A water tank of height 10 m, completely filled with water is placed on a level ground. It has two holes one at 3 m and the other at 7 m from its base. The water ejecting from :
A drop of water falls in air with terminal velocity \(v_0\). If 27 such drops combine to form a big drop. Then terminal velocity of this new big drop will be:
When 27 identical small drops of radius \(r\) coalesce, the radius \(R\) of the single large drop is \(R = 27^{1/3}r = 3r\). Since terminal velocity \(v \propto r^2\), the new terminal velocity is \(v' = 3^2 v_0 = 9v_0\).
A small ball of density \(rho\) is immersed in a liquid of density \(sigma (\sigma > \rho)\) to a depth h and released. The height above the surface of water upto which the ball jumps is:
Applying the work-energy theorem, work done by the buoyant force over depth \(h\) equals total work done against gravity: \(V \sigma g h = V \rho g (h + H)\). Solving for height \(H\) gives \(H = \left(\frac{\sigma}{\rho} - 1\right)h\).
A liquid is flowing in a cylindrical pipe of internal diameter \(4\text{ cm}\) with a velocity of \(5\text{ m/s}\). If this tube is joined with another tube of internal diameter \(2\text{ cm}\) then the velocity of flow of liquid in the smaller tube will be (in \(\text{ms}^{-1}\))
Using the equation of continuity, \(A_1 v_1 = A_2 v_2\), which ⇒ \(d_1^2 v_1 = d_2^2 v_2\). Substituting \(d_1 = 4\text{ cm}\), \(v_1 = 5\text{ m/s}\), and \(d_2 = 2\text{ cm}\) gives \(16 \times 5 = 4 \times v_2\), so \(v_2 = 20\text{ m/s}\).
Terminal velocity of iron ball of radius \(1\text{ mm}\) in glycerine is \(v_1\) and \(v_2\) is the terminal velocity of lead ball of radius \(2\text{ mm}\) in glycerine then \(v_1 : v_2\) is : [\(\rho_{\text{glycerine}} = 13.6\text{ g/cc}\), \(\sigma_{\text{Fe}} = 7.6\text{ g/cc}\), \(\sigma_{\text{Pb}} = 11.6\text{ g/cc}\)]
Terminal velocity \(v_T \propto r^2(\sigma - \rho)\). For iron, \(v_1 \propto 1^2(7.6 - 13.6) = -6\). For lead, \(v_2 \propto 2^2(11.6 - 13.6) = -8\). Thus, the ratio is \(v_1 : v_2 = -6 / -8 = 0.75\).
Consider the following statements:
Statement A: When a capillary tube is dipped into a liquid, if the liquid neither rises nor falls in the capillary, then the angle of contact must be zero.
Statement B: The working of a Venturi-meter is based on Bernoulli’s theorem.
Based upon the above information, pick the correct option:
If liquid neither rises nor falls, the net vertical surface tension force is zero, meaning the contact angle is \(90^\circ\). Venturi-meter works on Bernoulli's theorem, so only statement B is correct.
A small ball of mass \(m\) and density \(rho\) is dropped in a viscous liquid of density \(\rho_0\). After sometime, the ball falls with constant velocity. The net force acting on the ball after it attains constant velocity, will be:
When a body falls with constant velocity (terminal velocity), its acceleration is zero. Thus, by Newton's second law, the net force on it is zero.
Assertion (A): Bernoulli’s theorem is based on energy conservation.
Reason (R): Bernoulli’s theorem holds good for all liquids.
Bernoulli's theorem is based on energy conservation (Assertion is true). However, it only holds for ideal (non-viscous, incompressible) fluids, not all liquids (Reason is false).
Assertion (A): Bernoulli’s theorem is based on energy conservation.
Reason (R): Bernoulli’s theorem holds good for all liquids.
Bernoulli's theorem is derived from the law of conservation of energy (Assertion is true). It only holds good for ideal (non-viscous, incompressible) fluids, not all liquids (Reason is false).
Assertion (A): Blood pressure of heart is same whether you lie down or stand up.
Reason (R): Pressure varies with height in a fluid under gravity.
Assertion (A) is true due to physiological regulation of blood pressure at the heart level. Reason (R) is true as pressure in a fluid varies with depth, \(P = \rho gh\). However, (R) describes a physical effect that (A) counteracts, so (R) is not the correct explanation for (A).