement-I: When a capillary tube is dipped into a liquid, the liquid neither rises nor falls in the capillary. The contact angle may be 0°.
Statement-II: The contact angle between a solid and a liquid is a property of the material of the solid and liquid as well : In the light of above statement.
Choose the correct answer from the options given below. - A sphere of relative density \( \sigma \) and diameter \( D \) has a concentric cavity of diameter \( d \). The ratio of \( \frac{D}{d} \), if it just floats on water in a tank, is:
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.